| 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.