Why are lens and format measurements so absurd?

[Nicoleica]

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Among my Nikkors are a 17-35/2.8 (77mm filter thread and about 105mm long) and a 105/2.5 (52mm filter thread and about 70mm long). Which is the "Big" one?

 

That's easy. Both of them, as they are both bigger than the one that came with the camera. I didn't say that my friend's system was accurate, or even realistic. Just that it was simple.

[Guest mc_k]

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Thanks for the diagram and discussion. Can you give a source for this--am not questioning it. Interesting! I have seen both "angle of view" and "image distance" used differently, so I was glad to have the diagram.

[Guest mc_k]

To be quite nitpicking, you could argue that the viewing angle was determined by (a) the distance between the lens and the sensor plane and ( by the bounding frame of the sensor area.

 

In order to restrict the angle of view to the image distance, you'd have to add the restriction that a reasonably sharp image had to be produced in the sensor plane.

 

I thought I wanted the same...to just use the right half of the diagram, with the full sensor/film dimension shown, and replace "image distance" with the usual conjugate distance in the lens equation.

 

What do I call this angle?

[NZDavid]

A friend of mine has a much simpler system. She has 3 classifications for lenses. "Normal" is the lens that came with the camera. All others are designated as "Big" lenses or "Small" lenses. (She applies this same system to cameras too.)

 

Some photographers may think they have a long lens but they have tiny sensors.

 

"Is that a telephoto you have in your camera bag...?"

[adli]

Mmm, okay it's the same piece of hardware that it was when you had the M8, but now it works for you in a different manner, does it not? Because you have a larger sensor, the lens records a greater portion of the subject on it, assuming equal distances from the camera. So, now you have a (say) 84 degree diagonal field of view, whereas on the M8 you had (say) a 68 degree field of view. Which I think was really the point that David was getting at when he started this thread.

 

It's not we users who make life complex, its the degree of inter-operability and the variety of film/sensor formats that draws us into a state of confusion.

 

Yes, but my point was that if I had bought the lens as an 68 degree lens, I would be be very disappointed when buying the M9 and having to re-brand my lens. Imageing haveing to send all your lenses back to Leica to change the FoV inscription when you change from M8 to M9, Does not really make sense, does it, as long as the lens is the same?

[NZDavid]

We've just been showing a friend around. She has one of those Nikon superzooms. It has a 15X lens. That is the main number on the barrel. Now remove the lens cap and it says (from memory) 5.0-75mm, 3.5-5.

 

So she has a 75mm lens. A portrait lens, then? Semi-tele? Only on 35mm. With a small sensor camera it will be a huge telephoto. But on a medium format film camera it will be a standard lens. On a large format camera 75mm is wide angle. Of course you knew all that already. So did I and so will everybody else used to Leica cameras. But for many people starting off those numbers will mean very little. And 5mm? Ultra fish-eye? No, just semi wide angle on a small format digital camera.

 

From time to time, however, you have probably spotted questions on this forum asking which lens to buy: 21, 24, 28 wide angle, for example? The prospective buyer has wanted to know whether there is really much difference when the focal lengths are so close. Well, yes. A 21mm lens (with 24x36mm format sensor or film) covers 92º, 24mm covers 84º, and 28mm covers 76º. A few mm in focal length makes a quite a difference to field of view. Again, you probably knew that already, but did you know by how much?

 

A 75mm lens or a 90mm? Must be quite difference with 15mm more focal length? 31º for the 75mm, 27º for the 90mm. Hmm. About the same number of degrees separate these lenses as do the 24mm and 28mm at the wide angle end, where just a few millimeters matter more.

 

Again, if you know what the focal-length numbers mean already, no worries. (The M8 is superseded, and not really relevant to this question. Future Leica Ms will surely all be 24 x 35mm format.)

 

I am not advocating ditching using focal-length measurements entirely. But I think the relationship between focal length and field of view is quite fascinating and deserves more recognition!

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I am not advocating ditching using focal-length measurements entirely. But I think the relationship between focal length and field of view is quite fascinating and deserves more recognition!

 

For me, that's a bit like buying your petrol by the mile instead of by the gallon.

[scsambrook]

For me, that's a bit like buying your petrol by the mile instead of by the gallon.

 

Well, I think that's what I actually do - sometimes at least !

[Guest mc_k]

o.k., here is a problem related to angle of view for someone. How does one calculate the framelines error with increasing distance? Leica lists some figures in the "Technical Data" appendix to the M8, M8.2, and M9 instructions, along with the reference distance at which the framelines are correct:

 

FOCAL LENGTH, REFERENCE DISTANCE, MAX ERROR

 

(from M8 manual)

28mm, 0.7m, 9%

90mm, 1m, 23%

 

(from M8.2 manual)

24mm, 2m, 7.3%

90mm, 2m, 18%

 

(from M9 manual)

28mm, 1m, 7.3%

135mm, 1m, 18%

 

They don't say if the error is a % of area or something else. How does one get a formula for the error to get these figures? Obviously they are calculated. I have posted on this before, but never got any solutions.

[Guest mc_k]

...Obviously they are calculated...

 

I should say, obviously they can be calculated.

[giordano]

It's too late and I'm too tired to spell this out, but the basic approach is to use the familiar lens equation to calculate the magnification (u/v) at various distances with the lens in question. Then take the difference between the magnification at the distance where the framelines are correct and the magnification at the focused distance, and that gives you the error in the framelines as a ratio which you can convert into a percentage of the image width or image area as you choose.

