zip line
zip line
(OP)
I'm trying to caculate the stress on a trolley cable (zip line) 30deg also the anchor points.
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RE: zip line
RE: zip line
RE: zip line
there may be some dynamic effects though, maybe do a energy balance ... the strain energy in cable produced by the weight, the weight doing it's conversion of potential energy into kinetic ...
RE: zip line
You'd also need to consider initial tension, and how accurately it can be fixed when the cable is installed.
RE: zip line
BA
RE: zip line
The one time I've been on a set, the drop was barely a a few feet over 60 ft. After about 6 runs, we were only maybe 10 ft lower than the first platform.
In any case, I don't think you haven't given enough information to solve the problem. The sag and weight of the cable are integral to the solution.
TTFN
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RE: zip line
P = concentrated load
a = horizontal dimension from P to left support
b = horizontal dimension from P to right support
s = cable sag at load P
L = horizontal dimension between supports = a + b.
The tension in the cable is H/cos(alpha) where alpha is the angle the cable makes with the horizontal.
If the weight of cable is considered by itself, the curve is a catenary, which involves hyperbolic functions. The equations can be found by googling "catenary".
If the concentrated load is to be considered in addition to the weight of cable, the mathematics may become a bit complicated. An approximation may be made by assuming the cable is a uniform load per unit of horizontal span. The approximation becomes better as the sag decreases (or worse as the sag increases).
If the sag at the concentrated load is known or can be measured, the horizontal component of the cable for a point load acting anywhere on the span is P'ab/sL where P' is the load P plus half the weight of the cable.
Thirty degrees is much too steep for a zip line.
BA
RE: zip line
words fail, see sketch ?