Thus far, however, we have considered only the case where the two cone pulleys were exactly alike. Now although this case occurs much more frequently than the case in which the cone pulleys are unlike, it is nevertheless true that unlike cone pulleys occur with sufficient frequency to make it desirable that convenient means be established for obtaining the diameters of their steps rapidly and accurately, and Table III. was calculated by the writer for this purpose; its accuracy is more than sufficient for the requirements of practice, the numbers in the table being correct to within a unit of the fourth decimal place (i.e. within .0001). It should be noticed that the tabular quantities are not the diameters of the steps, but these diameters divided by the distance between the centres of the cone pulleys; in other words, the tabular quantities are the effective diameters of the steps only when the centres of the pulleys are a unit’s distance apart. By thus expressing the tabular quantities in terms of the distance apart of the axis, the table becomes applicable to all cone pulleys whatever their distance from each other, the effective diameters of the steps being obtained by multiplying the proper tabular quantities by the distance between the centres of the pulleys.

Before describing and applying the table, we will call attention to the term “effective” diameter. The effective radius—as is well known—extends from the centre of the pulley to the centre of the belt; the effective diameter, being twice this effective radius, must also equal the actual diameter plus thickness of belt.

The table is so arranged that the diameter (divided by distance between centres) of one step of a belted pair will always be found in the extreme right-hand column; while its companion step will be found on the same horizontal line, and in that vertical column of the table corresponding to the length of belt employed. For example, if column 14 of the table corresponded to the length of belt employed, some of the possible pairs of diameters would be as follows:

The upper row of this series of pairs being taken from column 14, and the lower row from the extreme right-hand column, the numbers in each pair being on the same horizontal line. If the distance between the centers of the pulleys were 60 ins. the effective diameters of the steps corresponding to the above pairs would be:

being obtained by multiplying the first series of pairs by 60; the length of belt which would be equally tight on each of these pairs would be 3.3195 × 60 ins. = 199.17 ins.

To get the actual diameters of these steps when thickness of belt = ⁷⁄₃₂ = 0.22 in., we have simply to subtract 0.22 in. from the effective diameters just given, thus:

would be the series of pairs of actual diameters.

In solving problems relating to the diameters of cone pulleys by means of the accompanying table, we must have, besides the distance between centres, sufficient data to determine the column representing the length of belt. The length of belt is seldom known because it is of small practical importance to know its exact length; but it may be estimated approximately, and then the determination of suitable diameters of the steps becomes an extremely simple matter, as may be seen from what has already preceded. When the length of the belt is not known, and has not been assumed, we indirectly prescribe the length of belt by assuming the effective diameters of the two steps of a belted pair; thus, in the following Figure (561), the length of belt is prescribed when the distance a b, and any one of the pairs of steps d₁d₁, d₂d₂, d₃d₃ and d₄d₄ are given. We will show in the following examples how the length of belt and its corresponding column of diameter may be found when a pair of steps (like d₁d₁), are given.

Since in this example the steps of the given pair are equal, we look for ¹⁵⁄₅₀ = 0.30, in the extreme right-hand column of table; we will find it in the 11th line from the top; now looking along this line for the diameter of the other step, = ¹⁵⁄₅₀ = 0.30, we will find it in column 10; consequently the numbers of this column may be taken as the diameters of the steps which are the companions or partners of those in the extreme right-hand column.

We can now easily determine the remaining members of the pairs to which 4.5 in., 9 in., and 21 in. steps respectively belong. To find the partner of the 4.5 step, we find ^4.5⁄₅₀ = 0.09 in the right-hand column, and look along the horizontal line on which 0.09 is placed till we come to column 10, in which we will find the number 0.4850; 0.4850 × 50 in. = 24.25 in. will be the effective diameter of the companion to the 4.5 in. step.

To find the partner to the 9 in. step, we proceed as before, looking for ⁹⁄₅₀ = 0.18 in the right-hand column, and then along the horizontal line of 0.18 to column 10, then will 0.4113 × 50 in. = 20.57 in. be the required companion to the 9 in. step of cone a.

