Pete Crosby writes:
>I have seen a couple of responses to this and, unless I am reading it
>wrong, the responders are reading it wrong. According to the above,
>the goat is tethered to an edge of the field and not to a post in the
>middle of the field. Therefore the length of the rope must be somewhat
>greater than the [radius] of the field in order for the goat to be able
>to cover half the surface.
Pete's dead right. And what's more worrisome is that this the second time in
two days that I've screwed up and significantly misread what someone has
written.
The "new" (and original) problem is schematicized below -- and is a little
bit harder to solve, but it can be broken up into a few component parts that
allow it to be solved in a few steps. Unfortunately, drawing a picture using
e-mail is difficult, so you'll have to use your imagination a bit to
interpret the drawing below.
The drawing is composed of two intersecting chords. The bottom chord is the
perfectly circular fence, R radius out from the center of the field. The top
chord is the radius of the tether, r radius out from the tether point on the
fence line.
limit of rope
. . . ____ y2 (y2 = r)
. . | field .
. . o center . . (y2-y1)
. |\ .____ y1
. . | \ .|.
. | \ R . | .
. r | \ . |
fence . | \ . | (y1-0) = y1
. | \ . |
. | \. |
. | . |
o--------------------- --- y = 0
tether point x1
The problem is to find the radius of the tether, r, such that the area
enclosed between the two chords is half the area of the circle specified by
the radius R. This will require the use of integral calculus, which is easy
enough, but we must perform two integrations over two distinct areas, a1 and
a2. The first area, a1, is bounded by the bottom chord and a horizontal line
at height y1 at its top. The upper area is bounded by the top chord and the
same horizontal line at height y1 at its bottom. The sum of the two areas, a1
+ a2, must equal half of pi*R^2, the area of the enclosed circle.
As Pete says, by inspection, we know that r must be greater than R.
Integrating an area is straightforward. At its simplest,
A = S da (using "S" as an integral symbol).
In our case, the differential area, da, is expressed by
da = x dy, where
x = 2 sqrt(R^2 - y^2) for the lower area, 0 <= y <= y1
and x = 2 sqrt(r^2 - y^2) for the upper area, y1 <= y <= y2
Note the difference between the two equations in the r's and R's. These
differences exist because the top and bottom chords are defined by different
radii. Rewriting the equations for x in the following manner allows us to
eliminate the x variable. Thus,
y1
a1 = 2 S sqrt(R^2 - y^2) dy
0
and
y2
a2 = 2 S sqrt(r^2 - y^2) dy
y1
Integrating, and substituting limits:
a1 = y1 sqrt(R^2 - y1^2) + R^2 arcsin(y1/R)
a2 = y2 sqrt(r^2 - y2^2) + r^2 arcsin(y2/r)
- y1 sqrt(r^2 - y1^2) - r^2 arcsin(y1/r)
Recognizing that r = y2, the second result simplifies to:
a2 = r^2*pi/2 - y1 sqrt(r^2 - y1^2) - r^2 arcsin(y1/r)
Thus, since
a1 + a2 = A/2 = R^2*pi/2
then
R^2*pi/2 = y1 sqrt(R^2 - y1^2) + R^2 arcsin(y1/R) + r^2*pi/2
- y1 sqrt(r^2 - y1^2) - r^2 arcsin(y1/r)
We've now got the answer, but it is unfortunately in terms of two variables.
We know R, the radius of the field, but we don't know r, the tether radius,
nor do we know y1, the height at which the two chords intersect. Thus we have
to develop a second independent equation. From the geometry of the situation,
we know these two equations to be true at the point (x1, y1) where the two
chords intersect:
R^2 = x1^2 + (r - y1)^2 (1)
r^2 = x1^2 + y1^2 (2)
Expanding the first equation leaves us with:
R^2 = x1^2 + y2^2 - 2*r*y1 + y1^2
Subtracting (2) from (1) gives us:
R^2 - r^2 = -2*r*y1
Isolating y1, we have:
y1 = (r^2 - R^2)/2r
We are now cooking with petroleum distillate.
Substituting y1, the governing equation looks like:
R^2*pi/2 = (r^2 - R^2)/(2r) sqrt(R^2 - ((r^2 -R^2)/(2r))^2)
+ R^2 arcsin((r^2 - R^2)/(2rR)) + r^2*pi/2
- (r^2 - R^2)/(2r) sqrt(r^2 - ((r^2 - R^2)/(2r))^2)
- r^2 arcsin((r^2 - R^2)/(2r^2))
We can make this a little simpler by normalizing the radius r into R units.
That is, we set R = 1. From this point on, r will be measured in units of R.
Doing this, the resulting equation is:
pi/2 = (r^2 - 1)/(2r) sqrt(1 - ((r^2 - 1)/(2r))^2)
+ arcsin((r^2 - 1)/(2r)) + r^2*pi/2
- (r^2 - 1)/(2r) sqrt(r^2 -((r^2 - 1)/(2r))^2)
- r^2 arcsin((r^2 - 1)/(2r^2))
We now have an equation in terms of only one variable, r. While there may be
a closed form solution to this, now is the time to do what Newton would have
done -- solve the equation by the numerical method of iterated approximation
rather than fuss with what will clearly explode into a mess.
That's exactly what I did. I used Equater, a set of scientific, financial,
programmer's and statistical calculators that were programmed up for the
HP3000 ten years ago (if you are an Adager, Robelle, or AICS customer, you
probably have a copy of Equater on your HP3000 somewhere).
By iterated approximation, I obtained the result that r = 1.057 R, a somewhat
surprising result in that r is only 6% bigger than R. However, the number
seems to check out -- and unless I've made a mistake somewhere, it's my entry
into the Grand Goat Sweepstakes.
My earlier comment about the actual length of the tether still holds,
however, if we're going to do this "perfectly." The actual length, r', must
be:
r' = r - width of goat's dentition
- distance of muzzle anchor to
one side of the goat's teeth
except that now, all goat head width's are measured in R units.
Wirt Atmar
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