I am writing a program in which I need to draw polygons of an arbitrary number of sides, each one being translated by a given formula which changes dynamically. There is some rather interesting mathematics involved but I am stuck on this probelm.
How can I calculate the coordinates of the vertices of a regular polygon (one in which all angles are equal), given only the number of sides, and ideally (but not neccessarily) having the origin at the centre?
For example: a hexagon might have the following points (all are float
s):
( 1.5 , 0.5 *Math.Sqrt(3) )
( 0 , 1 *Math.Sqrt(3) )
(-1.5 , 0.5 *Math.Sqrt(3) )
(-1.5 , -0.5 *Math.Sqrt(3) )
( 0 , -1 *Math.Sqrt(3) )
( 1.5 , -0.5 *Math.Sqrt(3) )
My method looks like this:
void InitPolygonVertexCoords(RegularPolygon poly)
and the coordinates need to be added to this (or something similar, like a list):
Point[] _polygonVertexPoints;
I'm interested mainly in the algorithm here but examples in C# would be useful. I don't even know where to start. How should I implement it? Is it even possible?!
Thank you.
for (i = 0; i < n; i++) {
printf("%f %f\n",r * Math.cos(2 * Math.PI * i / n), r * Math.sin(2 * Math.PI * i / n));
}
where r
is the radius of the circumsribing circle. Sorry for the wrong language No Habla C#.
Basically the angle between any two vertices is 2 pi / n and all the vertices are at distance r from the origin.
EDIT: If you want to have the center somewher other than the origin, say at (x,y)
for (i = 0; i < n; i++) {
printf("%f %f\n",x + r * Math.cos(2 * Math.PI * i / n), y + r * Math.sin(2 * Math.PI * i / n));
}