How do I calculate distance from subject if I know focal length and film width?

6

68degrees

Well-known
Local time
11:43 AM
Joined
Dec 11, 2012
Messages
882
Location
USA
View My Gallery
For example if a I want to photograph a subject that is 7 feet wide, and I want the subject to fill the entire width of the frame. The frame is about 60mm wide and the lens is an 80mm. What is the formula to calculate the distance of the camera from the subject? Does anyone here know how to calculate this on paper? Thanks.
 
There is very little that needs to be done in terms of calculation. You know the focal length, you know the image width (= the film width) and you know the subject width. The distance is just scaled focal length and so is equal (210/60)*80 = 280 or 2.8 meters. 210 is approx. 7 feet in centimetres.

This works because the angle under which the lens "sees" your film is the same (by definition) as the angle under which the lens sees the subject - so you have two triangles that differ only by a scaling factor.

Just imagine a 1:1 macro - in that case the focal length would be equal the distance to subject and the subject size would be equal the film size.

Does this help or should I make you a drawing?

EDIT: OK, here is the drawing:

11955598286_7280b9d3a8_c.jpg


I hope I did not make a mistake putting this quickly together 🙂
 
I think that's not quite correct, but a good approximation for small magnifications. 1:1 macro is exactly where it breaks down.
I'm too tired to set it right now, let me do it tomorrow...
 
Is there a particular reason that you can't determine this in situ? Take the camera with lens, mark out a 7 foot length on a wall, and then look through the viewfinder.
 
Yes the drawing was a great help. It doesnt get any simpler and faster than that. Thanks.

I made a few sketches myself and tried to look for a pattern but i couldnt come up with anything. So thanks. Now that I know the formula I can calculate in seconds any possible scenario while at the kitchen table so I can plan where I would have to be , how close, what lens, what camera etc etc.

I started thinking of this because I contemplated medium format such as 6x7. I thought. If my subject is 50 feet away and my focal length is 135mm and my subject is 6 feet tall, it doesnt matter what size the film is, the subject is still going to be the same size on the film whether its 35mm or 6x9. The differnce is just that there will be a lot of wasted space around the subject with the big film. If I want to fill the 6x9 frame with the subject I need a longer lens or must be closer, a distance tha would be much too close if I used the 35mm.
 
68degrees said:
Yes the drawing was a great help. It doesnt get any simpler and faster than that. Thanks.

I made a few sketches myself and tried to look for a pattern but i couldnt come up with anything. So thanks. Now that I know the formula I can calculate in seconds any possible scenario while at the kitchen table so I can plan where I would have to be , how close, what lens, what camera etc etc.

I started thinking of this because I contemplated medium format such as 6x7. I thought. If my subject is 50 feet away and my focal length is 135mm and my subject is 6 feet tall, it doesnt matter what size the film is, the subject is still going to be the same size on the film whether its 35mm or 6x9. The differnce is just that there will be a lot of wasted space around the subject with the big film. If I want to fill the 6x9 frame with the subject I need a longer lens or must be closer, a distance tha would be much too close if I used the 35mm.
Click to expand...

I must be dense because I don't understand this last paragraph at all, or the last sentence of the second paragraph. Maybe it's because I tend to do photography intuitively/by the seat of my pants, rather than forcing it to fit into a preconceived plan worked out at the kitchen table.
 
kossi008 said:
I think that's not quite correct, but a good approximation for small magnifications. 1:1 macro is exactly where it breaks down.
I'm too tired to set it right now, let me do it tomorrow...
Click to expand...

Yes, you are correct and I though of that, but that boils down whether the focusing is done simply by moving the whole lens - in that case the effective focal length doubles before you reach 1:1 magnification. But with a lens that has internal focusing (movie lenses do no 'zoom in' as you focus closer) may keep the same effective focal length as when focused to infinity.

Because of the additional complexity I kept it as simple as possible to make the point without more involved math (which I would first have to write down myself to be honest).
 
FrankS said:
I must be dense because I don't understand this last paragraph at all, or the last sentence of the second paragraph. Maybe it's because I tend to do photography intuitively/by the seat of my pants, rather than forcing it to fit into a preconceived plan worked out at the kitchen table.
Click to expand...

you aren't dense. maybe a little set in your ways but nothing wrong with that if you like your ways. Im still trying to develop my own way. You remind me of a great guitarist who doesnt know a note of music but plays by ear. Other musicians play by theory and what notes go with what chords and what scales etc. There are no rules in art.
 
There are no rules in art.

True enough. Good light to you sir!

Edit added after some thought about why I reacted to your post:

Just for the sake of photographic discussion, if you worked out that you would have to be say, 12 feet away from a subject with a certain camera and lens, when you actually got into that photographic situation, would you set up your subject and then measure/pace back 12 feet and take the picture? Wouldn't you adjust your position until things looked right in the viewfinder? I'm guessing you would. Again, not to be argumentative, what is the usefulness of the kitchen table calculations? (I hope I'm not coming across as a pest, and I apologize if I am. Just wondering.)
 
FrankS said:
Is there a particular reason that you can't determine this in situ? Take the camera with lens, mark out a 7 foot length on a wall, and then look through the viewfinder.
Click to expand...
This is good as long as you don't mind a margin. Many, if not all, viewfinders don't show 100% of what will be in the frame.
 
