International Library of Technology
374B
Applied Mechanics and Strength of Materials
197 ILLUSTRATIONS
Prepared Under Supervision of
A. B. CLEMENS
DIRECTOR, MECHANICAL SCHOOLS, INTER- NATIONAL CORRESPONDENCE SCHOOLS
LINK MECHANISMS
GEARING
GEAR TRAINS AND CAMS
PULLEYS AND BELTING
MATERIALS OF CONSTRUCTION
STRENGTH OF MATERIALS THE TESTING OF MATERIALS
Published by INTERNATIONAL TEXTBOOK COMPANY
SCRANTON, PA.
Link Mechanisms: Copyright. 1906, by INTERNATIONAL TEXTBOOK COMPANY. Gearing: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY. Gear Trains and Cams: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY. (Pulleys and Belting: Copyright, 1906, by INTERNATIONAL TEXTBOOK COMPANY.
Materials of Construction: Copyright, 1927AJJQ6, by INTERNATIONAL TEXTBOOK COMPANY. -**"-.*«»,.
Strength of Materials, Parts 1 and 2: Copyright, 1906, by INTERNATIONAL TEXT- BOOK COMPANY.
The Testing of Materials: Copyright, 1906, by INTERNATIONAL TEXTBOOK COM- PANY. •
Entered at Stationers' Hall, London.
All rights reserved
Printed in U. S, A.
6095 sty'
PRUSS OF
INTERNATIONAL TEXTBOOK COMPANY 94217
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PREFACE
The volumes of the International Library of Technology are made up of Instruction Papers, or Sections, comprising the various courses of instruction for students of the International Correspondence Schools. The original manuscripts are pre- pared ^ by persons thoroughly qualified both technically and by experience to write with authority, and in many cases they are regularly employed elsewhere in practical work as experts. The manuscripts are then carefully edited to make them suit- able for correspondence instruction. The Instruction Papers are written clearly and in the simplest language possible, so as to make them readily understood by all students. Necessary technical expressions are clearly explained when introduced.
The great majority of our students wish to prepare them- selves for advancement in their vocations or to qualify for more congenial occupations. Usually they are employed and able to devote only a few hours a day to study. Therefore every effort must be made to give them practical and accurate information in clear and concise form and to make this infor- mation include all of the essentials but none of the non- essentials. To make the text clear, illustrations are used freely. These illustrations are especially made by our own Illustrating Department in order to adapt them fully to the requirements of the text.
In the table of contents that immediately follows are given the titles of the Sections included in this volume, and under eacli title are listed the main topics discussed.
INTERNATIONAL TEXTBOOK COMPANY B
CONTENTS
NOTE. — This volume is made up of a number of separate parts, or sections, as indicated by their titles, and the page numbers of each usually begin with 1. In this list of contents the titles of the parts are given in the order in which they appear in the book, and tinder each title is a full synopsis of the subjects treated.
LINK MECHANISMS Pages
Relative Motions of Links 1-35
Introduction 1-2
General Kinematic Principles 3-8
Kinds of constrained motion; Plane motion of a rigid body; Relative motion.
Levers 9-35
Steam-engine mechanism; Quick-return motions; Straight-line motion; Universal joint.
GEARING
Toothed Gearing 1-46
Rolling Curves on Surfaces 1-4
Spur Gearing 5-32
General principles ; Involute systems ; Cycloidal, or rolled-curve, system; Proportions of gear-teeth; Con- struction of tooth profiles.
Bevel Gearing 33-40
Spiral and Worm-Gearing 41-46
GEAR TRAINS AND CAMS
Gear Trains !~29
Engine-Lathe Gear Trains 6-14
Back-gear train; Screw-cutting train.
Epicyclic Trains • 15-23
vi CONTENTS
GEAR TRAINS AND CAMS— (Continued)
Pages
Revolving Gear Trains 24-25
Reversing Mechanisms 26-29
Cams and Cam Trains 30-45
Rotary Cams 31-39
Sliding and Cylindrical Cams 40-41
Ratchet Mechanisms 42-45
PULLEYS AND BELTING
Belt Gearing , 1-33
Kinematics of Flexible Gearing 1-5
Length of Open and Crossed Belts 6-17
Power Transmission by Belt 18-22
Care and Use of Belting 23-26
Belt Connections for Non-Parallel Shafts 27-33
MATERIALS OF CONSTRUCTION
Metals 1-54
Iron 1-20
Manufacture and Properties of Iron 1-7
Forms of iron; Ores of iron; Separation of iron from its ores; Blast furnace; Pig iron.
Cast Iron 3-16
Cupola; Characteristics of cast iron; Carbon in cast iron; Silicon, sulphur, phosphorus, and manganese m cast iron ; Change of volume of cast iron ; Chilled castings ; Malle- able castings;' Case-hardening malleable castings.
Wrought Iron 16-20
Purity of wrought iron ; Puddling furnace ; Siemens regen- erative furnace; Puddling and rolling; Properties of wrought iron; Defects of wrought iron.
Steel 21-42
Classification 21-22
Open-Hearth Steel and Bessemer Steel 23-26
Blister Steel and Shear Steel 27
Crucible Steel 28-30
Superiority of crucible steel ; Tool steel.
Electric Furnace Steel 30-31
CONTENTS vii
MATERIALS OF .CONSTRUCTION— (Continued)
Pages
Ingots and Steel Castings ......................... 32-34
Ingots; Shaping of ingots; Defects in ingots; Steel castings.
Alloy Steels .................................... 35-42
Tungsten steel; Air-hardening, or self -hardening, steel; Manganese steel; Nickel steel; Chrome steel; Nickel- chrome steel; Chrome-vanadium steel; Stainless steel; Titanium steel.
Non-Ferrous Alloys ............................. 43-54
Brasses and Bronzes ............................. 43-47
Value of non-ferrous alloys; Brass; Bronze. Bronze and Brass Castings ................... • • • • 48
Miscellaneous Alloys ............................ 48-51
Copper-nickel alloys; German silver; Silicon bronze; Copper-manganese alloy; Babbitt metal; Solder.
Aluminum Alloys ............................... 52
Magnalium; Duralumin; Aluminum-zinc alloys; Alumi- num-copper alloy.
Special Alloys ..... ' ............................. 53~54
Die-casting alloys; Stellite. Non-Metals ..................................... 55-71
Mortars and Concrete ............................ 55-59
Portland cement; Portland-cement concrete; Portland- cement mortar; Difference between mortar and con- crete ; Cement ; Cement paste ; Proportioning of concrete ; Methods of mixing concrete.
Timber ..................................... ; • •
Characteristics of timber; Evergreen timber; Tropical timber ; Hardwood timber ; Wood preservatives.
Transmission Rope ............................... 67-68
Manufacture of rope; Mounting of hemp rope. Selection of Materials ........................... 69~71
STRENGTH OF MATERIALS, PART 1 Stress, Deformation, and Elasticity ................ 1-18
Stress and Deformation .......................... 1- 2
Elasticity ...................................... 3~ 5
6-7 Tension ........................................
Compression ....................................
viii CONTENTS
STRENGTH OF MATERIALS, PART I— (Continued')
Pages
Shear 9-10
Factors of Safety 11-13
Pipes and Cylinders 14-18
Elementary Graphic Statics 19-26
Force diagram and equilibrium polygon; Composition of moments; Graphic expressions for moments.
Beams 27-52
Definitions 27
Simple Beams 28-43
Reactions of supports; Vertical shear; Bending moment; Simple beam with uniform load; Simple beams with mixed loads.
Overhung Beams and Cantilevers 44-52
STRENGTH OF MATERIALS, PART 2
Beams 1_15
Strength of Beams 1-10
Deflection of Beams 11-13
Comparison of Strength and Stiffness of Beams 14-15
Columns 16-22
Torsion and Shafts 23-28
Ropes and Chains . . 29-34
THE TESTING OF MATERIALS
Methods and Appliances 1_39
Test Pieces 2-4
Apparatus for Testing Materials 5_14
Tension Test 15-17
Records of the Test 18-24
Compression Test 25-27
Method of making the test; Records of the test.
Transverse Test 28-31
Shearing and Torsion Tests 32
Miscellaneous Tests 32-39
LINK MECHANISMS
Serial 990
Edition 1
RELATIVE MOTIONS OF LINKS
INTRODUCTION
1. Mechanics.— The science that treats of the motions of bodies and of the forces that produce or tend to produce such motions is called mechanics.
Applied mechanics comprises the principles of pure mechanics as applied to the design and construction of machinery or to works of engineering The part of applied mechanics that relates to machinery is called the mechanics of machinery.
The mechanics of machinery may be divided into two. chief branches: kinematics of machinery, treating of the motions of machine parts without regard to the forces acting- dynamics of machinery, treating of the forces acting on machine parts and of the transmission of force from one part to another.
2. Free and Constrained Motion.— A body is said to be free when it may move in any direction in obedience to the forces acting on it. A body is constrained when the nature or direction of its motion is determined by its con- nection with other bodies.
Examples of free bodies are the moon, the sun, and other heavenly bodies. Examples of constrained bodies are seen m every machine. Thus, the crosshead of an engine is con- w™ If SUldeS' Md the Shaft ^ the b-ings in
COPVR1SHTKD BV ,NTBRNAT10NAL TBXTB
LINK MECHANISMS
JB-
|
H |
^>n a |
77 |
In constrained motion, every point of the constrained body is forced to move in a definite path, no matter what may IDC the direction of the force that causes the motion. Any lorce that tends to give the body some other motion is at once neutralized by an equal and opposite force developed in the constraining members. For example, the block a, Fig. 1, is enclosed by stationary guides b and c and its only possible motion therefore is a sliding motion along the line EF. Suppose that the acting force P has the direction of
the line M N. Then, if a were free, the force P would cause it to move along the line M N. Let P be resolved into the components H and respectively, acting parallel and perpendic- ular to EF. The force H causes a to move along EF; the force V, on the other hand, would cause it to move vertically downwards, but the downward push of a against the guide c develops a reaction R in the guide c that is equal and opposite to V. Hence, V and R are balanced, the net vertical force is zero, and as a result there can be no motion vertically. In this way, any force that tends to cause a to move in any direction except along the line EF creates a reaction in one of the restrain- ing bodies &, c that exactly neutralizes that tendency.
3. Definition of a Machine. — A machine is an
assemblage of .fixed and moving parts so arranged as to utilize energy derived from some external source for the purpose of doing work.
In the operation of machinery, motion and force are com- municated to one of the movable parts and transmitted to the part that does the work. During the transmission, both the motion and the force are modified in direction and amount, so as to be rendered suitable for the purpose to which they are to be applied.
LINK MECHANISMS 3
The moving parts are so arranged as to have certain definite motions relative to each other, the effect of which is to compel the part doing the work to move in the required way. The nature of these movements is independent of the amount of force transmitted; in other words, in a model of a machine operated by hand, the relative motions of the parts will be precisely the same as in the machine itself, although in the latter case a great amount of power may be trans- mitted and much work done.
GENERAL KINEMATIC PKINCIPLES
KINDS OF CONSTRAINED MOTION
4. Plane Motion.— All constrained motions of rigid bodies, however complicated, may be divided into three classes; viz., plane motion, spherical motion, and screw motion.
A body is said to have plane motion when all its points move in parallel planes. Nearly all the motions of machine parts belong to this class. For example, the motions of all parts of the steam engine, except the governor, are plane; all points of the piston, piston rod, and crosshead move in equal parallel straight lines; all points of the crank, shaft, and flywheel move in circles of various radii that lie in par- allel planes; and points of the connecting-rod describe oval curves, which likewise lie in parallel planes. There are two special cases of plane motion of much importance, namely, rotation and translation. A body is said to rotate about an axis when all points of it move in parallel circles whose centers all lie on the axis; this motion is very common. Thus, a flywheel rotates about the axis of the shaft, a pulley about the axis of the line shaft, etc. A body has a motion of translation when the direction of a straight line in that body is always parallel to or coincides with its original direction. The paths of any points in the body may be either straight or curved. The motions of the piston of a steam engine and the parallel rod of a locomotive are examples of trans- lation in straight and curved paths, respectively.
4 LINK MECHANISMS
5. Spherical Motion, — A body has spherical motion
when each point of it remains always at a definite distance from some fixed point so as to move in the surface of an imaginary sphere with the fixed point as a center. In machinery, there are few examples of spherical motion; the universal joint and the balls of the steam-engine flyball gov- ernor are two familiar examples of this motion.
6. Screw motion consists of a rotation about a fixed axis combined with a translation along the axis. An exam- ple is the motion of a nut on a bolt.
Of the three forms of motion, plane motion occurs most frequently in machinery. Unless the contrary is stated, it is assumed in the following pages that the motion of a machine part is plane.
PLANE MOTION OF A RIGID BODY
7. Point Paths. — During the motion of a rigid body, such as a machine part, each point of the body traces a line, which is called the path of the point. In plane motion, the path lies in a plane, and in the case of a machine part the path is usually a closed curve; that is, a path that, if fol- lowed continuously, will bring the body to its original position. The path of any point of a body rotating about a fixed axis is a circle, which is a closed curve. The direction in which a point is moving at any instant is the direction of the tangent to the path of motion at the given point.
Also, in plane motion, the motions of any two points of a body determine the motion of the body as a whole. Thus, in the case of a connecting-rod, if the motions of any two points of the rod, at any instant, are known, the motion of the entire rod for that instant is determined.
8. The Instantaneous Center. — In Fig. 2, A and B are two points of a rigid figure, which may be in any plane sec- tion of the rigid body parallel to the plane of motion, and m and n, respectively, are their paths. The direction of motion of A at a certain instant is the tangent a, at the point A, to
LINK MECHANISMS
the path m. Similarly, the tangent b to the path n shows the direction of motion of B at the same instant. Let the lines e and / be drawn through A and B perpendicular, respectively, to the tangents a and b, and let O be their inter- section. Suppose that some point be chosen, say E, on the line e, and that the figure be rotated about this point. For the sake of clearness, imagine the figure to be a disk of paper with a pin stuck through it at the point E. Evidently, when the disk is rotated, the direction of motion of A will be perpendicular to AE, that is, in the direction of the tan- gent a. Hence, a rotation about any point on the line e will cause A to move, for the in- stant, in the direction of the tangent a.
In the same way, a rota- tion about any point in the line / will cause B to move in the direction of the tan- gent b. Therefore, by choosing as the center of rotation the point <9, which lies on both e and /, both A and B will be caused to move for the instant in their F*«- 2
proper directions. But, in plane motion, the motions of two points of a body determine the motion of the whole body. Hence, the actual motion for the instant is a rotation about the point (9, which for this reason is called the instan- taneous center of the figure. Whatever may be the motion of a figure in a plane, it is possible to find a point about which a rotation will, for an instant, give the figure the same motion.
Having the center 0, it is easy to find the direction of motion of any third point, as C. Since, for the instant, the whole figure is rotating about 0, the point C is moving in a direction perpendicular to CO. In general, at a given instant, any point of the figure is moving in a direction per- Dendicular to the line joining it to the center O.
6
LINK MECHANISMS
9. The instantaneous center can always be found if the motions of two points are known. Through the two points lines are drawn perpendicular to the directions in which the points are moving, and the intersection of these two lines is the instantaneous center. In Fig. 3, for example, there are
two links a, b that turn about the fixed points A and B and are pinned at C and D to a body c. The center of the pin D is a point of c as well as of #, and it is moving in a direction at right angles to the link a; hence, a line drawn from D at right angles to this direction is the center line of the link. Likewise, the link b stands at right angles to the direction of motion of C and therefore the intersection P of the center lines of the links, extended, is the -instantaneous center for the motion of the body c. To find the direction of motion of any other point of c, as £, join E to P; then the direction of motion of E is perpendicular to PE.
In Fig. 4, the motion of the connecting-rod B C is consid- ered. All points in the crank A B describe circles about the fixed point A. The end B of the connecting-rod B C rotates about the fixed point A, while the end C moves to and fro in the straight line A X. Evi- dently, the perpendicular to the direction of motion of B is the crank A B, and the perpendicular to the motion of C is a line perpendicular to AX at C; therefore the instantaneous
FIG. 4
LINK MECHANISMS 7
center P lies at the intersection of the center line of the crank A £, extended, and a perpendicular to A X at the point C.
10. If two points A, B, of a body, Fig. 5, are moving- in the same direction, the perpendiculars m and n are parallel. In this case, the body has a motion of translation, and all points of it are moving in the same direction with the same speed.
11. Angular Telocity.
A body rotating about a fixed axis is said to have unit angular velocity when the distance moved through in 1 second by any point is equal to the distance between the point and the axis. Hence, if the point chosen is 1 foot from the axis, the angular velocity of the body is unity when this point travels 1 foot in 1 second, that is, when its speed is 1 foot per second; if the point has a speed of 5 feet per second, the angular velocity is 5; and so on.
Suppose that a point at a distance of r feet from the axis has a speed of v feet per second. Then, as the speeds are proportional to the radii, a point at a distance of 1 foot from the axis has a speed of 1 x v = ^ feet per second, and the
angular velocity is therefore -_ units. Let w denote the angu- lar velocity; then,
*> = - (1)
r
and v - rw
Formula 1 may be used to determine the angular velocity of a body rotating about an instantaneous, instead of a fixed, axis, or, what is the same thing, the angular velocity of a point of a plane figure rotating about an instantaneous center. Thus, in Fig. 3, suppose that the velocity of the point D is 20 feet per second and the distance from D to the
LINK MECHANISMS
instantaneous center P is 4 feet; then the angular velocity of the body c about the center of rotation P is, by formula 1-
v 20 -
^ _____ 5
r 4
Having the angular velocity of the body, the linear velocity of any other point, as Cor E, is easily found by formula 2;
thus,
linear velocity of C — angular velocity X P C linear velocity of E — angular velocity X PE
In the foregoing consideration of angular velocity, the unit is based on the angle subtended by an arc equal to the radius of the circle forming the path of the point in motion; this angle is called a radian. In other words, a radian is the angle subtended by an arc equal in length to its radius.
The length of the circumference of a circle, that is, the arc subtending the angle of one complete revolution, or 360°, is 2 nr. This angle, measured in radians, is there-
fore equal to = 2rc radians. It follows, therefore, that
r
one radian is equal to ~- = 57.296°.
12.
F<
RELATIVE MOTION
Two bodies, each of which is moving relative to some fixed body, have, in general, rela- tive motion. For example, the crank and the connecting-rod of an engine have each a certain motion relative to the frame; each has also a motion relative to the other, which motion is a turning about the axis of a crankpin. An illustration of the principle of relative motion is shown in Fig. 6. A block b is constrained to slide on a rod a, and this rod is pinned at A to a fixed body, so that the only possible motion of a is a rotation about A. Now, it a does not move, the block b simply slides along a; if,
FIG. 6
LINK MECHANISMS 9
however, a rotates while b slides along it, the actual motion of b will have the direction P G, but its motion relative to a is in the direction P E, just as though a were at rest. That is, the motion of b relative to a is not in the least affected by the motion of a. In general, the relative motion of two bodies is not affected by any motion they may have in common. This is a principle of great importance, and may be further illustrated by the following familiar examples: The relative motions of the parts of a marine engine are not influenced by the. rolling of the ship. The relative motions of the moving parts of a locomotive are not affected by the motion of the locomotive on the track.
From this principle, it follows that in studying the relative motions of two bodies, any motion common to both may be neglected; also, if desirable, a common motion may be given them without affecting their relative motions.
LEVERS
13. Use of Levers. — Levers are used in mechanisms to guide a moving point, as the end of a moving rod, or to transfer motion from one line to another. There are three kinds of levers: (1) Levers whose lines of motion are parallel; (2) levers whose lines of motion intersect; and (3) levers having arms whose center lines do not lie in the same plane.
In proportioning levers, the following points should in general be observed; they apply to all three cases just mentioned:
1. When in mid-position, the center lines of the arms should be perpendicular to the lines along which they give or take their motions, so that the lever will vibrate equally each way.
