# elliptical footprint of a round beam

• Apr 20th 2010, 02:45 AM
ilindema
elliptical footprint of a round beam
Hi,

I have some problems with a basic geometric problem (it is a long time ago I had some mathematics)
How do I calculate the surface area of a elliptical footprint of a beam that falls into the surface under a 45 degree angle? I only have the diameter D of the incident beam.
I got:
sqrt(2) pi (D/2)^2

is this correct?

thanks,
Ilindema
• Apr 20th 2010, 03:54 AM
HallsofIvy
Quote:

Originally Posted by ilindema
Hi,

I have some problems with a basic geometric problem (it is a long time ago I had some mathematics)
How do I calculate the surface area of a elliptical footprint of a beam that falls into the surface under a 45 degree angle? I only have the diameter D of the incident beam.
I got:
sqrt(2) pi (D/2)^2

is this correct?

thanks,
Ilindema

Start by drawing an isosceles right triangle The hypotenuse is the image of the beam and one of the legs is a diameter of the beam. That is, $c^2= D^2+ D^2= 2D^2$ so c, the major axis of the ellipse, is $D\sqrt{2}$. The minor axis is, of course, the diameter of the beam, D.

The area of an ellipse with major and minor semi-axes a and b is $\pi ab$ so, yes, the area of this ellipse is $\pi \left(\frac{D}{2}\right)\left(\frac{D}{2}\sqrt{2}\ right)$, exactly what you have.
• Apr 20th 2010, 08:19 AM
ilindema
thanks a lot! Now I have some peace (Bow)