| 4 | .5 | in. | 9 | in. | 15 | in. | 21 | in. | on cone | A, | |||
| 24 | .25 | 20 | .57 | 15 | in. | 8 | .5 | „ | B. |
If the thickness of belt employed were 0.25 in. the actual diameters of steps would be,
| 4.25 | 8.75 | 14.75 | 20.75 | on cone | A, |
| 24.00 | 20.32 | 14.75 | 8.25 | „ | B, |
and the length of belt would be 2.9425 × 50 = 147.125 in.
Example 2. Given the effective diameters
| 6 | in. | 12 | in. | 18 | in. | 24 | in. | on cone | A, |
| 30 | in. | — | — | — | „ | B, | |||
and the distance between centres = 40 in.
Required the unknown diameters on cone B.
We must, as before, first find the vertical column corresponding to the length of belt which joins the pair of steps 6 in/30 in. We find the number 6⁄40 = .15 in the right-hand column, and then look along its horizontal line for its partner 30⁄40 = 0.75. Since we do not find any number exactly equal to .7500, we must interpolate. For the benefit of those not familiar with the method of interpolation we will give in detail the method of finding intermediate columns of the table. On the aforesaid horizontal line we find in column 16 a number 0.7520, larger than the required 0.7500, and in column 15 a number 0.7014, smaller than 0.7500; evidently the intermediate column, containing the required 0.7500, must lie between columns 16 and 15. To find how far the required column is from column 16, we subtract as follows:
| 0.7520 | 0.7520 | |
| 0.7500 | 0.7014 | |
| .0020 | 0.0506 |