in which we place E, 2E, 3E, etc., in place of Y, thus calling the rods one hundred feet apart, we have
| Centre. | Rod 1. | Rod 2. | Rod 3. | Rod 4. | Rod 5. |
|---|---|---|---|---|---|
| 0 | E2 b2 × a | 4E2 b2 × a | 9E2 b2 × a | 16E2 b2 × a | 25E2 b2 × a |
| 0 | 1002 5002 × 80 | 2002 5002 × 80 | 3002 5002 × 80 | 4002 5002 × 80 | 5002 5002 × 80 |
| 0 | 3.20 | 12.80 | 28.80 | 51.20 | 80.00 |
Problem 3.
227. To find the angle E C G, fig. 103. The formula for the angle between the axis of the tower, and the tangent to the curve of the cable at the point of suspension is
tang a = E C G = 2a
b.
Span being one thousand feet, b is five hundred; and a being eighty feet, we have
tang E C G = 160
500 = log 160 – log 500:
or 2.204120 – 2.698970 = tang 9.505150 = 17° 45′ = E C G.
Also, 90° – 17° 45′ = 72° 15′ = angle G C A, or A C H.
When the points of suspension are not at the same elevation, we proceed in the same manner: only using G L, G E, in place of F L, F C, in fig. 103 A.