# How to get apparent linear diameter from angular diameter

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?

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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.

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

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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.
@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