Matus said:
The distance is just scaled focal length and so is equal (210/60)*80 = 280 or 2.8 meters. 210 is approx. 7 feet in centimetres.
Click to expand...

So (210cm / 60mm) * 80mm = 280 cm 2.8 meters away from a 7 feet wide subject will fill the frame on a 6x6 camera?

so with a longer lens like a 300mm, I would have to be farther away to fit it all in the frame. Lets see.

(210cm / 60mm) * 300mm = 1,050 or 10.5 meters away.

Lets try 6x9 Kodak Medalist
(210cm / 90mm) * 100mm = 2.3 meters.

What about 6x7 Pentax 6x7
(210cm / 70mm) * 300mm = 9 meters away.

Nikon FM
(210cm / 35mm) * 300mm = 18 meters


So if you were trying to get a photo of a bald eagle with a 7 feet wingspan, and you wanted a full frame wing tip to wing tip of the thing flying straight at you, or opening its wings on its nest.

If you have the Nikon FM you have to be only 18 meters away to get that shot.
If you bring the pentax 6x7 with 300mm , you have to be 9 meters away to get the same shot.

Sure you could shoot the pentax 6x7 from 18 meters but then the eagle would still only be as big as with the FM and you would have all that wasted film to crop. No advantage.

If you know that you wouldnt be able to get closer than 18 meters for some obstacle or state park regulation, there would be no point in bringing the 6x7 correct?
 
I divide the desired magnification, which is 213.4cm, or 2,134mm divided by 60mm, equals 355.6 times. That means that I need the subject to be 355.6 focal lengths from the lens's principal plane. 355.6 x 80mm = 2,845mm, or 2.845M, or 112 inches, or 13' 3".

Since few of us know where the principal plane is inside the lens, I just add one more focal length, to get 2,925mm = 2.925M = 115.2 inches, = 9 feet, 7.157 inches. And that is the distance from the subject to the film plane, which is usually marked on the camera body.
 
This is the same as the APS v 35mm lens focal length multiplication we had to do before full frame DSLRs.

Or equivalency for those of us who have used 120, 4x5 and larger formats.
Yes, the format is everything as a lens of a given focal length will be a wider field of view on the larger format camera.

That said, the reason to take such a camera, like a 120 body or a 4x5 is tonal range, enlargeability and subject separation/selective focus.

The subject projected on the film may, in theory, be exactly the same size as in the examples given but if you're doing field of view equivalency between formats, the enlargement factor of the larger piece of film has the advantage.
It's not all as simple though as a focal length and a piece of film. Depending upon your enlargement, your circle of confusion for viewing will change and as a result you can get a completely different look from one format to another given all variables between the two remain the same.

This is the reason we don't all shoot micro 4/3 cameras or smaller, in spite of their ability to reach out farther with a given focal length compared to 35mm.

It all goes back to reproduction ratio and the ability of the lens to resolve fine details in the subject. Then the questions of subject/environment separation come into play with regard to what field of view you want. If you're shooting 4x5 and want a wide field of view, you use a lens shorter than about 150mm.

The FOV doesn't really get wide until you get to about 120mm (127mm was a favorite of the old press photographers using Graflexes) and then it gets really wide when you go to a 90mm or even 65mm lens. But a 90mm lens still doesn't have a lot of depth of field in this case because of its focal length so you can get a wide shot with decent subject/environment separation. Lesser so with a wider lens such as a 65mm but it is still evident.

Personally, I love giant pieces of film and normal to wide lenses.

I get where you're going but an easier trick would be to walk around with a string and a wire frame. A "poor-man's director's viewfinder" so to speak. You set the wire frame to the aspect ratio of the film you're shooting then mark on a string or even a slide rule the different focal lengths on one scale and distances on another. Slide the frame out and you have your field of view.

After a few years of using a 35mm rangefinder, I could just look at something and know my field of view coverage with a 21, 35, 50, 90 or 105mm lens. It just becomes instinctual with practice.

Phil Forrest
 
68degrees said:
If you know that you wouldnt be able to get closer than 18 meters for some obstacle or state park regulation, there would be no point in bringing the 6x7 correct?
Click to expand...

If that was all you were going to photograph, yes.

I usually do the opposite thing, and try to make sure there is some extra space in the frame, for cropping purposes. You never know when a little thing like the parallax factor is going to ruin a shot (on non-SLR cameras).

I can see where this is a good thing to know when photographing static objects, such as in a museum setting. And it's also helpful in making your gear selection when shooting wildlife so that you don't have to get too close, thus spooking the subject.

PF
 
You must log in or register to reply here.

Similar threads

K
Guide numbers -- how to calculate?
Replies
12
Views
2K
KoNickon
K
A
If you could travel with just one focal length or lens, what would it be?
7 8 9
Replies
174
Views
10K
Russ@SVP
Russ@SVP
popavvakum
Kiev cameras and Jupiter-8Ms: working distance, lens resolution and focus
Replies
11
Views
1K
rfaspen
rfaspen
lxforrest
A DIY retro digital camera project - Blog #2 Rethinking of retro digital camera
Replies
0
Views
219
lxforrest
lxforrest
lxforrest
A DIY retro digital camera project - Blog #2 Rethinking of retro digital camera
Replies
4
Views
658
lxforrest
lxforrest
Back
Top Bottom