Calculating aperture diameter?
Moderators: Pau, rjlittlefield, ChrisR, Chris S.

 Posts: 215
 Joined: Sun Jul 22, 2012 2:30 pm
 Location: Bedford UK
Calculating aperture diameter?
Can anyone tell me how to calculate the diameter of the aperture at f8 for a given lens?
I have a rodagon 80mm f5.6 that has an illuminated aperture scale and a transparent plastic insert. What I am wanting to do is remove the iris and the plastic bits and fit an aperture disc that gives f8 with the lens reversed. I will then blank off the holes in the barrel. From what I have read, the calculation is not as straightforward as I thought.
The iris is broken so I can't accurately measure the diameter at f8.
Any help on this would be much appreciated.
Steve
I have a rodagon 80mm f5.6 that has an illuminated aperture scale and a transparent plastic insert. What I am wanting to do is remove the iris and the plastic bits and fit an aperture disc that gives f8 with the lens reversed. I will then blank off the holes in the barrel. From what I have read, the calculation is not as straightforward as I thought.
The iris is broken so I can't accurately measure the diameter at f8.
Any help on this would be much appreciated.
Steve
I was under the impression it's part of the definition of f stop. Ie an 80mm lens with an f stop of 8 has a pupil diameter of 10mm ( 80/8 ). I gather from wikipedia there are exceptions though, so you're best off asking someone who actually knows for a fact/has one.
My extrememacro.co.uk site, a learning site. Your comments and input there would be gratefully appreciated.
 rjlittlefield
 Site Admin
 Posts: 21395
 Joined: Tue Aug 01, 2006 8:34 am
 Location: Richland, Washington State, USA
 Contact:
Re: Calculating aperture diameter?
Correct, it's not straightforward. f/8 does not really mean that the aperture is 1/8 of the focal length. Instead, it means that the entrance pupil is 1/8 of the focal length. The entrance pupil is determined by the aperture, but there's a constant scaling factor caused by the magnifying or shrinking effect of whatever lens elements the aperture is seen through.steveminchington wrote:Can anyone tell me how to calculate the diameter of the aperture at f8 for a given lens?
I have a rodagon 80mm f5.6 that has an illuminated aperture scale and a transparent plastic insert. What I am wanting to do is remove the iris and the plastic bits and fit an aperture disc that gives f8 with the lens reversed. I will then blank off the holes in the barrel. From what I have read, the calculation is not as straightforward as I thought.
The simplest way to get the diameter exactly right is to use a twostep process. First make a prototype aperture that is large enough to be easily measured but small enough that for sure it's determining the entrance pupil. Stick that into the lens and measure the resulting entrance pupil. Then scale the entrance pupil to f/8 (so 10 mm diameter in your case), scale the aperture diameter by the same factor, and make the final aperture.
Since you want the lens to be f/8 when reversed, just be sure to make the measurement with the lens reversed also.
More information about measuring the pupil diameter can be found in the FAQ about pupil factor, HERE.
Rik
 rjlittlefield
 Site Admin
 Posts: 21395
 Joined: Tue Aug 01, 2006 8:34 am
 Location: Richland, Washington State, USA
 Contact:
Yes, but since the PMR is based on diameter, a reversed f/5.6 with PMR 0.5 becomes f/11.Does this mean that when you reverse a lens the apertures on the diaphragm scale are to be corrected by the pupil magnification ratio (if the PMR is 0.5 then 5.6 on the aperture ring means 8 with the lens reversed) ?
It's interesting how these numbers play out with reversed lenses.
When you reverse a lens and keep it at infinity focus, the fnumber gets adjusted by the PMR. When you extend a lens to make it focus closer, the PMR also plays into how the effective fnumber grows with extension. It turns out that these two effects partially compensate for each other, with the compensation becoming more complete at longer extensions and higher magnifications.
To take a hypothetical example, suppose you have a retrofocus lens with FL 20 mm, f/4, and PMR=2.
When you reverse the lens, it becomes f/2. "Great!" you say, "A wide aperture is just what I need for high magnification!"
But now extend that lens to provide say 5X magnification. Because of the PMR, the effective fnumber grows faster than the naive factor of magnification+1.
Taking PMR into account, the effective aperture of that reversed and extended lens is f/22. This is remarkably close to the f/24 that you would have gotten with the naive calculation of 4*(5+1) based on ignoring PMR altogether.