In like manner for the partner of the 21 in. step we get 0.1700 × 50 in. = 8.5 in. The effective diameter therefore will be,

If the thickness of belt employed were 0.25 in. the actual diameters of steps would be,

and the length of belt would be 2.9425 × 50 = 147.125 in.

Example 2. Given the effective diameters

and the distance between centres = 40 in.

Required the unknown diameters on cone B.

We must, as before, first find the vertical column corresponding to the length of belt which joins the pair of steps 6 in/30 in. We find the number ⁶⁄₄₀ = .15 in the right-hand column, and then look along its horizontal line for its partner ³⁰⁄₄₀ = 0.75. Since we do not find any number exactly equal to .7500, we must interpolate. For the benefit of those not familiar with the method of interpolation we will give in detail the method of finding intermediate columns of the table. On the aforesaid horizontal line we find in column 16 a number 0.7520, larger than the required 0.7500, and in column 15 a number 0.7014, smaller than 0.7500; evidently the intermediate column, containing the required 0.7500, must lie between columns 16 and 15. To find how far the required column is from column 16, we subtract as follows:

then the fraction ^.0020⁄_.00506 = 0.04 nearly will represent the position of the required intermediate column; namely, that its distance from column 16 is about ⁴⁄₁₀₀ of the distance between the adjacent columns, 15 and 16.

To find other numbers in this intermediate column we have only to multiply the difference between the adjacent numbers of columns 16 and 15 by 0.04, and subtract the product from the number in column 16. But it is not necessary to find as many numbers of the intermediate columns as are contained in either of the adjacent columns; it is only necessary to find as many numbers as there are steps in each of the cone pulleys. We will now illustrate what has preceded, by finding the partner to the 12 in. step of cone a. Find, as before, the horizontal line corresponding to ¹²⁄₄₀ = 0.30, then take the difference between the numbers 0.6413 and 0.5867 of columns 16 and 15, and multiply this difference, 0.0546, by 0.04; this product = 0.0022 subtracted from 0.6413, will give 0.6391, a number of the intermediate columns corresponding to the length of belt of the present problem. Multiplying by the distance between the axes = 40 in. we get 0.6391 × 40 = 25.56, for the diameter of the step of cone b which is partner to the 12 in. step of cone a.

To find the companion to the 18 in. step, we proceed in the same manner, looking for the horizontal line ¹⁸⁄₄₀ = 0.45, and interpolating as follows:

0.5094 - (0.5094 - 0.4500) × 0.04 = 0.5070.

Consequently, 0.5070 × 40 in. = 20.28 in. will be the required partner of the 18 in. step.

In like manner, for the 24 in. step, we have

0.3500 - (0.3500 - 0.2840) × 0.04 = 0.3474, and 0.3474 × 40 = 13.90.

The effective diameters are therefore

The actual diameters, when thickness of belt = 0.20 in., are:

And the length of belt will be:

[3.5080 - (3.5080 - 3.4137) × 0.04] × 40 in. = 140.17 in.

Example 3. Given the effective diameters:

and the distance between the centres = 60 in.

Required the remaining diameters on cone b.

The horizontal corresponding to ¹²⁄₆₀ = 0.20 lies ²⁄₃rd way between the horizontal line, corresponding to 0.18 and 0.21; the number ³³⁄₆₀ = 0.5500, corresponding to the companion of the 12 in. step, will therefore lie ²⁄₃rd way between the horizontal lines 0.18 and 0.21. We have now to find two numbers on this ²⁄₃rd line, of which one will be less and the other greater than 0.5500. An inspection of the table will show that these greater and less numbers must lie in columns 13 and 12. The numbers on the ²⁄₃rd line itself may now be found as follows:

In column 13, 0.5750 - ²⁄₃(0.5750 - 0.5513) = 0.5592.

In column 12, 0.5213 - ²⁄₃(0.5213 - 0.4967) = 0.5049.

