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View Full Version : Is there any way to guess a lens' coverage?

williaty
19-Dec-2017, 18:56
I'm on the hunt for a fast lens to use for ambrotypes. Obviously, this pretty much means projection lenses since I can't afford petzval prices and I've never seen a place to find coverage specs for projection lenses. Is there any way to guess how large the image circle will be based off the focal length, aperture, or anything else I can glean from an ebay auction?

Leigh
19-Dec-2017, 19:09
Not reliably.

Len Middleton
19-Dec-2017, 19:17
williaty,
Under the category of there is little new,
as you might find it an interesting start to your discussion.
Hope that helps,
Len

xkaes
19-Dec-2017, 19:24
IF you know the OPTICAL FOCAL LENGTH OF THE LENS and the ANGLE OF COVERAGE (AKA COVERING POWER) of a lens:

The image circle is roughly, approximately equal to the optical focal length of the lens times the tangent of the angle of coverage.

EX. Fujinon SWD 65mm with 105 degree angle of coverage = 65 * tan(105) = 169mm image circle.

Bob Salomon
19-Dec-2017, 19:27
IF you know the OPTICAL FOCAL LENGTH OF THE LENS and the ANGLE OF COVERAGE (AKA COVERING POWER) of a lens:

The image circle is approximately equal to the optical focal length of the lens times the tangent of the angle of coverage.

EX. Fujinon SWD 65mm with 105 degree angle of coverage = 65 * tan(105) = 169mm image circle.

Except a projection lens only covers the largest format it was designed to project. Not the size of the image that it projects onto the screen.

goamules
19-Dec-2017, 19:52
Basically, with Petzvals they are about a 20 degree lens. The faster ones (F3.1-F3.6) have less coverage than the slower ones (over F3.8).

You need a much longer one to cover than you would with more normal coverage designs, like Tessars and Planars. For example, a 300mm Tessar will cover 8x10. But a 300mm (11 inch) fast Petzval will not. If you tell us the size plate you want to cover, we'll tell you what focal length Petzval to look for.

Just in case you didn't know, you don't have to have a super fast lens for wetplate, if you are shooting outdoors. An F4.5 Tessar makes a good starter lens.

williaty
19-Dec-2017, 19:55
I'm shooting portraits indoors with strobes. Hence, fast lens. I have a 150mm f/2.8 Heidosmat that works well with the lights I have but I find it a little short. I'd rather have something in the 225-250mm region.

As I said, with Petzval prices so far out of control, I have no possibility of affording one.

Right now, I want to cover 4x5 because I'm using 4x5 as a proving ground to verify I'm actually serious about ambrotypes before committing to at least an 8x10 setup and, if a miracle occurs, something really large like 11x14 or even 16x20 in a few years.

renditiont
20-Dec-2017, 00:53
i recently mounted a gaumont kalee 241mm f/1.9 on graflex series D . pretty cool lens
https://c1.staticflickr.com/5/4733/38428545544_c3fc744809_c.jpg

renditiont
20-Dec-2017, 00:54
also a small portion of cheap lenses i found on fleabay. god bless fleabay. but to return to your question, yes you can gauge a lens coverage , often based on a combination of a. focal length b. configuration types, and c. rear's size
https://c1.staticflickr.com/5/4549/38335364684_8d4764a216_c.jpg

williaty
20-Dec-2017, 01:15
Good lord, that's an insane number of lenses!

I think you single handedly drove up the price of every lens that's ever been made! ;)

renditiont
20-Dec-2017, 01:43
Good lord, that's an insane number of lenses!

I think you single handedly drove up the price of every lens that's ever been made! ;)

Not really, but that black hole known as fleabay is fun to hunt for cheap lenses. for 4x5 projection lenses most people like 175mm f/2.5 or 200mm f/2.5 buhl and hektor (they have the same configuration). for 8x10 hektor 300mm f2.8 is pretty nice as starter.

Steven Tribe
20-Dec-2017, 03:30
Good lord, that's an insane number of lenses!

