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27 Oct 16
A farmer owns a circular field which has a radius of 20m.
He wishes to tether a goat to a post on the edge of the field (ie, the circumference of the circle) with a rope so that the goat will have access to exactly half of the field.
How long should he make the rope?
28 Oct 16
Originally posted by Blood On The TracksI've seen this before I think.
A farmer owns a circular field which has a radius of 20m.
He wishes to tether a goat to a post on the edge of the field (ie, the circumference of the circle) with a rope so that the goat will have access to exactly half of the field.
How long should he make the rope?
The divisions are going to be "half moon" shaped but my geometry knowledge is not good enough to apply to the problem
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28 Oct 16
Originally posted by Andrew KernHi
Im probably not right, bu in my minds eye it looks correct.
If the radius is A and the circumference is B, then wouldnt you need to do A+.5A=Half the circle?
I think you are saying, for my 20m radius field, that it should be 30m rope?
Afraid that isnt correct ~ would be too long, letting goat eat much more than half
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31 Oct 16
4 edits
Originally posted by Blood On The TracksThere probably is a cleaner way. ( to be honest I'm not going to find a numerical solution...because I'm lazy and I don't suppose you are after a numerical approximation, However, this should yield the answer if I followed through)
A farmer owns a circular field which has a radius of 20m.
He wishes to tether a goat to a post on the edge of the field (ie, the circumference of the circle) with a rope so that the goat will have access to exactly half of the field.
How long should he make the rope?
Let the radius of the field = R
Length of rope = L
The equation that needs to be solved:
1/2* Field Area = Area of Circular sector of radius "L" subtended by angle φ(L,R) + 2* Area Circular segment of Chord Length "L" subtended by angle ß(L,R)
1/2*π*R² = 1/2*arccos [L/( 2*R )]*L² + R²*( arcsin[ L/R*√(1 - ( L/( 2*R))² ) ] - L/R*√( 1 - ( L/( 2*R))² ))
Solve for L numerically ( maybe there is a compact analytic solution I'm missing?)
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01 Nov 16
1 edit
Originally posted by joe shmoWell, I went through with it. I found a small error in my equation when I tried to solve it.
There probably is a cleaner way. ( to be honest I'm not going to find a numerical solution...because I'm lazy and I don't suppose you are after a numerical approximation, However, this should yield the answer if I followed through)
Let the radius of the field = R
Length of rope = L
The equation that needs to be solved:
1/2* Field Area = Area of Ci ...[text shortened]... 2*R))² ))
Solve for L numerically ( maybe there is a compact analytic solution I'm missing?)
Corrected version:
1/2*π*R² = arccos [L/( 2*R )]*L² + R²*( arcsin[ L/R*√(1 - ( L/( 2*R))² ) ] - L/R*√( 1 - ( L/( 2*R))² ))
The answer is about 23.175 m.
23 Dec 16
Originally posted by joe shmodang! thats a lot. Neveer couldve figured that out nice work.
Well, I went through with it. I found a small error in my equation when I tried to solve it.
Corrected version:
1/2*π*R² = arccos [L/( 2*R )]*L² + R²*( arcsin[ L/R*√(1 - ( L/( 2*R))² ) ] - L/R*√( 1 - ( L/( 2*R))² ))
The answer is about 23.175 m.