Goat grazing inside a circular fence.

Fusion_power · 2003-09-05, 06:06

This is a similar problem to the one about the cow grazing beside the silo.

There is a goat that is inside a circular fence. But because the goat is always jumping the fence and visiting distant but lush pastures, the owner decided to tie him inside the pen so he can't leave. The fence is 100 ft in diameter The goats tether is exactly 60 feet long and is attached to a fence post. Presume the goats head is at the end of the tether for purposes of solving this equation. The tether is just enough of a nuisance that the goat does not jump the fence.

If the goat eats all the grass it can reach on the first day and the owner then ties him to a fence post exactly opposite (across the circle from) the one he was previously tied to, how much of the grass (area) will the goat be unable to graze over the two days?

Fusion

dweick · 2003-09-23, 23:04

I decided to solve it using software approximation.

Visualize a 100' radius circle within a box 100' on a side. Make the upper left hand corner of the square (0,0). The center of the grassy area is at (50,50). You have one goat rope tethered at (0,50) and a second tethered at (100,50).

For every point on the 100x100 grid compute the distance between the point at the three targets (above). If the distance is less than 50' to the center target and the distance to each goat tether point is greater than 60' then add 1 sq ft to an accumulator.

To make the software converge better I used a dx and dy to further divide the grid. Using 0.001 for dx and dy (so I'm determining whether each 0.00001 sq ft patch in the 100x100 grid is an area of interest I came up with 442.7485 sq ft. I'll round that off to 442.75 as my answer.

Takes 30 secs or so on my 700MHz PIII laptop to make a run at the above resolution. If I use 0.1ft dx, dy I get 442.59 sq ft.

Fusion_power · 2003-09-25, 04:40

Interesting solution.

The entire circle has an area of:
3.14 * 50 * 50 = 7850 square feet.

If the goat can't reach 443 square feet, then 7850 - 443 = 7407 square feet that the goat can graze.

Somewhere around here I have a pi calculator based on a similar principle to the method you used.

Fusion

2003-09-05, 06:06	#1
Fusion_power Aug 2003 Snicker, AL 7·137 Posts	Goat grazing inside a circular fence. This is a similar problem to the one about the cow grazing beside the silo. There is a goat that is inside a circular fence. But because the goat is always jumping the fence and visiting distant but lush pastures, the owner decided to tie him inside the pen so he can't leave. The fence is 100 ft in diameter The goats tether is exactly 60 feet long and is attached to a fence post. Presume the goats head is at the end of the tether for purposes of solving this equation. The tether is just enough of a nuisance that the goat does not jump the fence. If the goat eats all the grass it can reach on the first day and the owner then ties him to a fence post exactly opposite (across the circle from) the one he was previously tied to, how much of the grass (area) will the goat be unable to graze over the two days? Fusion

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2003-09-23, 23:04	#2
dweick Apr 2003 2×5 Posts	I decided to solve it using software approximation. Visualize a 100' radius circle within a box 100' on a side. Make the upper left hand corner of the square (0,0). The center of the grassy area is at (50,50). You have one goat rope tethered at (0,50) and a second tethered at (100,50). For every point on the 100x100 grid compute the distance between the point at the three targets (above). If the distance is less than 50' to the center target and the distance to each goat tether point is greater than 60' then add 1 sq ft to an accumulator. To make the software converge better I used a dx and dy to further divide the grid. Using 0.001 for dx and dy (so I'm determining whether each 0.00001 sq ft patch in the 100x100 grid is an area of interest I came up with 442.7485 sq ft. I'll round that off to 442.75 as my answer. Takes 30 secs or so on my 700MHz PIII laptop to make a run at the above resolution. If I use 0.1ft dx, dy I get 442.59 sq ft.

2003-09-25, 04:40	#3
Fusion_power Aug 2003 Snicker, AL 7×137 Posts	Interesting solution. The entire circle has an area of: 3.14 * 50 * 50 = 7850 square feet. If the goat can't reach 443 square feet, then 7850 - 443 = 7407 square feet that the goat can graze. Somewhere around here I have a pi calculator based on a similar principle to the method you used. Fusion