If you're into computation, that f/22 can be obtained in two different ways. Using the formulas from Lefkowitz:
1. Nominal f/4, PMR 2, using the "reversed" formula that f_e = f_r*(1/P)*(1+P*m). Then 4*(1/2)*(1+2*5) = 22.
2. Reversed "f/2", PMR 1/2, using the "front forward" formula that f_e = f_r*(m/P+1). Then 2*(5/(1/2)+1) = 22.
The key intuition, at least for me, is that the fnumber really specifies the angle of the cone of light. If a lens acts like f/4 when it's focused on a distant object and located close to the sensor, then "of course" it's also going to act like an f/4 lens when it's reversed and located close to the subject.
The quotation marks around "Of course" allude to an old joke about mathematicians. There's this math prof giving a lecture. He gets to a certain point in the derivation, says "Of course it's obvious that...", and stops. He stares at the blackboard, rubs his chin, scratches his head, paces a while, and eventually leaves the room. 5 minutes later he comes back, resumes his place at the blackboard, and continues the lecture. "Yes," he says, "it is obvious..."
I wrestled with PMR for a long time before eventually that bit of intuition about reversal and high magnification came to me.
Rik

 Posts: 215
 Joined: Sun Jul 22, 2012 2:30 pm
 Location: Bedford UK
Thanks Rik for your very clear explanation. I knew there were other factors at play but wasn't sure how to work it out.
I'm not sure I fully understand the second part you posted though  I will have to ponder on that. No doubt it will be useful for future reference.
Meanwhile, back to the Rodagon...
Thanks, Steve
I'm not sure I fully understand the second part you posted though  I will have to ponder on that. No doubt it will be useful for future reference.
Meanwhile, back to the Rodagon...
Thanks, Steve
I was considering the fixed aperture in a similar lens recently (used I believe for machine printing) and wondering how to make one with a "soft" edge. I'm not one to worry overmuch about bokeh, but it would be interesting.
What initially came to mind was a piece of b/w film with a hole, and appropriate exposure/development around it to make a circular grey grad. But a piece of clear acetate could be printed more easily.
What initially came to mind was a piece of b/w film with a hole, and appropriate exposure/development around it to make a circular grey grad. But a piece of clear acetate could be printed more easily.
Chris R, I've been thinking about something like this as well, following a conversation with Rik in which he brought up the concept of an optically softedged Waterhouse stop. After a bit of homework, I now use the term "apodizing stops" for such things (the phrase being wellestablished in some circles of photography, such as oldtime largeformat).ChrisR wrote:. . .aperture. . . wondering how to make one with a "soft" edge. I'm not one to worry overmuch about bokeh, but it would be interesting.
I think the benefit for us stackers wouldn't be in the bokeh per se, but in the possibility of creating softeredged outoffocus elements that might be easier for stacking algorithms to ignore and concentrate on subject detail. (While my thinking on this is muchinformed by Rik's comments in that conversation, any errors in this explanation are undoubtedly mine.)
When I get around to trying this, I think I'll make up a varied bunch of axial gradients around clear centers in Photoshop (should be quite easy), bunch them together on an 8x10" image, and send the file to a shop that does film printing. Should cost about $10 for a single piece of 8x10" film, which should permit a lot of experimenting. (What is the optimal clear diameter? How much margin should be used in gradient? How steep should that gradient be?) Then I can cut them out, insert in lenses, and see what is to be learned. I probably won't cut a center hole in the acetate, as in the film days, we projected or printed through acetate all the time without problem.
Of course, the experiment queue is long and full of elephantine projects ); it would be great if you do this first, in which case I can save time by learning from you.
Cheers mate,
Chris S.
Re: Calculating aperture diameter?
How do you find the entrance pupil if there isn't an existing iris? For example, any unmounted lens from Surplus Shed.rjlittlefield wrote: Stick that into the lens and measure the resulting entrance pupil.
@ChrisS  I had no idea you'd looked at this. I found a reference here
http://forums.dpreview.com/forums/read. ... e=33746072
which suggests some possibilities with external stops. I hadn't thought of softening of bokeh/"aperture images" being useful in stacking.
I wonder where I put that lens..
