rhill at siue.edu
Tue Mar 21 01:49:22 MDT 2017
Hi Jim and all,
I finally came up with an answer to this problem. Let's say you have
two points (e.g. opposite sides of a canyon). level with each other
and a distance D apart. You have a cable, rope, chain, etc. with a
constant mass per unit length, typically denoted by µ (the Greek
letter mu). It's not hard to see that the maximum tension in the
cable will be at the ends. How long a cable (L) should you use to
minimuze the tension at the ends, and how much will this tension (T)
The shape of the hanging cable is well-known (to those who know it
well :-) to be a catenary, or hyperbolic cosine (cosh). Without
going into the gory details of the solution, here is my answer:
Optimal length of cable: L = [1 / (z * sqrt(z^2 - 1))] * D
Tension at ends: T = (z/2) * µgL = [1 / (2 * sqrt(z^2 - 1))] * µgD
where z is obtained by solving the transcendental equation
tanh z = 1/z .
Note that µgL is the actual weight of the sagging cable, while µgD is
the weight of that amount of straight cable which would connect the
two end points if that were possible (g is the usual acceleration of
gravity which relates mass to weight).
Using a calculator, I found by iteration that
z = 1.19967864025773 approximately,
leading to the results
L = 1.25773645616809 * D
T = .599839320128867 * µgL = 0.75443978076916 * µgD .
Thus the tension at each end, in the optimal case, is about 6/10 the
weight of the sagging cable, or about 3/4 the weight of a straight
piece of cable connecting the two ends.
It also turns out that the angle (call it theta) of the cable from
the horizontal at the endpoints will be
theta = arc sin (1/z) = 56.46583512745° (*),
and the maximum amount of sag (at the center) will be
|y|_max = [(cosh z - 1)/2z] * D = 0.337661500968676 * D .
(*) This angle seems a bit larger than one would expect, but for
comparison, a V-shape cable with the same D and L would make an angle
of 37.337° ffom the horizontal, while a circular arc would make an
angle of 72.063°, so the above value is at least somewhere in
between. But please let me know if you find any errors.....
I'll leave it to you to put in actual numerical data.....
------------ On Jim Horn's message ------------
Did you ever come up with an answer to this? I started to dive into
it but never finished...
> Hi Jim and all,
> Hmm, this brings up an interesting physics/mathematics problem that
> I hadn't encountered before. Ignore for the time being any
> stiffness in the wire, wind effects, etc. It is well known that a
> flexible wire or chain suspended betwen two points will have the
> shape of a catenary (hyperbolic cosine), a problem in itself that I
> have used in our physics graduate courses. Now suppose the wire
> runs between two points at the same level and at a distance D apart.
> It's not hard to show that the maximum tension will be at the
> endpoints. The question is, what length L of wire will minimize
> tension at each end?
> If L is close to D (i.e. the wire almost straight), the tension will
> be very large as Jim points out, while if L is very large the
> tension will also be very large, having to support a large weight.
> Therefore there must be some value of L between D and infinity that
> gives a minimum tension. The answer for L will be some number
> D, such as 2D or pi*D, etc.
> Will have to work on that.......
> -- Roger
> ------------ On Jim Horn's message ------------
> The greatest challenge there may be preventing its breakage. With a
> limit strength of about 0.3 pounds (138g), it can support 10,000
> of itself. But a horizontal section can require many times its own
> weight to keep somewhat straight. And just overcoming the inertia of
> its spool as a drone stops and starts moving could exceed that. Not
> mention the effect of wind loads, transients from oscillation,
> induced fatigue, etc. Do you really need sub-MOA/milliradian diamter
> from only a foot away?
> Jim Horn, WB9SYN/7
>> Assume it is #40 magnet wire (very fine) and I want to string it
>> across a canyon. . .
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