PDA

View Full Version : Calculating f stop of Telephoto lens

IanG
26-Nov-2014, 08:31
My ex-Military 17" Dallon has no aperture scale but is marked as having a max aperture of f5.6, the diameter of the aperture blades is 42.8mm wide open min is 5.0mm, the lenses stop down to f45.

I'm assuming the normal calculation f stop = focal length (431,8mm)/diameter doesn't apply to telephoto lenses without an additional fixed factor. Having assumed there's a factor involved the calculated theoretical scale looks to be very close to the ones on 2 similar 17" f5.6 Dallons I've found photos of on the Internet (one's here (http://www.leicashop.com/vintage_en/dallmeyer-17-5-6-dallon-tele-anastigmat-f-leica-sku16342-60.html)).

Ian

Dan Fromm
26-Nov-2014, 09:21
Measure the entrance pupil's diameter with the diaphragm wide open. It should be 3 1/8".

IanG
26-Nov-2014, 10:25
The front element diameter is approx 77mmmm give or take 0.5mm, it's set back in the barrel. That calculates at around f5.6 which is marked on the lens cap and sleeve the lens came stored in. This indicates a fixed factor of approx 1.8 between the aperture calculated from the diameter of the aperture & focal lengthen.

It's fairly obvious that the effective aperture of a telephoto is larger than a standard lens of the same focal length due to the inverse square law, the satandard lens at the same actual aperture diameter is further from the film plane at Infinity (and other distances)

So I need to divide the theoretical aperture by approx 1.8 to get the effective aperture with this telephoto.

Ian

Dan Fromm
26-Nov-2014, 11:25
Ian, in this case measurement beats calculation.

Nodda Duma
26-Nov-2014, 11:37
F/# is ratio of the focal length to the entrance pupil. The entrance pupil is the aperture as imaged by the lenses in front of its position in the optical path.

In many designs the entrance pupil diameter is not the same as the physical aperture diameter: the only case where they are equal is when the aperture stop is on or in front of the first lens element (sounds like the case here when the lens is wide open).

But when stopped down, the entrance pupil will be set by the physical aperture diameter and not the front lens. Very interesting.

Dan Fromm
26-Nov-2014, 12:28
Jason, Ian's lens' aperture is somewhere in the middle of the lens. The entrance pupil, as you just said, is an image of the stop viewed through the front of the lens. We know what its diameter has to be with the lens wide open, and I've given Ian the number.

His problem is one many of have faced when we've switched a lens' cells from one shutter or barrel to another that has no aperture scale. We have to make a new aperture scale. SKGrimes does this routinely. There are standard aperture scales for Copal shutters for many common lenses. They're sold in long rolls. For common lenses put into Copal shutters, the machinist just snips a scale from the roll and attaches it to the shutter.

For uncommon lenses, SKGrimes' practice is to measure the entrance pupil with the aperture wide open, mark the barrel or shutter, stop down until the entrance pupil is one stop (1/sqrt(2)) smaller, mark, ... They've done this for me with, e.g., an aerial camera lens whose aperture scale was on a sleeve that had to be removed to mount the lens on a board.

ic-racer
26-Nov-2014, 12:42
It's fairly obvious that the effective aperture of a telephoto is larger than a standard lens of the same focal length due to the inverse square law,

1) Lens construction is not part of the equation
2) Inverse square law is not part of the equation

[Effective Aperture Calculation] Multiply the marked f-stop by (1 + M), where M is the magnification on the ground glass -- From the Large Format Photography Home Page

IanG
26-Nov-2014, 14:03
Jason, Ian's lens' aperture is somewhere in the middle of the lens. The entrance pupil, as you just said, is an image of the stop viewed through the front of the lens. We know what its diameter has to be with the lens wide open, and I've given Ian the number.

His problem is one many of have faced when we've switched a lens' cells from one shutter or barrel to another that has no aperture scale. We have to make a new aperture scale. SKGrimes does this routinely. There are standard aperture scales for Copal shutters for many common lenses. They're sold in long rolls. For common lenses put into Copal shutters, the machinist just snips a scale from the roll and attaches it to the shutter.

