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Then I asked: What if the goat is tethered to a round silo with a radius of 10 feet? It is not quite so straightforward, is it? No longer do you have definite lengths of rope at succeeding corners. Now you will find that after a half circle (yellow area) the rope will begin to shorten continually as it wraps around the silo. This curve is the involute of a circle. The area under this curve is obtained by adding up an infinite number of infinitesmally small sectors of a circle, with the radius decreasing a bit for each one, as shown in the diagram. There are a couple of other ways to look at it but all my sources, after doing the calculus to show how it is done, bring it down to one equation: Ai=L^3/6R. This is for just the one side involute. To get the whole area we double this and add the free semicircle. The resultant
equation is A=R^2*pi/2+2*(L^3/6R).


But what if the rope or tether is longer than piR? This makes it even more complicated. Just like the square barns, if the rope is longer than 1/2 the perimeter there will be an overlap.  Now the nice neat equation will not work. We will have to actually perform the integration. But we will have to stop the integration at the point of intersection to avoid the overlap, and do other calculations before we can obtain the total area the goat can graze. The problem then is to find the point of intersection mathematically. It was a matter of trial and error so I turned to BASIC, which I have not used for many years.

 As I have previously demonstrated the rope can describe a 1/2 circle freely , but as it begins to wrap around the silo it shortens. At each of several steps we can calculate the area of a portion of a circle by using the remaining rope at that point as the radius. The formula for a sector is: sector area = (pi * remaining^2)*step/360.

With BASIC I calculate as many as 180,000,000 sectors of decreasing radius, which is a pretty close to an "infinite number of infinitesimally small sectors of a circle." Close enough so that my answer agrees with all the professional mathematicians to the third decimal place, which is as far as anyone reports.
Having satisfied myself that my BASIC program produced comparable results by using a rope of 1/2 circumference, I began to add my special calculation to find the overlap area. The program calculates thousands of very smalIareas and accumulates them into a bigarea. I saw that the remaining rope as it wound around the silo always made a right angle with a radius. This gave me two sides of a right triangle, which gave me also the hypotenuse. I imagined that entire triangle rotating about the center of the silo with the rope and hypotenuse shortening and the angle A getting smaller and smaller. The program keeps track of that angle. When Angle A + rotation = 180 it knows my hypotenuse is coincident with the center line of the silo. so it stops the rotation and  figures the resultant triangle, adds that to the accumulated area and subtracts the sector of the silo within the triangle.

Once I found the rotation angle, I plugged that into the regular calculus formula and came up with the same answer as the professional calculator Wolfram's Alpha, for another proof that my system works. My BASIC program and comparison of results can be seen below

LET starttime=time
PRINT time$;" "; date$
print "pi = ";pi
INPUT prompt "radius of silo ":radius
INPUT prompt "length of rope ":rope
input prompt "lower limit ":lower
input prompt "upper limit ":upper

PRINT "working"
LET wrap = radius
LET bigarea = 0
LET supplement=0
LET degrees = 180/pi
LET step=.000001

FOR theta = lower to upper step step
LET usedrope = wrap * theta
LET remaining = rope - usedrope
LET smallarea = (remaining^2*pi)*step/(2*pi)

LET bigarea = bigarea + smallarea
LET involute=bigarea
LET supplement=pi-theta
LET angleA = atn(remaining/radius)
if abs(theta-tan(theta)-(rope/radius))<.00001 then exit for !theta-tantheta=L/R

NEXT theta

LET trianglearea=remaining*radius/2
LET sectorarea=radius^2*pi*supplement/(2*pi)
LET involutearea=involute+trianglearea-sectorarea
LET semicircle=rope^2*pi/2
LET totalarea =involutearea*2+semicircle
LET lineangle = theta + angleA
print remaining
print usedrope
print supplement
print radius
PRINT "rotation angle is"; theta
PRINT "Angle A = ";angleA
PRINT "Angle A + theta = ";lineangle
PRINT "involute to"; rotation;" = ";
PRINT USING ("####.####"): involute
PRINT "add final triangle area +";trianglearea
PRINT "subtract sector area -";sectorarea
PRINT "equals final involute =";involutearea
PRINT "double this ="; involutearea*2
PRINT "Add semicircle +";semicircle
PRINT "Total Area ="; totalarea
LET stoptime=time
PRINT "Program took";(stoptime-starttime)/60;" minutes"

Results for default rope at radius*pi
compared to Wolfram Alpha
My program
radius of silo = 10
length of rope = 31.415927
rotation angle is 179.6697
Angle A = .33030236
Angle A + rotation = 180.
involute to 179.6697 = 516.7713
Wolfram Alpha

double this = 1033.5426
Add semicircle + 1550.3139
Total Area = 2583.8565

My special program with breakout
radius of silo = 10
length of rope = 50
rotation angle is 107.777
Angle A = 72.223001
Angle A + rotation = 180.
involute to 107.777 = 1577.662

add final triangle area + 155.94683
subtract sector area - 63.026458
equals final involute = 1670.5824
double this = 3341.1647
Add semicircle + 3926.9908
Total Area = 7268.1555

When I submitted this program to Dr. Peterson, of Drexel University's Dr. Math program, he replied with the following equation to find the stopping point: theta-tan(theta)=L/R. This returns many values, but only one will fit since it must be between pi/2 and pi radians.

 Dr. Peterson suggested that my BASIC routine is unnecessary, but some sort of trial and error method must still be used with this equation.

The conclusion is that if the tether is longer than 1/2 the circumference of the circle the problem cannot be solved by just one equation.

To answer Phil's question go to Silo Demo









In my research I ran across student papers from a class by J. Wilson of the University of Georgia. He posed a combination of the square barn problem and the silo problem. Take a look, but be careful. On the silo portion most of them are right in principle, but they don't go far enough for a really good estimate.