0.5592 will be the number on the ²⁄₃rd line, which is greater than 0.5500, and 0.5049 will be the one which is less than 0.5500. The position of the intermediate column, corresponding to the length of belt of the present example, may now be found, as before, briefly. It is:

Consequently the required column lies nearest column 13, ¹⁷⁄₁₀₀th way between columns 13 and 12. To find any other number in the required column, we have only to multiply the difference between two adjacent numbers of columns 13 and 12 by ¹⁷⁄₁₀₀, and subtract the product from the number in column 13. For example, to find the diameter of the partner to the 18 in. step of cone a, we find the numbers 0.4750 and 0.4177 of columns 13 and 12, which lie on the horizontal line corresponding to ¹⁸⁄₆₀ = 0.30; the difference, 0.0573, between the two numbers is multiplied by 0.17, and the product, 0.0573 × 0.17 = 0.0097, subtracted from 0.4750. This last difference will equal 0.4653, and will be the number sought. If we now multiply by 60, we will get 27.92 in. as the effective diameter of that step on cone b which is the partner to the 18 in. step of cone a.

To find the companion of the 24 in. step, we proceed after the same fashion; the horizontal line ²⁴⁄₆₀ = 0.40 lies ¹⁄₃rd way between 0.39 and 0.42; hence,

In column 13, 0.3900 - ¹⁄₃(0.3900 - 0.3594) = 0.3798;

In column 12, 0.3294 - ¹⁄₃(0.3294 - 0.2975) = 0.3188;

And 0.3798 - (0.3798 - 0.3188) × 0.17 = 0.3694.

The required effective diameter of the step, which is partner to the 24 in. step, will therefore be 0.3694 × 60 = 22.16 in.

In like manner we obtain partner for the 30 in. step, thus:

In column 13, 0.2944 - ²⁄₃(0.2944 - 0.2600) = 0.2715.

In column 12, 0.2300 - ²⁄₃(0.2300 - 0.1940) = 0.2060.

Also 0.2715 - (0.2715 - 0.2060) × 0.17 = 0.2604, and 0.2604 × 60 in. = 15.62 in. = diam. of step belonging to the same belted pair as the 30 in. step of cone a.

The effective diameters will be:

and the actual diameters when belt is 0.22′′ thick:

and the length of belt is found to be:

[3.2252 - (3.2252 - 3.1310) × 0.17] × 60 in. = 192.55 in.

In all the preceding problems it should be noticed that we arbitrarily assumed all the steps on one cone, and one of the steps on the other cone. It will be found that all of the practical problems relating to cone-pulley diameters can finally be reduced to this form, and can consequently be solved according to the methods just given.

For those who find difficulty in interpolating, the following procedure will be found convenient: Estimate approximately the necessary length of belt, then divide this length by the distance between the centres of the cone pulleys; now find which one of the 33 lengths of belt (per unit’s distance apart of the centres) given in the table is most nearly equal to the quotient just obtained, and then take the vertical column, at the head of which it stands, for the companion to the right-hand column. Those numbers of these companion columns which are on the same horizontal line will be the companion steps of a belted pair. The table is so large, that in the great majority of cases not only exact, but otherwise satisfactory values can be obtained by this method, without any interpolation whatever.”

The teeth of the back gear should be accurately cut so that there is no lost motion between the teeth of one wheel, and the spaces of the other, because on account of the work being of large diameter or of hard metal (so as to require the slow speed), the strain of the cut is nearly always heavy when the back gear is in use, and the strain on the teeth is correspondingly great, causing a certain amount of spring or deflection in the live spindle and back gear spindle. Suppose then, that at certain parts of the work there is no cut, then when the tool again meets the cut the work will meet the tool and stand still until the lost motion in the gear teeth and the spring of the spindles is taken up, when the cut will proceed with a jump that will leave a mark on the work and very often break the tool. When the cut again leaves the tool a second jump also leaving a mark on the work will be made. If the teeth of the gears are cut at an angle to the axial line of the spindle, as is sometimes the case, this jumping from the play between the teeth will be magnified on account of a given amount of play, affording more back lash in such gears.

