Lens focal length/framing/distance math?
I am trying to decide right now between a 20mm prime or a 1835 zoom. I know the prime is a better lens but the 1835 has zoom capabilities. I am wanting to know how much it would actually affect my shooting and have wondered this in the past.
How can I solve for X? At 18mm my subject fills the frame when I am 10 feet away. Thus, at 35mm my subject will fill the frame when I am X feet away?
Thanks!
Jeremy

Nikon D700 w/grip, Nikon D300s, AF 50mm f/1.8D, AFS Nikkor 2485 f/3.54.5 ED VR, AFS NIKKOR 70200 f/2.8G ED VR II, AFS Nikkor 1870 f/3.54.5G ED, SB700, 2xYN565EX
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#1. "RE: Lens focal length/framing/distance math?"  In response to Reply # 0
Thu 30Aug12 08:04 PMPersonally, I use fCalc, which is a shareware program. The user manual for the program contains the actual equations.
Unfortunately, these equations can be hard to use as more than an approximation to answer questions such as yours. That's because a) the equations are simplified, and b) lenses like both the 1835mm and 20mm change focal length with focus distance. But you'd have to be just under twice the distance to your subject to fill the frame with a 35mm lens than an 18mm.
One of the biggest mistakes a photographer can make is to look at the real world and cling to the vain hope that next time his film will somehow bear a closer resemblance to it.  Galen Rowell

#2. "RE: Lens focal length/framing/distance math?"  In response to Reply # 0
MotoMannequin Registered since 11th Jan 2006Thu 30Aug12 08:45 PMThis is related to angle of view, which results in a fairly complicated trig function which I could lay out for you. However, unless you are shooting macro or otherwise a completely flat surface (everything in your frame is at the same distance), I believe it's not a worthwhile approach.
Changing your position and focal length so that your subject "fills the frame" will yield a radically different photo because your perspective will have changed. Even if the subject is the same size, its relation to the foreground and background will have changed. Therefore, I'd suggest you instead work with different focal lengths to get a feel for perspective, which is more exaggerated at wide focal lengths, until you develop an intuition and then know what focal length(s) you want to use as soon as you see the subject. Peterson's Learning to See Creatively has as set of exercises designed to teach exactly this. They are worthwhile to work through, as is the book.
That doesn't help answer the question of which lens. Looking at your profile, it looks like 50mm is your widest lens. In this situation, I believe the 20mm prime will leave some holes to fill, namely another prime at 28mm or 35mm. In that case you could cover the entire range nicely with an 1835. That really becomes a question of budget, and whether you're more comfortable with primes or zooms.
Larry  a Bay Area Nikonian
My Nikonians gallerywww.temperedlight.com 
#3. "RE: Lens focal length/framing/distance math?"  In response to Reply # 2
jeremy_c Registered since 14th Jul 2012Thu 30Aug12 08:56 PM>This is related to angle of view, which results in a fairly
>complicated trig function which I could lay out for you.
>However, unless you are shooting macro or otherwise a
>completely flat surface (everything in your frame is at the
>same distance), I believe it's not a worthwhile approach.
I do understand how focal length and point of view affects the resulting photograph and that my question may not be totally appropriate. At 18mm the photograph is going to be dramatically different than at 35mm, even if the framing remains exactly the same.
My question was thus a bit elementary. What I was really wanting to know is if I was stuck w/just the prime and had no other option, how far would I have to move forward or backward to achieve the same framing and simply taking whatever the perspective compression/expansion/distortion result would have been due to my choice of the prime over the zoom.
I eventually will be purchasing the 2470, but that is a few months off. I would love to just purchase the 1224, but the 2470 has to come before that one.Jeremy

