Differential Indexing

Workshop practice 2 : Differential indexing – brief write up of the process Differential indexing has to be employed when the number of divisions to be achieved ac hieved is not possible by simple indexing. You have learnt about simple indexing in the first year. It was employed to divide the circumference into equiangular divisions. The angular movement required was carried out on the indexing head. In order to increase accuracy, a magnified value of the angular movement is executed at the “index crank” and a mechanism is employed to reduce or demagnify this movement to the required amount at the “part”. If we try to directly rotate the part by, say, 1 degree – we might incur a considerable amount of error. But we would not incur the same percentage error if we rotated by 40 degrees. Hence, we rotate the index crank by 40 degrees which is reduced by mechanisms to 1 degree at the part. How is indexing carried out ? There is an index crank (handle) which wh ich rotates a worm on the same shaft. This worm in turn rotates the worm gear and the ratio of their rotations per minute is 40:1. There is an index plate mounted coaxially with the index crank shaft and it remains stationary during simple indexing. It has a number  of “hole circles” on its faces (equispaced holes which can accommodate the crank pin). If the required number of divisions on the part is 30, it has to be rotated by 1/30 rotations. Hence, to achieve that rotation on the part, the worm has to be rotated 40/30 times (unitary method). 40/30 is n othing but 1 and 1/3. Hence, if there is a 3-hole circle, we can easily do that by rotating the index crank once fully and then by 1 hole on the 3-hole circle.  Please note that the figure is slightly modified from actual setup. There are two spiral gears in the actual setup which are dropped here for simplicity.  Now, for differential indexing, the situation is that we don’t have the required number of hole ho le circle on the index plate. Example – say we have to achieve 31 divisions on the part and this means that the rotation to be achieved by the index crank is 40/31 – that is – 1 rotation and 9 holes on the 31 hole circle. But the problem is nearest hole circle and give the that 31 hole circle is not there on the index plate. Hence – what we do is: use the nearest hole index plate a definite amount of rotation to compensate for the mismatch. Hence, in case of differential indexing  – the index plate is not stationary. If you look at the index plate – you will notice that we have connected up a number of change gears “U” with it to connect it up with the worm gear shaft. That means – if the worm gear shaft rotates, the index plate will have to rotate. Look at the figure showing the angular rotations A and B carried out on the index plate. Suppose we have to get 71 divisions on the part but the 71 hole circle is not there. 70 hole ho le circle, however, is there. In that case – if we move the index crank from one hole to the next, we will have achieved 1/70 of a rotation on the index crank. If however (as shown in the figure), at the same time, if the index plate rotates back by a certain amount – we will achieve a rotation less than 1/70 of a rotation. The change gears are chosen in such a way that while the index crank moves by A (=1/71), the index plate rotates by 1/70 – 1/71 to provide the next hole at 1/71 position. Hence, even though the 71 hole circle circle is not there – the the second hole of the index plate rotates  backwards to provide one at the exact spot. How much should the value of the change gear ration U be in order to achieve that ? Shart from the index crack – it must be getting 1/71 rotation if the process works. The worm gets the same amount and the worm gear gets (1/40) X (1/71). The output of the gears U gets UX(1/40)X(1/71), where U is nothing but the gear ratio of the change gears (No/Ni). 1

Hence, in order to get a 1/70 – 1/71 movement at the index plate, U  ×

1 40

1 ×

71

1 =

70

1 −

71

Or U = 40/70 Questions : 1. Can you generalize the expression of the change gear ratio U ? 2. Can you check on the set up whether the ratio U has been correctly set on the machine ? 3. Self study – we are cutting gear teeth on the machine by differential indexing. Do you think that the gear  tooth shape is changing with the number of teeth on the gear ? Try to find out the relation of cutter  number with the number of teeth on a gear. 4. Find out the relations between different gear parameters and the module of a gear – a data sheet has been  provided.

Change gears U Index crank 

Worm gear  Your   part

Worm B A Index  plate

2