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zip line

zip line

zip line

(OP)
I'm trying to caculate the stress on a trolley cable (zip line) 30deg also the anchor points.    

RE: zip line

That's nice.  Now explain your problem.

RE: zip line

I believe that is a basic statics problem. You still got your old text book?

RE: zip line

on one level it is "just" a tension cable carrying a weight.

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

Assume a certain amount of strain in the cable, calculate the geometry based on the strained length, find corresponding force in the cable from statics, then repeat until that force matches the assumed strain.

You'd also need to consider initial tension, and how accurately it can be fixed when the cable is installed.

RE: zip line

I'm with Ron.  Explain your problem please.  A sketch wouldn't hurt.

BA

RE: zip line

If 30° is the unstressed slope, that's pretty sporting.  

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

If the weight of the cable is neglected, a cable subject to a concentrated load anywhere on the span will have a horizontal component of Pab/sL where:

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

picking up on the word "trolley", maybe this is more the opposite of a zip line ... more like a gondola cable ???

words fail, see sketch ?

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