2. If a vibrating link is connected to the lever, its point of attachment should be so located as to move equally on each side of the center line of motion of the link.
3. The lengths of the lever arms must be proportional to the distances through which they are to vibrate.
I L T 374B— 2
10
LINK MECHANISMS
14. Reversing Levers.— An example of a lever that illustrates the foregoing principles is shown in Fig. 7. The crank SR is the driver, and through the connecting-rod RH gives a motion to the point H approximately along the center line A JB, which is transferred by the lever E H to the line CD. The lever vibrates equally each way about its fulcrum or center 0, as indicated by the lines cb and da. When in mid-position, its center line EH is perpendicular to the lines of motion CD and A B. The horizontal distances traversed by the points E and H, respectively, are propor- tional to the arms E 0 and H O, or y : E 0 = x : HO. The
j>
I ^r I
|_ as M
FIG. 7
vibrating link EK connects the point E with the rod KD, which is constrained to move in a straight line by the o-uide g> and, in accordance with principle 2, the lever is so proportioned that the point E will be as far above the center line of motion CD, when in mid-position, as it will be below it in the two extreme positions; that is, the points c and d are as far below the line as the point E is above it.
At the bottom of the lever, where the rod HR connects with the crank, the same principle holds, the point H being as far below the line A B as the points a and b are above it. Frequently, the distance between the center lines CD and A B is given, and the extent of the motion along these lines,
LINK MECHANISMS
11
from which to proportion the lever. A correct solution to this problem is troublesome by calculation, because it is not known at the start how far above and below their respective lines of motion the points E and H should be.
A graphic solution is shown in Fig. 8. Draw the center lines of motion CD and A B> and a center line S T perpen- dicular to them. Draw ME parallel to ST at a distance from it equal to •§• y, or half the stroke along CD; also, the parallel line HN, on the other side of 5 71, and at a distance from it equal to i x, or half the stroke along A B. Connect points Mand. N by a straight line; where this line inter- sects 6* T, as at Os will be the center or ful- crum of the lever. With 0 as a center, find by trial the radius of an arc that will cut ST as far below the line AB as it does H N above this line, or so that the distance Pie- 8
n will be equal to the distance m. As an aid in determining the correct radius, describe an arc cutting S T, with O as a center and a radius ON. The distance n will be a little more than i /. Now, draw a straight line through points H and O. The part included between H N and ME determines the length of the lever. In this case, the length of the shorter arm is equal to 0 E, and that of the longer arm to OH.
15. Noii- Re versing Levers.— Fig. 9 shows the same construction applied to a lever in which the center O is at
12
LINK MECHANISMS
one end of the lever. This lever does not reverse the motion like the previous one, since, when the motion along A B is to the right or left, the motion along CD will be in the same
direction. The figure is lettered like the preceding one, so that the construc- tion will be easily under- stood.
16. Reducing Mo- tions.— It is often desir- able that a lever mechanism shall reproduce on a greater or smaller scale, along one line, the exact motion that occurs along another line; that is, for every change in the rate of the motion along one line a corresponding change shall be produced along the other line. Figs. 10 and 11 illustrate three indi- cator reducing motions that accomplish this.
In Fig. 10, the lower end of the lever attaches to the cross- head of the engine through the swinging link H R. The indi- cator string is fastened to the bar CD, which receives its mo- tion from the lever through the link EK, and slides through the guides g,g in a direction par- allel to the line of motion A B of the crosshead. In order that the bar CD shall have the same kind of motion as the cross- head, it is necessary that the lengths of the links EK and FlG'10
HR shall be proportional to their respective lever arms; thus OH: HR = OE\ Elf The pins must be so placed
LINK MECHANISMS
13
that the connecting; links will be parallel; if parallel at one point of the stroke, they will be so at all points. When the links JSKand HJt are parallel, OK\ OR = O£:OH,and the length of the indicator diagram will bear the same ratio to the length of the stroke as O E bears to OH.
It is to be observed that the pins O, K, and R are in one straight line, and, in general, it may be said that any arrange- ment of the lever that will keep these three pins in a straight line for all points of the stroke will be a correct one.
In Fig. 11, two such arrangements are shown. In the first, the pins K and R are fast to the slide and crosshead,
FIG. 11
respectively, and slide in slots in the lever. In the second, they are fast to the lever, the slots being in the slide and crosshead. In both, the pins K and R are in a straight line with the pin O during the whole stroke.
17. Bell-Crank Levers.— Levers whose lines of motion intersect are termed bell-crank levers; they are used very extensively in machine construction for changing the direc- tion and the amount of motion. The method of" laying out a lever of this kind to suit a given condition is as follows:
In Fig. 12, suppose the angle CAB, made by the lines of motion, to be given, and that the motion along A B is to be
14
LINK MECHANISMS
twice that along A C. Draw c d parallel to A C at any con- venient distance from A C. Draw a b parallel to A B and at a distance from A B equal to twice the distance otcd from A C.
FIG. 12
Through the intersection of these two lines and the apex A of the angle, draw the line A F. The center (9 of the bell- crank may be taken at any point on A F suited to the design
FIG. 13
of the machine. Having chosen point (9, draw the perpen- diculars C^and OH, which will be the center lines of the lever arms.
•LINK MECHANISMS
15
In Fig. 13, a construction is shown that may be employed when the two lines CD and A B do not intersect within the limits of the drawing. In Fig. 14, the same construction is
PIG. 14
applied to a non-reversing lever, in which the center O falls outside of the lines A B and CD. The figures are lettered alike, and the following explanation applies to both. Draw cd parallel to CD and a b parallel to A B, as before, so that the distance of cd from CD : distance of a b from A B = amount of mo- tion along CD : amount of motion along A B. Again, draw lines gh and ef in exactly the same way, but taking care to get their distances from CD and A B different from those of the lines just drawn. Thus, if FIG. 15
cd should be 6 inches from CD, make gh some other dis- tance, as 4 inches or 8 inches, and then draw <-/ at a propor- tionate distance from A B. Through the intersections of a b
16 LINK MECHANISMS
with cd and ef with ght draw the line IF, which will be the line of centers for the fulcrum 0.
Levers falling tinder the third classification are usually bell-crank levers, with their arms separated by a long hub, so as to lie in different planes. They introduce no new principle. Fig-. 15 shows the general construction of a lever of this kind.
18. Blow-Motion Mechanism. — A mechanism consist- ing of two connected levers, or of a crank and lever, is sometimes used to secure slow motion in one of the levers. Such an arrangement is shown in Fig. 16 where two levers,
PIG. 16
a and b, are arranged to turn on fixed centers and are con- nected by the rod r. The lever a is actuated by the handle h secured to the same shaft. If h and the lever a be turned counter-clockwise, the lever b will turn clockwise, but with decreasing velocity, which will become zero when the lever a and the rod r lie in the same line. Any further motion of a will cause b to return toward its first position, its motion being slow at first and then faster.
To obtain the greatest advantage, the lever b should be so placed that it will occupy a position perpendicular to the link r at the instant when a and r are in line. To lay out the motion, therefore, supposing the positions of the centers and lengths of the levers to be known, describe the arc B A
LINK MECHANISMS 17
about the center 0, with a radius equal to the length of the lever b. Through C, the center of the lever a, draw the line MN tangent to the arc just drawn; r and a must then be in line along the line MN, and the lever b must be perpendicu- lar to this line when in position b,. Generally, there will be a certain required amount of movement for the lever b. To secure this, draw b in its extreme left-hand position; then with C as a center and a radius equal to the length of a, draw the arc ED. Set the compasses to the length AD, and from B as a center draw an arc cutting the arc ED & E. CE will be the second position of the levers. In this combination, if the parts were proportioned to allow a to rotate like a crank, r and a would come into line twice during each revolution.
This mechanism has been used to operate platen printing presses, in which oscillation of the handle k moves the platen to and from the type through the lever b. It is also used to operate the exhaust valves in Corliss engines.
THE STEAM-ENGINE MECHANISM
19. Relative Positions of Crank and Crosshead.
ihe frame, connecting-rod, crosshead, and crank of a steam engine form what is called a four-link slider-crank mechanism. This mechanism is shown in diagrammatic form m Fig. 17, with the frame at «, the crank at b the connecting-rod at ,, and the crosshead at D. It is cus'tom- ary to speak of these parts as links, the frame a being termed the fixed link, since it is stationary, and the motions of the other links are considered with respect to the fixed link. The crank rotates about the point O with a uniform motion, and the crankpin E describes the crankpin circle The crosshead moves to and fro in a straight line, and the connecting-rod has a rather complex motion, since one end moves m a circle, while the other reciprocates in a straight me. The object of this mechanism is to transform Ihe tp-and-fro motion of the piston and crosshead into con- tinuous rotary motion of the crank and shaft, and thereby
18
LINK MECHANISMS
enable the energy of the steam to be utilized in doing useful work.
In the study of the steam-engine mechanism, there are two points to be investigated: (1) the position of the cross- head, or piston, in the case of the actual engine, for any crank position; and (2) the velocity of the crosshead or piston at any position of the crank. In Fig. 17, let m be the crankpin circle. From the dead centers F and G, with a radius equal to the length DE of the connecting-rod, draw arcs at H and K. Then H and K are the extreme
FIG. 17
positions of the crosshead pin. To find the position of the crosshead for any crank position, as, for example, OL, take L as a center, and with a radius equal to D E draw an arc cutting H K &\. IV. Then -N is the position of the crosshead when the crank is at 0 L. To find the position of the crank for a given crosshead position, as R, take R as a center, and with D E as a radius draw an arc cutting the crankpin circle at S and T. Then, when the crosshead is at R the crank may be at OS or at O T.
20. Relative Velocities of Crosshead and Crank- pin. — The crankpin moves with a practically uniform
LINK MECHANISMS 19
velocity, while the crosshead has a variable velocity Start- ing from rest, at the beginning of the stroke, the velocity of the crosshead increases to a maximum, and then decreases to zero at the end of the stroke before starting on the return When the velocity of the crankpin is known, the velocity of the crosshead can be easily determined for any crank position. In Fig. 17, produce the center line OE of the crank indefinitely, and at D erect a line perpendicular to HO Ihis perpendicular will intersect the center line of the crank at a point V, which, according to Art. 9, is the instantaneous center for the connecting-rod D E, since the points D and E are rotating, for the instant, about U. For the instant, then the Imear velocities of D and E are proportional to their distances from U, since the linear velocities of points in a rotating body are proportional to their radii. Hence,
E __ UE
_ __ the linear velocity of D ~ ~U~D
^ Extend the center line of the connecting-rod until it intersects the vertical line OL at V. By geometry, it may be proved that the triangle VOE is similar to the tri- angle D UE, and that their corresponding sides are pro- portional. That is,
O E _ UE _ the linear velocity of E O V UD the linear velocity of D Therefore, the velocity of the crankpin is to the -velocity of the crosshead as the length of the crank is to the intercept that the connecting-rod, or the connecting-rod produced, cuts off on the perpendicular through the center of the main bearing.
21. Velocity Diagrams for Crosshead.— The con- clusion reached in the last article may be used in constructing a diagram that will represent the velocities of the crosshead at different points of its travel. Such a diagram is shown by the curve 1-2' -3', etc., Fig. 18.
It is assumed that the crankpin of an engine moves with a constant velocity, and the length AB of the crank may be assumed to represent, to some scale, this velocity Then by placing the center of the crosshead pin in each of the
20
LINK MECHANISMS
positions 1, 2, 3, etc. and extending the center lines of the corresponding connecting-rod positions until they intersect the vertical line A C, the points 1", 2", 3", etc. are located. At the position 1, the velocity of the crosshead is zero, and at the position 2 its velocity is represented by the distance A 2". This distance is laid off vertically above 2, making 2-2' = A 2". Similarly, the distances 3-3f, 4-4', etc. are equal, respectively, to A3", A 4", etc., and represent the velocities of the crosshead at the points 3, 4, etc. By locating from eight to twelve crosshead positions, enough points will be obtained to permit a curve to be drawn through them, and this curve l-2f-3'-4', etc. is the velocity diagram of the crosshead. The height of any point on this curve above the horizontal line 1-10 gives the velocity of the crosshead, and
321
PIG. 18
hence of the piston also, for the corresponding position of the crank, to the same scale that the length of the crank represents the linear velocity of the crankpin.
Instead of laying off the velocities vertically above 1-10, they may be laid off on the corresponding crank positions. That is, in the position A D of the crank, the velocity A E of the crosshead is laid off equal to A-5"; on AF, the velocity A G is laid off equal to A-2"; and so on for each crank position. By taking several crank positions, a number of points, A,G,E, etc., are obtained, through which a smooth curve m may be drawn. This curve is then the dia- gram of crosshead velocities for the forward stroke.
To determine the crosshead velocity at any position of the crank, simply draw the crank in the desired position and
LINK MECHANISMS 21
note the distance from the center A to the point where the crank-line intersects the curve m. For example, if the crank is at AH \ AK represents the corresponding crosshead velocity. If, now, the crosshead velocities be determined for the positions 10 to 1 on the return stroke, they will be tound to be equal to the corresponding velocities on the for- ward stroke. Hence, if the curve n is plotted in the same manner as «, but below the center line, as shown, it will represent the velocity diagram for the return stroke More- over the two curves m and * will be exactly alike, and will be symmetrical with respect to the center line The curve l-4>-7'-io may also be used to find the crosshead velocity at any point on either stroke.
22. Crank and Slotted Crosshead.-If the connect- ing-rod in Fig. 17 be increased to a very great length, an arc drawn through L will be nearly a straight line coinci- ding with L O, and the horizontal movement of the crankpin will therefore, be practically the same as that of the cross- head. If the connecting-rod were increased to an infinite tength the two movements would be exactly the same -rig. 19 shows the crank and slotted crosshead mechanism by which this is ac- complished. Con- sider the crank o b as
the driver. The crank- pin b is a working fit in the block r, which is arranged to slide in PIG. 19
the slotted link /. The rods / and k are rigidly attached to the link, and are compelled to move in a straight line by the guides g,g>. As the crank rotates, the rods / and h are given a horizontal motion exactly equal to the horizontal motion of the pin b. If the erank rotates uniformly, the motion of the sliding rods is said to be harmonic, and the mechanism itself is often called the Harmonic-motion mechanism
22
LINK MECHANISMS
Harmonic motion may be defined as the motion of the foot of a perpendicular let fall on the diameter of a circle from a point moving with uniform velocity along the circumference.
23. The Toggle joint. — The togglejoint, shown in Fig. 20, is a. mechanism for producing a heavy pressure by the application of a small effort. It will be seen that it resembles the steam-engine mechanism with crank and con- necting-rod of about the same length. The effort F is applied at the joint B and the resistance P at the slide. The mechanical advantage, that is, the ratio of resistance to effort, depends on the angle between the links. It is evident that the point Q is the instantaneous center of
ri- _ 1H' o motion of the link
\—-~* A g Now> ag the
angle ABQ becomes smaller, that is, as the links approach the position in which they would form a straight horizontal line, the center Q moves toward the joint A, and the velocity of A relative to that of B grows less.
Since Q is the instantaneous center and QA is per- pendicular to the direction of the force /*, the product PX QA is the moment of P about Q. The moment of F about Q is FX. QC. Considering the mechanism to be, for the instant, in equilibrium, and neglecting friction, PX QA = FX QC, or
(1)
FIG . 20
This formula may be used to calculate the pressure P when the force acts vertically at B.
If the force acts at some other point than B, as, for example, F' at E, it becomes necessary to find the equiva- lent pressure at the joint B. Assuming that F' and F are parallel, and taking moments about O, F' X O D = F'X O G,
LINK MECHANISMS 23
or p = pt j TW OD •
OG' ' ~0~G = ~O& SmCe
and 0^ are similar. Hence, F= F> X ||. Substituting this value of fin formula 1,
OBY.QA
In case the force acts other than vertically, it is necessary to find its vertical component, which, when substituted for ForF' m the proper formula, will give the value of />. The distances QC and QA will vary according to the posi- tion of the toggle. However, by laying out the mechanism accurately to a fairly large scale, these distances may be measured with sufficient accuracy, so that the pressure /may be calculated for any. position.
*AM«JC i.-m Fig. 20, let ^equal 80 pounds; Q C, 24 inches- and , 5 inches; find the pressure P. . mcnes, and
SOLUTION.— Applying formula 1,
24 P = 80 X ~ = 384 Ib. Ans.
-ceX^isL322'~lQ a t0gglejoint like that shown in Pig. 20, the
d QA, 6 inches; what is the pressure P?' ' ' ln° SS>
SoLimoN.-Applying formula 2 and substituting the values given,
~ 19 v «" = 400 lb. Ans.
EXAMPLES FOR PRACTICE
In Fig. 20, if ^^ is 8 inches and Q C is 36 inches, what force be applied vertically at B to produce a pressure P of 180 pounds?
Ans. 40 Ib.
2 In a togglejoint similar to that in Fig. 20, the arms OB, AH and BQ are equal, each being 14 inches in length; the height B G* 4 inches; and the force F is 2,800 pounds. Find the
CQ equais
Ans. 4,695.74 Ib QC> 24 inches! ^^- 16
Ans. 462 Ib.
24
LINK MECHANISMS
4. In case a force F of 200 pounds, Fig. 20, should act at right angles to OB at JB, find the pressure P when Q C equals 32 inches; QA, 8 inches; and OQ, 60 inches, remembering that the angle that P makes with the vertical is equal to thp ingle Q O A.
Ans. 793 lb., nearly
5. A point revolves about an axis at a speed of 4,200 feet per minute; if the point is 5 feet from the axis, what is its angular velocity in radians per second? Ans. 14 radians per sec.
6. The angular velocity of a point is 25 radians per second and its distance from the axis of revolution is 8 feet; what is its linear velocity? •
Ans. 200 ft. per sec.
7. A flywheel having an outside diameter of 24 feet makes 56 revo- lutions per minute; find the angular velocity of the wheel, in radians per second. Ans- 5-86 radians per sec.
QUICK-RETURN MOTIONS
24. Vibrating-Link Motion. — Quick-return motions are used in shapers, Blotters, and other machines,
where all the useful
J JJ— LL s - — ~™ work is done during
the stroke of a recip- rocating piece in one direction. During the working stroke, the tool must move at a suitable cutting speed, while on the return stroke, when no work is performed, it is de- sirable that it should travel more rapidly. The mechanism shown in Fig. 21, known as a vibrating-link mo- tion, is applied to shaping machines operating on metal. Motion is received FlGl21 from the pinion pt
which drives the gear g. The pin b is fast to the gear, and
LINK MECHANISMS
25
pivoted to it is the block /, which is fitted to the slot of the oscillating link CD. As the gear rotates, the pin describes the circle bedc, the block slides in the slot of the link CD and causes CD to oscillate about the point D, as indicated by the arc CC', the path of the joint C The rod / connects the upper end of the link with the .tool slide, or ram, r, which is constrained by guides (not shown) to reciprocate in a straight horizontal line.
During the cutting stroke, the pin b travels over the arc deb, or around the greater arc included between the points of tangency of the center lines CD and CD During the return stroke, the pin passes over the shorter arc bed and as the gear f rotates with a uniform velocity, the time, of the forward and return strokes will be to each other as tha length of the arc deb is to the length of the arc bed The throw of the slotted link and the travel of the tool can be varied by the screw s, which moves the block / to and from the center of the gear. The rod /, instead of vibrating equally above and below a center line of motion is so arranged that the force moving the ram during the cutting stroke will always be downwards, causing it to rest firmly on the guides. y
25. To lay out the motion, proceed as follows: Draw he center line S T, Fig. 22, and parallel to it the line mn, he distance between the two being equal to one-half the ongest stroke of the tool. About 0, which is assumed to be the center of the gear, describe the circle bdc with a radius equal to the distance from the center of the pin b to the center of the gear #, Fig. 21, when set for the longest stroke. Divide the circumference of the circle at b and d into upper and lower arcs extending equally on each side of the center linear, and such that their ratio is equal to that of the times of the forward and return strokes. In this case the time of the forward stroke is double that of the return stroke and the circle is divided into three equal parts, as shown at 6,d, and c, thus making the arc db equal to one-half the arc deb. Draw the radial lines Ob and O d. Through b
I L T 374B-3
LINK MECHANISMS
Z? where it inter- point Cohere it
,tt c . the upper end of the slotted .eve. ™h C draw the horizontal line CO, making C'-E equal to CA. Draw C'A which will be tangent to the orde at rf. thus the other extreme position of the lever.
