Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a line $100$ $mm$ long and I want to draw an arc from endpoint to endpoint with a height of 3mm.

I use this formula to find the radius of the arc $$\frac{W . W}{ 8 H} + \frac{H} {2}= 418.1667.$$

But the CAD software I'm using, FreeCAD, uses radius and the angles for the start and endpoints to draw the arc. How do I find the start and end angles that will give me a 100mm straight line between them?

share|cite|improve this question
up vote 1 down vote accepted

See the picture below. From Pythagoras, $R^2=50^2+(R-3)^2,0=2500-6R+9,R=\frac {2509}6 $ confirming your value. Then $\sin \theta=\frac {300}{2509}, \theta = \arcsin \frac{300}{2509}\approx 6.867^\circ$. Since the tangent is perpendicular to the radius at the point of tangency, this is the angle between the tangent and the chord, which I suspect is the angle you are looking for.

enter image description here

share|cite|improve this answer
I don't recognize which formula this one comes from 0=2500−6R+9 and I don't follow where you got 300/2509. I was trying 90 - sin(418.1667/415.1667) = ~6.243, but that wasn't coming out right. – getSurreal Jul 22 '13 at 20:56
It helps if I do my formula correct in excel and understand when to use arcsin instead of sin. asin(418.1667/415.1667)*180/PI(). Now I get 6.867. – getSurreal Jul 22 '13 at 21:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.