5/6 (Not a homework question) You have a horizontal gap of length L and
\_ bullshit.
\_ double-bullshit. I made a paperclip chain and am curious how
to most easily drape it (and I'm a big nerd).
a rope/chain/cable with mass per unit length w, and there is vertical
gravity of acceleration g. What length of cord should you use to
bridge the gap to as to minimize the tension at the anchor points?
\_ A suspension like this is described by a hyperbolic cosine, the
least amount of tension is produced when the rope or chain is exactly
the length of L. Let me know if you need more proof. -scottyg
\- re: hyperbolic cos: also and probably more commonly known as
Cantenary arc/arch [like the st. louis object]. you can
google from "cantenary" ... somebody probably has the
solved integral that gives you a closed form for arc length.
this name actually comes from the idea of "hanging chain",
so appropos to you underlying physical model. and now lets
talk about the feynman sprinkler.
\_ And when you can't find the answer, search for
"catenary."
\_ Perhaps you misunderstood. The gap width is L. Trying to bridge
that with a cord onle L long would require infinite tension when
under gravity.
\_ it is a simple optimization problem that reduces down to the
more string you have, the greater the weight, and therefore the
greater the tension. L is the minimum amount of string one
would need to span the gap, therefore it is also the optimized
length for minimum tension. See the equations below for tens-
ion, T, as they are correct. BTW, I have a degree in Physics,
and not CS, if that gives me any more credit. -scottyg
\_ "assume the horse is a sphere"... sorry, nope. it doesn't.
\_ no it wouldn't.
\_ assuming an infinite tensile-strength cord, yes it would.
Take a reasonably heavy rope or chain and try to pull it
to be absolutely straight while not supported in the
middle. You can't do it.
\_ but that's not what we're trying to do. you can
straighten it by pushing in the middle. in any case
the answer is "as straight as you can".
\_ What I'm trying to do is find what length of paper
clip chain will bridge a gap L with minimum tension.
The answer is NOT L, because that has extremely
high tension which bends the paper clips all out of
shape. But the answer is not 100L either, because
that is way too much extra weight.
\_ you don't have to pull the ends to straighten
the chain. you can straighten it theoretically
without bending.
\_ Based on what little I remember about CE, scottyg is right.
A cable w/ uniform weight per unit length is described as
a cantenary. You need two equations to determine the length,
L, of the cable. Assuming that you know the separation between
the endpoints (X) and the "sag" (S) (the distance from the
horizontal passing through the end points and the lowest point
of the cable), you can use the following equations:
EQ1.1 The half length, L/2, equation:
EQ 1.1 The half length, L/2, equation:
y^2 - (L/2)^2 = c^2
EQ2.1 The parameter, c, equation:
EQ 2.1 The parameter, c, equation:
y = c cosh (X/c)
Since y = S + c you can simplify to:
EQ 1.2: L = 2(2c^2 - 2Sc - S^2)^(1/2)
EQ 2.2: S/c + 1 = cosh (X/c)
You can use excel or some such to solve for c in EQ 2.2 and
then it is simple to substitute into EQ 1.2 to get L. iirc,
in most cases it is simpler to just think of the thing as a
parabola.
then it is simple to substitute into EQ 1.2 to get L.
Now that I re-read your question, it seems like you want to
minimze S as well as L right? I don't remember enough calc
to help you out there, but I think that you probably need
to do this by minimizing the tension T at the end points.
The equation for T at the end points is:
EQ 3.1: T = w(c^2 + (L^2)/4)^(1/2)
Good luck!
\_ Hmm, what scottyg claims above is that tension T is minimum
when the string length L equals to the gap width X. Now,
as L approaches X, S approaches 0. Using your equations,
as X approaches L, S approaches 0. Using your equations,
we get this:
limit(L->X) S = 0;
limit(X->L) S = 0;
EQ 2.2: S/c + 1 = cosh (X/c)
limit(L->X) (S/c + 1) = limit(L->X) cosh (X/c)
1 = limit(L->X) cosh (X/c)
0 = limit(L->X) X/c
thus limit(L->X) c = infinity
limit(X->L) (S/c + 1) = limit(X->L) cosh (X/c)
1 = limit(X->L) cosh (X/c)
0 = limit(X->L) X/c
thus limit(X->L) c = infinity
EQ 3.1: T = w(c^2 + (L^2)/4)^(1/2)
limit(L->X) T = limit(L->X) (w(c^2 + (L^2)/4)
limit(X->L) T = limit(X->L) (w(c^2 + (L^2)/4)
^(1/2))
= infinity
I don't see how the tension is minimum in this case, using
your presumably correct equations. --- L&S CS major
your equations. --- L&S CS major
your presumably correct equations. So scottyg is wrong.
--- L&S CS major
\_ I think that what scottyg is getting at is that you
have to balance the tension at the end points against
the tensile strength of the material that makes up
the cable in order to find the min sag and length.
\_ Sorry, but this "L is the minimum amount of
string one would need to span the gap, therefore
it is also the optimized length for minimum
tension" seems to contradict that.
\_ Okay, I see your point. When I originally
posted the equations I was probably wrong
about the minimizing the tension at the
end points. Anyway, do you see anything
wrong w/ balancing the tensile strength
of the material w/ the tension at the
end points to get the optimal length
and sag?
\_ Not at all. That seems like a very reasonable
way to think about it. |
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