Lens design.

[nhemann]

I am slowly thinking my way through a homebuilt camera: I have a good source for loose lenses, good handle on how to make a bellows, plenty of skill with the tools, wood and etc. The question of the day is - how the heck do I determine what the image circle will be for a given focal length? Is it dependent on the lens diameter? There has to be an equation or something, right? Any light you LF or DIY-ers could shed would be most appreciated; reference material that you can direct me too? Even better. I'm planning on a single element on to start with - the engineer in me is dying to understand how to set up a multi-element as well - but first things first.
Thnks
neil

[Ian C]

The diameter of the image circle at infinity is a function of both the focal length and the projection angle of the lens. We cannot calculate coverage unless we know both of these parameters.

If the projection angle of the lens is θ and the focal length = f, then the diameter d of the image circle at infinity focus is

d = 2f*tan(θ/2)

[nhemann]

awesome, Ian - now we are getting somewhere. I assume that the projection angle is a design parameter from the OEM? I'm looking at a typical spec from the supplier and they look like this:

Manufacturer's part# 01 LAM 138, original unit price: $54.00. Made of SF8 glass, design wavelength is 546.1nm. Clear aperture of 36mm, center thickness is 7mm, edge thickness is 5.2mm. Surface quality is 60-40 scratch and dig, centration is 3 arc minutes. Edges are .25-.50 bevel. Manufacturer's companion doublet for this lens: 01 LAO 138. Aplanatic lenses are free of spherical aberration and coma.

Would it just be a matter of getting ahold of the OEM or can this be determined otherwise - assuming that I am working in the world of horse shoes, if a decent estimate determines that its going to cover 4x5, 8x10 etc with some margin I am in good shape, eh?

[Toffle]

The diameter of the image circle at infinity is a function of both the focal length and the projection angle of the lens. We cannot calculate coverage unless we know both of these parameters.

If the projection angle of the lens is θ and the focal length = f, then the diameter d of the image circle at infinity focus is

d = 2f*tan(θ/2)

Thanks for throwing that out there, Ian. I was just about to start making stuff up.
Is it safe to say, though that the projection angle is a product of the rear element/group, or can the other groups affect this as well?

Supplimental question.... if, as the OP indicates, he is dealing with loose lens elements, (I've got a box full, too) how do you calculate the projection angle?

(I suppose you could reverse the calculations from the focal length and image circle, but any time I've tried to measure an image circle, I have come up with very fuzzy numbers... about as fuzzy as the edge of my circle. The last thing I want to promote here is fuzzy math. You can get eaten by dragons here when you talk fuzzy math. )

[nhemann]

Its not nearly as dangerous as fuzzy logic, but just like The New Math, you need to be careful with the fuzzy stuff.

[Ian C]

The projection angle is determined by the lens design. If the supplier can’t provide the projection angle, then your only recourse is to test the lens when you get it to determine the angle and diameter of the image circle at infinity.

You’d have to set up the lens so that you focused a distant scene (infinity focus) onto a white surface and mark the boundaries of the projected circle with pencil.

This is most easily done by placing the lens and white surface in a darkened room and projecting a distant sunlight scene thorough a relatively small opening in a cardboard covering over the window.

Then you’d know the diameter of the circle and can calculate the projection angle (although by then you’d already know the diameter of the circle).

θ = 2*arctan(d/2f).

The circle not only has to cover the diagonal of the film holder, but you need to judge the quality of the projection as well. It might be that you judge, say, 70% of the diameter of the circle as usable, but the outer part might be too fuzzy or the light might falloff too much for a practial image.

[Ian C]

I don’t think that there’s an easy way for us to determine ahead of time what the result would be of assembling multiple elements into a compound-element design.

The lens making companies have that ability, of course, but that’s beyond what we can do with the simple formulas available to us.

[nhemann]

wow - I've just had a lot of floundering around the internets get prevented. Thank you so much. And boy, the neighbors are going to love watching as I crawl into a calibrated camera obscura to figure that one out. lol.

Ok then next issue: large vs small lens - as I see it from what you said above - given two lenses identical in quality and performance - I would pick a larger lens be able to keep my subsequent range of useful apetures larger i.e a low f of say 1.2 vs f4.5 or something? Is there a reasonable lower limit to f-stop in LF, I have seen a few pics here - Andrew Moxom's shot of that gun barrel comes to mind where where the DOF appears to almost literally be a plane.

[Chris Lange]

I've ground my own lens before (as part of a glassworking studio at my university), and I ended up with a fixed focus ~65mm/1.4 single element optic, mounted inside an M2 extension ring for my Nikon.

It wasn't horribly difficult, just a time consuming process. Granted, I can make nearly any type of element I may need, since we have a full blowing and casting shop at our disposal.

[Jim Jones]

I vaguely remember that the lowest practical f/ number for photographic lenses is around f/.7. As early as 1934 Leitz made a 75mm Summar f/0.85, but it was not used for traditional photography. Photographers had to be satisfied with the 73mm f/1.9 Hektor of 1931. I have one, and it is a very compact lens for that speed.