Edit  I've just found your (and a couple of other) posts containing "apodizing" in this forum  I 've been offline quite a bit recently.
http://forums.dpreview.com/forums/read. ... e=33746072
which suggests some possibilities with external stops. I hadn't thought of softening of bokeh/"aperture images" being useful in stacking.
I wonder where I put that lens..
Edit  I've just found your (and a couple of other) posts containing "apodizing" in this forum  I 've been offline quite a bit recently.
 rjlittlefield
 Site Admin
 Posts: 21395
 Joined: Tue Aug 01, 2006 8:34 am
 Location: Richland, Washington State, USA
 Contact:
Re: Calculating aperture diameter?
There's always something that limits the diameter of the light bundle. With an unmounted lens, it might be just the edge of the lens. Whatever it is, use that.elf wrote:How do you find the entrance pupil if there isn't an existing iris? For example, any unmounted lens from Surplus Shed.
Rik

 Posts: 215
 Joined: Sun Jul 22, 2012 2:30 pm
 Location: Bedford UK
So here's the theory in practice...
This is the lens I am working on
First step is to make a test aperture. I found a washer that was about the right size and ground down the outside diameter so that it would fit inside the lens. The hole was 12.1mm dia.
Then photograph the lens each way round focusing on the aperture. For this I put the lens on a small light box on my copy stand and used a 90f4 lens wide open to make a shallow DOF image.
If you have a copy stand you can measure the focus distance on the column.
Then photograph a ruler at the same focus distance so the scale is exactly the same and overlay it on the lens image to measure the aperture.
We can see that the aperture measures 17.0mm from the front and 17.6 from the rear of the lens.
Now divide these by the actual diameter of 12.1 and we have a magnification factor of 1.4 on the front and 1.45 on the rear.
Since I wanted an aperture of 10.0 to give me f8 when reverse mounted, I needed to make another aperture with a diameter of 10 / 1.45 = 6.9mm dia. So off to the washer box again...
The nearest I could find was 6.5 so I drilled it out to 7.0 which was close enough.
Then to repeat the procedure with the camera still set to the same focus distance.
So there we have it  the aperture from the front is now 9.75 and from the rear exactly 10.0mm dia.
So I can now make a proper fixed aperture out of a piece of aluminium sheet and paint it matt black and mark the lens as f8 for reverse mounting. Job done!
Thanks to Rik for explaining the method.
Steve
This is the lens I am working on
First step is to make a test aperture. I found a washer that was about the right size and ground down the outside diameter so that it would fit inside the lens. The hole was 12.1mm dia.
Then photograph the lens each way round focusing on the aperture. For this I put the lens on a small light box on my copy stand and used a 90f4 lens wide open to make a shallow DOF image.
If you have a copy stand you can measure the focus distance on the column.
Then photograph a ruler at the same focus distance so the scale is exactly the same and overlay it on the lens image to measure the aperture.
We can see that the aperture measures 17.0mm from the front and 17.6 from the rear of the lens.
Now divide these by the actual diameter of 12.1 and we have a magnification factor of 1.4 on the front and 1.45 on the rear.
Since I wanted an aperture of 10.0 to give me f8 when reverse mounted, I needed to make another aperture with a diameter of 10 / 1.45 = 6.9mm dia. So off to the washer box again...
The nearest I could find was 6.5 so I drilled it out to 7.0 which was close enough.
Then to repeat the procedure with the camera still set to the same focus distance.
So there we have it  the aperture from the front is now 9.75 and from the rear exactly 10.0mm dia.
So I can now make a proper fixed aperture out of a piece of aluminium sheet and paint it matt black and mark the lens as f8 for reverse mounting. Job done!
Thanks to Rik for explaining the method.
Steve
 rjlittlefield
 Site Admin
 Posts: 21395
 Joined: Tue Aug 01, 2006 8:34 am
 Location: Richland, Washington State, USA
 Contact:
Very nice! Thanks for the feedback and illustrations.
It's interesting to note that even though this lens has PMF not much different from 1, there is still a large difference between the pupil diameters and the physical aperture diameter. What we see here is typical for lenses that are fairly thick and have net positive elements on both sides of the aperture.
Rik
It's interesting to note that even though this lens has PMF not much different from 1, there is still a large difference between the pupil diameters and the physical aperture diameter. What we see here is typical for lenses that are fairly thick and have net positive elements on both sides of the aperture.
Rik