Posted by kay on .
Two poles, 60 feet tall and 20 feet tall, stand on opposite sides of a field. The poles are 80 feet apart. Support cables are placed from the top of each pole to the bottom of the opposite pole. How far above the ground is the intersection of the cables. What if the poles were 120 feet a part.
1--Let A and B = the two height of the two poles.
2--Let X = the height of the poles crossing.
3--Let C be the distance between the two poles.
4--From the figure, A/C = X/Y or AY = CX.
5--Similarly, B/C = X/(C - Y) or BY = BC - CX.
6--Y = CX/A = (BC - CX)/B from which X = AB/(A+B).
Note - X is actually one half the harmonic mean of the two dimensions A and B, the harmonic mean being 2AB/(A + B).
Therefore, the height of the crossing is totally independant of the distance between the two buildings.
The height h is given by 1/h = 1/60 + 1/20
h = 15'
This is true regardless of the width of the field.
THANK YOU TCHRWILL