Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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: $\displaystyle\sin\left(\frac{\alpha}{2}\right)=\frac{D/2}{r}$

share|improve this answer

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.

share|improve this answer
    
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 0:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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