# Horse Pen

• Jun 21st 2011, 04:49 AM
twr318
Horse Pen
Hello Folks,
Let me begin by stating I'm not good at mafamatics! I know this will be very simple for probably everyone on this forum. I just need to know.
I have 12 - 12' metal panels. They are straight. They will be connected end to end. What is the radius of this configuration?
• Jun 21st 2011, 05:32 AM
godelproof
Re: Horse Pen
• Jun 21st 2011, 06:12 AM
twr318
Re: Horse Pen
Thanks! Could you tell me how to calculate this for future reference?
• Jun 21st 2011, 06:50 AM
HallsofIvy
Re: Horse Pen
Strictly speaking the term "radius" only applies to circles. And one way to approximate this is to assume that you are talking about a circular pen with circumference 12(12)= 144 feet. For a circle the circumference is given by $\displaystyle 2\pi r$ so we must have $\displaystyle 2\pi r= 144$ so that $\displaystyle r= \frac{72}{\pi}= 22.91$ feet or 22 feet, 11 inches. That is 3 inches (about 1%) shorter than godelproof's answer.

What he did was imagine drawing lines from the center of the pen to the ends of each panel. That divides the 12 sided polygon into 12 isosceles triangles each with a base length of 12 feet and an angle of 360/12= 30 degrees. Now draw a line from the center of the pen to the center of each panel. We now have 24 right triangles with one leg of length 6 and opposite angle 30/2= 15 degrees. Since "sine" is "opposite side/ hypotenuse", taking the distance from the center of the pen to the end of each panel to be x, we have $\displaystyle \frac{6}{x}= sin(15)$ so that $\displaystyle x= \frac{6}{sin(15)}$. Using a calculator to find that sin(15)= 0.25882, we have $\displaystyle x= \frac{6}{0.25882}= 23.18$ feet of 23 feet and 2 inches as godelproof said.

As I said before, strictly speaking the term "radius" only applies to circles. What godelproof calculated was the "ex-radius", the radius of the circle that passes through the vertices of the polygon. One could just as well use the "in-radius", the radius of the circle inscribed in the polygon, that is tangent to each panel at its center. That radius is the other leg of the right triangle so is given by $\displaystyle tan(15)= \frac{6}{x}$ so that $\displaystyle x= \frac{6}{tan(15)}$. tan(15)= 0.26795 so that $\displaystyle x= \frac{6}{0.25795}= 22.4$ feet or 22 feet 4 inches.

For small angles, sine and tangent are almost the same so for polygons with many sides, the "in-radius" and "ex-radius" are almost the same- and my circle approximation is between them.
• Jun 21st 2011, 07:00 AM
godelproof
Re: Horse Pen
Sure. Suppose you have N panels, each M feet long, then your pen will have a radius R given by:

R=$\displaystyle \frac{M}{2sin(\pi/N)}$.
• Jun 21st 2011, 07:05 AM
godelproof
Re: Horse Pen
Quote:

Originally Posted by HallsofIvy
Strictly speaking the term "radius" only applies to circles. And one way to approximate this is to assume that you are talking about a circular pen with circumference 12(12)= 144 feet. For a circle the circumference is given by $\displaystyle 2\pi r$ so we must have $\displaystyle 2\pi r= 144$ so that $\displaystyle r= \frac{72}{\pi}= 22.91$ feet or 22 feet, 11 inches. That is 3 inches (about 1%) shorter than godelproof's answer.

What he did was imagine drawing lines from the center of the pen to the ends of each panel. That divides the 12 sided polygon into 12 isosceles triangles each with a base length of 12 feet and an angle of 360/12= 30 degrees. Now draw a line from the center of the pen to the center of each panel. We now have 24 right triangles with one leg of length 6 and opposite angle 30/2= 15 degrees. Since "sine" is "opposite side/ hypotenuse", taking the distance from the center of the pen to the end of each panel to be x, we have $\displaystyle \frac{6}{x}= sin(15)$ so that $\displaystyle x= \frac{6}{sin(15)}$. Using a calculator to find that sin(15)= 0.25882, we have $\displaystyle x= \frac{6}{0.25882}= 23.18$ feet of 23 feet and 2 inches as godelproof said.

As I said before, strictly speaking the term "radius" only applies to circles. What godelproof calculated was the "ex-radius", the radius of the circle that passes through the vertices of the polygon. One could just as well use the "in-radius", the radius of the circle inscribed in the polygon, that is tangent to each panel at its center. That radius is the other leg of the right triangle so is given by $\displaystyle tan(15)= \frac{6}{x}$ so that $\displaystyle x= \frac{6}{tan(15)}$. tan(15)= 0.26795 so that $\displaystyle x= \frac{6}{0.25795}= 22.4$ feet or 22 feet 4 inches.

For small angles, sine and tangent are almost the same so for polygons with many sides, the "in-radius" and "ex-radius" are almost the same- and my circle approximation is between them.

That's really all one can say about the problem!! Nice!