| As it happens, this just came through the USENET.
Topher
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From: [email protected] (Dave Boyd)
Newsgroups: sci.math
Subject: Re: Dog chained to a silo.
Message-ID: <[email protected]>
Date: 16 Mar 92 17:52:32 GMT
References: <[email protected]>
Organization: Hewlett-Packard, Greeley, CO
In sci.math, [email protected] (Dave Boyd) writes:
> L = length of the rope
> R = the _radius_ of the silo
> The area is then
> pi * L^2 L^3
> A = -------- + -----
> 2 3*R
> Dave Boyd
> Hewlett-Packard, Greeley, Colorado
> Standard Disclaimers Apply
Thanks to those who have pointed out that I solved the problem
only for the case that L <= pi*R. The original poster gave values
of L=10 and R=3.5, so I did solve his problem.
As L gets (very) large, A should approach pi * L^2. Maybe
someone would care to post a general solution?
Dave Boyd
Hewlett-Packard, Greeley, Colorado
Standard Disclaimers Apply
|
| I found that this wasn't too hard.
Below we see the silo. The cow is tied to the barn at the '*', and has
gone counter-clockwise, wrapping the leash around the barn.
Cow
\
\
/-----\
/ \
| |
( � *
| |
\ /
\-----/
We start the leash *tangent* to the silo, move it counter-clockwise
until the cow touches the silo, and consider the area covered.
Let 't' be the angle (around the center of the silo) from the '*' to
the point of tangency (between the leash and the silo), and let 'T'
be the angle at which the cow meets the silo. Let 'x(t)' and 'y(t)'
be the cow's coordinates as a function of 't'.
Recall that the silo's radius is R, and the leash has length L.
Then T = 2 pi L.
At angle t, t R of the leash is against the silo, and L - t R is in
the field, tangent to the silo.
The point of tangency is (R cos(t), R sin(t)).
The cow is at (R cos(t) - (L-tR) sin(t), R sin(t) + (L-tR) cos(t)).
We recognize this as (almost) the form: (cos(a+b),sin(a+b)); normalize
it first by multiplying and dividing by D = sqrt(R�+(L-tR)�); the cow
is at D (R/D cos(t) - (L-tR)/D sin(t), R/D sin(t) + (L-tR)/D cos(t)) =
D (cos(acos(R/D)+t), sin(acos(R/D)+t), and we see it's at distance D
from the center of the silo.
-
We wish to integrate / D�/2 du around the silo, where 'u' is the angle
to the COW; using the angle to the point of tangency would be incorrect.
Since u = acos(R/D)+t, du = (L-tR)�/( (L-tR)� + R� ) dt.
Int D�/2 du = Int (R�+(L-tR)�)(L-tR)�/( (L-tR)� + R� )/2 dt
= Int (L-tR)�/2 dt = (R�t� - 3LRt� + 3L�t)/6.
So plug in the bounds from 0 to T (T = 2 pi L) and multiply by 2,
or use the bounds of -2 pi L to +2 pi L, and you'll get L�/(3R).
Of course we need to add in the semicircle for which the leash isn't
tangent at all. That adds pi L�/2 to the area.
And I suspect the cow has no access to the interior of the silo.
Since the integral included it, we must subtract pi R� (L/(pi R)) = L R.
So the cow� grazes on an area of: L�/(3R) + pi L�/2 - L R.
Unless L > pi R, in which case we should only integrate to u = pi.
But t is our relevant variable; we find that u = pi when
(L-tR)/R = tan(t). C'est domage et bon chance.
� Unless of course it was a goat.
|
| re .6
> So plug in the bounds from 0 to T (T = 2 pi L) and multiply by 2,
> or use the bounds of -2 pi L to +2 pi L, and you'll get L�/(3R).
>
> Of course we need to add in the semicircle for which the leash isn't
> tangent at all. That adds pi L�/2 to the area.
>
> And I suspect the cow has no access to the interior of the silo.
> Since the integral included it, we must subtract pi R� (L/(pi R)) = L R.
>
> So the cow grazes on an area of: L�/(3R) + pi L�/2 - L R.
re .4
> L = length of the rope
> R = the _radius_ of the silo
>
> The area is then
>
> pi * L^2 L^3
> A = -------- + -----
> 2 3*R
There seems to be a conflict between these two answers.
FWIW I agree with the second answer (by algebra/calculus somewhat different
from that in .6).
I am suspicious of the answer of .6. Consider the case R=1, L small, say
0.01. The area will be negative???
|
| re .8
>The mistake in .6 is that the area in the triangular area formed by the center
>of the silo and the two endpoints of the semicircle was never added in. This
>area is 0.5 * 2*L * R = L*R which cancels the -L*R term in .6.
Yep, that's it.
re .6
>But t is our relevant variable; we find that u = pi when
>(L-tR)/R = tan(t). C'est dommage et bon chance.
Indeed this can't be solved for t in closed form using 'conventional'
trig functions. However it can be seen that as L->oo, t->pi/2 and this
makes the total area/(pi L�) -> 1 as is intuitively reasonable.
|