Calculating the distance between stars

[blobrana]

Hum,
To figure it roughly don't bother with Spherical trigonometry,
Just use standard 2d coordinates with Pythagoras theorem...

You need to know the distances of the two stars from the earth, and since that is a very rough estimate then that cuts down on the overall errors...


Perhaps if you imagine the one stars eclipsing the other star then it’s a simple case of subtracting one distance from another...

As you separate the two stars just use Pythagoras theorem (2d not even 3d coordinates) to work out the distance...
[Draw a line through both stars, it doesn’t matter the about alt or dec sime there is no up in space, and no one can hear you scream...]

[blobrana]

Hum,
but if you do want to mess up you head with Cartesian coordinates (you still need to convert the earth observed stars into an angle) check this



 :-)

[blobrana]

Hum,
yes, its better, but i find right-angles easier to work out in my head...
from those stellar distances you can really use the base angle to be almost 90 degrees...


The `right angle` is the same distance away as the closest star....

The `opposte` length is star2 - star1(closest)
So it`s a simple case to work out the hypotenuse

See!





























The Earth- X

[blobrana]

Hum,
OK, (to use the sine method)
Here’s an example of the blue and red stars of Orion...

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

RIGEL RA:   5h10m00.00s   DE:-08°00'00.0"

And remembering that 24h = 360º, 1h = 15º etc...

[The angle from the earth is about 18 degrees]
remember that 15 degrees = 1 hour

So 45minutes difference in the RA of the two stars = 45/60 x 15 = 11 degrees (thats one side)

The DEC is -8 - 7 = 15 degrees (the other side)

11sqr=121  15sqr=225   121+225=346 srqroot = 18.6degrees....


We have the angle!

Now all we need are the distances in light years to the two stars...



@Cymric
(to use the rough method)
the angle of the base line....(adjacent) the rightangle is the same distance from the earth as the first star...

the differance between alcor and mizar is a few seconds......the base line, in this case,would be so small  as to be forgotten about...(unless they were very close together)in which case

[blobrana]

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 ]

[blobrana]

hum
is that x²?

/¯1180723.5909957732  = 1086 light years ?

[blobrana]

Hum,
i cant check it using my head
but that is in the right magnatude and what i`d expect

[blobrana]

@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?

[blobrana]

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!

[blobrana]

Hum,

So i use a quick mental check to see if the onboard computer hasn't gone mad:
Let’s see...
We have the angle 18 degrees (near enough to 20 degrees and a third of 60 degrees , which is a  isosceles triangle  , with 3 equal sides...
Betelgeuse 427
Rigel 772

772-427=345 (so it must be bigger than that)

Now this is where you use the (other) rough Pythagoras/trig  calculation

One side of the triangle (at right angle) is 345^2 = 119025
The base line (guessed/work from the angle, a third of 427 ) is (say) 142 light years
142 *142 = 20164
119025 +20164 = 139189


Sqr139189 = 373 light years...
Which is pretty close to my spaceships odometer....

(er, only 19 light years out -  ok so you wouldn`t risk a hyperjump with it but its a rough guide)

[blobrana]

hum,
>> but falls all to hell

Yeah, using the rough mental method, it would, but it'll be within the right magnitude,
If you didn't have a calculator with cosine on it (er, like me), and just to verify that the cosine-rule calculation (see my other post) did produce a meaningful result...

The important thing is that it is all on one plane, so that the rough and accurate method work...




[Anyway I've got a spaceship that does that all for me...]
[go on, ask me the distance between two stars...any stars...]
 :-)

[blobrana]

Hum,
a nice link...

I'll peruse that one later...