grazing goat problem, how much of the area can the goat graze ?

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pauli relanderi 2021-10-16

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https://ww2.mathworks.cn/matlabcentral/answers/1565301-grazing-goat-problem-how-much-of-the-area-can-the-goat-graze

编辑： DGM 2021-10-16

采纳的回答： DGM

Hello !

Im just wondering does anyone know how would i go about solving this in matlab

How much of the fields area can it reach to eat ?

when

r is 0.8 %goats leash

R is 1 %the big circler radius

ive tried calculations i found online and got nowhere

This is what ive gotten

clear all

clc

r=0.8

R=1

d=R

alfa=acos(r./(2.*R))

A=(r.^2-(2.*R.^2)).*alfa+R.^2.*pi-R.^2.*sin(2*alfa)

Buts this gets me answer of 0.8318 which is too much it should be around 2-3, atleast i think it should

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Star Strider 2021-10-16

What is the problem statement, and the goat’s constraints (other than the leash length)?

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Answer 1

DGM 2021-10-16

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https://ww2.mathworks.cn/matlabcentral/answers/1565301-grazing-goat-problem-how-much-of-the-area-can-the-goat-graze#answer_810066

编辑：DGM 2021-10-16

在 MATLAB Online 中打开

Your answer is reasonable. As a point of verification:

rf = 1;
rg = 0.8;
% intersection point locations
% could also use circcirc()
xint = (2*rf^2 - rg^2)/(2*rf);
yint = sqrt(rf^2 - xint^2);
% sector areas
secareaf = rf^2 * asin(yint/rf);
secareag = rg^2 * asin(yint/rg);
% sum of triangle areas
triarea = yint*(xint + sqrt(rg^2 - rf^2 + xint^2));
% total area
totalarea = secareaf + secareag - triarea
totalarea = 0.8318

Furthermore, it's helpful to check things visually. Consider the simplified case where rf = 1 and rg = 1.

Given that the area of either circle = pi, then visually we know Aintersection < pi/2 ~ 1.5708

and considering the inscribed triangles, we know Aintersection > sqrt(3)/2 ~ 0.8660

Running the above code for this second case gives a result of 1.2284, which is indeed within those boundaries. Looking back at the original problem with the smaller rg, we should expect Aintersection to be correspondingly smaller than 1.2, even if the geometry becomes less obvious.