For uncommon lenses, SKGrimes' practice is to measure the entrance pupil with the aperture wide open, mark the barrel or shutter, stop down until the entrance pupil is one stop (1/sqrt(2)) smaller, mark, ... They've done this for me with, e.g., an aerial camera lens whose aperture scale was on a sleeve that had to be removed to mount the lens on a board.

I thought you'd grasp my points Dan, and coming from different perspectives we reach exactly the same conclusions.

I think the point Jason's missing is I know at max aperture it's an f5.6 lens so half the diameter is f11 and so on, and I can calculate the apertures between.

Ian

IanG
26-Nov-2014, 14:18
1) Lens construction is not part of the equation
2) Inverse square law is not part of the equation

Actually they are key in telephoto lens designs. A telephoto lens shortens the nodal point to film distance so for a given aperture mm or inch diameter there's a big change in exposure. I suppose a simplistic way of looking at it a 12"/300mm standard lens sits on the maximum extension on my Wista or Super Graphic, the 17" Dallon sits at the end of an unextended trackbed at Infinity, roughly half the extension, now equate that to the inverse square law, it's a no brainer.. . . . . . . . it's about 2 stops` difference.Use Maths not math :D the plural adds the missing gaps . . . . . . . .

Ian

Dan Fromm
26-Nov-2014, 15:09
A telephoto lens shortens the nodal point to film distance

Eh? Wot? It does no such thing. The rear node-to-film plane when the lens is focused to infinity is the same for all lenses of the same focal length. Tele lenses' rear nodes are way out in front ...

Ian, I'm not sure whether you're in the UK or Turkey but either way its past your bed time. Have a glass of warm milk, sleep well and reconsider in the morning.

Nodda Duma
26-Nov-2014, 15:32
Just to clarify, the rear nodal point is the exit pupil position. If the medium is the same in object space and image space (ie air) then that also is the Principal Point and gives the effective focal length.

ic-racer
26-Nov-2014, 15:49
Maybe you used the wrong words but you wrote "Effective Aperture" and that is defined as "marked F-stop times (1+ Magnification). The equation contains neither variables for lens construction or the inverse square law.

Dan Fromm
26-Nov-2014, 16:15
ic, look at post #11 again. I see "effective focal length."

IanG
27-Nov-2014, 05:44
Something simple is being missed, maybe I used the wrong terms (I've a bad head cold) but I'm talking about real world practice rather than theory so I can add an f stop scale.

Yes the f stop = focal length/aperture diameter equation is simplistic and doesn't work as is with telephoto lenses, but then, in post #1 I said there must be a fixed factor between the aperture calculated by the above equation and the actual effective f-stop as marked, I never mention variables.

Maybe you used the wrong words but you wrote "Effective Aperture" and that is defined as "marked F-stop times (1+ Magnification). The equation contains neither variables for lens construction or the inverse square law.

The terms actual and effective are probably misleading unless defined and I'm talking about marked apertures relative to actual aperture diameter and the physical aperture required is less than the theoretical from the f stop = FL/diameter by a fixed factor with a telephoto lens.

At infinity focus the front element of this 17" telephoto lens is only 14" from the film plane and the aperture diaphragm 11" So we have an actual focal length and an effective 17" focal length, and due to the inverse square law and the fact we need less bellows extension than a non telephoto design like a Symmar of the same focal length the actual physical aperture diameter is smaller than would be needed to achieve the same aperture with a standard lens like a Symmar etc.

In fact Schneider's faster 480mm Symmar S only has a maximum aperture of f8.4, the 360mm f6.8, the main restriction being the size of the Compur or Copal 3 shutters, the maximum aperture diameter is only (very) approx 55mm, a 17" f5.6 of similar design would require a max aperture diameter of 77mm and would be a huge lens.

Just to clarify, the rear nodal point is the exit pupil position. If the medium is the same in object space and image space (ie air) then that also is the Principal Point and gives the effective focal length.

I'll leave the theory to you :D You could probably explain what I'm saying a bit better. What FL do I use to calculate exposure factors with extension ? My guess is a nominal 240mm.

