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Determining focal length
So... I feel like this may be one of those really dumb questions with a simple answer. But I haven't figured it out yet..

Is there a way to determine what size object will fill the frame of my camera at a given focal length? For example, I am standing 30 feet from a 5 foot x 7 foot mural painted on a wall; what focal length will I need to fill the frame with the mural? Or, working in reverse, my lens has a max focal length of 200mm, how far away from the same mural will I have to stand to fill the frame? There has to be a simple equation for this. Maybe a concise explanation of how focal length relates to size/distance would help also..

I suppose I could probably do some math, but I am too lazy, and figured I'd just probe for someone who knows the answer already. Thanks in advance!Eric..

Comments (13)

Too lazy to answer....

E1 w/ grip, e510, e300 w/ grip, 8mm FE, 14-54mm, 35mm, 50mm, 40-150mm,50-200mm, fl-50, fuji - 6800..

Comment #1

What I meant was that I'm too lazy to go on a goose chase looking for the formula when I have a fantastic resource (mostly) full of helpful people right here.Eric..

Comment #2

JudgeBenWiles wrote:.

Is there a way to determine what size object will fill the frame ofmy camera at a given focal length? For example, I am standing 30 feetfrom a 5 foot x 7 foot mural painted on a wall; what focal lengthwill I need to fill the frame with the mural?.

How big is your camera's sensor? (Or film size?) Your profile says you're using a D80, so I'll assume that's correct..

The sensor size of a D80 is 23.6mm x 15.8mm..

30/5 = 6. 6x15.8= 95.30/7 = 4.3. 4.3x23.6=101..

So call it about a 100mm lens (a 5x7 won't fit precisely on a 3:2 sensor)..

Or, working in reverse, my lens has a max focal length of 200mm, howfar away from the same mural will I have to stand to fill the frame?.

200/15.8 = 12.7. 12.7x5 = 63.200/23.6 = 8.5. 8.5x7 = 59..

So call it about 60 feet..

There has to be a simple equation for this..

There is. Basically "the distance divided by the size" is the same both in front of (subject distance) and behind (focal length) the lens. A 100 foot object 100 feet away casts a 20mm image with a 20mm lens. a 500 foot object 5000 feet away casts a 20mm image with a 200mm lens..

This is provided that the lens is focused at infinity, which it usually isn't. But if the subject distance is significantly greater than the focal length, it works out to be fairly close. Also note that the focal lengths marked on lenses are only at infinity focus, so at closer distances the actual focal lengths are different...

Comment #3

Marvelous. I'm still processing all that info, but very helpful. Thanks!.

And yes, D80. And actually, my walkaround lens is 18-135 currently, but I can adapt any of this info to my equipment.Eric..

Comment #4

There probably is some equation out there somewhere but I've never come across it. If you get out there and take lots of pic's with primes you'll very quickly get to 'know' the focal length to use before putting the viewfinder to your eye ... HonestSimonhttp://www.simonstanmore.com..

Comment #5

Yes, I imagine so! Keep in mind this is coming from someone with very little experience in "real" hands-on photography (non-P&S). The idea behind my post was planning for a future shoot. I know what I am going to be shooting and about how far from the subject I will be, so I wanted to know how to calculate the focal range I would need in a lens. I'm sure as I shoot more, this will become second nature to me. Thanks again!Eric..

Comment #6

Simon Stanmore wrote:.

There probably is some equation out there somewhere but I've nevercome across it. If you get out there and take lots of pic's withprimes you'll very quickly get to 'know' the focal length to usebefore putting the viewfinder to your eye ... Honest.

Is trigonometry no longer taught in high school? Do people no longer graduate from high school? I don't expect everyone to remember their trig but it's quite simple to dig it up on the net..

And yes, you never need it in practice, just when answering hypothetical questions about how far you have to be to take in a certain view..

Leonard Migliore..

Comment #7

Well, I dooo hate trig, but can still rustle up a tangent when I need to. My problem is I don't have the photographic knowledge to base any mathematics on. I didn't know how focal length behind the lens related to subject distance in front..

As I mentioned, I could certainly research this for myself, but I found it more convenient to ask around here, where people generally are very forthcoming with helpful information regarding subjects about which they are already well-versed. Also, I can usally grasp a new concept better when it is explained to me in simple terms (which Doug kindly provided), and not in textbook tech-speak form. Apologies if it was stupid or lazy, but this IS the Beginners' forum.Eric..

Comment #8

JudgeBenWiles wrote:.

Well, I dooo hate trig, but can still rustle up a tangent when I needto. My problem is I don't have the photographic knowledge to base anymathematics on. I didn't know how focal length behind the lensrelated to subject distance in front..

As I mentioned, I could certainly research this for myself, but Ifound it more convenient to ask around here, where people generallyare very forthcoming with helpful information regarding subjectsabout which they are already well-versed. Also, I can usally grasp anew concept better when it is explained to me in simple terms (whichDoug kindly provided), and not in textbook tech-speak form. Apologiesif it was stupid or lazy, but this IS the Beginners' forum.Eric.

I agree that Doug outlined the solution for you most effectively. My response was to Simon's "there must be some equation out there". That's the similar triangles part. You have two isosceles triangles. One triangle has the focal length as it's height and the sensor size as it's base. The other triangle has the distance to the subject as it's height and the subject size as it's base..

Leonard Migliore..

Comment #9

Right you are. See? Now wasn't that easier than insulting my education? And a whole lot more informative for this beginner..

Just because I am a photographic noob doesn't mean I am unintelligent or incapable of doing a Google search. I am simply using a different resource..

Very good explanation, thank you!.

Edit: Fair enough, but I felt a touch insulted, since my question was the one being discussed. I do appreciate your explanation!Eric..

Comment #10

JudgeBenWiles wrote:.

Right you are. See? Now wasn't that easier than insulting myeducation? And a whole lot more informative for this beginner..

Just because I am a photographic noob doesn't mean I am unintelligentor incapable of doing a Google search. I am simply using a differentresource..

Very good explanation, thank you!.

Edit: Fair enough, but I felt a touch insulted, since my question wasthe one being discussed. I do appreciate your explanation!Eric.

You have been perfectly clear in your search, and I did not intend to denigrate it. I'm sorry that you feel insulted and apologize for it. My original gratuitous insult, such as it was, was directed at someone else (I probably should not have done that either, so I apologize to Simon, too.)..

Leonard Migliore..

Comment #11

Aw man, now I'm bummed that I picked a fight with a nice guy unintentionally. I wish you were an a-hole instead, because I was ready for a throwdown. Cheers, thanks for the help, sincerely. Eric..

Comment #12

Than the focal length, then.

F/S = D/FOV.

Where F = focal length in mmS = sensor size in mmD = focus distance in feetFOV = field of view at focus point in feet.

Obviously, S and FOV are either vertical or horizontal measurements.Leonhttp://homepage.mac.com/leonwittwer/landscapes.htm..

Comment #13

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This question was taken from a support group/message board and re-posted here so others can learn from it.

 

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