I Diameter of a Galaxy: Calculate Distance/Arc Min?

[NebulaBilly]

My last thread got lots of hits and was a success so I thought I would start another discussion based on the Galaxy I was speaking about, In my previous thread we found the distance to the galaxy, now I want to take this further and work at the diameter of the galaxy. If two edges of this galaxy are 4.0 arc minutes apart then to work out the diameter would it be Dia = Distance/Arc Min?. I'm not complete sure about Arc Mins and that formula may be totally incorrect.

 

[mfb]

Did you draw a sketch?

would it be Dia = Distance/Arc Min

That formula would suggest a galaxy with a larger angular diameter has a smaller actual diameter. Does that look realistic?
Also, why would you use arc minutes in such a formula? What is special about 1/60th of a degree?

 

[NebulaBilly]

The problem to solve states the edges are 4.0 arc minutes apart, its why I would include the arc minutes, like I say arc minutes I am not sure about. Also I think I wrote that incorrect Dia = Distance x Arc Minutes

 

[mfb]

Use the value of 4 arcminutes, sure. But keep the unit there. Don't work with "4". A formula cannot depend on our arbitrary definition of arcminutes.

Dia = Distance x Arc Minutes

Same question as before: Why arcminutes? Why not Dia = Distance x Arcseconds? Or Dia = Distance x Degrees? Clearly these three formulas will give completely different results, if you use the same angle of 4 arcminutes = 240 arcseconds = 1/15 degree.
If you see a situation like that, you know something has to be wrong. You cannot have "number of arcminutes" appearing in such a formula.

Did you draw a sketch? The correct formula is basic trigonometry.

 

[NebulaBilly]

Yes I mean I could use the basic trigonometry but would you not need the arc minutes in radians? or can it simply be
60.1xTan(8/60), I was trying to come up with a similar way and you are correct they way I am trying to do it I would need to convert arc minutes to seconds and the Mpc to Parsecs

 

[mfb]

but would you not need the arc minutes in radians?

How you calculate the tangent of 4 arcminutes is up to you. Converting it to radians is a good idea.

 

[NebulaBilly]

Ok so let's try and do this, so D = dtana if I was to use the numbers 60.1 Mpc and 4 arc minutes converted into radians then it would read D=60.1tan0.01, this looking good so far? that would be 0.01Mpc what if I wanted to used Kpc would = 10?

 

[stefan r]

Ok so let's try and do this, so D = dtana if I was to use the numbers 60.1 Mpc and 4 arc minutes converted into radians then it would read D=60.1tan0.01, this looking good so far? that would be 0.01Mpc what if I wanted to used Kpc would = 10?

We are not allowed to direct answer things that look like homework problems. mfb's answer looked deliberately vague.

There is an angle angle theorem in geometry. You have one angle. Astronomers assume 90° for the object you are looking at. That makes it the observed width not the actual width. The ratio of the sides of similar triangles are all the same. If the long leg is double size than the shorter leg in the similar triangle is also double size.

A parsec is derived from "parallax second". There are 60 seconds in a minute. 1/15 degree = 4 minutes = 240 seconds.
Mega, M = 1,000,000
kilo, K = 1000
Figure how wide the UFO would be if we knew it was 1 parsec away. Then use AA theorem to get the correct observed diameter.

 

[mfb]

4 arcmin = 1/15 degree = 1/15 * pi/180 rad, that is less than 0.01.

that would be 0.01Mpc what if I wanted to used Kpc would = 10?

That looks like the result of a rounding error, but apart from that it is not too far off.

 

[Chronos]

Without some knowledge of the inclination and actual geometry of the galaxy of interest, a direct measurement of apparent size has very limited value. To obtain a proper size it is also useful to know its proper distance.