Talk:Milliarcsecond

Distance of a tennis ball with a size of one milliarcsecond
A sphere with radius r at a distance D will have an angular size of &theta; if

$$D={r\over\tan(\theta/2)}.$$

Since a tennis ball has a diameter of about 6.5 cm, and one milliarcsecond is (1/3,600,000)&deg;,

$$D={3.25\,{\rm cm}\over\tan\,(1/7\,200\,000)^\circ}\approx13\,407\,{\rm km}.$$

&mdash;Bkell 00:40, 21 August 2005 (UTC)


 * Isn't one milliarcsecond 1/1,296,000,000&deg;? --LiamE 16:50, 16 June 2006 (UTC)


 * Ignore me - I've worked out my own mistake! ITS 1/1,296,000,000 of a circle not a degree. The original poster is correct. --LiamE 19:09, 16 June 2006 (UTC)


 * Rather than having to take arctans of things, you can use the small-angle approximation (which is an extremely good one for angles of milliarcsecond size);
 * $$D\approx {d \over\theta}$$ (&theta; is in radians, d is the diameter of the object, and D is the distance from the object).
 * 1 milliarcsecond = $$\frac{pi}{648000000}$$ radians.


 * Putting that into the approximation; $$D\approx\frac{648000000 d}{pi\times\theta}$$, with &theta; now in milliarcsec.


 * For a 6.5 cm object subtending 1 mas, we get $$D\approx {1.34\,\times\, 10^7\, m}$$ Richard B 02:47, 5 September 2006 (UTC)