Vascilli

12-Apr-2012, 22:06

So I put said lens on said camera and give it as much rise as I can. (Portrait orientation) How tall a building can I get in the frame without converging verticals from say, 100m?

View Full Version : Nikkor-SW 120 on 4x5: How much rise?

Vascilli

12-Apr-2012, 22:06

So I put said lens on said camera and give it as much rise as I can. (Portrait orientation) How tall a building can I get in the frame without converging verticals from say, 100m?

Leigh

13-Apr-2012, 03:29

The 120mm allow about 84mm rise. You need to learn the math.

- Leigh

- Leigh

Doremus Scudder

14-Apr-2012, 02:31

... You need to learn the math.

Possibly a bit more to help you understand :)

If you are infinitely far away, you can get an infinitely tall building in the frame. If you are really close, say two feet from the building, you're not going to be able to get an entire story in the frame.

It depends an angles and distance. If you don't want to do the math (I wouldn't either, my trig is not up to snuff these days, although I keep trying to find time for a brush-up), you can take your camera out, find a tall building and see for yourself.

Best,

Doremus

Possibly a bit more to help you understand :)

If you are infinitely far away, you can get an infinitely tall building in the frame. If you are really close, say two feet from the building, you're not going to be able to get an entire story in the frame.

It depends an angles and distance. If you don't want to do the math (I wouldn't either, my trig is not up to snuff these days, although I keep trying to find time for a brush-up), you can take your camera out, find a tall building and see for yourself.

Best,

Doremus

Leigh

14-Apr-2012, 09:31

No trigonometry required. It's a simple case of similar triangles.

I'll put the building 120m away rather than 100, for reasons that will be obvious:

Pick a point on the front of the lensboard for reference.

When focused at infinity, that point will be 120mm in front of the film.

Given that the building is 120m away, you have a ratio of 1000 to 1.

That same ratio holds for the heights of the image and the subject.

So 1" of image height equals 1000" (83 1/3 feet) of subject height.

If the building is 80 feet tall, its image on film would be just under one inch.

When you shift the lens up, the field of view at the subject changes by 1000 times the amount of shift.

While the specific numbers (120mm, 120m, 1000:1 ratio) only apply to this particular example,

the principle applies to all lenses and all formats.

- Leigh

nb. The explanation ignores minor factors like the distance between the two principal planes of the lens

and the difference between flange focal length and optical focal length.

The error introduced by this simplification is negligible.

I'll put the building 120m away rather than 100, for reasons that will be obvious:

Pick a point on the front of the lensboard for reference.

When focused at infinity, that point will be 120mm in front of the film.

Given that the building is 120m away, you have a ratio of 1000 to 1.

That same ratio holds for the heights of the image and the subject.

So 1" of image height equals 1000" (83 1/3 feet) of subject height.

If the building is 80 feet tall, its image on film would be just under one inch.

When you shift the lens up, the field of view at the subject changes by 1000 times the amount of shift.

While the specific numbers (120mm, 120m, 1000:1 ratio) only apply to this particular example,

the principle applies to all lenses and all formats.

- Leigh

nb. The explanation ignores minor factors like the distance between the two principal planes of the lens

and the difference between flange focal length and optical focal length.

The error introduced by this simplification is negligible.

Vascilli

14-Apr-2012, 13:40

No trigonometry required. It's a simple case of similar triangles.

I'll put the building 120m away rather than 100, for reasons that will be obvious:

Pick a point on the front of the lensboard for reference.

When focused at infinity, that point will be 120mm in front of the film.

Given that the building is 120m away, you have a ratio of 1000 to 1.

That same ratio holds for the heights of the image and the subject.

So 1" of image height equals 1000" (83 1/3 feet) of subject height.

If the building is 80 feet tall, its image on film would be just under one inch.

When you shift the lens up, the field of view at the subject changes by 1000 times the amount of shift.

While the specific numbers (120mm, 120m, 1000:1 ratio) only apply to this particular example,

the principle applies to all lenses and all formats.

- Leigh

nb. The explanation ignores minor factors like the distance between the two principal planes of the lens

and the difference between flange focal length and optical focal length.

The error introduced by this simplification is negligible.

Aha. I'm familiar with similar triangles for subject size and whatnot, I just couldn't relate it to shift. (Why, I don't know...) That helps.

I'll put the building 120m away rather than 100, for reasons that will be obvious:

Pick a point on the front of the lensboard for reference.

When focused at infinity, that point will be 120mm in front of the film.

Given that the building is 120m away, you have a ratio of 1000 to 1.

That same ratio holds for the heights of the image and the subject.

So 1" of image height equals 1000" (83 1/3 feet) of subject height.

If the building is 80 feet tall, its image on film would be just under one inch.

When you shift the lens up, the field of view at the subject changes by 1000 times the amount of shift.

While the specific numbers (120mm, 120m, 1000:1 ratio) only apply to this particular example,

the principle applies to all lenses and all formats.

- Leigh

nb. The explanation ignores minor factors like the distance between the two principal planes of the lens

and the difference between flange focal length and optical focal length.

The error introduced by this simplification is negligible.

Aha. I'm familiar with similar triangles for subject size and whatnot, I just couldn't relate it to shift. (Why, I don't know...) That helps.

David Lobato

15-Apr-2012, 10:19

My quick calculation shows you will have approx. 4 inches of front rise with the 120mm Nikkor on 4x5 film in the landscape (horizontal) format. So according to Leigh's figures you can photograph a building at least 320 feet tall at 120 meters distance. The next question is if your camera has 4 inches of front rise. Apologies for the mixed units btw.

Vascilli

15-Apr-2012, 13:00

The camera is no problem. It's a Toyo 45G.

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