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Finding the center of an n-gon (circle) based on angle and side-length

by STENDEC
Tags: angle, based, circle, ngon, sidelength
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STENDEC
#1
Jan7-13, 08:38 AM
P: 14
I hope this is self-evident to someone, i'm struggling.

I have a program that draws circles (n-gons really) of various sizes, but by translating-rotating-translating-rotating-..., not by x=sin/y=cos. That works as intended, but my wish is to offset the circle so that its center is (0,0) in the coordinate system. For that i need its center. Currently the circle itself originates from- and hence touches the (0,0) coordinates, so its center is somewhere above, in the y-axis.



Position of ? is sought after. A wider angle would result in ? rising for instance.

I found lots of tutorials on how to do it on paper using dividers and i also considered that it's a isosceles triangle, but it seems all textbook examples assume that one of the symmetric sides is already known.
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jedishrfu
#2
Jan7-13, 09:32 AM
P: 2,781
you could use similar triangles and some trig to get the radius along the y-axis.

Notice you can extend a perpendicular bisector from the first n-gon side which intersects the y-axis

so that 1/2 the n-gon side is the short edge the perpendicular creates the right angle and the y-axis is the hypotenuse.

This triangle is similar to the one formed by the n-gon edge and the x-axis.

So I get something like:

radius along y-axis = (1/2 n-gon side) / sin theta
STENDEC
#3
Jan7-13, 10:09 AM
P: 14
Quote Quote by jedishrfu View Post
you could use similar triangles and some trig to get the radius along the y-axis.
Yes, you're right. After some more reading and pondering i came to this solution:

[itex]\alpha =[/itex] angle in degrees
[itex]s =[/itex] segment length

To get the inner angle between the sides, we subtract from a half-circle. We then divide by two, to get the inner angle of the isosceles triangle:
[itex]\beta = (180 - \alpha) \div 2[/itex]

degrees to radians:
[itex]\phi = \beta\times\frac\pi{180}[/itex]

Distance to center point can then be gotten from [itex]s\div 2 * tan(\phi)[/itex].

Edit: Just saw you extended your reply, oh well :)

jedishrfu
#4
Jan7-13, 11:25 AM
P: 2,781
Finding the center of an n-gon (circle) based on angle and side-length

Quote Quote by STENDEC View Post
Yes, you're right. After some more reading and pondering i came to this solution:

[itex]\alpha =[/itex] angle in degrees
[itex]s =[/itex] segment length

To get the inner angle between the sides, we subtract from a half-circle. We then divide by two, to get the inner angle of the isosceles triangle:
[itex]\beta = (180 - \alpha) \div 2[/itex]

degrees to radians:
[itex]\phi = \beta\times\frac\pi{180}[/itex]

Distance to center point can then be gotten from [itex]s\div 2 * tan(\phi)[/itex].

Edit: Just saw you extended your reply, oh well :)
Glad you figured it out.


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