Calculate Focal Point

Focal point - Wikipedia, the free encyclopedia

r = 0, but it works only if the energy source is on the cylinder's axis.

The foci would that be the length way of the center (axis), because it is slice lengthways. if you need the formula follow lyn's link

I think this is what you are looking for but, it will not work with a cylinder segment. You need a parabolic surface to have a focal point.

Doesn't need to be parabolic. Can be a section of a sphere or an ellipsoid.

If I remember right r/2 can be used as an approximation for a spherical mirror but I think that only works for a small segment.

EDIT: Usbport posted while I was typing and stated it much better than I did.

If you're using the cylinder as a reflector, like a shiny metal tube sliced lengthwise, the focal length is 1/2 the radius of the cylinder - for light coming from 'infinity' or some distance much, much greater than the radius.

Note that the focal point will be a blur spot (or in this case a blurry line), since for light coming from infinity you need a parabolic shape, rather than a circular cylinder to get a sharp focus.

If the light is not coming from infinity, then use this equation to find I, the image distance:

1/f = 1/O + 1/I, where f is focal length, O is the distance to the object and I is the image distance. Again, f is one-half the radius. You need to solve for I.

If the cylinder is a clear glass or plastic cylinder used as a lens, then you need to use the lens-makers equation. In this case the simple form is:

1/f = (n-1)·[1/R1].

Here f is focal length, R1 is the radius of the cylinder, and n is the index of refraction of the glass or plastic.

This has everything one could want.

Colors, circles and arrows. Focus (geometry),

Point F is a focus point for the red ellipse, green parabola and blue hyperbola.

... except for the geometry for a circular section of a cylinder, where e = 0. Which is what the Anon Poster described.

I don't care. I think it's homework anyway.

What about the Petzval Field Curvature? Maybe the AP needs to know that, too.

This is not homework.I found a piece of corrugated metal pipe,made from aluminum, and think it would make a good heat reflector for a camp fire, to direct the heat in a preferred direction.I am trying to calculate the focal distance.It is a lengthwise 1/2 slice of what appears to have been a 4 foot diameter pipe.I realize the corrugations will tend to scatter the reflection and complicate the calculations, but a general rule is adequate for this purpose.The 1/2 Radius I think will suffice.I wish it were a parabola,but it is what it is, and I think it would be very difficult to reform into a parabola.

Thanks to everyone for their helpful links and suggestions.