| 2.32 | ) | 96 | .00 | ( | 41 |
| 92 | 8 | ||||
| 3 | 20 | ||||
| 2 | 32 | ||||
| 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 smallest wheel. | Number of cutters in the set. | |||
| 12 | × | 16 | = | 192 |
Then
| Number of cutters in the set. | Number of the cutter. | |||
| 16 | - | 15 | = | 1 |
And
| Number of teeth in smallest wheel. | The number of the cutter. | The number of the teeth in the largest wheel. | ||
| 12 | × | 15 | ÷ | 300 |