Thread: Present uncertainty in relative positions of Alpha and Proxima Centauri

1. kzb
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Originally Posted by Hornblower
I just crunched the equation in my trusty Texas Instruments calculator, which can do trig functions to ten figures. A 1" uncertainty in the angular separation had less than 1% of the effect of the published distance uncertainties.
That's good, it means is negligible in an uncertainty estimation. Especially since the real uncertainty in the angular separation is certainly less than 1".

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Originally Posted by kzb
Tell us the exact calculation algorithm and then we can sort out how to propagate the uncertainties through it.
See my above post:
1,3 mpc (270 AU) uncertainty of one leg of right triangle map to uncertainty of the hypotenuse, seeing how the other leg has little uncertainty, dominated by the uncertainty of distance to Sun (under 10 AU)?
My estimate would be 270 AU*38mpc/63 mpc=160 AU uncertainty of Alpha-Proxima distance.
I use small angle approximation twice.
Designating as A the point behind Proxima, I approximate the triangle A-Proxima-Alpha as a right triangle. Not an exact shape, because the Alpha-Sun-Proxima angle is nonzero. But close enough for the estimation of errors.
Now, taking the external error of Sun-Proxima distance relative to Alpha, I designate as B the point behind Proxima which is the furthest possible possible location of Proxima.
Then I draw a normal from B to Alpha-Proxima line and designate the crossing as C.
Since I above approximated Alpha-Proxima-A as a right triangle (right angle at A), note that Proxima-B-C is a similar right triangle (sharing the angle A-Proxima-Alpha, and with right triangles at A and C). Therefore, I can use the above established distances Proxima-Alpha, Proxima-A, A-Alpha, and the uncertainty distance Proxima-B, to compute the remaining sides Proxima-C (and B-C)
Using small angle approach for the second time, I approximate Alpha-B as equal to Alpha-C. Therefore the distance Proxima-C is the uncertainty of distance Alpha-Proxima.

3. kzb
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Originally Posted by chornedsnorkack
See my above post:

I use small angle approximation twice.
Designating as A the point behind Proxima, I approximate the triangle A-Proxima-Alpha as a right triangle. Not an exact shape, because the Alpha-Sun-Proxima angle is nonzero. But close enough for the estimation of errors.
Now, taking the external error of Sun-Proxima distance relative to Alpha, I designate as B the point behind Proxima which is the furthest possible possible location of Proxima.
Then I draw a normal from B to Alpha-Proxima line and designate the crossing as C.
Since I above approximated Alpha-Proxima-A as a right triangle (right angle at A), note that Proxima-B-C is a similar right triangle (sharing the angle A-Proxima-Alpha, and with right triangles at A and C). Therefore, I can use the above established distances Proxima-Alpha, Proxima-A, A-Alpha, and the uncertainty distance Proxima-B, to compute the remaining sides Proxima-C (and B-C)
Using small angle approach for the second time, I approximate Alpha-B as equal to Alpha-C. Therefore the distance Proxima-C is the uncertainty of distance Alpha-Proxima.
I can't quite follow what you mean. It may be you are calculating the AB to Proxima distance acceptably well, but from the sound of it you are not propagating the uncertainties correctly.

See this:

http://www.geol.lsu.edu/jlorenzo/geo...tiespart2.html

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Originally Posted by kzb
I can't quite follow what you mean. It may be you are calculating the AB to Proxima distance acceptably well,
I did not actually calculate the distance.

5. Originally Posted by kzb
The next problem is finding the cosine of very small angles !

The standard uncertainty of 2 degrees 11' 06":

As reported, the arcsecond number could be between 5.5 and 6.5 arcseconds.

The standard uncertainty (assuming all values between these limits are equally probable) is 0.5/SQRT(3) = 0.289 arcseconds

0.289 arcseconds is 0.0000802 degrees.

How do you find the cosine of that?
Wolfram Alpha, Octave, Maxima, Maple, .....

6. kzb
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Originally Posted by chornedsnorkack
I did not actually calculate the distance.
I think we need to. It's the only way for ordinary mortals to see how to propagate the uncertainties correctly.

There is that formula I used, but if you have a simpler calculation method, that would make the uncertainty propagation easier to do.

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