In practice the scale I've calculated and marked in pencil on the unpainted lens board matches the scales on photographs of the same lens which I know has an f5.6 to f45 range (it will stop down a touch past f45).

Ian

Dan Fromm
27-Nov-2014, 06:13
What FL do I use to calculate exposure factors with extension ?

The lens' focal length. 17.5", I think you said it was. Same as always. And you measure extension from the infinity position. Same as always. If you need rear node-to-film distance, it is focal length plus extension from the infinity position. Same as always.

Nodda Duma
27-Nov-2014, 06:35
Something simple is being missed, maybe I used the wrong terms (I've a bad head cold) but I'm talking about real world practice rather than theory so I can add an f stop scale.

Yes the f stop = focal length/aperture diameter equation is simplistic and doesn't work as is with telephoto lenses, but then, in post #1 I said there must be a fixed factor between the aperture calculated by the above equation and the actual effective f-stop as marked, I never mention variables.

The terms actual and effective are probably misleading unless defined and I'm talking about marked apertures relative to actual aperture diameter and the physical aperture required is less than the theoretical from the f stop = FL/diameter by a fixed factor with a telephoto lens.

At infinity focus the front element of this 17" telephoto lens is only 14" from the film plane and the aperture diaphragm 11" So we have an actual focal length and an effective 17" focal length, and due to the inverse square law and the fact we need less bellows extension than a non telephoto design like a Symmar of the same focal length the actual physical aperture diameter is smaller than would be needed to achieve the same aperture with a standard lens like a Symmar etc.

In fact Schneider's faster 480mm Symmar S only has a maximum aperture of f8.4, the 360mm f6.8, the main restriction being the size of the Compur or Copal 3 shutters, the maximum aperture diameter is only (very) approx 55mm, a 17" f5.6 of similar design would require a max aperture diameter of 77mm and would be a huge lens.

I'll leave the theory to you :D You could probably explain what I'm saying a bit better. What FL do I use to calculate exposure factors with extension ? My guess is a nominal 240mm.

In practice the scale I've calculated and marked in pencil on the unpainted lens board matches the scales on photographs of the same lens which I know has an f5.6 to f45 range (it will stop down a touch past f45).

Ian

It's thanksgiving. My theoretical work today involves the most efficient way to consume a desired amount of turkey and beer.

While I know it (or know where to look), I'm not too much a fan of pure theory..I came into lens design as an engineer so I'd rather get my hands dirty. I'll leave the base theory to the physicists (if you think this math is bad..trust me it's only approximations to the full blown derivations). No negative thickness lenses for me! In my professional opinion if you have a system that works in the field then stick with it....focus on the how and don't worry about the why unless you're taking a quiz :)

That said, I would suggest a copy of "Lens Design Fundamentals" by Kingslake. I think the relationship you're looking for in regards to telephotos is in there and it's a good reference if you are interested at all in the technicalities of optics. Another real good one is the book and iphone app called "Field Guide to Geometrical Optics" by John Grievencamp. Good reference...Grievencamp was a professor of mine at U of A.

How's that for a long-winded way of saying I'm too lazy to go look it up? ;)

IanG
27-Nov-2014, 09:14
The lens' focal length. 17.5", I think you said it was. Same as always. And you measure extension from the infinity position. Same as always. If you need rear node-to-film distance, it is focal length plus extension from the infinity position. Same as always.

The extensions factors for asymmetrical lenses are far greater with telephotos than with symmetrical lenses, this is because the actual FL is less than the effective FL, it's the other way around with retro-focus lenses but the differences are very are much smaller. With the 17" lens on my Speed Graphic film plane to aperture ring is 25cm at Infinity and 34cm at closest focus possible which is 226cm.

You could probably explain what I'm saying a bit better. What FL do I use to calculate exposure factors with extension ? My guess is a nominal 240mm.

Are you saying use the Infinity extension as the "base focal length" so rather than divide the Overall extension by the 17" focal length, divide by the Infinity focus extension, because that's the same as I asked and assumed. Makes perfect sense, I found a reference that you need to add a factor if you use the marked focal length in bellows extension factors, 0.57 for a 360mm Tele Arton. it's 0.56 for my lens, The 1 cm difference is arbitrary as I had to chose a point to measure from and chose the aperture ring, front of the lens standard to film plane is 24cmm which is also the approx actual FL.

