Calculating triple convertible length

[rrankin]

I got an old triple convertible today with a camera. It has the aperture scale for the two lenses, plus their combined length, but doesn't have the focal length of the third 'combined' length listed. The front lens is 23", rear is 17" and the combined is... Well, maybe it DID have the combined focal length but that is where the hole was drilled to mount the scale plate.

Obviously I can put it on a camera and focus at infinity with the combined lenses.

I can also tell because the max aperture opening is 1" and the max for the combined lenses is f8.

I can tell by looking at the scale and seeing that about f8.x on the combined aperture scale is about f16 on the 17" scale, so it's about half.


Since I don't have my Stroebel book in the USA to research this, and I was curious, I thought someone here will know this off the top of their head - given 2 arbitrary lenses in front and rear, how is the combined focal length calculated?

Or conversely, given that you want X focal length, how do you calculate what focal lengths to use in the front and rear?

I did try to search for an old thread, but didn't get anything useful.

Thanks, Richard

 

[Alan Davenport]

Here is some information I got from the Newsgroup: rec.photo.equipment.large-format. It is from August of 2003.

"The typical formula used for figuring out the combined focal length is 1/f = 1/f1 + 1/f2 - [d/f1 * f2] where f is focal length of combination, f1 and f2 are focal lengths of individual cells and d is the distance between the nodal points of the individual cells." The author of this article is Richard Knoppow.
I do not know how to measure the nodal distance "d" or whether it can simply be ignored.

Alan

 

[Lee L]

For the basics, see: http://en.wikipedia.org/wiki/Lens_(optics)

Look for the section on compound lenses for formulas that should get you in the ballpark.

Lee

 

[rrankin]

Thanks. The Wiki article explained the 'd' part, which did turn out to be sort of insignificant but did add about a half inch to the final calculation.
Cheers,
Richard