! ;)

Not really! I would guess that it is about average for anyone who gets interested in the optics side of photography.
You can learn a lot from reading, but there is nothing better than "hands-on" experience with real lenses! With lots of web market places, it is comparitively easy to both buy and re-sell bargain items.

All Petzvals have the same optic design at a specific designated makers speed ( perhaps apart from the Dallmeyer version, which I think is actually identical in reality). I think it is possible to draw up a table of Petzvals coverage (image circle) based on the combination of just the speed and focal length.

Pere Casals
20-Dec-2017, 04:57
this pretty much means projection lenses since I can't afford petzval prices and I've never seen a place to find coverage specs for projection lenses.

I'd just ask seller if you want to know the actual illumination circle, the actual format it was projecting with the projector will give you a clue of what is the sharp circle, and seller also can measure what illumination circle has, he can do that just projecting in the wall with lens focused at infinite.

Jimi
20-Dec-2017, 06:01
I'm shooting portraits indoors with strobes. Hence, fast lens. I have a 150mm f/2.8 Heidosmat that works well with the lights I have but I find it a little short. I'd rather have something in the 225-250mm region.

Leitz made a few projection lenses too - there is for example a Hektor 200/2.5. Also remember that you are doing closer than infinity stuff so the coverage gets better at portrait distances.

Bernard_L
20-Dec-2017, 08:44
Coverage from degradation of IQ: not really.
Coverage from light falloff: somehow, approximately. Look through the lens from the front against a uniformly lighted background. Turn the lens; You should at some point see that two circular apertures that the light must go through are intersecting, something like: (). Note the angle from boresight. That is the on-scene angular coverage, the value generally given in datasheets. This is of course a soft limit, depending on how much loss (1 stop?) is acceptable. Ditto from the rear: gives you the angle of illumination of the film by the lens. I did this with a Petzval and could see that the coverage was indeed small.

goamules
20-Dec-2017, 09:54
I'm shooting portraits indoors with strobes. Hence, fast lens. I have a 150mm f/2.8 Heidosmat that works well with the lights I have but I find it a little short. I'd rather have something in the 225-250mm region.

As I said, with Petzval prices so far out of control, I have no possibility of affording one.

Right now, I want to cover 4x5 because I'm using 4x5 as a proving ground to verify I'm actually serious about ambrotypes before committing to at least an 8x10 setup and, if a miracle occurs, something really large like 11x14 or even 16x20 in a few years.

Well, the nice thing is you're starting with a small plate size. So many hear about wetplate, then decide immediately they want a 1) Fast, 2) Large, and 3) Cheap lens (pick any TWO!). You're giving yourself lots of options. You can also look for a triplet, many were F3.5. Or just add some light, my friend Mark will probably weigh in, he uses lots of lenses with strobes for wetplate.

On projector lenses, many are Petzvals, most others are Triplets (some other designs I'm sure). There is a Kodak brand proj lens that is fast and about the right size, but I can't recall the name. Big silver things....

renditiont
20-Dec-2017, 17:41
see how many buhl 240/2.5 you can spot
https://c1.staticflickr.com/5/4732/24324576747_21223f20b4_b.jpg

Pere Casals
21-Dec-2017, 04:34
also a small portion of cheap lenses i found on fleabay. god bless fleabay. but to return to your question, yes you can gauge a lens coverage , often based on a combination of a. focal length b. configuration types, and c. rear's size
https://c1.staticflickr.com/5/4549/38335364684_8d4764a216_c.jpg

Wow

goamules
22-Dec-2017, 16:40
You could always use the How Many Leicas fit inside method. This one is a 1200mm, or 47" focal length. Its front element area would hold about 1.5 Leica II cameras. 47 divided by 1.5 = 31.33 (the diagonal of the size of the plate that would be covered. Let's make it 31.24 to make the math easier. Using the Pythagorean Theorem, you can now determine the rectangles (or square) covered. We can now determine this one covers 20x24 inch. *

https://c2.staticflickr.com/6/5479/9615750377_6542e9c0a9_c.jpg

*calculations may be suspect