Drone comments

richard nineteenfortyone richard1941 at gmail.com
Wed Mar 22 19:48:08 MDT 2017


Why not use the Lambert W function?  It's included in the WP-34s!   Just
solve z = 1/(asinh(1/z)+W(1/ln(z)) for n.
K6YVL

On Tue, Mar 21, 2017 at 12:49 AM, Roger Hill <rhill at siue.edu> wrote:

> 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)
> be?
>
> 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.....
>
> -- Roger
>
> ------------ On Jim Horn's message ------------
>
> Roger,
>
> Did you ever come up with an answer to this? I started to dive into
> it but never finished...
>
> Jim
>
> > 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
> > the
> > 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
> > times
> > 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
> feet
> > 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
> to
> > mention the effect of wind loads, transients from oscillation,
> flexing
> > 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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