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Fiery Phoenix
2010-Jun-09, 06:37 AM
OK, I'm going to try and write my question in a slightly different way so it can be easily understood. I REALLY need to figure this out (sci-fi story purposes).

First of all, consider the following picture of a galaxy:
http://www.astronomytrek.com/_/rsrc/1263665193156/the-milky-way-galaxy/spiral%20galaxy.jpg

Imagine yourself aboard a spaceship and heading towards the galaxy in the picture above. How the galaxy is viewed in the picture is EXACLTY how you are peceiving it as a pilot/passenger. That is to say, the galaxy in the picture is with respect to your perspective. Moreover, assume the galaxy has the same exact diameter as the Milky Way (100,000 ly).

This is the core question:
Given the above information, is there a way for you as a person on that spaceship to manually determine your approximate current distance to the galaxy?

I've tried looking on Wikipedia but it's too complicated. I'd GREATLY appreciate any help. I just need a formula to figure out the distance, given the above info.

Thanks in advance. Awaiting a response...

agingjb
2010-Jun-09, 06:52 AM
If it looks, out of a window, roughly as it looks on my screen, then perhaps 300,000 ly. (It's eight inches across; my viewing distance is two feet; -ish in each case.)

How many reservations do want? Galaxies don't have exact sizes. Galaxies probably don't look that bright. Other people have smaller or larger screens, or sit closer or further away.

Fiery Phoenix
2010-Jun-09, 07:00 AM
If it looks, out of a window, roughly as it looks on my screen, then perhaps 300,000 ly. (It's eight inches across; my viewing distance is two feet; -ish in each case.)

How many reservations do want? Galaxies don't have exact sizes. Galaxies probably don't look that bright. Other people have smaller or larger screens, or sit closer or further away.

Thank you for your response. I want as many reservations as possible; disregard how different people would view it from different angles, the galaxy's brightness, etc. All I'm asking for is the galaxy's overall distance from my ship. Sorry, I should've mentioned that in the OP.

I only need a formula or something allowing me to calculate the current distance from any galaxy in a situation like this. And yes, assume how it is visilbe on the screen is how you'd see it from the ship.

Jens
2010-Jun-09, 07:00 AM
I think the simple answer is no, you can't.

One other difficulty, you say that it looks like that, but that doesn't really tell us much. You would need to specify how much angle of the sky it takes up. How much of the sky is your picture covering? It can't be half because the perspective would be strange. In other words, how big is the "window" that you are looking through.

Fiery Phoenix
2010-Jun-09, 07:05 AM
I think the simple answer is no, you can't.

One other difficulty, you say that it looks like that, but that doesn't really tell us much. You would need to specify how much angle of the sky it takes up. How much of the sky is your picture covering? It can't be half because the perspective would be strange. In other words, how big is the "window" that you are looking through.

Hey Jens!

I did say in my reply to Agin to assume the size of the window is the size of the picture. That is, what you're seeing from the ship is just that... the picture. There's a reason I had to assume this, and it's what you brought up. I'm aware of the inconsistencies that will arise if you don't make these assumptions and get it out of the way.

Jens
2010-Jun-09, 07:33 AM
OK, but another important factor is how far away you are from the screen. I sometimes sit about 40 cm away, and sometimes may 80 cm away. The galaxy might be half or twice the size depending on the distance.

Fiery Phoenix
2010-Jun-09, 07:40 AM
That won't matter. I'd imagine the ship to have a single pilot viewing it as if it were a constant image; just gets bigger as you approach it.

I know there are lots of variables, but regardless, isn't there any way AT ALL to realisticallly calculate the distance in such a case?

Strange
2010-Jun-09, 08:41 AM
Given the above information, is there a way for you as a person on that spaceship to manually determine your approximate current distance to the galaxy?

Can't see your picture for some reason. But isn't this just a matter of simple trigonometry? If the person on board knows the size of the galxy (you imply they do) then they can measure the apparent (angular) width and calcuate the distance. I don't see the problem. Unless no one on board has studied basic geometry because they have computers to do all that stuff for them. Even then, someone smart could work it out from first principles fairly easily using the idea of ratios and similar triangles...

And if you are travelling fast enough to see a change of size, you can obviously calculate your velocity relative to the galaxy as well.

Jens
2010-Jun-09, 08:45 AM
If the person on board knows the size of the galxy (you imply they do) then they can measure the apparent (angular) width and calcuate the distance. I don't see the problem.

Stupidly I missed that part about assuming it is the same size as the Milky Way. In that case, as Strange said, it would be a simple problem. It could be done manually, as long as the person had a pencil.

Van Rijn
2010-Jun-09, 08:50 AM
Hey Jens!

I did say in my reply to Agin to assume the size of the window is the size of the picture. That is, what you're seeing from the ship is just that... the picture. There's a reason I had to assume this, and it's what you brought up. I'm aware of the inconsistencies that will arise if you don't make these assumptions and get it out of the way.

