T.R | Title | User | Personal Name | Date | Lines |
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889.1 | Why not untide the rope and use a yard stick :-) | HPSTEK::XIA | | Mon Jun 20 1988 16:28 | 7 |
| I think that is as good as you can get. I believe any further fine
tuning of that model will not necessarily improve the accuracy (From
an engineer's point of view, of course). The problem is that
to further improve the model, the shape of the rope has to be taken
into consideration. I mean the "inner side" of the rope covers
a shorter distance than the "outer side".
Eugene
|
889.2 | The Tape Reel Problem | DRUMS::FEHSKENS | | Mon Jun 20 1988 19:24 | 16 |
| I once solved this problem sort of in general, trying to find the
fraction of the total length as a function of the fraction of the
(maximum radius - minimum radius). The application was to be able
to estimate the length (and hence elapsed time) of the partially
played part of a cassette based on the relative size of the source
and destination "pancakes". I believe approaching the problem this
way is independent of the thickness of the layers, but I may be
wrong. Anyway, I used exactly the method you propose, letting the
layer thickness be dt and integrating suitably. The answer was
surprisingly simple, but I've forgotten it anyway, and am too lazy
to repeat the exercise. Note, however, that I *assumed* you started
out knowing the length you are looking for, but maybe you could
work backwards from this approach.
len.
|
889.3 | It's not really a spiral | AKQJ10::YARBROUGH | Why is computing so labor intensive? | Mon Jun 20 1988 20:50 | 11 |
| Near the end of each loop the rope must "reach out" to 1/2" beyond the
radius of the last loop. Call the arc length of the "reach" s, and the
radius of the current loop r; call the angle at which the rope leaves the
previous outer edge theta. Then we have
s = sqrt((r+1/2)^2-r^2)
theta = arctan(s/r)
then s-r*theta is the difference between the length of the "reaching" rope
and its circular counterpart. The total length of all the loops will be
2*Pi*3+(sum[i/2,i=1..30]) + sum[(s-r*theta), r=3(.5)18].
|
889.4 | No, a REAL spiral was intended | POOL::HALLYB | The smart money was on Goliath | Tue Jun 21 1988 13:52 | 17 |
| Re: .3 -< It's not really a spiral >-
.0> If you were to view this from above you would see a nice neat spiral,
.0> assuming for simplicity that someone has managed to lay out the rope
.0> in a smooth fashion.
By the above statement back in .0 I meant to imply that in fact
a perfect spiral was laid out, not a circle-bump-circle as .3 intends.
It actually would be kind of interesting to pursue this from the
rope's point of view (.1 fails to consider that the rope is a helix
of strands, therefore has no "inside"/"outside"), but I would really
prefer to know if there is a general formula for length of a spiral.
I'm not even sure I know how to parameterize a spiral -- the rope
notion was just to make sure we all knew what a spiral was.
John
|
889.5 | | CLT::GILBERT | | Tue Jun 21 1988 19:31 | 33 |
| It sounds like you want equal distances between successive 'loops'
of the spiral. Using polar coordinates, you'd have:
r(theta+2pi) = w + r(theta) (1)
A possible solution to this recursion is:
r(theta) = theta*w/(2pi) (2)
Then, you probably want to compute the *length* of the spiral, from
theta_0 to theta_1. That is,
theta_1
integral r(theta) d theta (3)
theta_0
Using (2), this equals:
w/(4pi) * ( theta_1^2 - theta_0^2 ) (4)
Now if we're just interested in N loops 'starting' at radius R,
we have theta_0 * w/(2pi) = R, and theta_1 = theta_0 + N * 2pi,
so substituting these into (4) gives the length as:
w/(4pi) * ( (theta_0+N*2pi)^2 - theta_0^2 )
= w/(4pi) * ( 4*N*pi * theta_0 + 4*N^2*pi^2 )
= w/(4pi) * ( 4*N*pi * 2*pi*R/w + 4*N^2*pi^2 )
= N * pi * ( 2*R + N*w )
Does this help?
|
889.6 | End of Part I | CHALK::HALLYB | The smart money was on Goliath | Wed Jun 22 1988 13:47 | 8 |
| That looks like the right answer, all right. Polar coordinates
sure make things easier, don't they.
For the special case N = 1, we have L = 2 * pi * (R + w/2);
as w --> 0 the spiral approaches a circle and the length approaches
the formula for circumference of a circle.
So the rope is about 165 feet long. Thanks.
|
889.7 | Arc-length is More Difficult | VAXRT::BRIDGEWATER | | Wed Jun 22 1988 15:54 | 47 |
| Re: .5
I think the problem of finding the arc-length of a spiral is a
bit more difficult. In what follows, I've used a somewhat
more compact form to make the equations manageable.
t = angle in radians
r(t) = radial length from pole (origin) to point.
Then the polar equation of a spiral is:
r(t) = c*t [ c = w/(2pi) where w is thickness of the rope ]
Now the arc-length, L, from t1 to t2 is:
t2
L = integral sqrt[r(t)^2 + r'(t)^2 ] dt
t1
where r'(t) represents the first derivative of r wrt t. So,
t2
L = integral sqrt[c^2*t^2 + c^2] dt or
t1
t2
L = [w/(2pi)] * integral sqrt[1 + t^2] dt
t1
| t=t2
L = [w/(4pi)] * [t*sqrt(1+t^2) + ln(t+sqrt(1+t^2)) |
| t=t1
If you want to measure the length of rope starting on the spiral
at the pole, then t1=0 and:
L = [w/(4pi)] * [t2*sqrt(1+t2^2) + ln(t2+sqrt(1+t2^2))]
If you want to express this in terms of number of revolutions of the
spiral, n, then make the substitution:
t2 = 2pi*n
- Don
|
889.8 | | CLT::GILBERT | | Wed Jun 22 1988 18:20 | 1 |
| Thanks. Yes, .5 is wrong (nice try, though).
|
889.9 | Solution to Original Problem | VAXRT::BRIDGEWATER | | Wed Jun 22 1988 23:47 | 23 |
| Re: .0 using equations in .7:
I didn't reread .0 so some of my assumptions in .7 were wrong.
Namely, the starting and ending angles of the spiral, t1 and t2,
should be specified by the following equations:
w = 0.5 inches
r(t1) = w/2pi*t1 = 3 inches
t2 = t1 + 2pi*30
Solving for t1 and t2 gives:
t1 = 12pi
t2 = 72pi
| t=t2
L = [w/(4pi)] * [t*sqrt(1+t^2) + ln(t+sqrt(1+t^2)) |
| t=t1
Evaluating yields approximately: 1979.27 inches
or: 164.94 feet
- Don
|