Modern Machine-Shop Practice Part 43

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

5.8 11.8 17.8 23.8 on cone A.

29.8 25.36 20.08 13.7 " B.

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:

12 in. 18 in. 24 in. 30 in. on cone A, 33 in. -- -- -- " B,

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:

0.5592 - 0.5500 = 0.0092 = 0.17.

0.5592 - 0.5049 = 0.0543

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:

12 in. 18 in. 24 in. 30 in. on cone A, 33 27.92 22.16 15.62 " B,

and the actual diameters when belt is 0.22" thick:

11.78 17.78 23.78 29.78 in.

32.78 27.70 21.94 15.40

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 a.s.sumed _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."

[Ill.u.s.tration: Fig. 562.]

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 s.p.a.ces 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 pa.s.sing 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 pa.s.s to the back of the lathe to operate the device.

[Ill.u.s.tration: Fig. 563.]

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.

[Ill.u.s.tration: Fig. 564.]

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.

[Ill.u.s.tration: Fig. 565.]

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 ill.u.s.trations 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.

[Ill.u.s.tration: Fig. 566.]

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.

[Ill.u.s.tration: Fig. 567.]