Let me apologize for the waste of bandwidth, but checking over my work I
found a semi-serious error that was put into the equations almost from
beginning. Because the two arcs face each other, their respective radii are
indexed negative to each other (that is, as one increases, the other
decreases). That was the mistake. I had both integration indices proceeding
the same direction.
The mistake was only semi-serious in effect because the differentical effect
was small in this case because r is approximately R. However, the world
should be kept safe for mathematics. With these new calculations, r is found
to be 15.8% bigger than R, rather than 5.7% as reported before.
Clearly, you should not try to solve these kinds of problems during working
hours (but it was enjoyable nonetheless). I've repeated all of the text
below, now double-checked for errors.
Wirt Atmar
==================================
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 arcs. The bottom arc is the
perfectly circular fence, R radius out from the center of the field. The top
arc 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 arcs 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 arc and a horizontal chord
at height y1 at its top. The upper area is bounded by the top arc 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 - (R-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 arcs 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 - (R-y)^2) dy
0
and
y2
a2 = 2 S sqrt(r^2 - y^2) dy
y1
Integrating, and substituting limits:
a1 = R^2*pi/2 - (R - y1) sqrt(2Ry1 - y1^2)
- R^2 arcsin((R - 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 = R^2*pi/2 - (R - y1) sqrt(2Ry1 - y1^2)
- R^2 arcsin((R - 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 arcs 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
arcs 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 + R^2 - 2*R*y1 + y1^2
Subtracting (2) from (1) gives us:
-r^2 = -2*R*y1
Isolating y1, we have:
y1 = r^2/2R
We are now cooking with petroleum distillate.
Substituting y1, the governing equation looks like:
R^2*pi/2 = R^2*pi/2 - (R - r^2/2R) sqrt(r^2 - (r^2/2R)^2)
- R^2 arcsin((R - r^2/2R)/R) + r^2*pi/2
- r^2/2R sqrt(r^2 - (r^2/2R)^2) - r^2 arcsin(r^2/2R/r)
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 = pi/2 - (1 - r^2/2) sqrt(r^2 - (r^2/2)^2)
- arcsin(1 - r^2/2) + r^2*pi/2
- r^2/2 sqrt(r^2 - (r^2/2)^2)
- r^2 arcsin(r^2/2r)
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 = 1.15873 R, a
seemingly reasonable result in that r is 16% bigger than R. The number seems
to check out -- and unless I've made a mistake somewhere again, it's my final
entry into the Grand "You Got My 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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