Rule.--Multiply the number of teeth in the smallest wheel of the train by the number of cutters it is proposed to have in the set, and divide the amount so obtained by a sum obtained as follows:--

From the number of cutters in the set subtract the number of the cutter, and to the remainder add the sum obtained by multiplying the number of the teeth in the smallest wheel of the set or train by the number of the cutter and dividing the product by the number of teeth in the largest wheel of the set or train.

Example.--I require to find how many wheels each cutter should cut, there being 8 cutters and the smallest wheel having 12 teeth, while the largest has 300.

Number of teeth in Number of cutters smallest wheel. in the set.

12 8 = 96

Then

Number of cutters Number of in set. cutter.

8 - 7 = 1

Then

Number of teeth in The number of the The number of the teeth smallest wheel. cutter. in largest wheel.

12 8 300

12 8 --- 300 ) 960 ( 0.32 900 --- 600 600

Now add the 1 to the .32 and we have 1.32, which we must divide into the 96 first obtained.

Thus

1.32 ) 96.00 ( 72 924 ---- 360 264 --- 96

Hence No. 8 cutter may be used for all wheels that have between 72 teeth and 300 teeth.

To find the range of wheels to be cut by the next cutter, which we will call No. 7, proceed again as before, but using 7 instead of 8 as the number of the cutter.

Thus

Number of teeth in Number of cutters in smallest wheel. the set.

12 8 = 96

Then

Number of cutters Number of in the set. cutters.

8 - 6 = 2

And

Number of teeth in The number of the The number of teeth smallest wheel. cutter. in the largest wheel.

12 8 300

Here

12 8 --- 300 ) 960 ( 0.32 900 --- 600 600

Add the 2 to the .32 and we have 2.32 to divide into the 96.

Thus

2.32 ) 96.00 ( 41 928 --- 320 232 --- 88

Hence this cutter will cut all wheels having not less than the 41 teeth, and up to the 72 teeth where the other cutter begins. For the range of the next cutter proceed the same, using 6 as the number of the cutter, and so on.

By this rule we obtain the lowest number of teeth in a wheel for which the cutter should be used, and it follows that its range will continue upwards to the smallest wheel cut by the cutter above it.

Having by this means found the range of wheels for each cutter, it remains to find for what particular number of teeth within that range the cutter teeth should be made correct, in order to have whatever error there may be equal in amount on the largest and smallest wheel of its range. This is done by using precisely the same rule, but supposing there to be twice as many cutters as there actually are, and then taking the intermediate numbers as those to be used.

Applying this plan to the first of the two previous examples we have--

Number of teeth in the Number of cutters in smallest wheel. the set.

12 16 = 192

Then

Number of cutters Number of the in the set. cutter.

16 - 15 = 1

And

Number of teeth in The number of the The number of the teeth in smallest wheel. cutter. the largest wheel.

12 15 300

12 15 --- 60 12 ----- 300 ) 180.0 ( 0.6 1800

Then add the 1 to the .6 = 1.6, and this divided into 192 = 120.

By continuing this process for each of the 16 cutters we obtain the following table:--

Number of Number of Cutter. Teeth.

1 12 *2 13 3 14 *4 15 5 17 *6 18 7 20.61 *8 23 9 26 *10 30 11 35 *12 42 13 54 *14 75 15 120 *16 300

Suppose now we take for our 8 cutters those marked by an asterisk, and use cutter 2 for all wheels having either 12, 13, or 14 teeth, then the next cutter would be that numbered 4, cutting 14, 15, or 16 toothed wheels, and so on.

A similar table in which 8 cutters are required, but 16 are used in the calculation, the largest wheel having 200 teeth in the set, is given below.

Number of Number of Cutter. Teeth.

1 12.7 2 13.5 3 14.5 4 15.6 5 16.9 6 18 7 21 8 23.5 9 26.5 10 29 11 35 12 40.6 13 52.9 14 67.6 15 101 16 200

To a.s.sist in the selections as to what wheels in a given set the determined number of cutters should be made correct for, so as to obtain the least limit of error, Professor Willis has calculated the following table, by means of which cutters may be selected that will give the same difference of form between any two consecutive numbers, and this table he terms the table of equidistant value of cutters.

TABLE OF EQUIDISTANT VALUE OF CUTTERS.

Number of Teeth.

Rack--300, 150, 100, 76, 60, 50, 43, 38, 34, 30, 27, 25, 23, 21, 20, 19, 17, 16, 15, 14, 13, 12.