The teeth of the wheels should always be of involute and not of epicycloidal form, for the following reasons. The transmission of motion by epicycloidal teeth is exactly uniform only when their pitch circles exactly coincide, and this may not be the case in time because of wear in the parts as in the live spindle journals and the bearings, and the back gear spindle and its bearings, and every variation of speed in the cut, however slight it may be, produces a corresponding mark upon the work. In involute teeth the motion transmitted will be smooth and equal whether the pitch lines of the wheels coincide or not, hence the wear of the journals and bearings does not impair their action.

The object of cutting the teeth at an angle is to have the point of contact move or roll as it were from one end to the other of the teeth, and thus preserve a more conterminous contact on the line of centres of the two wheels, the supposition being that this would remove the marks on the work produced by the tremor of the back gear. But such tremor is due to errors in the form of the teeth, and also in the case of epicycloidal teeth from the pitch lines of the teeth not exactly coinciding when in gear.

The pitch of the teeth should be as fine as the requisite strength, with the usual allowance of margin for wear and safety will allow, so as to have as many teeth in continuous contact as possible.

Various methods of moving the back gear into and out of gear with the cone spindle gears are employed. The object is to place the back gears into gear to the exact proper depth to hold them securely in position, and to enable the operator to operate the gears without passing to the back of the lathe. Sometimes a sliding bearing box, such as shown in Fig. 562, is employed; a is the back gear spindle, b its bearing box, and d a pin which when on the side shown holds b in position, when the back gear is in action. To throw it out of action d is removed, b pushed back, and d inserted in a hole on the right hand of b; the objection is that there is no means of taking up the wear of b, and it is necessary to pass to the back of the lathe to operate the device.

Another plan is to let the back gear move endwise and bush its bearing holes with hardened steel bushes. This possesses the advantage that the gear is sure, if made right, to keep so, but it has some decided disadvantages: first, the pinion a, Fig. 563, must be enough larger than the smallest cone-step b to give room between b and c for the belt, and this necessitates that d also be larger than otherwise; secondly, the gear-spindle f projects through the bearing at f, and this often comes in the way of the bolt-heads used for chucking work to the face plate. The method of securing the spindle from end motion is as follows: On the back of the head is pivoted at i, a catch g, and on the gear shaft f are two grooves. As shown in the sketch, g is in one of these grooves while h is the other, but when the back gear is in, g would be in h.

Sometimes a simple eccentric bush and pin is used as in Fig. 564, in which a is the spindle journal, b a bush having bearing in the lathe head, and d a taper pin to secure b in its adjusted position.

In large heavy lathes having many changes of speed, there are various other constructions, as will be seen upon the lathes themselves in the various illustrations concerning the methods of throwing the back gear in and out. The eccentric motion shown in Fig. 573 of the Putnam lathe, is far preferable to any means in which the back-gear spindle moves endways, because, as before stated, the end of the back-gear spindle often comes in the way of the bolts used to fasten work to the large face plate. This occurs mainly in chucked work of the largest diameter within the capacity of the lathe.

In many American lathes the construction of the gearing that conveys motion from the live spindle is such that facility is afforded to throw the change gears out of action when the lathe is running fast, as for polishing purposes, so as to save the teeth from wear. Means are also provided to reverse the direction of lead screw or feed screw revolution. An example of a common construction of this kind is shown in Fig. 565, in which the driving wheel a is on the inner side of the back bearing as shown. It drives (when in gear) a pair of gears, one only of which is seen in the figure at b, which drives c, and through r, d, i, and s, the lead screw. A side view of the wheel a and the mechanism in connection therewith is shown in Fig. 566, in which s represents the live spindle and r is a spindle or shaft corresponding to r in Fig. 565. l is a lever pivoted upon r and carrying two pinions b and e; pinion b is of larger diameter than e, so that b gears with both c and e (c corresponding to wheel c in Fig. 565), while e gears with b only.

