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Author Topic: Calculating the distance between stars  (Read 10917 times)

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Offline KennyRTopic starter

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Re: Calculating the distance between stars
« Reply #14 on: July 21, 2004, 05:59:23 PM »
Thanks for your replies. It seems to me that working out the cartesian coords first and then doing pythagoras on them, like Cymric said, would be the best way to do it. At least, it would be the most accurate way. I'm not convinced that 2D math is any good on a sphere.

But...! Why don't we all test our theories? Blobrana gave us two stars, with coordinates:

Betelgeuse RA: 5h55m00.00s DE:+07°00'00.0"

RIGEL RA: 5h10m00.00s DE:-08°00'00.0"

Betelgeuse is 650 ly from Earth, Rigel is 1400 ly. Now, what is the distance between them?
 

Offline Karlos

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Re: Calculating the distance between stars
« Reply #15 on: July 21, 2004, 06:49:14 PM »
Quote

KennyR wrote:
I'm not convinced that 2D math is any good on a sphere.


That isn't quite what anybody was implying. The 2D argument comes from the observation that you have only 3 points to consider and any three distinct points always lie in a 2D plane (basically defining a perfectly ordinary 2D triangle within that plane). I should point out that I'm not convinced that the simple pythagoras right angle triangle stuff applies, but if you know the length of any two sides of a triangle and angle subtended between them, you can work out the other angles and side length.
 
Of course the difficulty is that you still have to determine the angle subtended between the lines connecting star1 and star2 to your point of reference.

Ultimately, I'm sure it would be the same complexity as cartesian conversion of the stars locations relative to earth and the resulting vector length.

I'm a tad busy but I will give the vector length and angle subtended versions a go when I have some time :-)
int p; // A
 

Offline blobrana

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Re: Calculating the distance between stars
« Reply #16 on: July 21, 2004, 06:50:16 PM »
Hum,
have you a calculator?

The two sides of the  triangle are  a and b, and the angle in between, C, - it`s `easy` to find the remaining side x.  by using the cosine formula:
x²  = a²  + b²  - 2 a b cos C

x²  = 1400²  + 650 ²  - 2* 1400 *650 * cos 18

=? (some one work it out for me)


[we know the angle C from as 18 degrees, the distances are already `known`]

[And it all is on a 2d plane, yeah, really ]

Offline Karlos

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Re: Calculating the distance between stars
« Reply #17 on: July 21, 2004, 06:53:32 PM »
@blob

That's what I mean :-)

x = 807 light years.
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Offline KennyRTopic starter

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Re: Calculating the distance between stars
« Reply #18 on: July 21, 2004, 06:53:52 PM »
Blob, the answer to that equation you printed is:

1180723.5909957732

Thats way, way off what I got. :-o
 

Offline Karlos

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Re: Calculating the distance between stars
« Reply #19 on: July 21, 2004, 06:57:55 PM »
@Kenny

Remember order of precedence

x = sqrt((1400*1400)+(650*650)-(2*1400*650*cos(18 deg)))
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Offline blobrana

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Re: Calculating the distance between stars
« Reply #20 on: July 21, 2004, 07:00:07 PM »
hum
is that x²?

/¯1180723.5909957732  = 1086 light years ?


Offline KennyRTopic starter

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Re: Calculating the distance between stars
« Reply #21 on: July 21, 2004, 07:00:39 PM »
I did it with cartesians, and I got:

798.949 light years.
 

Offline KennyRTopic starter

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Re: Calculating the distance between stars
« Reply #22 on: July 21, 2004, 07:03:02 PM »
@Karlos

I just put it in c:ev as-is (replacing ² for ^2 of course). It's usually good at precedences. Not this time though. :)

Ok, so Blobrana was very close with the 2D trick. Seems her maths works too.
 

Offline blobrana

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Re: Calculating the distance between stars
« Reply #23 on: July 21, 2004, 07:03:32 PM »
Hum,
i cant check it using my head
but that is in the right magnatude and what i`d expect

Offline Karlos

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Re: Calculating the distance between stars
« Reply #24 on: July 21, 2004, 07:03:51 PM »
Note that blobrana's solution is *very* sensetive to the accuracy of the angle subtended (since it is a factor in the last term of the expression).

I think, within the accuracy of measure 799 and 807 are in close enough agreement.
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Offline blobrana

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Re: Calculating the distance between stars
« Reply #25 on: July 21, 2004, 08:14:17 PM »
@kennyR

Ahhh!
error

>>Betelgeuse is 650 ly from Earth, Rigel is 1400 ly.

Where on earth did you get those figures?


Well, i`ve just `flow` there, and the real distances are
Betelgeuse 427.480ly
Rigel      772.907ly

And from my spaceship porthole, at Rigel, i can see that Betelgeuse  is 392 ly away....


Is this the reason that the invasion fleet hasn`t arrived here KennyR? Hum?

Offline Karlos

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Re: Calculating the distance between stars
« Reply #26 on: July 21, 2004, 08:32:22 PM »
Well the figures may be wrong but the agreement between the methods is what matters here :-D
int p; // A
 

Offline KennyRTopic starter

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Re: Calculating the distance between stars
« Reply #27 on: July 21, 2004, 08:32:45 PM »
Quote
Blobrana wrote:
Where on earth did you get those figures?


An advanced scientific method. First I put the text "betelguese distance" into google and took the first distance that appeared. In this case it was

Betelguese > Earth
Betelgeuse is ~1400 light years from Earth, Betelgeuse. BV Color. Common Names.
Constellation. Coordinates. Distance from Earth. Harvard Revised Number. ...

And for Rigel, more of the same.

Try it yourself. =)
 

Offline blobrana

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Re: Calculating the distance between stars
« Reply #28 on: July 21, 2004, 08:36:13 PM »
HHum,
google is obviously hiding something...the truth...

trust me,
i`ve been there...
have a deeper look...

Perhaps goofle has been `infected`?

----------------------------------------
i `m off to check --->


hum, yea...
try the revised distance
see!

Offline Cymric

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Re: Calculating the distance between stars
« Reply #29 from previous page: July 21, 2004, 11:12:56 PM »
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.
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