FIG. 22
on the
of intersection on me ate c c
IL
the line
,.
h oyn intersections so that they will
ine R L and numoei LU »-
the
LINK MECHANISMS
27
point 2 to 3, etc., on the forward stroke. On the return stroke, from point 9 to 10, 11, 12, and 1, the motion is much less uniform.
A property of this motion is that, as the radius Ob is diminished to shorten the stroke, the return becomes less rapid, as can be seen from the figure by comparing the motion when Ob is the radius with the motion when Ob' is the radius.
26. Whitworth Quick-Return Motion.— This mech- anism is shown in principle in Fig-. 23. The pin d, inserted in the side of the gear /, gives motion to the slotted
8
PIG. 23
link CD, as in the vibrating link motion. This motion closely resembles the previous one, the difference being that the center D of the slotted link lies within the circle described by the pin 6, while in the previous case it lies without it To accomplish this result, a pin^ is provided for the gear to turn on, and is made large enough to include another pin D placed eccentrically within it, which acts as the center for the link CD. With this arrangement, the slotted link, instead of oscillating, follows the crankpin during the com- plete revolution, and thus becomes a crank. The stroke line RL passes through the center D, which is below the center O of the pin p. The forward or working stroke occurs
LINK MECHANISMS
28
-
pin describes the shorter
that it will divide
n r fr P IIILU uvvw t*iv<vj, Ui c g A
times of the forward 'and return strokes. The point D where Us line cuts the center line ST is the position for the center of the slotted crank. The motion is plotted as m F> 24 Divide the crankpin circle dcge into a number o equal parts. From the center of the slotted crank, draw radtaTlines through these points to the outer circle, which represents the path of pin C, Fig. 23, using the latter point, of
8
13
FIG. 24
intersection as centers, and, with a radius equal to the length- of the connecting-rod, strike off points on the stroke line which will show the movement of the tool for equal
amounts of rotation of the driving gear.
Fie 25 shows the mechanism as practically constructed. The gear .g is driven with a uniform velocity in the direction of the arrow by the pinion h. It rotates on the large pin A which is a part of the frame of the machine, and carries the pin b which turns in the block k. This block is capable of sliding in a radial slot in the piece /, as is shown m the sectional view. This piece / is supported by the shaft d, which turns in a bearing extending throu-gh the lower part
LINK MECHANISMS 29
of the large pin. The line R L drawn through the center of d is the line of motion of the tool slide. The connecting-rod that actuates the tool slide is pinned to the stud t, which in turn is clamped to the piece /.
28. The radial slot t, in Fig. 25, holds the stud c, to which the connecting-rod is attached; the slot provides for the adjustment of the length of the stroke, and if the point
PIG. 25
of Attachment of the rod to the tool slide is also made adjustable, the position of the stroke, as well as its length, can be changed. Thus, in the case of a shaping machine, it is not only desirable to regulate the distance passed over by the tool, but to have the stroke extend exactly to a certain point. These two adjustments are often required in mechanisms where reciprocating pieces are employed.
STRAIGHT-LINE MOTIONS
29. Parallel Motion.— A parallel motion, more properlv called a straight-line motion, is a link mecha- nism designed to guide a reciprocating part, as a piston rod in a straight line. In the early days of the steam engine,' parallel motions were extensively used to guide the pump and piston rods; but they are now seldom met with, except on steam-engine indicators, where they are employed to give a straight-line motion to the pencil. Very few parallel
v:
6095
30
motions produce an absolutely straight line, and it is custom- ary to design them so that the middle and two extreme posi- tions of the guided point will be in line.
30. Tlie Watt Parallel Motion.— The best-known motion is the one shown in Fig. 26, which was invented by James Watt in 1784. The links AB and CD turn on fixed centers A and D. The other ends B, C are connected
by the link CB, which has the point 0 so chosen that it will pass through three points 0,, O, 0, in the straight line 55 perpendicular to the links CD and A B when in their middle positions. When the point 0 is at the upper extremity of its motion at Olt the linkage assumes the position AB, C^D; at the lower extremity it assumes the position A B, C,D.
Having given the length 0, 03 of the stroke, the middle position 0 of the guided point, the center of one lever A, and the perpendicular distance between the levers when in
LINK MECHANISMS
31
mid-position, the motion may be laid out as follows: Let SSbe the path of the guided point, 0 its middle position, A the given center, and A B and CD indefinite parallel lines, representing: the middle positions of the levers. From m, where AB intersects 6" 5, lay off on 55 the distance ma, equal to one-fourth of the stroke. Join A with a and draw an indefinite line aB, perpendicular to A a. The point £}, where a B intersects line A B, is the right-hand extremity of lever AB, and the lower extremity of the link CJB. The point C is obtained by drawing an indefinite line through B and O; this intersects the line CD in the required point. To find the center D, lay off nb equal to one-fourth stroke; connect Cand t, and from b draw an indefinite line perpendic- ular to Cb\ the center will be at its intersection with CD.
If the positions of both centers should be known, mark points a and b as before. Draw A a and bD, and through b and a draw perpendiculars to these lines; the points B and C, where they intersect the center lines of the levers, are the extremities of these levers. Join B and C by the link B C, and the point Ot where the center line of this link cuts the line of motion 55, is the position of the guided point O on the link B C.
UNIVERSAL, JOINT
31. Hooke's coupling, or the universal joint, shown in Fig. 27, is used to connect two shafts, the center lines of
PIG. 27
which are in the same plane, but which make an angle with each other. It is generally constructed in the following manner: Forks /, / are fastened to the ends of the shafts «
32
LINK MECHANISMS
and 5, and have bosses c,c tapped out to receive the screws s,s. The ends of these screws are cylindrical, and are a working fit in corresponding beatings in the ring r. The details of construction may be seen in the right-hand part of this figure. In heavy machinery, the forks are forged and welded to the shafts. The ring r furnishes an example of spherical motion, and the point of intersection of the two axes is the center. The motion transmitted by the universal joint is not uniform; that is, if the driving shaft turns uni- formly, the driven shaft will make the same number of revolutions, but it will have alternately greater and less angular velocities at different points of a single revolution, and there are only four positions of the joint in one revolu- tion in which the two shafts have the same speed.
32. The maximum and minimum speeds may be found as follows: In Figs. 28 and 29, let m be the driving shaft and n the driven shaft, the center lines of both shafts being in the plane of the drawing; also, let the center lines of the two shafts intersect at O, and denote by e the acute angle between them. In Fig. 28, the fork of the shaft n is in the plane of the two shafts, and in this position it is evident that the
PIG. 28
plane of the ring connecting the two forks is perpendicular to the shaft n, for in this position both arms S S' and R R' are perpendicular to n.
From formula 2, Art. 11, v = rw; that is, the linear velocity of a rotating point equals the radius times its angular velocity. Let v denote the linear velocity of the point S, and wn the angular velocity of the shaft n\ then, since ,5*
LINK MECHANISMS 33
is at a distance OS from the shaft n, the linear velocity equals OS times the angular velocity of n, or v = OS X w Now, when the ring: is in this position, it is moving tor the instant, as if it were rigidly attached to shaft m and hence J> may be considered for this instant as a part of m Prom S, therefore, let the perpendicular 5 T be dropped on the shaft m, and since the velocity of 5 is v, the relation of the Imear velocity to the angular velocity wm of the shaft m is v - S Tx wm. But, as the linear velocity v of the point ^ is the same in both cases,
OS X wn = STX wm< or wn = wm V ~
OS But in the triangle O S T, ||= cos O S T, and the angle
OST keeps changing as the shafts revolve. Let x denote this variable angle OST. Then, = cos *, and
wn — wm X cos x (1) When the fork on m is in the plane of the shafts, as in A the same course of reasoning will apply. Hence,
FIG. 29
in this position, v = OR x wm and v = R T X w There fore, RTXWn = O R x wm, or
1
wn =
x
= Wm x
cos^r
These two formulas give the extremes of angular velocity of the driven shaft, formula 1 being for the minimum, and tormula A for the maximum, angular velocity of n
34
LINK MECHANISMS
33. From the foregoing- formulas, it is apparent rhat when the shaft m rotates at a uniform speed, the shaft n is given a variable speed, which is alternately less and greater than the speed of the driving shaft. However, there are four positions in each revolution at which the angular velocities of the driving and driven shafts are equal, and these are the positions in which the distances of the points 6" or R from the two axes are exactly equal. In the positions shown in Figs. 28 and 29, the angles e and x are equal, and the driven shaft n has its least and greatest angular velocities, respect- ively. If the driving shaft rotates uniformly, the least speed of the driven shaft is equal to the speed of the driving shaft multiplied by the cosine of the angle between the two axes, and the greatest speed is equal to the speed of the driver 1
multiplied by
Thus, if the shafts revolve at the rate
cos x
of 100 revolutions per minute, and the angle e between them is 30°, the least speed of the driven shaft will momentarily be 100 X cos 30° = 100 X .866 = 86.6 revolutions per min- ute, and the greatest speed, — = — — - = 115.47 revolu- tions per minute.
34. To obviate the variation in angular velocity, the double universal joint is used, as shown in Fig. 30. Let
FIG. 30
a and b be the two shafts to be connected. Draw their center lines, intersecting at 0 and bisect the angle A OB by
LINK MECHANISMS 35
the line O C. The center line MN of the connecting shaft d must now be drawn perpendicular to 0 C. Care must be taken that the forks on the intermediate shaft He in the same plane Thus constructed, a uniform motion of a will result in a uniform motion of the driven shaft b, though the inter- mediate shaft will have a variable motion. Let x = angle between a and d = angle between d and 6, and denote the three angular velocities by wa, wd, and w6. Then, with the forks as shown in the figure, --* = cos x, and ^ = 1 Multiplying, Wa ' Wd cosx'
"Wd .. Wh 1 „„
-X- = «»*X— , or^ = 1, whence «,, = «,.
It can be proved that this law holds good for all positions.
//, therefore, the forks on the intermediate shaft are in the same plane, and the intermediate shaft makes the same angle with the driving and driven shafts, the angular velocity of the driven shaft is equal to that of the driving shaft.
This arrangement is often employed to connect parallel shafts, as would be the case in Fig-. 30 if OB took the direc- tion 5 T. In such a case it makes no difference what the angle x is, except that, if the joints are expected to wear well, it should not be too great. If the forks of the inter- mediate link d are at right angles, the variation of motion of the shaft b is greater than in the single coupling. In this case the greatest and least angular velocities of B would be,' respectively, wa X -~- and wa X cos2*
COS X
Serial 991
GEARING
TOOTHED GEARING
Edition 1
ROLLING CURVES AND SURFACES 1. Direct-Contact Transmission.-^ the various ways of transmitting motion from one machine Ja to another that by direct contact is the most common Thlt is, a surface of one of the parts is in contact with a similar surface of another part, and the motion of the first ca™ es
the motion of the second. In Fig. 1 for examole th,
parts a and b are in con- example, the
tact at P and the rotation
of a in the direction shown
by the arrow will cause a
rotation of b in the same
direction.
2. Condition of
Rolling.— The relative motion of the two bodies in contact may be: (a] sliding motion; (b} roll- ing motion; (,) combined sliding and rolling motion. When the point of contact P is not situated on the line joining the
": 2r7r°n M^ * there wm be Slidin^ but <*£
as n Fig. 2, /Mies in the same line with M and N, the bodies will roll on each other without sliding. The condition for rolling, therefore, is that the point of contact shall lie in the line of centers.
PIG. 1
COPYR.GHTED BY .NTERNAT.ONAL TEXTBOOK COMPANY,
ALL RIGHTS RESERVED
2 GEARING
3. Natt-re ol Boiling Curves.-In order that two curves may rotate about fixed points and roll together, the known as rolling curves, must be .pecully con- to ive the
Bthese are not closed curves; that is, it is impossible to start from any point on the curve and, by following its outline in one direc- tion, return to the start- ing point. Hence, the curve CPE, Fig. 2, can- not be used to transmit motion continuously to the curve £>/>^when revolving in one direction, but it may reciprocate through a partial revolution, and the^two curves will remain in contact. On the other band, a pair of closed curves, as the circles shown in Fig. 3, will roll in contac and may turn continuously in one direction. Two equal ellipses Fio-. 4, will roll together if the distance between the axes of 'rotation is equal to the length of the major axis.
FIG. 2
FIG. 3
PIG. 4
4. Angular Velocity Ratio.— In Fig. 1, let a line n be drawn through P at right angles to the common tangent t to the two curves. A line like this, perpendicular to the tan- gent, is called the normal to the curve at the given point,
GEARING S
Let Q denote the poipi in which this normal n cuts the line through the centers M and N; then it can be proved, by a rather difficult process, that the segments M Q and N Q are to each other inversely as the angular velocities of a and b;
that is,
angular velocity of b _ M_Q /-j\ angular velocity of a N Q
This is a general rule that applies to all cases of transmis- sion by direct contact.
In Figs. 2, 3, and 4, the normal must cut the line M N in the point P itself; hence, for cases of pure rolling,
angular velocity of b __ M P /o\ angular velocity of a N P
When the radii M P and N P remain constant throughout the entire revolution, as in the ease of circular gears, and the motion of the gears is uniform, the angular velocities may be measured in revolutions per minute.
Let Na — angular velocity of a, in revolutions per minute;
and Nt = angular velocity of b, in revolutions per minute.
Then formula 2 becomes
Nj, = M_P (3)
Na ' NP
When P lies between M and N, as in Figs. 2, 3, and 4, the curves a and b rotate in opposite directions. If Q or P lies outside of M N, as in Figs. 1 and 5, they rotate in the same direction.
When the rolling curves are circles, as in Figs. 3 and 5, the point of contact P will always be at the same distance from the centers M and N; but for any other forms of curves, the position of P will change. It follows that the velocity F*- 5
ratio, that is, the fraction — ?, is constant for rolling circles,
Na
but is variable in other cases. Thus, in Fig. 4, suppose a
GEARING
to rotate clockwise about M with a constant number of rev- olutions per minute. As the rotation proceeds, the point P
moves toward N, the ratio — — increases, and in consequence
the angular velocity of b increases and attains its maximum
value when P is nearest to IV.
5. Boiling Cones. — While the bodies represented in
Figs. 1 to 5 are shown as plane figures rotating about fixed points, they are, in reality, representa- tions of solids with definite thicknesses rotating- about parallel axes. Cases of parallel axes of rotation are the most 0 common, but frequently non-parallel axes must also be used. Let OM and ON, Fig. 6, be two axes of rotation in- tersecting in the point 0; if on these axes are constructed two frustums of cones, a and If, respectively, these cones will. roll together and one will transmit motion to the other. Assuming that there is no slipping, the velocity ratio may
be found in the same manner as for rolling cylinders. At E,
for example, the circles forming the bases of the frustums
Nb EA
are in contact, and the velocity ratio is
EB'
since the
angular velocities are inversely proportional to the radii. Similarly, at any other point, as F, the velocity ratio is — -
J. V a
FC_ FD'
But since triangle O CFis similar to triangle OAE,
and triangle ODF is similar to triangle QBE, EA : FC = OE: OF, and EB\FD = 0 E : OF, so that /?/4 : FC ~ E B : FD; from which, by transposing, E A : E B
FC
= FC'.FD.
which means that, for a
pair of cones running in contact without slipping, the velocity ratio is constant and is equal to the inverse ratio of the corresponding radii at any point.
GEARING
SPUR GEARING
GENERAL PRINCIPLES
6. Positive Driving.— When motion is transmitted by friction gearing, that is, by cylinders or cones that roll in contact, there is more or less slipping of the contact surfaces; hence, when it is necessary to transmit motion with a definite and unvarying velocity ratio, friction wheels cannot be used and some means must be employed to insure positive dri- ving. In the case of rolling wheels or cylinders, the circum- ferences may be provided with projections a, a, Fig. 7, and
PIG. 7
with corresponding grooves b, bt so that the projections of one wheel fit into the grooves of the other. Evidently the same number of these projections pass the line of centers in a given time on wheel B as on wheel A, so that a given number of revolutions of the driver A causes a definite number of revolutions of the follower B.
7. Condition of Constant Telocity Ratio.— The pro- jections on the rolling surfaces insure that the number of revolutions of the follower per minute or per hour is the same that would be obtained by rolling without slipping, but
1 L T 374B— 4
GEARING
the angular velocity from projection to projection will not be constant unless the bounding curves of the projections, or teeth, as they are called in gear-work, have certain forms. The shape of the tooth depends on the principle stated in Art. 4, namely, that when the driver and follower have direct contact, the angular velocities are inversely as the segments into which the line of centers is cut by the common normal at the point of contact.
In Fig. 8, a pair of teeth m and n are attached to the parts a and b and are in contact at C. The common normal NN at C cuts the line of centers A B in the point P\ then if Va and Vb denote the angular velocities of a and b, respectively,
Va = BJP
Vb AP
In order, therefore, that this ratio of the velocities may be strictly constant, the point P must always lie in the same position on the line of centers. Hence the following law:
Law.—- />z order that toothed wheels may have a constant -velocity ratio, the common normal to the tooth curves must .always pass through a fixed point on the line of centers.
8. Pitch. Surfaces and Pitch lanes. — In Fig. 8 let the circles e and /, with centers A and B, be the outlines of
two rolling cylinders in contact at P. Let the tooth curves m and n be so constructed that their common normal shall always pass through P. The veloc- ity ratio produced by the rolling of these cylinders is precisely the same as that caused by the teeth m and n\ for in the case of rolling cylinders the velocities are inversely proportional to the radii, that is,
V, Nb AP
GEARING 7
The bounding surfaces e and / are called pitch surfaces, and the lines e and / are called pitch lines or pitch curves, or, in the case of circular gears, pitch circles.
The point of tangency P of the pitch lines is called the pitch point, and it must lie on the line of centers. This is the meaning of the term as generally used by designers, and is the meaning intended wherever the term pitch point is mentioned in this treatise on gearing. In the machine shop, however, this term is frequently used in a different sense, being there considered as any point in which the tooth out- line intersects the pitch circle, as indicated in Fig. 9.
In the case of circular pitch^lines, the pitch point P lies in a fixed position; but if the pitch lines are non-circular, as in the case of rolling ellipses, Fig. 4, the pitch point moves along the line of centers. In order to include non- circular wheels, the law of Art. 7 may be made general, as follows:
Law. — In order that the motion prod^lced by tooth driving shall be equivalent to the rolling of two pitch surfaces, the common normal to the tooth curves must at all times pass through the pitch point.
The object, then, in designing the teeth of gear-wheels is to so shape them that the motion transmitted will be exactly the same as with a corresponding pair of wheels or cylinders without teeth, which run in contact without slipping. In actual work, two general systems of gear-teeth are used. The one is known as the involute system and the other as the cycloidal system, both of which will be discussed in succeeding paragraphs.
9. Definitions. — Referring to Fig. 9, which shows part of a circular gear-wheel, the following definitions apply to the lines and parts of the tooth.
The circle drawn through the outer ends of the teeth is called the addendum circle; that drawn at the bottoms of the spaces is called the root circle. In the case of non- circular gears, these would be called, respectively, the addendum line and the root line.
GEARING
The addendum is the distance between the pitch circle and the addendum circle, measured along a radial line. The root is the distance between the pitch circle and the root circle, measured along a radial line. The term addendum is also frequently applied to that portion of a tooth lying between the pitch and addendum circles, and the term root to that portion of the tooth lying between the pitch and root circles.
FIG. 9
The working surface of the addendum, that is, the part of the working surface outside of the pitch circle, is called the face of the tooth. The working surface of the root is called the flank of the tooth.