With the lever l in the position shown, neither b nor e engages with a, hence they are at rest; but if lever l be raised as in Fig. 567, b will gear with wheels a and c, and motion will be conveyed from a to c, wheel e running as an idle wheel, thus c will revolve in the same direction as the lathe spindle.

But if lever l be lowered as in fig. 568, then wheel e will gear with and receive motion from a, which it will convey to b, and c will revolve in the opposite direction to that in which the lathe spindle runs. To secure lever l in position, a pin f passes through it and into holes as i, j, provided in the lathe head. Lever l is sometimes placed inside the head, and sometimes outside as in Fig. 569, and it will be obvious that it may be used to cut left-hand threads without the use of an extra intermediate change gear, which is necessary in the construction shown in Fig. 570, in order to reverse the direction of lead screw revolution.

Sometimes the pin f is operated by a small spring lever attached to l, so that the hand grasps the end of l and the spring lever simultaneously, removing f from the hole in h, and therefore freeing l, so as to permit its operation. By relaxing the pressure on the small spring lever pin f finds its own way into the necessary hole in h, when opposite to it, without requiring any hand manipulation.

In larger lathes the lever l is generally attached to its stud outside the end bearing of the head h.

It is preferable, however, that the device for changing the direction of feed traverse be operative from the lathe carriage as in the Sellers lathe, so that the operator need not leave it when it is necessary to reverse the direction of traverse.

The swing frame, when the driving gear d is outside of the back bearing (as it is in Fig. 570), is swung from the axis of the lead screw as a centre of motion, and has two slots for receiving studs for change wheels. But when the driving gear is inside the back bearing as in Fig. 571, the swing frame may be suspended from the spindle (r, Fig. 565) that passes through the lathe head, which may also carry the cone for the independent feed as shown in Fig. 571, no matter on which side of the lathe the lead screw and feed rod are. This affords the convenience that when both lead screw and feed rod are in front of the lathe, the feed may be changed from the screw cutting to the rod feed, or vice versâ, by suitable mechanism in the apron, without requiring any change to be made in the driving gears.

In the lathe shown in Fig. 572, which is from the design of S. W. Putnam, of the Putnam Tool Company of Fitchburg, Massachusetts, the cone pinion for the back gear, and that for driving the feed motion, are of the same diameter and pitch, so that the gear-wheel l in Fig. 573 may (by means of a lever shown dotted in) be caused to engage with either of them. When the latter is used in single gear it would obviously make no difference which wheel drives l, but when the back gear is put in and l is engaged with the cone pinion, its speed corresponds to that of the cone, which being nine times faster than the live spindle, enables the cutting of threads nine times as coarse as if the back gear was not in use. This affords very great advantages for cutting worms and threads of coarse pitches.

An excellent method of changing the direction of feed motion, and of starting or stopping the same, is shown in Fig. 574, which represents the design of the Ames Manufacturing Company’s lathe.

In the figure, a is the small step of the lathe cone, b the pinion to drive the back gear, c a pinion to drive the feed gear, giving motion to d, which drives e, the latter being fast to g and rotating freely upon the shaft f, g drives h, which in turn drives i. The clutch j has a featherway into which fits the feather c, on the shaft f, so that when the clutch rotates it rotates j through the medium of c; k is a circular fork in a groove in j, and operated by a lever operated by a rod running along the front of the lathe bed. This rod is splined so that a lever carried by the apron or feed-table, having a hub and enveloping the rod, may by means of a feather filling into the spline operate the rod by partly rotating it, and hence operate k. Suppose now that this lever stands horizontal, then the clutch j would stand in the position shown in the cut, and d, e, g, h, and i, would rotate, while f would remain stationary. By lifting the lever, however, j would be moved laterally on f (by means of k) and the lug a on j would engage with lug b on g, and g would drive j, which through c would drive f, on which is placed a change gear at l, thus traversing the carriage forward. To traverse it backward the lever would be lowered or depressed below the horizontal level moving k, and therefore j, to the right, so that lug a would engage with lug b on i, hence f would be driven by i, whose motion is in an opposite direction to g, as is denoted by the respective arrows.