Nikon D700 w/grip, Nikon D300s, AF 50mm f/1.8D, AFS Nikkor 2485 f/3.54.5 ED VR, AFS NIKKOR 70200 f/2.8G ED VR II, AFS Nikkor 1870 f/3.54.5G ED, SB700, 2xYN565EXVisit my Nikonians gallery.
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#4. "RE: Lens focal length/framing/distance math?"  In response to Reply # 3
MotoMannequin Registered since 11th Jan 2006Thu 30Aug12 09:14 PM  edited Thu 30Aug12 09:26 PM by MotoMannequinWell, I just ran through the math and it turns out what I said isn't exactly true  all the trig functions cancel each other and the resulting formula is pretty simple.
Let:
a = angle of view
D = sensor diagonal
f = focal length
d = subject distance
w = subject width
we have:
a/2 = arctan(D/(2*f)) (formula for angle of view)
w/(2*d) = tan(a/2) (definition of tangent)
Combining these yields:
w/(2*d) = D/(2*f)
So, if we have 2 different focal lengths, and 2 different subject distances, but want the same subject width:
w/(2*d1) = D/(2*f1)
w/(2*d2) = D/(2*f2)
or...
w = 2*d2*D/(2*f2) = 2*d1*D/(2*f1)
d2/f2 = d1/f1
d2 = d1*(f2/f1)
or, simply, if you double your focal length, double your subject distance. Of course this is simplified because lenses actually change angle of view with focus distance, but this should work as a general rule, just take into account that the focal length reading on your lens might not be accurate
Anybody feel free to correct me...
Larry  a Bay Area Nikonian
My Nikonians gallerywww.temperedlight.com 
#7. "RE: Lens focal length/framing/distance math?"  In response to Reply # 4
gah Registered since 10th Oct 2012Wed 10Oct12 10:45 PM>Well, I just ran through the math and it turns out what I
>said isn't exactly true  all the trig functions cancel each
>other :+ and the resulting formula is pretty simple.
>a = angle of view
>D = sensor diagonal
I would have used the long side of the sensor, instead of the diagonal,
but maybe this is the official definition. I don't usually frame for
the diagonal.
>f = focal length
>d = subject distance
>w = subject width
>a/2 = arctan(D/(2*f)) (formula for angle of view)
>w/(2*d) = tan(a/2) (definition of tangent)
>Combining these yields:
>w/(2*d) = D/(2*f)
Yes, you can also get it directly from similar triangles.
>So, if we have 2 different focal lengths, and 2 different
>subject distances, but want the same subject width:
(snip)
>d2/f2 = d1/f1
(snip)
>or, simply, if you double your focal length, double your
>subject distance. Of course this is simplified because lenses
>actually change angle of view with focus distance, but this
>should work as a general rule, just take into account that the
>focal length reading on your lens might not be accurate ;)
The difference normally only needs to be considered for closeup
or macro photography, unless you measure really carefully.
 glen


#5. "RE: Lens focal length/framing/distance math?"  In response to Reply # 3
MotoMannequin Registered since 11th Jan 2006Thu 30Aug12 09:17 PM>I eventually will be purchasing the 2470, but that is a few
>months off. I would love to just purchase the 1224, but the
>2470 has to come before that one.
Since you're shooting an FX camera, I guess you mean the 1424 not 1224?
Larry  a Bay Area Nikonian
My Nikonians gallerywww.temperedlight.com 
#6. "RE: Lens focal length/framing/distance math?"  In response to Reply # 5
jeremy_c Registered since 14th Jul 2012Thu 30Aug12 10:02 PM>>I eventually will be purchasing the 2470, but that is a
>few
>>months off. I would love to just purchase the 1224, but
>the
>>2470 has to come before that one.
>
>Since you're shooting an FX camera, I guess you mean the 1424
>not 1224?
>
Opps, yes. I recently upgraded from a D7000, getting my lenses mixed.
JeremyJeremy

Nikon D700 w/grip, Nikon D300s, AF 50mm f/1.8D, AFS Nikkor 2485 f/3.54.5 ED VR, AFS NIKKOR 70200 f/2.8G ED VR II, AFS Nikkor 1870 f/3.54.5G ED, SB700, 2xYN565EXVisit my Nikonians gallery.
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