The diameter of the pitch circle is called the pitch diam- eter. When the word diameter is applied to gears, it is always understood to mean the pitch diameter unless other- wise specially stated— as outside diameter, or diameter at the root.
The distance from a point on one tooth to a corresponding- point on the next tooth, measured along- the pitch circle, is the circular pitch.
The fillet is a curve of small radius joining- the flank of the tooth with the root circle, thus avoiding the weakening effect of a sharp corner.
THE INVOLUTE SYSTEM '
10. Production of the Involute Curve. — In general, the involute of any curve may be defined as the curve that is described by a point in a cord as it is unwound from the original curve, keeping the unwound portion of the cord straight. Thus, suppose that a cord is wound around the curve a, Fig. 10, and let P be any point on the cord. Then, as the cord is unwound, the point P will describe a curve m that is an involute of the curve position of the cord shown.
In the case of an involute to a circle, it is convenient to conceive the curve described as follows: Suppose a, Fig. 11,
FIG. 10
Pi A being the last
FIG. 11
to be a circular pulley having a cord wound around it, and let the pulley be pinned at a fixed point 0. At the end P cf the cord a pencil is attached, and there is a fixed groove
10
GEARING
or guide e tangent to the pulley at P. Suppose also that the pulley has attached to it a sheet of paper or cardboard c. Now take hold of the pencil and pull it along the groove e. The pulley and paper will thus be caused to rotate about O, and the pencil will trace on the moving paper the involute Pi Pt, or m. If now the pulley is turned backwards, so that the string is wound up, the pencil will move from P, to P and will retrace the curve PsPi on the moving paper, the
point Pl moving to P, its original position,
Let t be a tangent to the curve m at the point P,. From the manner in which the curve is pro- duced, it is evident that the tangent is perpendic- ular to P P*. Similarly, the tangent tt at any other point <2i is perpendicular to Qi Q, and in general any tangent to the cir- cle a is a normal to the involute m.
11. Involute Tooth Curves. — Let 0 and 0', Fig. 12, be the centers of two cylinders that are a short distance apart, and D E a cord that has been wound several times around them in opposite directions. If the cylinder D Df be turned in the direction of the full arrow. the cord D E will cause the cylinder E E' to turn in the oppO' site direction, as shown by the full arrow, and points on the cord D E will describe portions of the involute curve. In order to better comprehend this, imagine a piece of paper to be attached to the bottom of each cylinder, as shown, and that the width of each piece is the same as the distance between
12
GEARING
11
the cylinders. Now, suppose that a pencil is attached to the cord at E, the point of tangency of the line DE with the cylin- der EE', in such a manner that it can trace a line on the piece of paper attached to the cylinder EEf if a proper motion be given to the cord D E. Turn the cylinder D D' in the direc- tion of the full arrow, that is, rotate it counter-clockwise. The point E will travel toward the cylinder D D' in the straight line ED, and will gradually move away from the cylinder EEf. During this movement, the pencil attached at E will trace on the piece of paper the involute curve m. In the same manner, if the pencil be attached at D and the cylin- der EE' be rotated in the direction of the dotted arrow, the involute mf will be traced on the piece of paper attached to the cylinder D D' . Suppose that those parts of the pieces of paper to the right of the curve m and to the left of the curve m' be lemoved, and that E E' be rotated until curve m takes the position ed\ also, that D D' be rotated until m1 takes the position gh, the two curves being in contact at k on the line D E. The cord D E, being tangent to the circle D D', is normal to the involute gh, and being also tangent to the circle EE', it is normal to the involute de. This would be true for any positions in which the curves de and gh could be moved into contact by rotating the cylinders. Moreover, this line DE always passes through a fixed point C on the line of centers. Therefore, de and gh are two curves whose common normal at the point of contact always passes through one fixed point C so long as the curves remain in contact. If, now, the pitch circles A A' and B B' be drawn through this point C, the motion transmitted by the involutes de and gh used as the outlines of teeth will be precisely equivalent to that produced by the rolling of these circles without slipping.
12. Properties of Involute Teeth. — The circles D D' and EE', Fig. 12, to which DE is tangent, are called base circles. It will be noted as a characteristic of involute teeth that the whole of the face and the part of the flank between the pitch and base circles is a continuous curve.
12 GEARING
For this reason, involute gears are sometimes called single- crurve gears.
If the centers 0. and O', Fig. 12, be moved apart so that the pitch circles do not touch, the relative velocities of D D' and BE' evidently remain unchanged, since the cylinders are connected by the cord D E. The curves described by the point k are also the same, because they are still involutes of the same circles. From this, it follows that the distance between the centers of involute gears may be varied with- out disturbing the velocity ratio or the action of the teeth— a property peculiar to the involute system.
In Fig. 12 let TT be drawn through C at right angles to the line'of centers O 0'\ then TT is the common tangent to the pitch circles A A' and B B> . As already stated, the line ED is the common normal of the tooth curves, and is therefore the line of action of the pressure between the teeth. The angle between the common normal to the tooth surfaces and the common tangent to the pitch circles is called the angle of obliquity. In the case of involute teeth, this angle is constant.
13. Angle and Arc of Action. — The angle through which a wheel turns from, the time when one of its teeth comes in contact with a tooth of the other wheel until the point of contact has reached the line of centers is the angle of approach; the angle through which it turns from the instant the point of contact leaves the line of centers until the teeth are no longer in contact is the angle of recess. The sum of these two angles forms the angle of action. The arcs of the pitch circles that measure these angles are called the arcs of approach, recess, and action, respectively.
In order that one pair of teeth shall be in contact until the next pair begin to act, the arc of action must be at least equal to the circular pitch.
The path of contact is the line described by the point of contact of two engaging teeth. In the case of involute gears the path of contact is part of the common tangent to the base
GEARING
13
circles. The arc of action depends on the addendum, or what amounts to the same thing, on the length of the tooth. With short teeth, the arc of action must necessarily be small; and if a long arc is desired, the teeth must be made long.
14. Standard Interchangeable Gears. — In order that two gears may run properly together, two conditions must be satisfied: (1) They must have the same circular pitch, and (2) they must have the same obliquity. If, therefore, all involute gears were made of the same obliquity, any pair of wheels having teeth of the same pitch would work properly together, and such gears would be said to be interchangeable.
The tooth selected for the standard is one having an angle of obliquity of 15°; that is, in Fig. 12, angle T C E — angle CO' E = 15°. With this obliquity, then, in the triangle 0' EC, 0' E = O'C cos CO1 E = O1 C cos 15° = .966 O> C; that is, the radius of the base circle equals .966 times the radius of the pitch circle. The distance between the base circle and pitch circle is thus about one-sixtieth of the pitch diameter.
In the interchangeable series of standard gears, the small- est number of teeth that a gear may have is twelve, for with a smaller number the arc of contact will be smaller than the cir- cular pitch, in which case one pair of teeth will separate before the next pair comes into contact, and the gears will not run.
15. The Involute Hack.
teeth described on a straight pitch line. It is usually a metal bar, in which the teeth are cut, al- though they may be cast. A rack, there- fore, may be consid- ered as a portion of „ the circumference of a gear-wheel whose radius is infinitely long, and whose sequentlv be regarded as straight.
-A rack is a series of gear-
E
FIG. IS
pitch line may con-
14 GEARING
In the involute rack, Fig. 13, the sides of the teeth are straight lines making an angle of 90° - 15° = 75° with the pitch line EF. Thus, on the contact side, the tooth outlines are perpendicular to the line of action N ' N. To avoid inter- ference, the ends of the teeth should be rounded to run with the 12-tooth pinion. A pinion is a small gear meshing with a rack or with a larger gear.
16. Involute Internal Gears. — An annular, or internal, gear is one having teeth cut on the inside of the rim. The pitch circles of an annular gear and its pinion have internal contact, as shown in Fig. 5.
FIG. 14
The construction of an annular gear with involute teeth is shown in Fig. 14. The obliquity of 15° is shown by the angle TCN, and the base circles EE' and D D1 are drawn tangent to the line of action NN't with 0 and O', respect- ively, as centers. The addendum circle for the internal gear should be drawn through F, the intersection of the path of contact NN' with the perpendicular OF drawn from the center of the pinion. The teeth will then be nearly or quite without faces, and the teeth of the pinion, to correspond,
GEARING
15
may be without flanks. If the two wheels are nearly of the same size, points c and d will interfere; this interference may be avoided by rounding the corners of the teeth.
. 15
THE CYCLOIDAL OR ROLLED-CURVE SYSTEM
17. Tlie Cycloid. — The name cycloid is given to the curve traced by a point on the circumference of a circle as it rolls on a straight line. Thus, in Fig. 15 the circle m rolls on the line A B, and the point P on the cir- cumference traces the path A CB, which is a cycloid. The roll- ing circle is called the generating cir- cle, the point P the tracing point, and the line AB the base line.
If the base line AB is an arc of a circle, the curve traced by the point P is called an epicycloid when the generating circle rolls on the convex side, and a hypocycloid when it rolls on the concave side. Fig. 16 shows an epicycloid, and
Fig. 17 a hypocycloid. The one property of cycloidal curves that makes them peculiarly suitable for tooth outlines is that the common normal through the point of contact will always pass through the pitch point. In Figs. 15, 16, and 17, P is the tracing point in each case and E is the point of contact of the generating circle and the line on which it rolls. Now, it is clear that the generating circle m is for the instant turning about E as a center, and the pQfflt P is therefore moving in a
Pm'16
16
GEARING
direction at right angles to PE. In other words, the tangent to the curve at P is perpendicular to PE, or PE is normal
to the curve.
A particular form of the hypocycloid is i* shown in Fig. 18. The generating circle m has a diameter equal to the radius of the circle n. In this case, a point P of m describes a straight line ACB, which passes through the cen- ter C of n and is therefore a diameter.
18. Generation of Tooth Outlines. — In Fig. 19 let a and «! be two pitch circles in contact at E and suppose a third circle m to be in contact with them at E. If the circle m rolls on the outside of the
circle a, the tracing point P describes the epicycloid If, however, it rolls on the inside of circle al, P describes the hypocycloid C P d. Now let an arc of the curve B P B± be taken as the outline of the face of a tooth on a, and let an arc of the hypocycloid CPd be taken as the outline of the flank of a tooth on a-,. These curves are in contact at P, and, more- over, PE is normal to each of them. Hence, the common normal through the point of contact passes through the pitch point, and the curves satisfy the condition required of tooth outlines.
The one rolling circle m generates the faces of the teeth of a and the flanks of the teeth of a*. A second circle mit rolling on the inside of a and on the outside of a^ may be used to generate the faces of gear-teeth for a^ and the flanks for a. While not necessary, it is customary that the generating cir- cles shall have the same diameter. If a series of gears of the same pitch have their tooth outlines all generated by rolling
FIG. 18
GEARING
17
circles of the same diameter, any gear will run with any other gear of the series; that is, the gears are interchangeable.
PIG. 19
19. Size of Generating Circle. — In Figs. 20 to 22 is shown the effect of different sizes of generating circles on the flanks of
FIG. 20 FIG. 21 FIG. 22
the teeth. In the first, the generating circle has a diameter
18 GEARING
equal to the radius of the pitch circle, the hypocycloid is a straight line, and the flanks described are radial. In the second, with a smaller circle, the flanks curve away from the radius, giving a strong tooth, and in the third, with a larger circle, the flanks curve inwards, giving a weak tooth, and one difficult to cut. It would seem, therefore, that a suitable diameter for the generating circle would be one-half the pitch diameter of the smallest wheel of the set, or one-half the diameter of a 12-tooth pinion, which, by com- mon consent, is taken as the smallest wheel of any set.
It has been found, however, that a circle of five-eighths the diameter of the pitch circle will give flanks nearly parallel, so that teeth described with this circle can be cut with a milling cutter. For this reason, some gear-cutters are made to cut teeth based on a generating circle of five- eighths the diameter of a 12-tooth pinion, or one-half the diameter of a 15-tooth pinion.
It is more common practice to take the diameter of the generating circle equal to one-half the diameter of a 12- tooth pinion rather than one-half the diameter of a 15-tooth pinion; this size, therefore, is taken in this discussion of the subject of gearing.
20. Obliquity of Action. — Neglecting friction, the pressure between two gear-teeth always has the direction of the common normal. In the case of involute teeth, this direction is always the same, being the line of action ED, Fig. 12. With cycloidal teeth, however, the direction ot the normal is constantly changing, and hence the direction of pressure between the teeth is likewise variable. In Fig. 19, let the tooth curves be just coming into contact at the point P. The direction of the pressure between them at that instant is PE, since P ' E is the common normal at P. As the gears rotate, however, the point of contact P approaches E, moving along the arc PPi E, and the common normal PE, therefore, makes a greater angle with O^ 0, until, when P reaches E, the common normal is at right angles to 0, 0.
GEARING
19
Beyond the point E, the point of contact follows the arc EQ until the teeth leave contact, and the direction of the common tangent gradually changes from a position at right angles to Ol 0 to the position EQ. Hence, with cycloidal teeth, the obliquity of the tooth pressure is greatest when the teeth first come in contact, decreases to zero at the pitch point, and increases again to a maximum at the final point of contact. As a rule, the greatest obliquity should not exceed 30°.
-In Fig. 23 is shown a rack
FIG. 23
21. Rack and Wheel.
and 12-tooth pinion. Tooth outlines for the rack in the cycloidal system are obtained by rolling the gener- ating circles oo' on the straight pitch line B B; the curves are therefore cycloids. The generating cir- cle o passes through the center O of the pinion and rolls on the pitch circle A A, and therefore describes radial flanks.
22. Epicycloidal Annular Gears.— Annular gears, or internal gears, as already explained, are those having teeth cut on the inside of the rim. The width of space of an internal gear is the same as the width of tooth of a spur gear. Two generating circles are used, as before, and if they are of equal diameter, the gear will interchange with spur wheels for which the same generating circles are used.
In Fig. 24 is shown an internal gear with pitch circle A A, inside of which is the pinion with pitch circle B B. The generating circle o, rolling inside of B B, will describe the flanks of the teeth for the pinion, and rolling inside of A A, the faces of the teeth, for the annular wheel. Similarly, the corresponding faces and flanks will be described by o'. The
20
GEARING
only special rule to be observed in regard to epicycloidal internal gears is that the difference between the diameters
FIG. 24
of the pitch circles must be at least as great as the sum of
the diameters of the generating circles.
This is illustrated by Fig. 25. a is the pitch circle of an in- ternal gear and b that of the pinion. Then, for correct ac- tion, the difference D — d of the diameters must be at least as great as c, the sum of the diam- eters of the generating circles. To take a limiting case, sup- pose a to have 36 teeth and b 24 teeth. A wheel with a diam- eter equal to D — d, as shown FlG- 25 dotted at <?, would, therefore,
have 36 — 24 = 12 teeth. In the 12-tooth interchangeable
system, this latter would be the smallest wheel of the series,
GEARING 21
and the generating circles would be half its diameter. From this, it follows that, if D-d, the diameter of <?, is not to be exceeded by the sum c of the diameters . of the genera- ting circles, b is the largest wheel that can be used with a. Hence, when the interchangeable system is used, the' num- ber of teeth in the two wheels must differ by at least the number in the smallest wheel of the set. If it is desired, for example, to have 18 and 24 teeth, and the gears are to interchange, generating circles of half the diameter of a 6-tooth pinion will be used, this being taken as the smallest wheel.
23. Formerly, the cycloidal system was used almost exclusively, but in later years the involute system has largely taken its place. The distinctive features of the involute system lie in the great strength of the tooth and in the fact that the distance between the centers of two meshing gears may be varied somewhat without affecting the uniformity of the velocity ratio. The chief objection that has been raised against involute teeth is the obliquity of action, which causes increased pressure on the bearings. If the obliquity does not exceed 15°, however, this objection is not serious.
PROPORTIONS OF GEAR-TEETH
24. Circular Pitch.— The circular pitch of a gear has been denned as the distance between corresponding points on adjacent teeth measured along the pitch circle. It is obtained by dividing the circumference of the pitch circle by the number of teeth in the gear. If the circular pitch is taken in even dimensions, as 1 inch, f inch, etc., the circum- ference of the pitch circle will also be expressed in even dimensions; but the diameter will be a dimension that can only be expressed by a fraction or a decimal. Thus, if a gear has 40 teeth and a 1-inch circular pitch, the pitch circle has a circumference of 40 inches and a diameter of 12.732 inches.
25. Diametral Pitch. — Ordinarily it is more conve- nient to have the pitch diameter a dimension easily measured,
I L T 374 B—S
22
GEARING
and for this reason a new pitch, called the diametral pitcli, has been devised. This pitch is not a measurement like the
circular pitch, but a ratio. It is the ratio of the num- ber of teeth in the gear to the number of inches in the diameter; or it is the number of teeth on the circumference of the gear for 1 inch of the pitch-circle diameter. It is obtained by dividing the number of teeth by the pitch diameter, in inches.
For example, take a gear that has 60 teeth and is 10 inches in diameter. The diametral pitch is the
ratio of 60 to 10 = -~
= 6, and the gear is called a 6-pitch gear. From the definition, it follows that teeth of any particular diametral pitch are of the same size, and have the same width on the pitch line, whatever the diameter of the gear. Thus if a 12-inch gear had 48 teeth, it would be a 4-pitch gear. A 24-inch gear, to have teeth of the same size, would have twice 48 or 96 teeth, and 96 -r- 24 = 4, the same diametral pitch as before.
Fig. 26 shows the sizes of teeth of various diametral pitches.
lOPitch
FIG. 20
GEARING
26. Relation of Diametral to Circular Pitch.— The relation between the diametral and circular pitches of a gear may be readily obtained.
Let p = diametral pitch;
A = circular pitch, in inches; n = number of teeth; d = diameter of pitch circle, in inches.
TABJLE I CIRCULAR AND DIAMETRAL PITCHES
|
Diametral Pitch |
Circular ' Pitch Inches |
Circular Pitch Inches |
Diametral Pitch |
|
2 1 |
I-57I |
2 |
I-57I |
|
21 1 |
1.396 |
. I* |
1.676 |
|
a |
1.257 |
if |
1-795 |
|
23 |
1.142 |
if |
1-933 |
|
3 |
1.047 |
I 2" |
2.094 |
|
« |
.898 |
irV |
2.185 |
|
4 |
.785 |
il |
2.285 |
|
5 |
.628 |
T 5 ITS" |
2-394 |
|
6 |
.524 |
ii |
2.513 |
|
7 |
•449 |
IT\- |
2.646 |
|
8 |
•393 |
ii |
2-793 |
|
9 |
•349 |
irV |
2-957 |
|
10 |
•314 |
i |
3-142 |
|
ii |
.286 |
JL5 1 6 |
3-351 |
|
12 |
.262 |
I 8 |
3-590 |
|
14 |
.224 |
Te |
3-867 |
|
16 |
.196 |
a 4 |
4.189 |
|
18 |
-175 |
Tg |
4-570 |
|
20 |
< C *7 |
5_ |
|
|
8 |
5-027 |
The length of the circumference of the pitch circle is TT d inches; hence, from the definition of circular pitch,
(1)
24
GEARING
From the definition of diametral pitch,
p = ± (2)
d
Multiplying these equations together, member by member,
~ =7i = 3.1416 d
(3)
Then,
A = £, and p =
A
(4)
that is, the product of. the numbers expressing" the two pitches is 7t, and when one pitch is known, the other may be found by dividing re by the known pitch.
EXAMPLE— If the circular pitch is 2 inches, the diametral pitch is 3. 1416 -r- 2 = 1.571, nearly. If the diametral pitch is 4, the circular pitch is 3.1416 -s- 4 = .7854 inch.
Table I gives in the first two columns values of the circular pitch corresponding to common values of the diametral pitch, and in the last two columns values of the diametral pitch corresponding to common circular-pitch values.
27. Backlash and Clearance.— Referring to Fig, 27, / and j denote, respectively, the thickness of tooth and width of space measured on the pitch line. Evidently, t -{• s - A>
the circular pitch, and s must be at least equal to /, in order that the tooth on the other gear may enter the space without binding. In practice, s is made somewhat greater than t, to provide for irregularities in the form or spacing of the teeth, and the difference between s and t is called the backlash.
