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Say I have an object, whose actual size is 10 units in diameter, and it is 100 units away.

I can find the angular diameter as such: $2\arctan(5/100) = 5.725\ $ radians.

Can I use this angular diameter to find the apparent linear size (that is, the size it appears to be to my eye) in generic units at any given distance?

cmal
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2 Answers2

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It appears you are using the wrong angular units: $2\;\tan^{-1}\left(\frac{5}{100}\right)=5.7248$ degrees $=0.099917$ radians.

The formula you cite above is valid for a flat object perpendicular to the line of sight. If your object is a sphere, the angular diameter is given by $2\;\sin^{-1}\left(\frac{5}{100}\right)=5.7320$ degrees $=0.100042$ radians.

Usually, the angular size is referred to as the apparent size. Perhaps you want to find the actual size of the object which has the same apparent size but lies at a different distance. In that case, as joriki says, just multiply the actual distance by $\frac{10}{100}$ to get the actual diameter. This is a result of the "similar triangles" rule used in geometry proofs.

Update: In a comment to joriki's answer, the questioner clarified that what they want is to know how the apparent size varies with distance. enter image description here

The formulae for the angular size comes the diagram above:

for the flat object: $\displaystyle\tan\left(\frac{\alpha}{2}\right)=\frac{D/2}{r}$; for the spherical object: $\displaystyle\sin\left(\frac{\alpha}{2}\right)=\frac{D/2}{r}$

robjohn
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Yes you can, but it's much easier to just use the original values. The ratio of the apparent size to the distance is constant, so in your case it's $1/10$, and you just multiply the distance by that to get the apparent size.

joriki
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  • I must be confused or not seeing things correctly: 1/10 x distance results in a larger number as the distance increases. Shouldn't the visual size get smaller as the distance increases? – cmal Oct 14 '11 at 21:59
  • I think I know what you're missing: 10 units is not the apparent visual size, it is the ACTUAL size. I need to know what the apparent visual size will be at any given distance. – cmal Oct 14 '11 at 22:11
  • @cmal: I, too, had a hard time figuring out what the question was. The angular diameter does indeed change as the distance changes. As the distance increases, the apparent size decreases. It is does not vary exactly as the inverse of the distance, but as the distance gets much larger than the actual size, it varies very close to the inverse of the distance. If the object is flat, the angular size would be $2;\tan^{-1}\left(\frac{5}{\text{distance}}\right)$. If spherical, $2;\sin^{-1}\left(\frac{5}{\text{distance}}\right)$. – robjohn Oct 14 '11 at 22:59
  • I understand what you're saying, but that's not what I'm looking for. I know how to find the angular diameter. What I need to find is the visual diameter (the size it appears from the viewpoint), measured in units, at any given distance. I appreciate you sticking with me on this, I hope that makes things more clear. – cmal Oct 14 '11 at 23:07
  • @cmal: would you please explain the difference between the angular diameter and the visual diameter? As far as I can tell, they are the same thing. Also, what units do you want? – robjohn Oct 15 '11 at 00:46
  • Now I'm completely confused :-) I don't understand what cmal is saying and I don't understand what robjohn is doing. @robjohn: How is it that you added an update to your answer two hours ago that sounds as if you now understand what the question is and then added a comment one hour ago that sounds as if you don't? – joriki Oct 15 '11 at 03:04
  • @cmal: I think you need to define what you mean by "the size it appears from the viewpoint, measured in units, at a given distance". The only two interpretations of that that I can come up with are robjohn's and mine, but it seems these are both not what you have in mind. – joriki Oct 15 '11 at 03:08
  • Angular diameter what you've provided. For "visual diameter", I want to know how to turn angular diameter into the size something will appear to be, to my eye, at any given distance away from me, measured in any linear unit (mm, cm, generic "units", etc). For instance, let's say I take a photo of an oncoming car, and I know that the car is 10 meters wide and 100 meters away. I want to know how wide the car will appear in a photo taken when it is 90 meters, 80 meters, or any other distance away from the camera. If possible ignore camera-specific variables. – cmal Oct 15 '11 at 03:11
  • @joriki: It sounded from a comment that cmal made: I need to know what the apparent visual size will be at any given distance that he wanted what I added in my update. Then he posted a comment that seemed very confusing, so I am trying to determine what he wants. cmal seems to think that visual diameter is different than angular diameter. – robjohn Oct 15 '11 at 03:14
  • @cmal: You've got quite a wide car there :-). The gist is in that last sentence: You can't ignore camera-specific values. How could there be an answer to that question? You could just print out the photo at twice the size. That's if by "how wide the car will appear in a photo" you mean the length of the image of the car in the physical photo. If that's not what you mean, then I think you still haven't clearly defined what you mean by "how wide it will appear". – joriki Oct 15 '11 at 03:17
  • Part of the problem is its a difficult question and I'm having trouble phrasing it. I apologize. Hopefully my last comment will clarify something. – cmal Oct 15 '11 at 03:19
  • Yes, it is an extremely wide car! Can you answer the questions if you DO assume camera specific values, or if you use the human eye instead of a camera? Yes, you are correct about what I am trying to determine "the length of the image of the car in the physical photo". – cmal Oct 15 '11 at 03:23
  • @cmal: If by your last comment you mean the one with the car and the photo that I replied to, unfortunately it doesn't. I think you should consider the possibility that the problem isn't so much that it's a difficult question but that perhaps it's actually not a well-defined question. Sometimes one thinks one has a clear concept in one's mind and after further thought it turns out that there was actually nothing there, just the illusion of something coherent. – joriki Oct 15 '11 at 03:23
  • @cmal: Well, if you do assume camera-specific values, then it depends on details of the lens (is it a wide-angle lens? you might need one for that car :-), but to first order it will simply be some constant times the angular diameter, where that constant is determined not only by the camera but also by the resolution at which you print out the picture. – joriki Oct 15 '11 at 03:26
  • If there is no amount of specific information I could provide that would allow you to help me, please let me know so we can both stop wasting our time.

    Otherwise: It is a 50mm f/1.8 lens and the photo is always being printed out 1 meter wide.

     – cmal Oct 15 '11 at 03:42
  • @cmal: I don't know enough about cameras to be of much help with those values. From what I know, I believe you'd need not only the focal length of the lens and the size of the print-out but also the film size, but I may be wrong. What I would do, if this is an option, is measure the size of the object in some photo, calculate its angular diameter for that photo, and then assume that the size in the photos is roughly proportional to the angular diameter. I think that should be a good approximation in most situations. – joriki Oct 15 '11 at 04:12
  • I'm not sure I know enough about cameras either, but as long as I know that the change in size is a constant that should be enough to go on. Sorry for the runaround. I'll let you know how it works out. – cmal Oct 15 '11 at 14:35
  • Well explained, easy question. But nobody understood it. – DR01D Jan 31 '18 at 01:14