Workshop Filling the frame calculation

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Martin
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Is there a calculation I can use to find at what distance from something I would need to be to fill the frame. For example, let's say I am using a 400mm lens on full-frame and my subject is 300mm tall, at what distance will I need to be from the subject for the top be at the top of the frame and the bottom...well, you get it?
 
This is easy to explain with diagrams, but I'll try. It uses only simple arithmetic and algebra.

If the subject is a distance u from the lens (warning: lenses are not a plane, so this should be distance from the front nodal plane, which isn't marked...) and the lens to image plane distance is v, then the magniication is the ratio of u and v.

If the object has a height h and the image a height j, then the magnification if j/h which is therefore equal to v/u.

The standard lens equation is 1/f = 1/u +1/v

From j/h = u/v, 1/v = j/hu

Substituting
1/f = 1/u + j/hu

For your example,

1/400 = 1/u + 24 (full frame height, landscape,or 36 if portrait, both approx)/300u

Assuming no slip in my algebra, rearrange to simplify and solve

300/400 = 300/u + 24/u

3/4 = 324/u

u = (324 × 4) / 3

u = 432mm

I sincerely hope someone will double check this - I'm rather tired and did this on screen rather than on paper...
 
Maybe I’m being dim here (quite likely :( ) but if you really meant 300mm (ie 12 inches) isn’t trial & error easier? Is there a 400mm lens that focuses to 17inches?
 
This is easy to explain with diagrams, but I'll try. It uses only simple arithmetic and algebra.

If the subject is a distance u from the lens (warning: lenses are not a plane, so this should be distance from the front nodal plane, which isn't marked...) and the lens to image plane distance is v, then the magniication is the ratio of u and v.

If the object has a height h and the image a height j, then the magnification if j/h which is therefore equal to v/u.

The standard lens equation is 1/f = 1/u +1/v

From j/h = u/v, 1/v = j/hu

Substituting
1/f = 1/u + j/hu

For your example,

1/400 = 1/u + 24 (full frame height, landscape,or 36 if portrait, both approx)/300u

Assuming no slip in my algebra, rearrange to simplify and solve

300/400 = 300/u + 24/u

3/4 = 324/u

u = (324 × 4) / 3

u = 432mm

I sincerely hope someone will double check this - I'm rather tired and did this on screen rather than on paper...

I'll have a look at this in detail later but I can see, from what sphexx says below, that I was a bit silly picking an object with height 300mm and would have done better with something like 20m to put things into the real world. However I shall substitute some better figures and see if that works out.
Maybe I’m being dim here (quite likely :( ) but if you really meant 300mm (ie 12 inches) isn’t trial & error easier? Is there a 400mm lens that focuses to 17inches?

Not being dim, as I have commented above, my object is a bit too small for a 400mm lens.

Loads of lens calculators on the web - mainly CCTV though

try this one

https://www.pointsinfocus.com/tools/depth-of-field-and-equivalent-lens-calculator/#{"c":[{"f":13,"av":"8","fl":50,"d":3048,"cm":"0"}],"m":0}

My, that's a site to make your eyes sore. I shall investigate.
 
There is another gotcha, apart from the "what the Dickens is the front nodal plane and where do I find it". I'm a medium and large format film photographer, and don't use autofocus zoom lenses. Autofocus means changing the focus for you, and, as implied by the lens formula, that means racking the lens out for close focus. Modern zoom lenses are big and heavy, and moving a big heavy object takes a lot of power. Which means short battery life.

As a result, it's better from a battery life point of view to simply move a few lens elements rather than the whole lens. And this usually (often, always?) changes the lens focal length. I have a couple of example photos taken with a 50mm prime lens and a zoom set at 50mm of an object at close distance. The image size us not the same (the zoom went more wide angle to reduce the amount it had to shift to focus).

So, if you calculate the distance, and it's a close up, and you use a zoom, be prepared to not fill the frame.
 
I'll have a look at this in detail later but I can see, from what sphexx says below, that I was a bit silly picking an object with height 300mm and would have done better with something like 20m to put things into the real world. However I shall substitute some better figures and see if that works out.

Thanks for clearing that up. I think I first read your 300mm as metres, ie something architectural maybe given the 400mm lens :) and then started scratching my head. Sometimes one can get bogged down in the figures and not see the larger picture :). I’m not much good at figures but I used to use a slide rule a lot which gets you judging whether a result is of reasonable magnitude or not ;).
 
Yep, that's what I want. Tried the formula with a few different variables and it checks out, or at least it gives intuitively reasonable figures. I'll do some real world tests tomorrow but it is showing me that if I want to take frame-filling pictures of greenfinches at 6m I'm going to require the Hubble Space Telescope -- only kidding, turns out to be about 1300mm; I won't be buying one.

Thanks a lot for that.
 
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