The root r is also made greater than the addendum a, so that there is a clearance between the point of the tooth of one wheel and the bottom of the space of the engaging wheel. This space is shown at c, Fig. 27 and is called
GEARING 25
bottom clearance, or simply clearance. When the wheels have equal addenda and equal roots, as in the interchange- able system, then c = r — a.
The sum r + a is the whole depth of the tooth and the working depth is e = 2 a, double the addendum.
28. Tooth Proportions for Cut Gears, Using Diam- etral Pitch. — Cut gears of small and medium size, such as are used on machine tools, are. invariably based on the diametral pitch system. For gears of this character the following proportions are used:
p = diametral pitch = - d
a = addendum = - inches P
9 • 1 r = root = — — inches
8^
17
a + r = whole depth = -— inches
8fi
1 f?rr t = thickness of tooth = =^- inches
P
c = clearance = — inches »'/
EXAMPLE.— The diametral pitch of a cut gear is 4; what are the tooth dimensions?
SOLUTION— Addendum = i = \ in.; root = •£- = j- in.- clear- P 4 Q p 32 '
ance = 87 = 32 in'; whole dePthof tooth = j^ = j|j in.; thickness of
tooth = ^ = .39+ in. Ans. P
29. Tooth Proportions for Gears Using Circular
Pitch.— With gears of large size, and usually with cast gears of all sizes, the circular-pitch system is used. In these cases, it is usual to make the addendum, whole depth, and thickness of tooth conform to arbitrary rules based on the circular pitch. However, none of these rules can be con- sidered absolute. Machine-molded gears require less clear- ance and backlash than hand-molded, and very large gears should have less, proportionately, than smaller ones. Table II
26
GEARING
gives the proportions that have been used successfully and will serve as an aid in deciding on suitable dimensions. Column 1 is for ordinary cast gears and column 2 is for very large gears having cut teeth. Circular pitch = pt.
TABLE II PROPORTIONS FOR GEAR-TEETH
|
i |
2 |
|
|
Addendum |
•3°A |
•3% |
|
Root . |
•40/>i |
•3S/i |
|
Whole depth |
•?oA |
• 65A |
|
Thickness of tooth |
•48A |
•495 A |
|
Width of space |
• 52 A |
•5°5A |
30. Geai*-Blanks. — A gear-blank is a disk of metal along the circumference of which gear-teeth are to be cut. The blank is turned in a lathe, and its diameter is made equal to the outside diameter of the desired gear. In making a gear-blank, therefore, the outside diameter rather than the pitch diameter is required.
Let D = outside diameter;
d = pitch diameter; a = addendum.
Then for any gear,
(1)
n 1
But, in the case of diametral pitch, d — - and a = -.
P P
Substituting these values,
n
D =
P
(2)
which may be used to find the outside diameter when the number of teeth and the diametral pitch are known.
EXAMPLE 1. — A wheel is to have 48 teeth, 6 pitch; to what diameter must the blank be turned?
SOLUTION.— By formula 2, D
— - — = — - — = 8.33 in. Ans. p h
GEARING 27
EXAMPLE 2.— A gear-blank measures 10i inches in diameter and is to be cut 4-pitch; how many teeth should the gear-cutter be set to space?
SOLUTION.— From formula 2, D = n + 2, orn = £>Xp-2 = 10| X 4 - 2 = 42 - 2 = 40 teeth. Ans. P
EXAMPLES FOR PRACTICE
1. (a) How many teeth has a 2^-pitch gear, 4 feet in diameter? (d) What is the circular pitch of this gear? A / (a) 120 teeth
AnM(£) 1.257 in.
2. What is the outside diameter of a gear-blank from which a wheel is to be cut having 50 teeth 4-pitch? Ans. 13 in.
3. The pitch diameter of a gear is 25 inches; what is its outside diameter, supposing it to be 6-pitch? Ans. 25.333 in.
4. A gear-blank measures 10.2 inches in diameter and is to be cut 10-pitch; how many teeth should the gear-cutter be set to space?
Ans. 100 teeth
CONSTRUCTION OF TOOTH PROFILES
31. Approximate Methods. — The exact construction of the outlines of gear-teeth involves a great amount of work, and hence is seldom employed except where great accuracy is required. In laying out the patterns of cast gears, especially if the pitch is small, it is not necessary to employ exact constructions, since the cast teeth will depart somewhat from the true form, however carefully the pattern may have been made. The part of the curve, whether involute, epicycloid, or hypocyclpid, used as the tooth outline is so short that for practical purposes one or more circular arcs closely approxi- mating the true curve may be used for the profile.
Many systems of drawing tooth profiles by means of arcs of circles have been proposed, among which may be mentioned especially the odontograph and the odontograph table. The odoiitograph. is a templet for drawing tooth outlines. Its form is generally determined by a set of radii that give a curve that closely approximates the correct tooth form. Tables of radii that give tooth outlines that approxi- mate very closely to the correct forms are also used, and these are called odontograph tables.
TABLE III
INVOLUTE ODONTOGRAPH TABLE FOR STANDARD INTERCHANGEABLE GEARS
|
Number of |
Divide by the Diametral Pitch |
Multiply by the Circular Pitch |
||
|
Teeth |
||||
|
Face Radiu |
Plank Radius |
3 Face Radiu |
Flank Radius |
|
|
10 |
2.28 |
.69 |
•73 |
.22 |
|
ii |
2.40 |
.83 |
•76 |
.27 |
|
12 |
2.51 |
.96 |
.80 |
|
|
13 |
2.62 |
•1.09 |
•83 |
•34 |
|
14 IS 16 |
2.72 2.82 2.92 |
1.22 1-34 1.46 |
' .87 .90 •93 |
•39 •43 •47 |
|
17 18 |
3.02 3-12 |
1.58 1.69 |
.96 .99 |
•So •54 |
|
19 |
3-22 |
1.79 |
1-03 |
• 57 |
|
20 |
3-32 |
1.89 |
1. 06 |
.60 |
|
21 |
3.41 |
1.98 |
1.09 |
•63 |
|
22 |
3-49 |
2.06 |
I. II |
.66 |
|
23 24 |
3-57 3-64 |
2.15 2. '2 4 |
1.13 1.16 |
.69 •7i |
|
26 |
3 7i 3.78 |
2.33 2.42 |
1.18 1.20 |
•74 77 |
|
27 o |
3.85 |
2.50 |
1.23 |
•/ / .80 |
|
28 |
3-92 |
2.59 |
1.25 |
.82 |
|
29 30 |
3-99 4.06 |
2.67 2.76 |
1.27 1.29 |
•85 .88 |
|
31 |
4-13 |
2.85 |
I.3I |
.91 |
|
32 |
4.20 |
2.93 |
i-34 |
•93 |
|
33 34 35 36 |
4.27 4-33 4-39 4-45 |
3-01 3-09 3.16 3-23 |
1.36 1.38 i-39 1.41 |
.96 •99 1. 01 1.03 |
|
37-40 |
4.20 |
4 |
||
|
41-45 |
4.63 |
T . Q |
||
|
46-51 52-60 61-70 71-90 91-120 |
S-o6 5-74 6.52 7-72 9-78 |
I .48 1.61 1.83 2.07 2.46 3T T |
||
|
121-180 181-360 ~ — |
13-38 21.62 |
.1 I 4.26 6.88 |
28
GEARING
29
32. The Involute Odontograpli Table.— Table III is an involute odontograph table that gives a series of radii for constructing arcs of circles for approximate tooth out^ lines. The centers of these arcs all lie on the base circle. For gears with from 10 to 36 teeth, the tooth profile consists of two circular arcs of different radii, one for the face of the tooth and one for the part of the profile between the pitch circle and base circle. For gears with more than 36 teeth, a single arc is used for the curve between base and addendum circles. The flanks from the base line to the bottoms of the
A
PIG. 28
spaces are straight radial lines, but at the bottom fillets should be put in to avoid sharp corners.
The construction of tooth outlines by the odontograph table is illustrated in Fig. 28. Suppose that it is required to draw a gear with 15 teeth, 3 diametral pitch. Then, as previously
explained, pitch diameter = - = -^ = 5 inches; addendum
= - = -g- = .333 inch; root = P
9
8 x 3
= .375 inch; by
Art. 14, radius of base circle = .966 X 2.5 = 2.415 inches; circular pitch = -— - = 1.047 inches; and thickness of
u
tooth = ----- = .523 inch. With these data, the pitch, base,
222—3
30
GEARING
addendum, and root circles are readily drawn, as shown at H> K, M, and Nt Fig. 28.
Let P be some pitch point. Referring to the odontograph table, the numbers in the columns for the face and flank radii are, respectively, 2.82 inches and 1.34 inches; but these must be divided by the diametral pitch 3, giving .94 inch and .45 inch, respectively, as the radii. Take points O and O, on the base circle, so that OP = .94 inch and O,P = .45 inch. With 0 as a center, strike the arc PA, and with <9t as a center, strike the arc PB. From B draw the radial line 3C and finish the outline with the fillet*. The radius of this fillet differs in practice. Some authorities recommend a radius equal to one-seventh the width of the space between two teeth, measured along the addendum circle, while others recommend a smaller radius. When the
form of the tooth is such that the root is liable to be weak, the radius should be made as large as possible. Knowing the circular pitch and thickness of tooth, the pitch points for other teeth may be laid off and the pro- files constructed in a similar manner.
The construction of the involute rack is shown in Fig. 29. The side of the tooth is a straight line A B inclined at an angle of 15° with a line at right angles to the pitch line L. To correct for interference, the points of the teeth are cut away as follows: From a point B midway between the pitch line L and the addendum line F, strike a circular arc B C from a center O on the pitch line with a radius 2.10
FIG. 29
OB =
P
inches, or .67 A inch.
33. The Cycloidal Odontograph Table. — For the
cycloidal system of interchangeable gears Table IV gives the radii of circular arcs for faces and flanks, and also the
M 3
|
O s |
0) o d J3 |
^0 SSRS^c^^S^^S^ |
|
|
3 CJ |
•a |
S |
M Ol CO M " |
|
g O 03 |
fa |
(/) •r-l |
|
|
12 |
1 |
||
|
Numbe |
8 a |
^ *"* W CVJ CO ^f T^ 1O IO '^O t^h 00 O"\ O M OOOOOQOOOOOOOMM |
|
|
CD en 03 EH |
03 O ci |
3 |
|
|
*4-J 3 |
fa |
.1 |
N ro-3-u-itxoo o N ^r^o ONroo co-^r |
|
a |
* |
||
|
o £ |
03 O a fl |
§ o ^^ o {^ ^^ {^^ ^ ^oo ^ |
|
|
"ctf 03 |
^ |
3 |
TT 'O c^ ro (» M M |
|
cS 5 |
fa |
en .H |
8£ S^^^aS^^^^-g^ |
|
•4-1 |
aJ tf |
I M M 1 1 |
|
|
umbers |
03 O a cd |
||
|
03 en 03 |
03 O |
3 |
|
|
03 |
" |
en .3 |
O\ O i-f "^"O TfO^O fOOCO O rooi^O |
|
5 |
a) |
M!N<MCNNCNN(M<M<NiM04NCNN |
|
|
Number of Teeth |
M ^ O co |
32 GEARING
distances, from the pitch circle, of the circles on which the centers lie.
For the 12-tooth gear, as previously stated, the flanks are radial; hence, the infinite values in Table IV, denoted by °o .
The manner of using the odonto graph table is shown by tin example in Fig;. 30, which shows the layout of the teeth
FIG. 30
of a 16-tooth gear, 2 diametral pitch. For tnis gear, the fol- lowing- data are readily obtained: Pitch diameter = — = ir
P
— 8 inches; addendum = — = 5 inch; root = -— = A inch.
P %P
From these dimensions, the addendum, pitch, and root circles can be drawn. From Table IV, the distance for faces is .09 inch, and that for flanks is 3.46 inches, which divided by 2, the diametral pitch, gives, respectively, .045 inch and 1.73 inches; hence, the circle of face centers is drawn .045 inch inside the pitch circle, and the circle for flank centers 1.73 inches outside the pitch circle. The radii in Table, IV are 2,10 and 7.86, and these divided by 2 give 1.05 inches and 3.93 inches, respectively, for face radius and flank radius. Let P be chosen as a pitch point. Then with OP = 1.05 inches, locate the center 0 on the circle of face centers and describe the arc PA; and with 0, P = 3.93 inches locate Oi on the circle of flank centers and describe the arc P B. At the bottom of the space put in a fillet, as usual.
The name three-point odontograph is sometimes given to this system, from the fact that the arc constructed by it passes .through three points on the true cycloidal curve.
GEARING
33
BEVEL, GEARING
34. Relation of Bevel to Spur Gearing-. — It has been shown that motion may be transmitted by rolling cylinders when the axes are parallel, or by rolling cones when the axes intersect. If rolling cylinders are used as pitch surfaces and are provided with teeth, spur gears are the result; if frustums of rolling cones are used as pitch surfaces and provided with teeth, the gears so formed are called bevel gears. Bevel gears are used to transmit motion from one shaft to another making an angle with the first. In case the shafts are at right angles, and the gears have equal pitch diameters, they are 'known as miter gears.
35. Generation of Bevel-Gear Teeth. — In the gen- eration of cycloidal tooth outlines for spur gears, generating circles were caused to roll on pitch circles. The tooth sur- faces, however, are produced by the rolling of generating cylinders on pitch cylinders. The generation of tooth sur- faces for bevel gears is accomplished, in a sim- ilar manner, by the roll- ing of a generating cone in contact with the pitch cone.
In Fig. 31, let COB represent a pitch cone, the part CDEB being the pitch surface of a bevel gear, and let A OC be the generating cone. If the genera- ting cone is supposed to describe the tooth sur- face M N G P by rolling on the pitch cone, the line NG representing the outer edge of the tooth will lie on the surface of a sphere whose radius is ON. The point N that describes this line is always at a fixed distance from
FIG. 31
34
GEARING
the center 0; hence, every point in the line NG is equally distant from O, and, as all points in a spherical surface are equally distant from a point within called the center, it follows that NG must lie on a spherical surface. The curve N G, therefore, may be called a spherical epi- cycloid, and by rolling the generating cone on the concave surface of the pitch cone a spherical liypocycloid is obtained. A spherical involute may be obtained b> unwrapping an imaginary flexible sheet from a base cone lying within the pitch cone. A point on the edge of this
PIG. 32
sheet will describe a curve on the surface of a sphere, and this curve will be a spherical involute.
To be theoretically exact, therefore, the tooth curves for a bevel gear should be traced on the surface of a sphere, as shown in Fig. 32. This method is not a practical one, how- ever, and in practice is replaced by what is known as Tred- gold's approximation, which is much simpler and is universally used.
By this method, the tooth curves are drawn on cones tan- gent to the spheres at the pitch lines of the gears, as shown
GEARING
35
in Fig. 33. The process is simply to develop or unwrap the surfaces of the cones, the unwrapped surfaces being- rep- resented by ABC and CDE in the figure. The length of the arc A B C is equal to the length 'of the pitch circle A1 C, and the length of the arc CDE is equal to that of the pitch circle CE'. The gear-teeth are then drawn on the
FIG. 33
unwrapped surfaces, precisely as for spur gears of the same pitch and diameter.
The teeth, as laid out by Tredgold's method, will vary somewhat from the shape of the spherical teeth. But although the tooth curves may not be exactly the same as the curves on the sphere, the difference is so slight that it has no appreciable effect on the uniformity of the motion transmitted.
86
GEARING
GEARING
37
36. Laying Out Bevel Gears. — Let OA and OB,
Fig1. 34, be the axes of the pitch cones, intersecting at O, and assume O C to be the line of contact of the two cones. Then, if CD is the radius of one cone, CE is the radius of the other,
and the velocity ratio is the ratio CD : CE.
Through C a line is drawn at right angles to 0 C) cutting the axes in the points A and B. With A and B as cen- ters and A C and B C as radii, circular arcs CG and CH are drawn. Suppose, now, that CJis chosen as the length of the rolling conical frus- tum, or, what is the same thing, the length of the face of a tooth. Through J draw a line parallel to ABt cutting OA and OB in K and/,, respectively. Then, with A as a center, and A M = K J as a radius, draw the arc MR, and with B as a center, and B N = L J as a radius, draw the arc NS. The four arcs CG, CH, MR, and NS are then considered as pitch circles of spur gears, and the tooth outlines are drawn according to the methods already given. Let na — the number of teeth in the gear whose axis is OA; then, as in spur gearing, the diametral pitch for the outside diameter of the cone is
PIG. 35
p =
diameter
I L T 374B— 6
38
GEARING
From this value of p, the addendum, the root, and the radii for the circular arcs of the tooth outline are obtained.
Similarly, for the inside radius, p = ^77:; and from this
2 J Q
value of p, the teeth on the pitch circles MR and N S are constructed. In making wooden patterns for cast gears, the teeth may be laid out on strips of thin sheet metal, and these strips may then be fastened to the ends of the rolling- frustums. The spaces in the metal strips will form guides at the ends of the teeth, so that the wood in the spaces may be cut away accurately. In Fig. 35 is shown a section of the gears laid out in Fig. 34.
37. Bevel-Gear Angles. — In Fig. 36, O C is an element of the pitch cone, OF marks the top and O E the bottom of the working portion of a tooth. The angles «, b, c, e, f, andg are named as follows: a is the center angle; b, the face angle; c, the cutting angle; e, the cutting decrement, or angle
of bottom; /, the face increment, or angle of top; g, the angle of edge.
In turning up a gear-blank to size, it is necessary to know the outside diameter, face angle, and, also, the angle of edge The face angle b is found by subtracting FrG- 36 the sum of the angles
a and / from 90°. The angle of edge g is equal to the center angle a. The center angle is determined from the ratio between the outside pitch radii CD and 0 D of the two gears. Thus, CD and O D represent the outside pitch radii
CD
of the two gears, and tan a = —
But CD and OD are proportional to the number of teeth in their respective gears1,
GEARING 39
hence, the number of teeth may be substituted for the pitch radii. Having: the value of the tangent, the corresponding angle may easily be found from a table of natural tangents. The cutting angle c is found by subtracting the angle e from the angle a. The angles e and /may be obtained by trigo- nometry, as follows: Let r be the outside pitch radius CD,
Fio- 36 Then tan / - --— - v
- ' tan f ~ o c - CD x oc' CD
CD
and — - = sin a. Substituting these values,
tan/ = —-sin a (1)
r
Now, CF is the addendum, and when diametral pitch is
1 2 r
used, CF =- = --, where n equals the number of teeth. p n
Hence, for diametral pitch, formula 1 reduces to
n 2
tan / = — sin a = - sin a (2)
r • n
In terms of the circular pitch, the addendum CF = .3 A and consequently, when the circular pitch is used, formula 1 becomes
tan / = :M sin a (3)
r
Having determined the value of tan /, the corresponding angle may readily be found by reference to a table of natural tangents.
In cutting bevel gears, the clearance is made constant for the entire length of the tooth, and the cutting angle is there- fore taken at the working depth of the tooth, allowance for clearance being made on the cutter. Then C E = CF, and it follows that the angle e must equal the angle /.
The outside diameter is equal to 2 B D. But,
ZBD = 2(J5C + CD) (4)
The angle B CF = angle £• = angle a. Then — - = cos
C F
BCF = cos a, or B C = CF cos a. But, CF is the
40 GEARING
2 r
addendum which rs equal to — for diametral pitch, and .3/>l
n
2 r
for circular pitch. Hence, for diametral pitch, B C = — cos a\
n
and for circular pitch, B C = .8/1 cos a. The radius CD = r. Substituting these values, formula 4 becomes, for diametml pitch,
outside diameter = 2 ( — cos a +r\ (5)
and for circular pitch,
outside diameter = 2 (.3 p^ cos a + r) (6)
The following example will show how to calculate the various angles:
EXAMPLE.— ^-Suppose that two bevel gears have 70 teeth and 30 teeth, respectively, and that the diametral pitch at the outer ends of the teeth is 3; find the center, face, and cutting angles, the angles of top, bottom, and edge, and the outside diameter, for each gear.
