Timeline for Is it exactly true that doubling the focal length makes everything look twice as big?
Current License: CC BY-SA 3.0
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| Jun 12, 2017 at 14:56 | comment | added | mattdm | Okay, but: that is in objective conflict with empirical evidence. Try it: take a picture of a tape measure so your distance is about half the total width, and then measure half the length in the center compared to the whole length. It'll be off a bit from perfect because it's a real-world lens not an ideal one, but unless you're using a lens with a non-rectilinear projection, it'll be really close to "something twice as big in the real world is twice as big in the frame". It certainly will be nothing like 18% off as suggested by 1.69 vs 2. | |
| Jun 12, 2017 at 14:12 | comment | added | Brandon Dube | @mattdm yes, it is | |
| Jun 12, 2017 at 10:53 | comment | added | mattdm | At infinity, I can definitely say that no camera will be ever able to record it. But, let's stick to finite distances, where what you're saying seems to me to break down much, much more quickly than that. Let's put the 1m and 2m boxes at 1m and then 10m. As noted before, at 1m, the 1m box has an angular measurement of 53.1° and the 2m box 90°. At 10m, it's 5.7° and 11.4°, respectively. From what you are saying, in the 1m-distant case, the 2m box will appear 1.69× (90°÷53.1°) wider than the 1m box, and in the 10m-distant case, it will appear 1.99× (11.4°÷5.7°) wider. Is that right? | |
| Jun 12, 2017 at 5:09 | comment | added | Brandon Dube | @mattdm angles work at all distances. What if you put your 1m box infinitely far away? | |
| Jun 12, 2017 at 3:45 | comment | added | mattdm | I'm really not sure where you are going with that. The point is, something twice as big at the same distance will be twice as big on the sensor, even though the angle isn't halved. Because angles and the size of things do not have that relationship. | |
| Jun 12, 2017 at 3:41 | comment | added | Brandon Dube | @mattdm and if you move the boxes to a different distance the angular relationship is different. Angles work at all distances. Distances do not work at all angles. | |
| Jun 12, 2017 at 3:27 | comment | added | mattdm | We don't even need to bring cameras and lenses into it. Imagine you have a box that's 1m square and 1m away from you. The angle from your nose to the edges of that box is 53.1°. Now, we replace the 1m box with a 2m box. Now, it's 90°. Obviously, 53.1° is not half of 90° — but who cares? Half of 2m is definitely 1m. | |
| Jun 12, 2017 at 3:12 | comment | added | mattdm | I'm not seeing where that matters. You are using the same model for your angular field of view numbers, aren't you? | |
| Jun 12, 2017 at 1:48 | comment | added | Brandon Dube | You should not use thin lens theory for thick lenses at finite conjugates. | |
| Jun 12, 2017 at 1:28 | comment | added | mattdm | @BrandonDube Clearly it isn't. :) Particularly, with your 1000mm, 2000mm, 10mm, 20mm lens examples — assuming everything to be perfect, at 10m distant, the horizontal distance viewable in the frame would be 36cm, 18cm, 36m, and 18m respectively — not just the 2× difference between each set, but also a simple 100× difference between your long lenses and the wide angle ones. Sure, that's 2.1°, 1°, 121.9°, and 84° (also respectively), but... who cares? | |
| Jun 12, 2017 at 0:26 | comment | added | Brandon Dube | @mattdm is the ratio of angular FoV not precisely the difference in the apparent size of something? | |
| Jun 12, 2017 at 0:03 | comment | added | mattdm | PS Not my downvote. Looks like someone downvoted the question and all of its answers. | |
| Jun 11, 2017 at 23:36 | comment | added | mattdm | This is all true, but I don't think very practically interesting. Is there a case where the ratio of angular FoV is really meaningful? | |
| Jun 11, 2017 at 22:53 | comment | added | Brandon Dube | The relationship you reference is described in the equation I gave. Consider the case of a 90-degree full field of view, a 21mm lens has a half field of view of 45.86deg. A 42mm lens has an HFOV of 27.26deg. These are not exactly double each other. Consider further one half of 21mm, this is 64.12deg which is certainly not 2x 46 deg. Your commentary on the entrance pupil is not valid -- for any focal length you can place the entrance pupil at any distance (even infinity). | |
| Jun 11, 2017 at 16:57 | comment | added | Michael C | The relationship between AoV and object height in an image is also non-linear. As matt's answer and his most recent comment point out, the frame of reference of the angles is the center of the entrance pupil and it is exactly non-linear in a way that is reciprocal to the the non-linear relationship of focal length and angle of view. Which means the non-linearity of each relationship cancels the other. If a 100mm lens focused at infinity projects an object at infinity as 10mm high on the sensor, a 200mm lens focused at infinity will project the same object onto the sensor at a height of 20mm. | |
| Jun 11, 2017 at 16:19 | history | answered | Brandon Dube | CC BY-SA 3.0 |