To throw all the feed motion out of gear, to run the lathe at its quickest for polishing, &c., the operation is as follows.

m is tubular and fast in n and affords journal bearing to wheel d. Through m passes stud o, having a knob handle at p. At the end of the hub of d is a cap fast in d, the latter being held endways between the shoulder shown on o and the washer and nut t. If then p be pulled outwards o will slide through m, and through the medium of t will cause d to slide over m, in the direction of the arrow, and pass out of gear from c, motion therefore ceasing at c.

q is the swing frame for the studs to carry the change wheels, and r a bolt for securing q in its adjusted position. s is a journal and bearing for h.

If it be considered sufficient the feed motion on small lathes (instead of feeding in both directions on the lateral and cross feeds as in the Putnam Lathe), may feed in the direction from the dead to the live centre, and in one direction only on the cross slide.

An example of a feed motion of this kind is shown in Figs. 575 and 576; f f is the feed spindle splined and through the medium of a feather driving the bevel pinion a having journal bearing in b. Pinion a drives the bevel gear c, which is in one piece with pinion d. The latter drives gear f, which drives pinion k, which is carried on a lever l, pivoted on the stud which carries f, so that by operating l, pinion k is brought into gear with pinion p, which is fast upon the cross-feed screw, and therefore rotates it to effect the automatic cross feed.

As shown in the cut, the lever l is in such a position as to throw k out of gear with p, and the cross feed screw is free to be operated by the handle by hand. At m is a slot in l in which operates a cam or eccentric, one end of which projects into l, while at the other end is the round handle r, Fig. 575, which is rotated to raise or lower that end of l so as to operate k. To operate the saddle or carriage the motion is continued as follows:—at the centre of f is a pinion gear g which operates a gear h, which is in one piece with the pinion i, and the latter is in gear with the rack running along the lathe bed.

If the motion from a to i was continuous, the carriage feed or traverse would be continuous, but means are provided whereby motion from f to i may be discontinued, as follows:—A hand traverse or feed is provided. j, Figs. 575 and 576, is carried by a stud having journal bearing in a hub on x and receiving the handle q; hence by operating q, j is rotated, operating the gear h, upon which is the pinion i, which is in gear with the rack running along the lathe bed.

To lock the carriage in a fixed position, as is necessary when operating the cross feed on large radial surfaces, the following device is provided:—n is a stud fixed in a hub on x, and having a head which overlaps the rim of h, as shown in figure. On the other side of that rim is a washer z on the same stud having a radial face also overlapping the rim of h, but its back face is bevelled to a corresponding bevel on the radial face on the hub of lever o (the hub of o being pivoted on the same stud). When therefore o is depressed the two-bevel face of the hub of o forces the washer z against the face of the wheel h, whose radial faces at the rim are therefore gripped between the face of the collar n and that of the washer z, hence h is locked fast. By raising the end of lever o, z is released and h is free to rotate.

Both the carriage feed and cross feed can only be traversed in one direction so far as these gears and levers are concerned, but means are provided on the lathe headstock for reversing the direction of motion of the feed spindle f so as to reverse the direction of the feeds. It will be observed that so long as f rotates, a, c, d, and f rotate, the remaining motions only operating when s is screwed up.

In order to obtain a delicate tool motion from the handle q it is necessary to reduce the motion between j and i as much as possible, a point in which a great many lathes as at present constructed are deficient, because q, although used to simply traverse the carriage along the bed, in which case rapid motion of the latter is desirable, is also used to feed the tool into corners when the lathe has no compound rest to put on light cuts on radial faces, hence it should be capable of giving a delicate tool motion.

On account of the deficiency referred to it is often necessary to put on a fine radial cut by putting the feed traverse in gear, and, throwing the feed screw gear out of contact with the other change wheels, pull it around by hand to put on the cut. In compound slide rests these remarks do not apply, because the upper part of the rest may be used instead of the handle q.

Many small lathes are provided with a tool rest known as the elevating rest, or weighted lathe.