Calculating the distance between stars

[Cymric]

Not being hindered by any knowledge on astronomy whatsoever (well, not a lot ;-)), isn't this just a problem of converting the spherical coordinates (altitude, azimuth, distance) to Cartesian ones and then applying Pythagoras' theorem? In fact, I'm sure that if I haul out my textbook on differential geometry I could apply Pythagoras directly to the spherical coordinates, but since that involves some messy mathematics with Jacobians and what-not, I prefer to use the workaround I understand.

In other words, assuming a right-handed (x, y, z)-coordinate system, with f the angle {P_x}{O}{P_y} (= azimuth), g the angle {P}{O}{P_z} (= 90 degrees minus altitude), and r the distance to the Sun, with P the position of the star in space, and a subscript indicating its projection on either one of the three coordinate axes, my math lecture notes state that

   P_x = r cos(f) sin(g)
   P_y = r sin(f) cos(g)
   P_z = r cos(g)

Of course these formulas do not take into account the lattitude of the observer, nor the tilt of Earth's axis, nor the angle of the ecliptic. But somehow, I feel that those will simply fall out of the equation if azimuth and altitude are given relative to the constant observer's location: of course the absolute distance between stars should not depend  on the position of the observer on Earth. Continuing, therefore, the position of the second star R is given b y a similar set of equations, resulting in a distance D of

    D^2 = (r cos(f) sin(g) - r' cos(f') sin(g'))^2 + (r sin(f) cos(g) - r' sin(f') cos(g'))^2 + (r cos(g) - r' cos(g'))^2

Of course the above is more or less an educated guess, so take it with a pinch of salt, and try to obtain verification from people who did astronomy rather than chemical engineering ;-).

Edit: corrected the angle for the altitude.

[Cymric]

Yeah, okay, granted, you can reduce the 3D problem to a 2D one. But I think it doesn't solve anything, as you now are faced with the problem of calculating the angle subtended between the two stars, as it does not trivially follow from the two angles associated with each star's position. (Try it for yourself.) Look at it this way: three distinct points indeed lie in a plane, and it is trivial to construct the equation for that plane provided the coordinates are Cartesian. Unfortunately, the are not, they are spherical. Things are much more complex when you try to do it with them buggers. I have a strong hunch that if you come up with a formula to compute the required subtended angle, and then used 2D-Pythagoras or the sine rule, you'd be doing exactly the same amount of work as I did, but via a tricky and not really necessary 3D->2D transformation.

Things are of course completely different if you get the subtended angle as a function of time straight from the beginning. But that is not what Kenny gave us to work with.

[Cymric]

blobrana wrote:
from those stellar distances you can really use the base angle to be almost 90 degrees...

Which angle is almost 90 degrees? Say you're dealing with Alcor and Mizar (the famous double in the tail of the Great Bear), I cannot make out any angle close to this value. Could you please explain?

[Cymric]

Okay, looks I made a few errors---it is a well-known fact engineers cannot calculate things properly ;-). First of all, Blobrana's method of Pyhtagoras-ing the angles and then applying the sine rule is simple and valid, of course. Second---and much more serious---I made an error in the conversion formula: in the expression for P_y the cos(g) should of course read sin(g). (It was very nice to enter a few numbers which mysteriously did not lie a distance 1 away from the origin in Cartesian coordinates, even though that is what I put them at in spherical ones...) Apologies all around. Third correction is that in order to use the astronomical values as given by Blobrana, you need to convert the value of f as well: negate the value before use. Astronomers use a left-handed system *sigh*. Also nice to track down.

So, to give a number example to calculate the angle subtended the hard way: Betelgeuse and Rigel are reported at 88.75 RA, 7 DE; and 77.5 RA, -8 DE; convert that to -88.75, 83; and -77.5, 98 degrees and plug those values into the corrected formulas. You end up with Cartesian coordinates (0.02165, -0.9923, 0.1218) for Betelgeuse and (0.2143, -0.9668, -0.1392) for Rigel---note that for now we do not need to know their distances to the Sun, as we're interested in the angle only, and that is independent of the distance. We can assume those to be 1 for convenience. Vector algebra will then tell you that the inner product of these two vectors equals the cosine of the angle subtended (times the length of both vectors, which is 1, hurray), and if I do the math, I end up with an angle of 18.7 degrees. Which is exactly equal to what you end up with following Blobrana's Pythagoras-approach.

Now I know why we are not taught spherical geometry :-D.

[Cymric]

With the cosine rule

    a^2 = b^2 + c^2 - 2 b c cos \alpha

with b = 427 ly, c = 772 ly, and \alpha 18.7 degrees, I calculate the distance a to be 392.2 ly. Funny how a constellation turns out not to be 'flat' even though we perceive it as such.

In any case, this solves Kenny's problem.