Fig. 561.
Example 1. Given the effective diameters
| 4.5 | in. | 9 | in. | 15 | in. | 21 | in. | on cone | A, |
| — | — | 15 | in. | — | „ | B, | |||
and the distance between centres equal to 50 inches.
Required the remaining diameters on cone b.
Since in this example the steps of the given pair are equal, we look for 15⁄50 = 0.30, in the extreme right-hand column of [table]; we will find it in the 11th line from the top; now looking along this line for the diameter of the other step, = 15⁄50 = 0.30, we will find it in column 10; consequently the numbers of this column may be taken as the diameters of the steps which are the companions or partners of those in the extreme right-hand column.
We can now easily determine the remaining members of the pairs to which 4.5 in., 9 in., and 21 in. steps respectively belong. To find the partner of the 4.5 step, we find 4.5⁄50 = 0.09 in the right-hand column, and look along the horizontal line on which 0.09 is placed till we come to column 10, in which we will find the number 0.4850; 0.4850 × 50 in. = 24.25 in. will be the effective diameter of the companion to the 4.5 in. step.
To find the partner to the 9 in. step, we proceed as before, looking for 9⁄50 = 0.18 in the right-hand column, and then along the horizontal line of 0.18 to column 10, then will 0.4113 × 50 in. = 20.57 in. be the required companion to the 9 in. step of cone a.
In like manner for the partner of the 21 in. step we get 0.1700 × 50 in. = 8.5 in. The effective diameter therefore will be,