Arc center point
Author
If I have the start and end points of an arc is there a way to find the center point, which is the center of rotation of the arc? Thanks.
There are infinite solutions to a problem like that. Don't you have anything more than a starting point and an end point for the arc?
Author
Oh right. Actually I know the plane in which the arc is going to be drawn, I have all the vectors that establish the user coordinate system and normal of the arc. I am just having trouble finding the center point of rotation on the same plane that has the start/end points of the arc. That should only have a single solution right?
A fun problem :)
a = start spoint
b = end point
c = center point
A triangle (a, b, c). One side has the length same as the distance between a and b. The other two sides have the length equal to the radius. The direction of the side from a to c is the same as the normal for the arc.
If you draw this on paper you should probably be able to solve it.
a = start spoint
b = end point
c = center point
A triangle (a, b, c). One side has the length same as the distance between a and b. The other two sides have the length equal to the radius. The direction of the side from a to c is the same as the normal for the arc.
If you draw this on paper you should probably be able to solve it.
Author
Not really sure what you mean by...
"The direction of the side from a to c is the same as the normal for the arc."
When you say direction of the side from a to c, are you talking about a vector? If you are then how is it the same as the normal, when the normal of the arc is a vector perpendicular to the plane that contains the arc.
Ok, I never heard of a normal to an arc before so I thought that it was the gradient for the start point. Well then we are bavk to the original state.
Consider the triangle in my last post and place it in a plane that has the normal as specified. With the info given, the radius which is equal to the length of side ac and the length of side bc can be any positive value bigger than half of the length of side ab and hence there are infinite number of solutions.
Consider the triangle in my last post and place it in a plane that has the normal as specified. With the info given, the radius which is equal to the length of side ac and the length of side bc can be any positive value bigger than half of the length of side ab and hence there are infinite number of solutions.
Author
Well actually bc would be the radius also and so you would end up with only one solution, not infinite.
Author
What if you were to set the condition that bc and ac had to be the radius, would that still result in infinitely many solutions?
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