Ian

Dan Fromm
27-Nov-2014, 09:38
Ian, focal length is focal length. You are confusing back focus -- the distance from the lens' rear vertex to the film plane when the lens is focused to infinity -- with focal length -- the distance from the lens' rear node to the film plane when the lens is focused to infinity.

I don't know what you mean by "extensions factors." If you're thinking of calculating effective aperture given magnification and aperture set, then the magic formula is "effective aperture" = "aperture set" * (1 + m). This is an approximation, there is a further correction when pupillary magnification is not 1.0. I have to go out, will write more later.

ic-racer
27-Nov-2014, 09:42
ic, look at post #11 again. I see "effective focal length."

A few posts got in the way, numbed 12 was in response to number 9.

IanG
27-Nov-2014, 11:01
Ian, focal length is focal length. You are confusing back focus -- the distance from the lens' rear vertex to the film plane when the lens is focused to infinity -- with focal length -- the distance from the lens' rear node to the film plane when the lens is focused to infinity.

Neither of those is the focal length of a telephoto that's the dilemma, yes I may be using the wrong term for that back focus point that's 240mm for my 17" lens.

[/QUOTE]I don't know what you mean by "extensions factors." If you're thinking of calculating effective aperture given magnification and aperture set, then the magic formula is "effective aperture" = "aperture set" * (1 + m). This is an approximation, there is a further correction when pupillary magnification is not 1.0. I have to go out, will write more later.[/QUOTE]

Extension is often expressed as a factor for a specific lens, it is the amount of additional exposure needed with increased extension to compensate for the change in effective aperture you can express it as a figure ie x4, x2 etc or as +2 stops, +1 stop etc.

What ever terminology we use I've been able to determine the f stop scale and can now make a sticker for the Speed Graphic to indicate the increased exposure needed with the 17" Dallon at closer distances.

Ian

IanG
27-Nov-2014, 11:39
This is an approximation, there is a further correction when pupillary magnification is not 1.0. I have to go out, will write more later.

I found the thread where Emmanuel Bigler describes the Pupillary magnification, it's the same factor I arrived at comparing the actual full aperture diameter compared to the theoretical full f5.6 aperture for the same FL symmetrical lens, I found a another PDF he doesn't link to as well obviously from the same French article (ebsite).

Ian

jb7
27-Nov-2014, 13:12
My ex-Military 17" Dallon has no aperture scale but is marked as having a max aperture of f5.6, the diameter of the aperture blades is 42.8mm wide open min is 5.0mm, the lenses stop down to f45.

I'm assuming the normal calculation f stop = focal length (431,8mm)/diameter doesn't apply to telephoto lenses without an additional fixed factor. Having assumed there's a factor involved the calculated theoretical scale looks to be very close to the ones on 2 similar 17" f5.6 Dallons I've found photos of on the Internet (one's here (http://www.leicashop.com/vintage_en/dallmeyer-17-5-6-dallon-tele-anastigmat-f-leica-sku16342-60.html)).

Ian

Assuming your purpose is to calibrate f stops for the lens, and assuming you already know the max aperture is f/5.6, which measures at 42.8mm, then why not just halve the area of that aperture, so that you can measure an aperture of 30.26mm to give you the position for f/8, and so on, and so forth ...

IanG
27-Nov-2014, 13:56
Assuming your purpose is to calibrate f stops for the lens, and assuming you already know the max aperture is f/5.6, which measures at 42.8mm, then why not just halve the area of that aperture, so that you can measure an aperture of 30.26mm to give you the position for f/8, and so on, and so forth ...

That's exactly what I did and it works perfectly. Using a spreadsheet made it very easy, (I already had a spreadsheet to calculate f-stops - that's why it was easy to find the factor needed as well - the "Pupillary magnification".

The internet searches bring up a lot of fallacies and very little is written about the questions I've asked.

