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I would like to know how to get a specific point on the circumference of a circle, given an angle. The diameter of the circle is 1, and the center point of the circle is { X: 0.5, Y: 0.5 }.

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You can work this out using basic trigonometry. http://www.freehomeworkmathhelp.com/Trigonometry/Trigonometry_Introduction/trigonometry.html

Tan(angle) = Opposite / Adjacent
Sin(angle) = Opposite / Hypotenuse
Cos(angle) = Adjacent / Hypotenuse

I always remember the above as

The Old Arab
Sat On His
Camel And Howled

The above means if we have the angle and one length of a right-angled triangle we can work out the lengths of the other sides. Luckly your problem can be thought of as calculating the length of triangle sides:

Circle Triangle Image

Above, r is the hypotenuse, x is the adjacent and y is the opposite.

So for x:

Cos(a) = x / r
Cos(a) * r = x
x = Cos(a) * r

And for y:

Sin(a) = y / r
Sin(a) * r = y
y = Sin(a) * r

This is assuming a circle at (0, 0), so we just add on the circle's center.

radius = 1;
CenterX = 0.5;
CenterY = 0.5;

x = Cos(angle) * radius + CenterX;
Y = Sin(angle) * radius + CenterY;

Note: The C# Math functions use angles in radians, so if you have degrees convert them first:

radians = degrees * Math.PI / 180
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    \$\begingroup\$ Keep in mind that the angle should be in radians. \$\endgroup\$
    – bummzack
    Oct 10 '11 at 11:21
  • \$\begingroup\$ Depends on the math library, but good point, XNA/.Net uses radians. \$\endgroup\$ Oct 10 '11 at 11:27
  • \$\begingroup\$ Why the old Arab was to howl? Here we are not all English speakers so a conceptual memonic rule is far better than a mnemonic sentence \$\endgroup\$
    – FxIII
    Oct 10 '11 at 12:38
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    \$\begingroup\$ @FxIII: True, it's not the most popular memonic, however i've always found it easy to remember than SOHCAHTOA as it's unusual. \$\endgroup\$ Oct 10 '11 at 13:01
  • \$\begingroup\$ Or you could just look it up, once you are familiar with the concept, the once or twice a year that you need it. \$\endgroup\$
    – Tetrad
    Oct 10 '11 at 15:29

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