Is it moving slowly with respect to the galaxy? The issue is that relativity can dramatically affect the view.

Fiery Phoenix
2010-Jun-09, 08:51 AM
Okay, so how do I "measure" the angular width? And what is the unit I should use for that?

Bear with me a little; I'm sure I'll do this eventually! ;)

Fiery Phoenix
2010-Jun-09, 08:53 AM
@Van, yes.

EDIT: Sorry for double posting...

Strange
2010-Jun-09, 08:58 AM
Okay, so how do I "measure" the angular width? And what is the unit I should use for that?

Ah, so its not the person on the ship that has a problem with trig, then :)

OK. Sit in a seat where the galaxy exactly fits the width of your window. Measure the distance from the window and the width of the window. Now divide the known width of the galaxy by the width of the window: call this the scale. Now multiply the distance from the seat to the window by scale. That gives you the distance to the galaxy.

Actually, it is probably more tractable to take the ratio of the width of the window to the distance to the window; and multiply this by the width of the galxy....

In other words, if the window is half as wide as the distance the observer is from it, then the galaxy is half as wide as the distance the observer is from it.

Fiery Phoenix
2010-Jun-09, 09:18 AM
Ah, so its not the person on the ship that has a problem with trig, then :)

OK. Sit in a seat where the galaxy exactly fits the width of your window. Measure the distance from the window and the width of the window. Now divide the known width of the galaxy by the width of the window: call this the scale. Now multiply the distance from the seat to the window by scale. That gives you the distance to the galaxy.

Actually, it is probably more tractable to take the ratio of the width of the window to the distance to the window; and multiply this by the width of the galxy....

In other words, if the window is half as wide as the distance the observer is from it, then the galaxy is half as wide as the distance the observer is from it.
Is that supposed to be how wide the galaxy is from the window?

Strange
2010-Jun-09, 09:26 AM
Is that supposed to be how wide the galaxy is from the window?

Not sure I understand that....

When I say "width of the window" I mean the "apparent" width of the galaxy (which is just filling the window).

Perhaps my final sentence would have been better as: if the distance of the observer from the window is twice the width of the window, then the galaxy is twice as far away as its actual width. So, if the window is 10 inches wide and your eye has to be 20 inches away from the window to make the galaxy just fill the window - then the galaxy is twice as far away as it is wide (200,000 ly).

Does that make more sense?

Fiery Phoenix
2010-Jun-09, 09:35 AM
Perfect. Thank you very much.

astromark
2010-Jun-09, 09:44 AM
Thankyou 'Strange'... That little explanation took me back to a schoolroom in south China and,
a very excitable maths teacher drawing triangles on a blackboard... I was the only European student... It was funny., and still is... :o:
The obvious problem with this method is just that. You do need to establish a scale.
If the target galaxy filling your window was not Andromeda but, some unknown... you have a problem.

Strange
2010-Jun-09, 09:51 AM
If the target galaxy filling your window was not Andromeda but, some unknown... you have a problem.

Indeed. I suppose if you were travelling fast enough, you could use the galaxy's blue shift to determine velocity relative to the galxay and then take two width measurements and figure out the change in width and hence ... (I can't do that off the top of my head though!)

astromark
2010-Jun-09, 09:53 AM
I did see that you only wanted a method... not a real problem then... Captain Picard always had Data as Kirk had Spock...
When all they needed was a rule of thumb, and a calculator...

Strange
2010-Jun-09, 09:58 AM
When all they needed was a rule of thumb, and a calculator...

Or even a slide rule, for us oldies ...

Fiery Phoenix
2010-Jun-09, 10:01 AM
Okay, this is a really stupid question, but I got close to 48 inches for the distance of the galaxy. The width of the window and the distance to it are 11 and 48 inches respectively. Since the galaxy is JUST filling the window, I came up with 48 inches as its distance. Does that sound plausibe?

Strange
2010-Jun-09, 10:07 AM
Er... no. How about 436,000 ly ?

(distance to window) / (width of window) = 4.36

(width of galaxy) * 4.36 = 436,000

Fiery Phoenix
2010-Jun-09, 10:14 AM
Er... no. How about 436,000 ly ?

(distance to window) / (width of window) = 4.36

(width of galaxy) * 4.36 = 436,000

Well, it looks like I mixed things up with centimeters and inches. Conversion error on my part.

I just re-did the calculation and it appears you're correct. I also did it again using centimeters rather than inches and it yielded the same result.

I guess that's it then. Sorry for being so confused; ever since I majored in computer science I forgot how to deal with the "real" math. lol

Thanks again, Strange. You're no longer strange to me. :D

Strange
2010-Jun-09, 10:15 AM
Thanks again, Strange. You're no longer strange to me. :D

I may have to change my name again then...

Your welcome.