 

Unfortunately this approach isn't valid with lenses that focus closer by shortening their focal length rather than by bodily moving away from the sensor (i.e. the WATE and virtually all zoom and internal focusing lenses, I'm not sure about the MATE) because their magnification doesn't vary with focus distance in the same way as simple lenses.

[AlanG]

I think if you are serious enough about photography to buy a $7,000 camera and lenses that may run into the thousands, this should either be very basic to you or you should learn it fast.

[Guest mc_k]

I think if you are serious enough about photography to buy a $7,000 camera and lenses that may run into the thousands, this should either be very basic to you or you should learn it fast.

 

thanks but I didn't ask for framing help, I asked how to calculate the framing errors in the Leica instructions. Seems like one of the most obvious questions one can ask, if you want to know all about a rangefinder. I have no idea why I can't just look this up somewhere.

 

If you can do the problem yourself, great, I would appreciate the info.

[Guest mc_k]

It's too late and I'm too tired to spell this out, but the basic approach is to use the familiar lens equation to calculate the magnification (u/v) at various distances with the lens in question. Then take the difference between the magnification at the distance where the framelines are correct and the magnification at the focused distance, and that gives you the error in the framelines as a ratio which you can convert into a percentage of the image width or image area as you choose.

 

Unfortunately this approach isn't valid with lenses that focus closer by shortening their focal length rather than by bodily moving away from the sensor (i.e. the WATE and virtually all zoom and internal focusing lenses, I'm not sure about the MATE) because their magnification doesn't vary with focus distance in the same way as simple lenses.

 

Thanks for the details. That is how I tried to do it before, except the only numbers that came out right were the ones from the M8 instructions....even if I used the exact focal lengths from the brochures, or other adjustments. It's probably just me, but I was wondering if anyone had given it a try before. I know the lens has to focus as a unit, but I don't really know the focusing details for any of those lenses: 24's, 28's, 90's, and 135. Thank you for the help.

[giordano]

Since the distance factor is only one part of framing error, maybe the thing to do is to make a series of exposures of a static subject at different distances, each time carefully comparing the image captured with the image in the finder and measuring the differences.

 

But if precise framing is really important it makes more sense to use a camera with a through-the-lens finder.

[AlanG]

thanks but I didn't ask for framing help, I asked how to calculate the framing errors in the Leica instructions. Seems like one of the most obvious questions one can ask, if you want to know all about a rangefinder. I have no idea why I can't just look this up somewhere.

 

If you can do the problem yourself, great, I would appreciate the info.

 

Sorry, but my post referred to the original point and had nothing to do with framing errors.

[Guest mc_k]

...

 

my mistake, sorry.

[UliWer]

o.k., here is a problem related to angle of view for someone. How does one calculate the framelines error with increasing distance? Leica lists some figures in the "Technical Data" appendix to the M8, M8.2, and M9 instructions, along with the reference distance at which the framelines are correct:

 

FOCAL LENGTH, REFERENCE DISTANCE, MAX ERROR

 

(from M8 manual)

28mm, 0.7m, 9%

90mm, 1m, 23%

 

(from M8.2 manual)

24mm, 2m, 7.3%

90mm, 2m, 18%

 

(from M9 manual)

28mm, 1m, 7.3%

135mm, 1m, 18%

 

They don't say if the error is a % of area or something else. How does one get a formula for the error to get these figures? Obviously they are calculated. I have posted on this before, but never got any solutions.

 

Those calculations are not in any way related to the lenses' angles of view nor did they depend on sensor size.

 

They just want to say the follwoing:

 

With the M8u the frames for the different focal lengthes in the viewfinder were exact at the shortest focussing distance (usually 0.7m, 1m for some lenses like a 90mm). Focussing the lens to infinity led to a larger picture in reality than you saw in the frames. The percentage indicates the difference: with a 90 mm lens and the frames in the viewfinder calibrated on 1m focal distance, you get a picture 23% larger than the frames, if you focus on infinity.

 

The calibration of the M8u was the traditional one used for all M cameras since the M2. But with digital opportunities to see the results at once, users started to complain: frames are to small and not precise, make them bigger! So they calibrated the frames for the M 8.2 on 2m , which led to less difference when focussed on infinity - but you get less on your picture than you saw in the frame, when you use the closest focus.

 

For the M9 they compromised on 1 m in general to make some concessions to those who found the traditional frames too small and as well to those who criticized the loss for close focussing caused by the calibration for the M8.2. So with the M9 you loose less of your frame view when you focus close, and gain less when you focus on infinity.

[Guest mc_k]

Those calculations are not in any way related to the lenses' angles of view nor did they depend on sensor size.

...

 

They are related to the angle of view. The % difference is calculated from the angular field of view--or whatever you want to call the angle--at infinity and at the reference distance. I never said they depend on format; I gave the values from the different instruction manuals because the different manuals have figures for different focal lengths.

 

What I suggested is that you can get a simple formula for the % error in terms of the reference distance and focal length. Give it a try and see what you get. Won't help your framing any, but if you have a curiosity about how framelines work, or want to adapt or make a finder, etc., this kind of thing should be interesting. As John said it would not work for all lenses.

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[UliWer]

They are related to the angle of view.....

 

Yes. When I said, they were "not in any way related..." I was wrong. Naturally the frames must be related to the lense's angle of view!

 

Though when the question is about the percentage the frames for the two versions of the M8 differ, we can see, that this percentage is not related to the lens, but to the calibration of the frame for a certain focal length.