SOLUTION. — Since the shafts are at right angles, use the above
CD formulas, referring also to Fig. 36. Tan'fl = —., = f{y = 2.33338, from
which « = 66° 48' for the larger gear. For the smaller gear, a = 1)0°
- 66° 48' = 23° 12'. Ans.
By formula 2, tan / = ,-% X sin 66° 48' = -& X .91914 = .02626; from which / = 1° 30'. Hence, e = 1° 30', also, and these angles are the same for both bevel gears. Ans.
For the large gear, a + f = 66° 48' -4- 1° 30' = 68° 18'. Hence, b for this gear is equal to 90° - 68° 18' = 21° 42'; also, c = a - e = 66° 48'
- 1° 30' = 65° 18', while g = a = 66° 48'. Ans.
For the small gear, a + f = 23° 12' + 1° 30' = 24° 42'. Hence, b for this gear is equal to 90° - 24° 42' = 65° 18'; also, c - a - e = 23° 12' - 1° 30' = 21° 42', and^' = a = 23° 12'. Ans.
/2 r \
For larger gear, using formula 5, outside diameter = 2 — cosa-f-r)
Now, r = Hj = | (^) = ^; n =* 70; cos a - cos 66° 48' = .39394.
(2 ^ 70 1 70\
-^ — X ™ X .39394 + ~r = 23.6 in 6 70 6/
nearly. Ans.
ln\
For the smaller gear, r = i - = £ (4£) = 5; n = 30; cos a = cos
\PI
23° 12' = .91914. Then, outside diameter = 2 l^~- X .91914 +
V oO <= 10.61 in. Ans.
41
SPIRAL AND WORM-GEARING
38. Spiral Gearing. — The name spiral gearing is
given to a system of gearing in which the wheels have
cylindrical pitch surfaces, as in
.spur gears, but in which the teeth
are not parallel to the axes. Each
tooth winds helically like a screw
thread, hence spiral gearing is
sometimes called screw gearing. In Fig. 37, suppose a and b to T
be two cylinders having axes A A A
and B B, respectively. The cylin- ders may be made to touch at a
single point P. Let a helix .y be
traced on the surface of b, passing
through P, and let TP T be the
tangent to the helix at P. Now,
through P, let a helical line be
traced on the cylinder a, having such an angle that its tan- gent is likewise T P T. If, now the helix on b be replaced by a projection similar to a screw thread, and that on a by a groove made to fit this projection, a rotation of b about the axis B B will cause a rotation of a about the axis A A. In order that the motion may be continuous, succes- sive grooves on a must be so spaced as to engage properly with the' thread on b.
•JB
FIG. 37
39. Relation of Spiral to Worm- Gearing. — The cylinder b may be pro- vided with only one thread or with several threads. Thus, in Fig. 37, the helix 6- may be used alone, or the helixes sltsa, and s3 may be added. If from one to three helixes are employed, the projections are called threads rather than teeth; the
FIG. 38
42
GEARING
cylinder and thread (or threads) together are called a worm and the other cylinder is called the worm-wheel. "
When each cylinder has several helixes, as shown i Fig. 38, the term tooth is used instead of thread; that is, i is said to be a spiral gear, having a certain number o teeth. It must be understood, however, that the numbe of teeth is the number of helixes wound around the cylindrics pitch surface, and that a worm with a single thread is i: reality a spiral gear with one tooth.
40. Pitches in Spiral Gearing. — The distance betwee corresponding points on two consecutive teeth, measure- around the cylindrical pitch surface at right angles to th
axis, is the circular pitch, a shown at A B, Fig. 39. This i also sometimes called the cir cumferential pitch. The dis tance measured at right angle -I — to the teeth, as A C, is called th normal pitch. The distanc measured parallel to the axis, a A D, is the axial pitch. Le these pitches be denoted by p^ p, and pa, respectively. If m denote the angle between the helix an a line on the pitch surface parallel to the axis, the m = B AC, and A C = A B cos BA C; that is,
pn = p^ cos m (1)
The number of teeth n on the gear is equal to the circurr ference of the pitch circle divided by the circular pitch; hence
2rrr %n r cos m /o\
n = — — = — — — (A)
Pi Pn
where r denotes the pitch radius.
In a pair of gears that run together, the normal pitche must be the same in both, but the circular pitches may c may not be the same. If the axes of the two gears are £ right angles, the circular pitch of one must be equal to th axial pitch of the other, and vice versa.
FIG. 39
GEARING
43
41. Computation of Spiral Gears for Given Teloc- ity Ratio.
Let na and nb — number of teeth in a and b, Fig. 40,
respectively;
ra and rb = radii of pitch cylinders o£ a and b] Na and N» = speeds of a and b, in revolutions per
minute;
ma = angle of teeth of a with axis of a] mb = angle of teeth of b with axis of b; e — angle between axes of a and b; h = distance between axes of a and b.
In the case of spiral gears, the revolutions per minute of the shafts are inversely as the num- bers of teeth, that is,
N6 na •
The radii ra and r,n however, have no direct influence on the velocity ratio of the gears, and within certain limits may be made of any length.
From Fig. 40, angle ma -\- angle mi, = angle e. The relative magni- tudes of angles ma and mb depend on the location of the tangent TPT. Having fixed this tangent and the angles ma and mb, the radii ra and rb may be obtained as follows:
From formula 2 of Art. 40, 2 TT ra cos ma
and
2 Ttrb cos mb
ffl . rrz
Dividing,
nb
~na
rb cos mb ra cos ma
(1) (2)
(3)
It will be seen from this formula FlG- 40
that the values of ra and rb may be assumed if desired and
44 GEARING
the corresponding; values of ma and m/, calculated. In prac- tice, however, the method given above is generally preferred.
The speed ratio — -? being given, and angles ma and mi,
Jvt
having been decided on, the ratio — can be found. Further,
ra
the sum ra + r/, = /z, the distance between the axes is known, and from the two equations thus at hand the pitch radii ra and rt may be calculated. An example will show the method.
EXAMPLE. — A pair of spiral gears with axes at 75° are to transmit a velocity ratio of 2| to 1, and the distance between their axes is 14 inches; find the pitch radii
SOLUTION. — Any number of solutions may be obtained, since it is possible to divide 75° into two angles in an infinite number of ways.
Suppose, first, ma = mt = 37|°; then ^ = 2.5 = ~- °°S ^!f-°, or — *
AV ra cos 37-2- ^a
— 2.5. But, ra + rt, — 14"; whence rb = 10" and ra = 4". Ans. For a second solution, let ma = 30° and mt = 45°; then
2.5 = = .817 , or - = 3.06.
ra cos 30° ra ' ra .817
Solving, ra = 3.45" and r6 = 10.55". Ans.
WORM-GEARS
42. Threads of Worms and Teeth of Worm-Wheels.
When the angle m of the helix, Fig. 39, is so great that the tooth becomes a thread winding entirely around the pitch cylinder, the combination is called a worm and worm- wheel. Usually, the axes of worm-gears are at right angles. The. screw may have a single thread or two or three threads With a single thread, the velocity ratio of the gears is equal to the number of teeth in the worm-wheel; thus, if the wheel has 40 teeth, 40 turns of the worm-shaft are required for 1 turn of the wheel.
Fig. 41 shows a worm and worm-wheel. It will be noticed that in the longitudinal section, taken through the worm, the threads appear to be like involute rack teeth. The worm is usually made in a screw-cutting lathe, and as it is easier to turn the threads with straight sides, it is better that they
GEARING
45
should be of the involute form. Involute teeth, then, should be used on the wheel, and should be of a pitch to correspond with the threads on the worm.
The circular-pitch system is used for worm-gearing because lathes are seldom provided with the correct change gears for cutting diametral pitches. The circular pitch is not so incon- venient, however, in the case of worm-gearing as with
FIG. 41
spur gearing. If the diameter of the worm-wheel should come in awkward figures, the diameter of the worm can be made such that the distance between centers will be the dimension desired. The circular pitch of the gear must be equal to the axial pitch of the worm.
43. Close-Fitting Worm and Wheel. — To make a close-fitting wheel, a worm is made of tool steel and then
46 GEARING
fluted and hardened like a tap. It is almost a duplicate of the worm to be used, being of a slightly larger diameter to allow for clearance. This cutter, or hob, is placed in mesh with the worm-wheel, on the face of which notches have been cut deep enough to receive the points of the teeth of the hob. The hob is then made to drive the wheel and is dropped deeper into it at each revolution of the latter until the teeth are finished.
Fig. 41 represents a close-fitting worm and wheel. The pitch circles are in contact at P. The outside diameter D of the worm may be made four or five times the pitch. The arcs C JC and E F are drawn about O and limit the addendum and root of the wheel teeth, the distance between them being the whole depth of the teeth. Clearance is allowed just as in spur gearing. The angle a is generally taken as either 60° or 90°. The whole diameter of the wheel blank can be obtained by measuring the drawing.
The object of nobbing a wheel is to get more of the bearing surface of the teeth on the worm-thread, making the outline of the teeth something like the thread of a nut.
GEAR TRAINS AND CAMS
Serial 992 Edition 1
GEAR TRAINS
1. Use of Trains. — Motion may be transmitted between two parallel shafts by a single pair of gear-wheels or by a -single pair of pulleys and a belt. However, it is often pref- erable to replace the single pair by a series of such gears or pulleys, especially if the required velocity ratio is great. Thus, in Fig. 1, it would be possible to transmit motion from shaft A to shaft F by means of two pulleys only, but to give the desired velocity ratio, pul- ley a would be very large and pulley / very small. By inter- posing the intermedi- ate shafts B and E and the pulleys b, c, d, and e, the velocity ratio is obtained with pulleys that do not vary greatly in size. In the same way, motion may be transmitted by a single pair of gears, but it is often better to introduce intermediate gears to avoid inconvenient sizes.
In general, a series of pulleys, gear-wheels, worms and worm-wheels, etc., interposed between the driving and driven shafts, is called a train of mechanism. If all the members are spur gears, the train is called a gear train or train of
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PIG. 1
GEAR TRAINS AND CAMS
gears; with bevel gears for members, it is called a bevel- gear train.
2. Velocity Katio of Two Wheels. — In Fig. 2, let a and b be two gear-wheels turning on the shafts A and B, and let P be the pitch point or point of contact of the roll- ing pitch circles. Evidently, since these pitch circles roll together, points on their circumferences have the same velocity. Let ra denote the radius of the wheel a, and Na the speed of «, in revolutions per minute, frequently written R. P. M. Then a point on the circumference of a travels,
FIG. 2
in 1 minute, ^^r^Na. feet, and a point on the circumference of b travels ^icrtNt feet. Hence, InraNa — 2 n rb Nb, or raNa = rbNb (1)
whence ^ = ^ (2)
Nb ra
That is, the revolutions per minute of two wheels running together are inversely as their radii.
The radii of the wheels are directly proportional to the diameters and to the number of teeth; hence it is allowable to use the ratio of either the radii, diameters, or numbers of teeth, as is most convenient. Furthermore, in any one
GEAR TRAINS AND CAMS 3
train, the radii of one pair, the diameters of another pair, and the numbers of teeth in a third pair, may be used.
3. Velocity Ratio of a Train of Gears. — Suppose that there is a train, as shown in Fig. 3, in which the pairs are a and b, c and d, and e and /. From formula 2 of Art. 2,
—? = -* = __*, where rand D denote, respectively, the radius 2V i, ra JJ
FIG. 3
and the diameter of the gear. For the second pair
= ^ = ^; and for the third pair, - * = ^ = ±2. ^ A: TV/ re A
ing together the left-hand and the right-hand members, respectively, of these three equations, and placing the prod- ucts equal to one another,
A/" y TV v /V TL v r^ v •»-•, 71 v D v /")
-* v a /*x -t v ^ /\ -t v e A ^ /\ r d /\ / / JLs & /^ /-/^ /\ -^-^/
TV* V A/J V /Vf ^VrVy yOv/lvA)
^VAA.-iVar/\YV/ r a S\ f( /\ re J-J a A /-/<: A -t-'e
But gears <5 and c are fixed to the same shaft, as are gears d and e, hence A7* = Nc and TVrf = A7",; therefore the first frac- tion reduces to — ?, which is the ratio between the speeds of Nf
4 GEAR TRAINS AND CAMS
the first and last shafts. Therefore the preceding formula
may be expressed thus:
product of radii, diameters, or number of teeth of all the
R. P. M. of the driving- shaft followers
R. P. M. of the driven shaft product of radii, diameters,
or number of teeth of all the drivers.
EXAMPLE. — In Fig. 4 is shown a train of six wheels on four shafts. Four of the wheels are gear-wheels, and two are pulleys connected by a belt. Let d1} d,, etc. denote the drivers, and /i, /"a, etc. the fol- lowers, A" the number of revolutions of di, and n the number of revolutions of /3. Suppose that the pulley A is 40 inches in diameter,
FIG. 4
and /! 35 inches; dz has 54 teeth, and /2 60 teeth; d3 is 1 foot in diam- eter, and /3 2 feet. What is the speed of fa if N = 100 revolutions per minute?
SOLUTION.— The product of the drivers X the speed of the driving shaft = the product of the followers X speed of the driven shaft; that is, 100 X 40 X 54 X 1 = « X 35 X 60 X 2. Hence,
The drivers and followers may be arranged in any order without changing the result; that is, they may be interchanged among them- selves, It should be noticed, however, that if the diameter of one driver be given in inches, the diameter of its follower must also be given in inches.
4. [Direction oi notation. — Axes connected by gear- wheels rotate in opposite directions. Hence, in a train consisting solely of external gear-wheels, if the number of axes be odd, the first and last wheels will revolve in the
GEAR TRAINS AND CAMS
same direction; if the number be even, they will revolve in opposite directions.
It is evident that in the case of a pinion working in an internal gear the two wheels will turn in the same direction.
5. Idlers.— In the train in Fig. 3, shaft B carries two gears, one being driven by gear a and the other driving gear d. Sometimes, however, only one intermediate gear is used, serving both as a driver and a follower. Such a wheel is called an idler, or idle wheel, and while it affects the relative direction of rotation of the wheels it is placed between, it does not affect their velocity ratio.
The following examples will illustrate this statement, and show a few ways in which idlers are used. In Fig. 5 is shown one method of arranging the change gears on the end of an engine lathe for changing the speed of the lead screw, which is used to feed the tool in screw cutting. The driver d^ re- ceives motion from the lathe spindle, and the follower f, drives the lead screw. The middle wheel, which is an idler, acts both as a driver and follower, or as dz and flf
Let N! and N* represent, respectively, the number of revolutions of dl and /„ and let the letters d,, d*, /,, and f, be used to denote either the radii, diameters, or numbers of teech of the wheels, as well as the wheels themselves. Then,
~ = -7 ~- But, as da and /j represent the same wheel,
•L V i I\ /\ /2
they have the same value, and — = — , or Na — — — . That
is, the speed of /, is exactly the same as though no idler were used.
To change the speed of the lead screw, a different size of wheel is put on in place of rfx or /„ or both are changed. The idler turns on a stud clamped in the slot in the arm a.
PIG- 5
GEAR TRAINS AND CAMS
This stud can be moved in the slot to accommodate the dif- ferent sizes of wheels on /. To bring the wheel in contact with the gear on s, the arm is swung about / until in the right position, when it is clamped to the frame by the bolt b. The way in which an idler changes the direction of rotation is shown in Fig. 6, which is a reversing mechanism, some- times placed on the headstock of a lathe for reversing- the
feed. Here the idler / is in contact with gears d and/, making three axes, an odd number, so that d and / turn in the same direction. Wheels /and /', however, are pinned to the plate p, which can swing about the axis of /, and they re- main always in contact. Moreover, as plate p swings about its center gear, 7 must necessarily remain in contact with gear /. If the lower end of the handle // be drawn upwards, the plate will be turned clockwise about the shaft of /, by means of the pin working in the slot in the lever /. The two idlers will take the dotted positions shown, and d will drive / through both of them. The number of axes will then be even, so that d and / will turn in opposite directions.
PIG. 6
ENGINE-LATHE GEAR TRAINS
THE BACK-GEAR TRAIN
6. Some of the best examples of gears in trains are to be found in engine lathes. Fig. 7 shows the headstock of an engine lathe, the spindle S turning in bearings, as shown, and having a face plate p and a center on the right end for placing the work. The lead screw /, used in screw cutting, is connected with the spindle by the train of gears on the
GEAR TRAINS AND CAMS 7
left, which will be described later. The back gears F^ and Da, on the shaft m n, have been drawn above the spindle for convenience of illustration, instead of back of it, where they are really placed.
It is important to keep the cutting speed within limits that the tool will safely stand. For turning work of different diameters and materials, therefore, the spindle must be driven at different speeds. This is accomplished by means of the cone pulley c driven by a similar pulley on the countershaft by means of a belt, and by means of the back gears.
The gear F, is fastened to the spindle S* and always turns
FIG. 7
with it. The cone <r, however, is loose on the spindle, but can be made to turn the latter by means of a lug, or catch, operated by a nut under the rim of F,. When the nut is moved out from the center, the lug engages with a slot in the large end of the cone. The cone will then turn with the spindle, and as many changes of speed may be had as there are pulleys on the cone. As ordinarily constructed, however, the cone alone does not give a range of speed great enough to include all classes of work, nor is the belt power sufficient for the larger work and heavier cuts. It makes the lathe more com- pact and satisfactory to construct the cone for the higher
8 GEAR TRAINS AND CAMS
speeds and lighter work, and to obtain the speeds for the heavier work by means of back gears. Referring to the figure, it will be seen that the back gears are connected by the sleeve s, and so form one rigid piece. The gear F, meshes with A, and A, which is loose on the spindle but fastened to the cone, meshes with • F>. To get the slower speeds, the nut mentioned before is moved in toward the center of Flt disengaging the gear from the cone, which is now free to turn on the spindle. Hence, if the back gears are in mesh with the gears on the spindle, the belt will drive the spindle at a slower speed through the cone and the
train A, Fl} A, F*.
The back gears cannot remain in gear when the cone and gear F, are connected; if they do, the lathe will not start, or teeth will be broken out of the wheels. To provide for throwing them in and out of gear, as required, the rod on which are the back gears and sleeve j is provided with eccen- tric ends at m and n, fitting in bearings in the frame. By turning the rod part way round by means of the handle /*, the gears can be thrown either in or out of gear.
A lathe is spoken of as running back-geared when the back gears are in, and as being in single gear when they are out of gear. This arrangement, or a modification of it, is used on upright drills, boring mills, milling machines, .and other machine tools.
EXAMPLE.— The four steps of the cone pulley of a lathe, as in Fig. 7, have diameters of 15, 12, 9, and 6 inches, respectively, and the cone on the countershaft is the same as that on the lathe. Suppose the gears A, Flt &„ and Ft to have, respectively, 30, 108, 24, and 84 teeth. The countershaft makes 105 revolutions per minute. Deter- mine the eight speeds of the spindle that may be obtained.
SOLUTION.— With the back gear thrown out, the following speeds are obtained:
1. With the belt on the first pair of steps, the spindle speed = ^ X 105 = 262| R. P. M.
2. With the belt on the second pair, the speed = ir X 105 = 140 R. P. M.
3. With belt on third pair, the speed = '& X 105 = 78f R. P. M.
4. With belt on fourth pair, the speed = & X 105 = 42 R. P. M.
GEAR TRAINS AND CAMS 9
30 X 24 5 The velocity ratio of the train is 7^5-77-57 = p^. Hence, with the
IUo X O* DO
back gear thrown in, the speeds for the various pairs of steps will be those given above multiplied by i^3. Thus,
5. 262| X & = 20f R. P. M.
6. 140 X <fe = 1H R- P. M.
7. 78f X A = 6i R- P. M.
8. 42 X A = 3* R. P. M.
