Info
이 질문은 마감되었습니다. 편집하거나 답변을 올리려면 질문을 다시 여십시오.
The code below generates random points inside a regular hexagon. I need the second coordinate y of the points to be an angle from 0:2pi. How can I modify the code please?
조회 수: 1 (최근 30일)
이전 댓글 표시
numEdges = 6;
R = 8;
xVertex = R * cos((0:6)*pi/3);
yVertex = R * sin((0:6)*pi/3);
xVertex = [xVertex , xVertex(1)];
yVertex = [yVertex , yVertex(1)];
requiredPoints = 20;
plot(xVertex, yVertex, 'b+-', 'LineWidth', 3);
grid on;
numPointsIn = 1;
while numPointsIn < requiredPoints
testx = 2 * R * rand(1) - R;
testy = 2 * R * rand(1) - R;
if inpolygon(testx, testy, xVertex, yVertex)
x(numPointsIn) = testx;
y(numPointsIn) = testy;
numPointsIn = numPointsIn + 1;
end
end
hold on;
plot(x,y,'r+', 'MarkerSize', 10, 'LineWidth', 2);
댓글 수: 0
답변 (1개)
Image Analyst
2018년 1월 21일
Something like
yAngle = atan2(y, x);
댓글 수: 7
Image Analyst
2018년 1월 21일
yAngle is in the range 0-2*pi, not -8 to 8. y is in that range, but not the angles.
To have the angle show up on the graph, use the sprintf() and text() functions.
str = sprintf('yAngle(1) = %f', yAngle(1));
text(1., .2, str);
Greg
2018년 1월 22일
I think the desired output is
yAngle = atan2(y, x) + pi;
plot(x,yAngle,'r+', 'MarkerSize', 10, 'LineWidth', 2);
이 질문은 마감되었습니다.
참고 항목
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!