Ian

Mark Sawyer
27-Nov-2014, 18:34
I think it could all be summed up in "yep, you figure it the same way as any other lens..." :rolleyes:

Dan Fromm
27-Nov-2014, 19:01
I found the thread where Emmanuel Bigler describes the Pupillary magnification, it's the same factor I arrived at comparing the actual full aperture diameter compared to the theoretical full f5.6 aperture for the same FL symmetrical lens

Ian

Wrong again, Ian. Here's the law, with definitions:

The relationship between extension (lens’ rear node to film) and magnification is e = f*(m + 1) where e is extension, f is the lens’ focal length, * is the multiplication operator and m is magnification. This is true for all lenses, regardless of pupillary magnification. When the lens is focused at infinity, m = 0.

For a symmetrical lens, with entrance and exit pupils the same size, i.e., pupillary magnification = 1, effective f/ number is f/ set * (m + 1) and the exposure increase factor is (m + 1)^2. Read f/ as “f over.” ^ is the exponentiation operator. So with a lens whose aperture is set to, say, f/4 the effective aperture when magnification = 1 is f/8 and, equivalently, the exposure increase factor is 4.

For a lens with pupillary magnification <> 1 mounted front forward, effective f/ number is f/ number set * ((m/p) + 1 ) where p is the pupillary magnification. With the lens reversed, effective f/ number is f/ number set * (1/p)*(1 + pm).

Pupillary magnification is (diameter of exit pupil)/(diameter of entrance pupil).

These relationships apply to all lenses regardless of construction. That's what Mark said in post #24.

Nodda Duma
27-Nov-2014, 20:59
You guys are talking the relationship between image space f/# and working f/# (the formal term for what you're calling effective f/#).

Dan one thing to point out is in your second paragraph, magnification is still very small (m = - image height / object height) for large object distances, and you still approximately have an f/4 lens even with a telephoto where p < 1. It's only really in macro photography conditions that you'll see that magnification be a significant factor in affecting the light cone.

Working f/# = (1-m)* f/# = (1-m)* (fe/Dep)

fe is effective focal length and Dep is diameter of the entrance pupil. Yes that assumes a 1:1 pupil ratio and the ratio is included for designs with different sized pupils. (For a telephoto p < 1).

working f/# describes the image forming cone for finite conjugates..ie for short distances.

m is just the (negative) image height / object height. Don't forget images are inverted in camera lenses and have a negative sign...that's why lens designers (led by U of A and Rochester) went to a different sign convention than what you're talking. In your convention magnification is a negative number. So a photograph of a 2m tall person may only have an image height of -0.02 m, giving a magnification of -0.01 in yours, +0.01 per the design community.

Note that for all but macro photography even m/p will still be a very small number and the working f/# will still be similar to the image space f/#. The exit and entrance pupil diameters of a telephoto are not vastly different.

Of course this post is not meant in any way to bring clarity to the discussion. :)

Tin Can
28-Nov-2014, 00:47
LOL

Got that right!

Dan Fromm
28-Nov-2014, 02:00
Jason, you're right. I brought all that in because Ian mentioned some correction or other for working at near distances. I think he wants to shoot portraits with his tele lens and is concerned about getting good exposure when he does. The scary thing -- I'm not sure I should mention it, if I do another discussion will open -- is that unless he's shooting reversal film his film's exposure latitude will save him.

These discussions tend to wander. He started out wanting to know how to calibrate his lens' unmarked aperture scale. I think that's been answered. I think.

Mark Sawyer
28-Nov-2014, 10:16
These discussions tend to wander. He started out wanting to know how to calibrate his lens' unmarked aperture scale. I think that's been answered. I think.

Nonsense! We can make it more complicated and confusing if we try! :rolleyes:

Dan Fromm
28-Nov-2014, 11:10
Nonsense! We can make it more complicated and confusing if we try! :rolleyes:

Right! Fuzzy images, fuzzy thinking, that's us. Including me, of course.

Mark Sawyer
28-Nov-2014, 13:03
Personally, I'm working on a formula that includes f/stop, bellows extension factor, depth of field, reciprocity failure, development time, the Scheimpflug principle, and, if the model is attractive, her telephone number...

Tin Can
28-Nov-2014, 13:28
Lol