THE SCREW-CUTTING TRAIN
7. Screw Cutting. — The engine lathe is .frequently used for screw cutting. Referring to Fig. 7, the screw-cutting mechanism is driven from the gear dlt which is fastened to the spindle; this connects with the lead screw /, through the gears /n da, the idler /, and the gear /2. To cut a screw thread, the work is placed between the centers of the lathe and made to turn with the face plate by the dog b clamped to the end of the work. The spindle runs toward the operator, or clockwise, as looked at from the outer end of the headstock, and the carriage, tool post, and tool, all of which are here represented by the pointer, are moved by the lead screw along the lathe bed parallel to the axis of the spindle.
In cutting a right-hand thread, the tool must move from right to left, and from left to right in cutting a left-hand thread.
Suppose that the lead screw has a left-hand thread, as shown in Fig. 7; then, to move the tool from right to left, the lead screw must turn counter-clockwise, regarded from the outer end 'of the headstock; to move the tool from left to right, it must turn clockwise. Hence, to cut a right-hand thread, the spindle and lead screw must turn in opposite directions and the number of axes in the train connecting them must be even; while to cut a left-hand screw, the spindle and lead screw turn in the same direction and the number of axes in the train must be odd. If the lead screw has a right-hand thread, these movements must be reversed.
In Fig. 7, the lead screw has a left-hand thread, and there are four axes in the train. To cut a left-hand screw, another
10
axis must be added and with this form of lathe provision is made for a second fixed stud, on which is placed a second idler between wheels d, and /.
An arrangement frequently adopted is shown in Fig. 8, The driving gear d1 on the spindle and the follower /t on the stud do not mesh directly together, but the idlers /i and /
PIG. 8
are interposed as explained 4n connection with Fig. 6. These two gears are always in mesh, and rotate on studs carried by an arm of the lever a, which is pivoted on the same stud as d- and /,. Otherwise, the train is the same as that of Fig. 7. When d, drives A -through the single idler 7t there are five axes in the train, but when the lever a is moved downwards ^ drives 7a, 7a drives Ilt and A drives /,;
GEAR TRAINS AND CAMS
ir
there are thus six axes in the train. Hence, by merely shifting the lever a, the number of axes may be made odd or even as desired. A right-hand lead screw is generally used with this train; hence, to cut a right-hand thread, /., is thrown out of gear, and to cut a left-hand thread it is thrown in gear. With a left-hand lead screw, the reverse would be true.
8. Gears for Cutting Thread of Given Pitch. — The
pitch of the thread cut depends on two things, namely, the pitch of the lead screw and the velocity ratio of the train connecting the lead screw and the spindle.
Let ,y = number of threads per inch on the lead screw;
t = number of threads per inch to be cut. Then, while the carriage moves 1 inch, the spindle must make / turns and the lead screw £ turns; hence, for a given travel of the carriage,
the turns of spindle _ t the turns of lead screw .y
Since the spindle and lead screw are, respectively, the driving and driven shafts, and following the rule of Art. 3,
/ _ product of numbers of teeth of all followers s product of numbers of teeth of all drivers With the train of Fig. 7 or that of Fig. 8, therefore,
£ d± X d*
When the lathe has a stud, as in Fig. 7, gears d1 and /, remain unchanged, and the different pitches required are obtained by different combinations of the gears d3 and /„.
The constant ratio -1 may be denoted by K; then
or
t_ =
s
f, =
(1) (2)
EXAMPLE.— -If the lead screw, Fig. 7, has six threads per inch, dl has 30 teeth, and A 60 teeth, find proper gears d, and /a to cut four, five, six, seven, eight, nine, and ten threads per inch.
12 GEAR TRAINS AND CAMS
SOLUTION.— K = £ = ~ = 2; s = 6. Then, using formula 2, «i oU
4 For four threads, /2 = S-TTH ^2 = A^a
•" xN 0
For five threads, /2 = A ^2
For six threads, /a = TS da, etc.
On the stud, place a gear 0fa having a number of teeth that is some
multiple of 12; say, 72 teeth; then /, = -^ d, = g-^-g X 72 = 6 '•
Hence, the number of teeth in fa, the gear to be placed on the lead screw, is six times the number of threads per inch to be cut. To cut four threads, a24-tooth gear is required on the screw; for rive threads, a 30-tooth gear; for six threads a 36-tooth gear, etc.
A number of other arrangements are possible. Thus, with an
• 84 84-tooth gear on the stud, the gear on the lead screw must have ^-- /
Qfi = 7 t teeth, and with a 96-tooth gear on the stud, f* = g-^-g t = 8 t.
Aus.
In designing- a set of gears for a lathe, the number of sepa- rate gear-wheels should be as small as can be found to cut the desired range of threads. The change gears should have involute teeth, so that the distances between centers may vary slightly without affecting the constancy of the velocity ratio.
9. Compound Gearing. — Lathes are frequently designed so that the screw-cutting train can be com- pounded. The single idler /., Fig. 8, is replaced by two gears d, and /„, Fig. 9, which are fastened together and turn about the axis s. Gear /3 is driven by da, and da acts as the driver for /a, the gear on the lead screw. The general formula
t f X / X /
for this arrangement is - = — ^ ^ 7, or
s di X d, X d*
t _ V- v *' v ^ f T)
~ — •**- /^ ~T ^ ~j \ -*- /
s a3 dt
The ratio — is usually kept constant for several pitches, da
the gears /2 and da being changed. Denoting this ratio by
or d.
• GEAR TRAINS AND CAMS
13
The ratio K •= — is usually either 2 or 1, and the ratio d-,.
Ki = ~ is usually either 4 or 2, or - or -, according1 as /3 or da
&$ ~£ &
is the larger of the two gears.
The following example illustrates the application of these principles:
EXAMPLE. — In a lathe with compound gearing, gears di and /"j are equal, gear d3 has 30 teeth, gear /3 has 36 teeth, and the lead screw has five threads per inch. Find a set of change gears, the smallest to have 24 teeth and the largest 64 teeth, to cut from four to six- teen threads per inch, using as small a number of gears in the set as possible.
14 GEAR TRAINS AND CAMS
/ Qft / / Qf\
SOLUTION. — From formula' 1, ^ = 1 X SK X -p or 5 = on Hence, /. = = X ™ X ^2 = 7; rfo. That is, to cut four threads per inch,
5 00 0
/a = !<£*; to cut five threads, fa = |^2; to cut six threads, /2 = f^2, and so on for the remaining threads. Since da is always multiplied by a fraction whose denominator is 6, let da be chosen as some multiple of 6, say 48. Then,
For four threads, /, = $ X 48 = 32 teeth
For five threads, /2 = I X 48 = 40 teeth
For six threads, /2 = f X 48 = 48 teeth
For seven threads, f, = &X 48 = 56 teeth
For eight threads, /a = I X 48 = 64 teeth
It is evident that for nine threads, with a 48-toqth gear on the stud, there would need to be a 72-tooth gear on the lead screw. But the largest gear is to have only 64 teeth. Hence, the size of d2 is changed to 24 teeth, and then,
For nine threads, /2 = | X 24 = 36 teeth
For ten threads, /, = ¥ X 24 = 40 teeth
For eleven threads, /„ = ¥ X 24 = 44 teeth
For twelve threads, /2 = nr X 24 = 48 teeth
For thirteen threads, /, = ¥ X 24 = 52 teeth
For fourteen threads, /2 = ¥ X 24 = 56 teeth
For fifteen threads, /, = ¥ X 24 = 60 teeth
For sixteen threads, /„ = ^ X 24 = 64 teeth
Therefore, the change gears required are those having 24, 32, 36, 40, 44, 48, 52, 56, 60, and 64 teeth, respectively, or eleven gears in all, since there must be two 48-tooth gears in cutting six threads per inch.
EXAMPLES VOU PRACTICE
1. In a lathe geared as in Fig. 7, the wheels dlt /i, and d^ have 18, 24,. and 16 teeth, respectively, and the lead screw has six threads per inch; what must be the number of teeth in /2 to cut eight threads per inch? Ans. 16 teeth
2. In a lathe like that in Fig. 7, /2 has 20 teeth, ds 15 teeth, and the ratio 1C has a value of f; how many threads per inch must the lead screw have to cut twelve threads per inch? Ans. 6 threads
3. In the lathe of Fig. 9, suppose that the values of K and Kl are 2 and 1, respectively, and that da has 24 teeth; find how many teeth f? must have, in order to cut ten threads per inch, assuming the lead Screw to have six threads per inch. Ans. 20 teeth
GEAR TRAINS AND CAMS
15
BPICYCLIC TRAINS
10. In the trains discussed thus far, all the wheels turn on fixed axes, that is, axes forming a part of the frame or fixed link of the machine. There are trains, however, in which awheel (or several wheels) turns on an axis that itself moves relative to the frame. In particular, there are cases in which the axis revolves about another fixed axis; a train containing such a combination is called an epicyclic train.
-In Fig. 10,
11. Principle of Combined Rotations.-
a wheel w is pinned at B to an arm a that rotates about the fixed point A. The wheel w turns about the pin B, which moves in the circular path s. With the arm and wheel in the first position, a line B C of the wheel coincides with A B, the center line of the arm. Suppose that the arm moves to a new posi- tion A B'; then, if the wheel were rigidly fastened to the arm, the line B C would be at B' O in the line AB'. If, however, the wheel has in the meantime turned on the pin B, B C will have taken some new posi- tion B' C". Let B' C" be prolonged to meet B C in E\ then the angle BAB1 = m± = angle through which the arm has turned; angle C' B' C" = mt = angle through which the wheel has turned on pin B, that is, relative to the arm; angle C" E C = m3 = angle through which the wheel has turned from its original position.
The angle m2 is an exterior angle of the triangle A E B' ', and it can be proved by geometry that angle m* = angle m>
FIG. 10
16 GEAR TRAINS AND CAMS
+ angle w,. That is, the angle through which the wheel has turned relative to the fixed frame, to which the arm a is attached at A, is equal to the sum of the angle through which the arm has turned and the angle through which the wheel has turned relative to the arm.
It is necessary to pay attention to the direction of the rotation; the above rule holds good only when the arm and wheel turn in the same direction. If, however, a counter- clockwise rotation is considered as + and a clockwise rota- tion as -, and the angles are given their proper signs, the rule holds good in all cases. The following examples illus- trate this principle:
1. The arm turns through 40° counter-clockwise, and the wheel turns on its pin through 60°, also counter-clockwise. Then mz = 40° + 60° = 100°, the angle through which the wheel has turned in a fixed plane.
2. The arm turns 110° counter-clockwise, and the wheel turns on the pin 45° .clockwise. Then, the turning of the wheel in a fixed plane is m, = + 110° - 45° = + 65°.
3. While the arm makes a complete turn (i. e., ml = 360° ) , the wheel makes three and one-half complete turns relative to the arm, both turns being counter-clockwise. Then, relative to the fixed link, the wheel makes + 1 -f 3| = + 4|
turns.
4. The arm makes two turns clockwise, and at the same time the wheel makes seven turns counter-clockwise relative to the arms. Relative to the fixed link, then, the wheel makes N = — 2-j-7 = +5 turns, that is, five turns counter- clockwise.
5. The arm makes one turn, and the wheel is fixed to the arm so that it cannot turn on the pin B. In this case the wheel makes one revolution relative to the fixed plane, though it does not turn on its axis. Referring to Fig. 10, if the wheel is fixed to the arm, m, — 0 and ?«3 = me, that is, the wheel turns through the same angle as the arm.
6. The arm makes one turn counter-clockwise, and at the same time the wheel makes one turn clockwise relative to the arm. Relative to the fixed plane, the wheel makes
GEAR TRAINS AND CAMS
17
FIG. 11
+ 1 — 1 =0 turns; that is, it has no motion of rotation, but has only a circular translation.
12. Train of Two Wheels. /
In Fig. 11, an arm a joins the two / gears b and c and holds them in \ mesh. Suppose that gear b is fixed and the arm a is turned counter- clockwise about the axis E. Then the gear c will turn on its axis; that is, relative to the arm, and the case is similar to that shown in Fig. 10. According to the principle stated in Art. 11, turns of wheel c = turns of arm a -\- turns
o£ c relative to a If the arm is fixed and the wheel b is given a complete
turn, the wheel c will make - turns, where b and c denote the
c
number of teeth, the radii, or the diameters of the wheels ,/; and c. It follows that when b is held fixed and the arm
•.makes one turn, the wheel c makes - turns relative to the
c
arm, just as it did when the arm was fixed. If the arm turns counter-clockwise, c turns counter-clockwise on its axis; hence, the total number of turns of c relative, to the fixed plane
is 1 + *. c
A convenient way of arriving at this result is the following: First, assume the arm to be held fixed and give the wheel b one negative turn, that is, a clockwise turn. As a result, the
wheels is given- positive turns. Second, assume the wheels c
and arm to be locked together so as to form a rigid body and give the whole system one positive turn; the first step gives wheel b one negative turn, the second a positive turn, and as a result wheel b is brought back to its first position. As regards its arm, the result of the two steps is to give it one
positive turr; and, as regards the wheel c, it is given -f- - turns
18 GEAR TRAINS AND CAMS
by the first, and -f one turn by the second step, or 1 + - turns
c
as a whole. It appears, therefore, that these separate motions give the same result as the single counter-clockwise turn of ivi a with b fixed. The results may be tabulated as follows"
|
Arm fixed .... |
WHEEL b ARM a . . -1 0 |
WHEEL |
|
Wheels locked . . |
. . +1 +1 |
c + 1 |
|
Result |
. . 0 (fixed) 4- 1 |
1 + * |
The algebraic sums in the third line give the total motion of each link, and the result may be inter- preted thus: With wheel b fixed, one turn
of arm a causes 1 -f - turns of wheel c. c
13. Annular Two-Wheel Train.
Suppose that the fixed gear is annular, as shown in Fig. 12. In this case, if the arm is fixed, a clockwise rotation of b about
its axis E will cause - clockwise turns of c about its axis F. c
Applying the method of successive motions,
|
Arm fixed . . . |
WHEEL b . . - 1 |
ARM a o |
WHEEL c b |
|
Wheels locked |
. . +1 |
+ 1 |
c 4-1 |
|
Result |
. . 0 (fixed |
) 4-1 |
1 -b |
c
That is, with the annular wheel b fixed, one counter-clock- wise turn of the arm causes 1 clockwise turns of wheel c
c
By the same method, it may be found that one clockwise
turn of the arm causes 1 counter-clockwise turns of the
wheel c.
GEAR TRAINS AND CAMS
19
EXAMPLE 1.— In Fig. 11, suppose that 6 has 60 teeth an I c 24 teeth If the arm a makes 1 counter-clockwise turn, how manj turns does c make, and in what direction?
ARM a 0
+ 1
SOLUTION. — Arm fixed . . Wheels locked
Wheel c makes + 3| turns; that is, 3* counter-clockwise turns!' Ans.
EXAMPLE 2.— In Fig. 12, let b and c have 60 and 24 teeth respectively; if the arm is given 1 clockwise turn, how many turns' will c make?
SOLUTION. — b
Arm fixed -f 1
Wheels locked ... — 1 _____
Wheel c makes + \\ turns. Ans.
14. Higher Epi- cyclic Trains. — In
Fig. 13, the arm a car- ries an idler c and the wheel d. By the addi- tion of the idler, the wheel d is caused to
>
|
ARM a 0 - 1 |
£ ,1 60 "~T 2~4~ - I |
|
|
1 |
M-l |
(= 1.5) |
|
/"" |
i ~""\ |
|
|
/ | |
|m \ >^ \ |
|
|
1 V |
r j |
|
|
m |
' < ,s \r^' |
|
|
i ,-',-.: 1 It * 1 |
H |
|
|
V » •>--' ' 'V x |
i |
|
|
^*ft*w |
. '" --^ |
turn in a direction / opposite to that of the | arm, as in the annular \
wheel train, Fig-. 12. \
Applying the usual method of analysis,
b Arm fixed . . . — 1
PIG. 13
ARM a
0
Wheels locked . + 1 +1 Result. . . .
c
+ 1
0 +1 1 +
d
With the arm fixed, one negative turn of b gives - positive
c
20
GEAR TRAINS AND CAMS
turns of the idler c, and for each turn of c, wheel d makes - turns in the opposite direction, that is, clockwise; hence,
while b makes one turn and c makes - turns, d must make
c
'l x - = - turns. The idler, therefore, has no effect on the c d d
number of turns of wheel d,
An interesting1 result is obtained when b and d are of th*
same size. In this case, 1-^=1-1=0, which means
a
that wheel d does not rotate at all but has merely a motion of translation. Thus, as shown in the figure, a ver- tical line m remains vertical for all positions of the wheel.
15. In Fig. 14, the arm a carries the two wheels c and d, which are fixed to the same shaft, and the wheel e. Wheels b and c mesh together, as do wheels d and e. With the arm
FIG. 14
fixed, let wheel b be given a clockwise turn. Then wheel c, and wheel d which is fast to it, must make - counter-clock- wise turns. For each turn of d, e makes - turns; hence,
e
while d makes - turns, e makes - X - turns. The motions c c e
when b is fixed and the arm is turned are;
GEAR TRAINS AND CAMS
21
|
b |
ARM a |
C AND d |
e |
|
_1 |
0 |
+ - |
-fix?} |
|
C |
\c e i |
||
|
+ 1 |
+ 1 |
+ 1 |
+ 1 |
Arm fixed . . . Wheels locked .
Result .... 0 +1 1+- 1 - - X
c c e
The same analysis may be applied to a train with any number of wheels. Thus, suppose that a second wheel / is fixed to wheel <?, Fig. 14, and that this gears with a wheel g that turns about a third moving axis G. For this arrange- ment, the analysis gives,
b ARM a c AND d e AND / &•• '
g
\ c/ c e c e g
16. Reverted Trains. — In Fig. 15 is shown a train in which the last wheel e turns on the stationary axis of the fixed wheel b, and is driven through the gears c and d, of
|
Arm 1 j |
, b b ., d \b \/ d |
|
|
fixed J |
c c c c e |
|
|
Wheels] 1 locked J |
+ 1 |
+1 +1 +1 |
|
Result 0 |
+ 1 |
+ fl + -^ i_^x^ 1 + ^ X - |
FIG. 15
which c engages with b and d with e. Such a train is called a reverted train. The analysis is as follows: Keeping the arm a fixed, a clockwise rotation of b causes - counter-clockwise turns of c and d (c and d are fixed
together), and for each turn of d, e makes - clockwise turns,
e
or - X - turns for one turn of b. Heiice, c e
22
GEAR TRAINS AND CAMS ARM a b c AND d
Arm fixed . . . Wheels locked . Result . . .
|
0 |
_ 1 |
_l_ _ |
\/ _ |
|
c |
c e |
||
|
+ 1 |
+ 1 |
+ 1 |
+ 1 |
|
+ 1 |
0 |
1 b c |
i-b-xd c e |
FIG. 16
That is, for each positive turn of the arm, wheel e makes
1 X - turns. If 1 is greater than - X -, <? turns in
c e c e
the same direction as the arm, but if 1 is less than - X
c e
it turns in the opposite direction. This reverted train may be used to give a slow motion to the wheel e. Suppose that the wheels have the following: numbers of teeth:
GEAR TRAINS AND CAMS 23
b, 41; c, 39; d, 40; e, 40. Then 1 - - X - = 1 - ^^~-
c e 39 X 40
= 1 — IT = — A. That is, 39 turns of the arm a are required to give 2 turns of the wheel <?.
17. Annular Reverted Train. — In Fig. 16 is shown the application of an annular reverted train. The annular wheel b forms part of the fixed frame of the machine. A frame a has pinned to it three pair of gears, cd, cd, cd, of which c, c, c engage the fixed annular wheel b, and d, d, d engage with the small pinion e, which is the driver of the train. Several revolutions of the pinion e are required to produce one turn of the frame a, and the train thus gives the desired mechanical advantage.
