0
\$\begingroup\$

I am trying to figure out what focal length lens I would need to get a specific object to completely fill the image from a camera. Assuming I have two RAW files from the same camera body, shot from the same position, is there a way to create a ratio or equation that would help map the necessary focal length to fill the frame with a given object, such that I could then say "that street light at 100mm is 50x50 pixels, so if I change to an 800mm lens, the ratio is y, so that object is now 200x200 pixels."

Furthermore, once that ratio is calculated, will it always apply to all lenses on that body, or would the ratio change based on the lens? For instance, if I have a 70-200 zoom lens, would changing to a 100-500 yield the same equation or does individual lens optics play a part in the calculation?

I'm considering investing in a longer lens to try out wildlife/birding photography but I'm trying to wrap my head around "800mm would make distant-object-x 'this big' in the frame." I'm trying to figure out whether mathematics can be used to help answer that question definitively.

Improve this question
\$\endgroup\$
1
  • \$\begingroup\$ When I go on bird walks the most common rig I see is a 100-400 zoom on an APS-C body, giving an effective focal length of 640mm. I have read about 150-600 zooms from Sigma and Tamron but have not seen them. I do a lot of bird photography and would not use anything less than 600mm effective. Most of the time I use a Nikon P900 with its 2000mm effective focal length. The whopping zoom range means the lens is not as crisp as the shorter zooms, but with practice one can find the bird and hand hold it. \$\endgroup\$
    – Ross Millikan
    Feb 17, 2021 at 16:12

2 Answers 2

Reset to default
1
\$\begingroup\$

The actual formula involves tangents and arc tangents, but for long lenses the angles are small and the tangents/arctangent can be approximated as a linear function so the results you get with a linear interpolation are going to be accurate enough (the actual focal length is not exactly what is advertised, once you take in account the focusing).

This said, there are limits to the usability of long lenses. A 800mm lens that you can carry around with have a rather small aperture (f/5.6 at best or even f/8) and in anything but very bright conditions you won't be able to use sufficiently short exposures to compensate for the increased sensitivity to camera shake. In addition such lenses are expensive (if good enough) and difficult to master (try shooting birds in flight...). Another limit is... your eyesight: you can't shoot a bird you didn't spot first, so it has to be close enough for you to see it. Your best zoom is proper behavior/technique and proper attire, a $20 ghillie suit may be a better investment than a $2000 lens.

If you want a lens to shoot birds, see what other people are using, it's sometime better to have a small bird in the picture than no bird at all.

Improve this answer
\$\endgroup\$
1
\$\begingroup\$

On a full frame/35mm camera; from about 50mm every time you double the focal length the FOV is approximately halved, and the size something is recorded at doubles (L&W). So for your example beginning from 100mm/50px; 100-200-400-800mm lenses result in 1x-2x-4x-8x or 50-100-200-400px.

The 50mm approximation for the full frame starting point correlates to the format's "normal lens" and an ~ 45* horizontal FOV. e.g. 30mm APS, 45mm FF, 150mm 4x5, 300mm 8x10; it does not work for wider focal lengths.

Improve this answer
\$\endgroup\$
2
  • 1
    \$\begingroup\$ It might be helpful to mention that this is only an approximation that gets closer to the actual case as the focal lengths increase. As the focal lengths decrease the approximation can get ridiculously inaccurate by several orders of magnitude. And the 50mm "line" only works for FF or smaller imaging areas. As format size increases, the minimum focal lengths where the approximation and the actual computation reasonably approach one another also increases. For an 8x10 view camera, the approximation doesn't begin to be close to reality until about 350mm and longer. \$\endgroup\$
    – Michael C
    Feb 16, 2021 at 23:18
  • \$\begingroup\$ @MichaelC, updated... \$\endgroup\$
    – Steven Kersting
    Feb 17, 2021 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.