EXAMPLE.— In Fig. 16, let b have 42; c, 11; d, 29; and e, 12 teeth. What is the velocity ratio, that is, how many turns of the driving shaft are required for 1 turn of the frame al
SOLUTION. — Applying the usual analysis,
FRAME a b c AND d e
r, c , n , b 42 b d 42 29
Frame a fixed . . 0 — 1 = +-v"= — X —
c 11 ^ c e 11 ^ 12
Wheels locked . . + 1 +1 +1 +1
Result ....
|
1 |
0 |
42 ~ II |
1- |
, 42 29 hllX12 |
Velocity ratio = l + ffxfi = W = 10.23; that is, 10.23 turns of the shaft are required for 1 turn of a. Ans.
EXAMPLES FOR PRACTICE
1. In Fig. 12, suppose that b has 60 teeth and c 12 teeth. If the arm a makes 5 clockwise turns, how many turns does c make, and in what direction? Ans. 20 turns, counter-clockwise
2. Let b and d, Fig. 13, each have 30 teeth. If a makes 2 clock- wise revolutions, and b 1 counter-clockwise turn, how many turns will d make, and in what direction? Ans. 1 turn, counter-clockwise
3. Suppose that the gears b, c, d, and e, Fig. 14, have 80, 40, 50, and 20 teeth, respectively. If a makes 50 turns counter-clockwise, how many turns does e make, arid in what direction?
Ans. 200 turns, clockwise
4. In Fig. 15, let b, c, d, and e have 45, 15, 35 and 25 teeth, respect- ively. If a turns 30 times clockwise, how many turns does e make, and in what direction? Ans. 96 turns, counter-clockwise
I L T 374B— 8
24
GEAR TRAINS AND CAMS
REVOLVING GEAR TRAINS
18. Differential Back Gears. — Upright drills for metal work are sometimes provided with arrangements for increasing the range of the speeds and driving power that are different from the back gears explained in connection with the engine-lathe train. Fig. 17 shows one arrangement for this purpose. The cone pulley c is loose on the shaft s. A casting d, also loose on the shaft, has teeth on the inside, thus forming an annular gear. A plate p carrying the small gear, or pinion, / is fast to the shaft. On the left-hand end of the pulley hub is another gear /, which is fast to the hub of the cone. The action is as follows: A pin, on which there
PIG. 17
is a collar and nut, is clamped in the slot / in the plate d. The pin projects through d, so that when it is placed in the inner end of the slot it will engage with a corresponding slot in the plate p. When it is lowered, however, the pin disengages from the plate, but the collar can be made to fall in a slot in the arm a, which is a part of the frame. With the pin in the former position, gears d, I, and / are locked together, so that the shaft must turn with the cone. With the pin in the lattei position, gear d is locked to the frame and cannot turn, while gear / still turns with the cone. Therefore, the plate p is driven by the train consisting of the wheels /, /, and d.
GEAR TRAINS AND CAMS
25
Consequently, p and the shaft s attached to it turn more slowly than the cone.
19. Automobile Gearing. — In Fig. 18 is shown a revolving bevel-gear train as applied to automobile construc- tion. The engine shaft s carries the bevel gear b and the flywheel w, both of which are keyed to it. The second bevel wheel d can turn on shaft .? and carries two sheaves, / and g. By means of a band brake k, the wheel d may be held sta- tionary; or by means of a clutch h, the sheave g may be fast- ened to wheel w so that d is fast to the shaft s. A yoke m attached to the sprocket wheel n turns freely on the shaft s
FIG. 18
and carries the intermediate bevel gears c and e. The shaft runs in bearings in the fixed frame a. The following combi- nations can be obtained:
1. Neither brake k nor clutch h is applied. The engine drives shaft s and wheel d, wheels c and e act as idlers, and d turns with the same speed as b, but in the opposite direction. No motion is imparted to the sprocket wheel n.
2. The brake k is applied, thus holding wheel d fixed. In this case, the yoke m turns one-half as fast as the wheel b driven by the shaft; that is, two revolutions of the shaft are required for one revolution of the sprocket wheel. With this combination, the automobile is driven at low speed.
26
GEAR TRAINS AND CAMS
3. The brake k is released and the clutch g is applied. The wheel d is thus locked to the shaft, and therefore d,&, yoke m, and sprocket wheel n all turn as one rig-id piece keyed to the shaft. Each turn of the shaft gives a revolution of the sprocket wheel n, and the automobile is driven at high speed.
4. With the clutch h in gear, the brake k is applied, thus locking the wheel w with the frame a; this stops the automobile.
REVERSING MECHANISMS
20. Clutch. Gearing-. — A reversing mechanism often used, especially when the reversal must take place auto- matically, is shown in Fig. 19. It consists of three bevel
FIG. 19
gears, of which the driver d is fast on the driving shaft t. The gears f, and L are continually in contact with d, and are loose on shaft m n. A sleeve c can move endwise on m n, but is ^ compelled to turn with it by a key a in the shaft.' When in the position shown, the notches in c engage with /„ and m n turns; when the sleeve is in mid-position, m n will not turn; ^ and when thrown to the right, it engages with /, and the direction of rotation is reversed.
If the reversal is to be automatic, some provision must be made to insure that after sleeve c has been pushed out of contact on one side, it will be thrown in contact with the other gear. One way of doing this is shown in the figure. The lever / is pivoted at o, and at the other end is forked to
GEAR TRAINS AND CAMS
27
embrace c, a roller on each prong running in the groove shown. On the rod r, which is pivoted to / at one end and slides in a guide g at the other end, are two collars b and e, held in place by setscrews. Helical springs are also placed on r against the inside of the collars. Suppose that the tappet t, which is free to slide on r, is given a motion to the right through mechanism connected with £, but not shown in the figure. When it reaches the spring, it will compress it until the pressure of the spring on e is sufficient to over- come the resistance of the clutch. Further movement of t will move £ to the right until free of flt when the energy stored in the spring will suddenly throw c in contact with fa. The time at which reversal occurs can be adjusted by chan- ging the positions of the collars b and e.
21. Quick-Return Clutch.
Gear. — Sometimes it is desirable to have a slow motion in one direc- tion with a quick return. Fig. 20 shows a method that may be used for this purpose. The driver d1 is made cup-shaped so as to allow a smaller driver d3 to be placed inside. For the slow motion, d, drives /a, and, for the quick return, d1 drives /a. The change of direction is effected by shifting the clutch on the shaft so as to engage alternately with f, and /,.
22. Reversing by Belts. — In many machines, espe- cially those running at high speed, a reversal of the motion is effected by belts rather than gears. Two belts are gener- ally employed — one open and one crossed. Sometimes, these belts are made to shift alternately from a tight to a loose pulley, and in other cases they are arranged to drive two pulleys on the same shaft in opposite directions, either of the pulleys being thrown in or out by means of a friction clutch, as, for example, in lathe countershafts.
23. Planer Reversing Mechanism. — Of the first class, just mentioned, Fig. 21 affords an illustration as applied to a
FIG. 20
28 GEAR TRAINS AND CAMS
planer operating on metal. The table h is driven forwards and backwards by the rack and gearing shown. The work to be planed is clamped to the table /z, and during the for- ward, or cutting, stroke a stationary tool removes the metal. As no cutting takes place on the return stroke, the table is made to run back from two to five times as fast as when moving forwards. There are three pulleys on shaft n, of which t is the tight or driving pulley, A and /„, shown in the top view, being loose on the shaft. There are two belt shifters g
PIG. 21
and k, in the shape of bell-cranks, pivoted at a and b, respect- ively. A crossed belt, moving in the direction shown by the arrow, runs from a small pulley on the countershaft, and is guided by the shifter .f. This belt drives the table during the cutting stroke. The other belt, which is open, runs over a large pulley on the countershaft, and is guided by the shifter k. The short arms of the shifters carry small rollers working in a slot in a cam-plate c, which is pivoted at o. The ends of this slot are concentric about o, but one end has a greater radius than the other. As shown in the figure, the two
GEAR TRAINS AND CAST^ X.'-1—''
• /,.» N
shifters are in mid-position and the belts a^-pn^»|he«*loose _. pulleys. Suppose, however, the rod e to be p'uJIed/Jtjzj ^hg^ left. Shifter k will not move, because its roller will con- tinue to be in the same end of the slot, which is concentric about o; but the roller on g will be pulled to the left by passing from the end of the slot that is of larger radius to the end of the smaller radius. The crossed belt will, there- fore', be shifted to the tight pulley, which will cause the table to run forwards until the dog d strikes the rocker s, thus throwing the cam in the other direction. The effect of this will be to first shift the crossed belt from the tight to the loose pulley, and then to shift the open belt to the tight pulley; one motion follows the other, so as to decrease the wear and tear of the belts. The length of the stroke of the table can be varied by changing the position of the dogs, which are bolted to a T slot on the edge of the table.
The table is driven as follows: The belt pulleys drive gear d1} which is keyed to the same shaft; dj. drives /1( which in turn drives dat keyed to the same shaft; da drives gear /, which in turn drives the table by means of a rack underneath it. The circumferential speeds of / and da are evidently the same, and the speed of the table is the same as the cir- cumferential speed of the gear da. The velocity ratio of the gearing is generally expressed in terms of the number of feet traveled by the belt to 1 foot passed over by the table.
Suppose the belt pulleys to be 24 inches in diameter, and to make N revolutions while the table travels 1 foot. Let the diameter of dl be 3 inches; of /,, 26 inches, and of da, 4 inches. The circumference of d3 is 12k inches, nearly, so that for every foot traveled by the table, da and /,. will turn
12 24 . — - = — times. The number of turns made by the belt
12~s 25
pulleys for each foot passed over by the table is,
therefore, N X 3 = II X 26, or N = ?i^l§ = 8.32. Hence,
75
for 1 foot of travel of the table, the belt travel is 8.32 X ft X 3.1416 = 52i feet. That is, the planer is geared to run 52i to 1.
30
GEAR TRAINS AND CAMS
CAMS AND CAM TRAINS 24. Cams. — A cam may be defined as a machine part so
shaped that by its motion it imparts a definite motion to
some other part with which it is in direct contact. The cam is the driving part and is used to impart a motion that could not easily be obtained by other means, such as by link work or gear trains. Cams have the disadvantage of small contact surface and correspondingly rapid wear, and as a general rule are to be avoided if other devices can be obtained that will do the work as well.
A cam train, is usually a mechanism of three links, the fixed link, the cam, and the follower. An example is seen in the stamp mill for ore crushing, as shown in Fig. 22. The rotating shaft a carries the cam b> which acts on the piece c. This piece is attached to the rod d that carries the stamp e. The rod is guided by a framework, not shown
in the figure. It will be seen that the train has three
links — the fixed framework, the shaft and attached cam, and
the stamp.
FIG. 22
GEAR TRAINS AND CAMS
31
ROTAKY CAMS
25. Fig. 23 represents a common application of the plate cam. The cam c is supposed to turn clockwise, as indicated by the arrow, about the axis b, and to transmit a variable motion through the roller or follower / and the lever / to the rod p. The lever swings on the axis O0 and
.-<&i^.
FIG. 23
the roller moves up or down on the arc 0-6, as the cam revolves. The roller is held in contact with the cam by its own weight and that of the lever and the rod.
Suppose that the locations of the cam shaft b and the rod p are known, and that the cam shaft is to rotate uniformly clockwise and impart motion to the rod, so that during the first half of every revolution it will move uniformly
32 GEAR TRAINS AND CAMS
downwards during the next quarter-turn remain stationary, "rWthe last; quarter return to its former position
with, a uniform motion.
n order to determine the outline of the cam on which the circumference of the roller bears, it is necessary to find an outoe that will give the center of the roller the reared To on. Then, by placing the point of the compasses a different places on this outline, and strike arcs ms.de of ft wth radii equal to the radius of the roller, the curve or the actual cam can be drawn, the curve torn* tangent to the arcs and parallel to the first outline.
In this case, the roller / moves in an arc directly over the center of *. Knowing the distance the rod /is to move, Tis necessary to choose the point 0, and the throw of the cam, "halts! the distance that /is to move, so that the move- m-nt of * will be to the movement of / as r is to a.
Now, with O. as a center and a radius equal to . descr.be the arc 0-5-6, in which the center of the roller is to move, and mark the highest and !owest points 0 and 6 of the ro ler The lower point should not be near enough the shaft to allow the roller to strike the hub of the cam.
It evidently makes no difference with the relative motions of the cam and roller whether the cam turns clockwise, and the lever remains with its axis at O., or whether the cam ,s aTsumed to be stationary, and the lever and roller to move couTe Clockwise in a circle about the center *. This latter assumption will be adopted in drawing: cams.
With 4 as the center, draw a circle through O. and space it into a number of equal parts, say twelve, and number the divisions around to the left. Now, assume the lever / to move around the axis i in a counter-clockwise direction It wil, take positions 0<-1, 0.-2, O.-3, etc. Hence using these several points on the outer circle as centers, and with ladu equal to a, the length of the lever, describe a series of arcs corresponding to the original arc 0-3-6 Number these arcs 1 2 3 etc., to correspond with the numbers on the outel circle ' During the first half-turn of the cam, or, what is the same thing, while the lever is moving from its position at O.
33
to Oe on the outer circle, the center of the roller must move uniformly from its outer to its inner position. Hence, draw the chord of the original arc 0-3-6, and divide it into six equal parts, numbering them toward the center as shown. Then, with b as a center, describe an arc through point 1, intersecting arc 1 in point Jd. Now, sweep arcs through the other points, getting 2^,3^4^, etc., which are all points in the path of the center of the roller. From O6 to 0B on the original circle, the center of the roller remains at a constant distance from b; hence, 6, and 9, must be connected by a circular arc. From Oe to O1Z, the points are found as before by dividing the chord 0-6 into three equal parts and numbering them as shown, the numbers running outwards.
Finally, draw the path of the center of the roller through points 11,21,31,41, etc.; then draw the outline for the cam itself parallel to it, as explained at first. This is easily done by setting the dividers to describe a circle whose radius shall be the same as that of the roller /. Next, with various points on the curve 0-1,-2,-S, . . . 11, as centers (the more, the better), describe short arcs as shown. By aid of an irregular curve, draw a curve that will be tangent to the series of short arcs; it will be the .required outline of the cam, and will be parallel to the curve O-li-2,-3^ . . . 11^
The question sometimes arises in designing cams of this nature, whether it is the chord 0-6 or the arc 0-6 that should be divided to give the roller the proper outward and inward motions. For all practical purposes either way is sufficiently exact, but neither is quite correct, though it is better to space the chord. The exact way would be to draw the rod p and the roller in the different positions desired, and then design the cam to meet the roller at these points.
26. Quick-Drop Cam. — The cam shown in Fig. 24 differs in principle from the preceding one only in that the roller moves in a straight line, passing to one side of the center b of the shaft. Let it be required to design a cam of this nature to turn clockwise, as indicated by the arrow, which will cause the roller a and rod p to rise with a uniform.
84
GEAR TRAINS AND CAMS
motion to a distance h during two-thirds of a revolution. When the roller reaches its highest point, it is to drop at once to its original position, and to remain there during the remainder of the revolution. Assume the distance from the center b to the center line of p to be equal to r. With r as a radius, describe a circle about d, as shown. The center line of the rod will be tangent to this circle in all positions. With the same center and a radius equal to &A0, Aa being the extreme outward position of the roller, describe the out- side circle A0-4-8-A0. Divide this circle into some convenient number of equal parts, the number depending on the fraction of a revolu- tion required for the different periods of motion. Since the roller is to rise during two-thirds of a revo- lution, it is well to use twelve divisions as before, thus giv- ing I X 12 = 8 whole divisions for the first period.
Now, proceeding as before, by assuming the rod to move about the cam to the left, its positions when at the points of division At,Aat etc., will be represented by drawing lines through these points, and tangent to the inner circle, whose radius is r. The points a and A0 are the two extreme positions of the roller. Divide the line aA0 into eight equal parts, numbering them from the inside outwards, since the first movement of the roller is outwards. With b as a center, draw concentric arcs through these points, intersecting the tangents at 1,, 2l} 3ly etc. At point 8 on the outer circle,
PIG. 24
GEAR TRAINS AND CAMS
35
\
the roller drops along the line 8-81, the point 81 being deter- mined by drawing an arc about b with a radius b a. From point 8, back to a, the rod is at rest. The true cam outline is now to be found, as was done in the last example, by striking small arcs from points on the curve a~!1-21-31-41-51 -6i-71-8-8l as centers, and radii equal to the radius of the cam-roller a. This cam can rotate clockwise only in driving the rod p.
27. Cam for Stamp Mill. — The cam shown in Fig. 25 is a modification of that in Fig. 24, and is a form that is generally used in stamp mills where ore is crushed. The axis O of the cam lies at a distance r from the axis B C of the stamp rod. To construct the cam outline, a circle c is first drawn about O as a center with a radius r. This circle is evidently tan- gent to B C at G. The vertical move- ment or stroke of the stamp rod isA£, A being the lowest and B the highest point of contact with the cam. This length of stroke AB is made equal to one-half the length of the circumference of the circle <:, and the point A is located at a distance above G equal to one- fourth the circumference of c, so that B is at a distance from G equal to three-fourths the circum- ference of c.
Through the point A, then, the curve m is constructed as an involute of the circle c. The final point B' is found by drawing a circle, with 0 as a center and OB as a radius, cutting the curve m at B'. It will be seen that B' is verti- cally below D and that B' D is equal to three-fourths the circumference of the circle c. For the radius GA of the curve m is equal to one-fourth the circumference of c, that is, equal to the arc Gff, since the curve m is formed by the unwinding of a string from the circle c, starting at H. Hence, when the string has been unwound from the
FIG. 25
36
GEAR TRAINS AND CAMS
or three-fourths the circumference, the radius DB't repre- senting the length of the string at that instant, must be equal to the arc HGD.
The inner curve n may be chosen at random, but it must lie wholly within the space between m and B1 D. For an involute curve the tangent to the base circle c is normal to the curve m, and it follows that in any position of the cam the tangent at the point lying on B C above G is perpen- dicular to B C, Hence, the point of contact of the cam and the collar / always lies on B C. As actually constructed, two of these cams are placed opposite each other, as shown in Fig. 22.
28. Cams With Tangential Followers.— -It is some- times necessary to construct a cam to act tangentially on a
FIG. 26
flat surface as a follower. The follower may have a motion of translation, as EF moving. along X Fin Fig. 26, or it may turn about a fixed axis, as CD swinging about C in Fig. 27. For the first case, the method of designing the cam is as follows: Let the line EF represent the flat surface of the
GEAR TRAINS AND CAMS 37
follower, O the center of rotation of the cam, and XY the line of motion of the follower. Suppose that the follower is to take the positions 2,3,4, etc. when the radii OAS, OA3, OA<, etc. of the cam coincide, respectively, with the line 0 Y. With 0 as a center and O2,O3,O4, etc. as radii, strike arcs . cutting corresponding: radii in the points A2,A3,A*, etc. Through A, draw a line ft, making the same angle with OAa as the follower EF makes with OX. In the figure, EF\$ at right angles to OX; hence /, is at right angles to OA*. This is the usual arrangement. Likewise, draw lines 4, ^, /6, etc. through At, At, At, etc. and perpendicular, respectively,
FIG. 27
to OAa, OA*, OAS, etc. Draw a smooth curve tangent to the lines 4, /3, /4, etc., which will be the desired cam outline.
Sometimes it will be found that the otitline cannot be drawn without making some part of it concave. In this case, the conditions required cannot be met and some of them must be changed. All parts of the outline of a cam acting tangentially on a straight follower must be convex, for a concave part cannot become tangent to the follower.
In Fig. 27 is shown the construction for a follower that oscillates about a fixed axis. This axis is denoted by C, and Cl, C2, C3, and C4 are follower positions corresponding to the radii OAltOA2, etc. of the cam. That is, when O A,
38
GEAR TRAINS AND CAMS
coincides with the line OB, the follower will be at C2, and so on. With 0 as a center and radius 02 draw an arc cutting the first radius at As, and through the point Az draw the line ta, making the angle mz with OAS, equal to the angle that C2 makes with OB. Likewise,