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#1
24-09-2016, 09:51 AM
 DavidLJ (David) Registered User Join Date: Jun 2011 Location: Sydney Posts: 120
Measuring size

Probably a silly question but would be good to get an opinion.

If you are given both the angular size of, and the distance to, a celestial object simple math should allow the object's linear size to be calculated. For example, in the attached grossly exaggerated diagram assume that you are at position O and that the short arc represents a segment of your night sky. You observe a galaxy that spans locations A and B on the sky and determine that it subtends an angle of L. Source references state that the distance to the galaxy is d. It is easy to calculate the linear size of the line A-B. It is also easy to calculate the length of the arc A-C-B. But which of the two is the correct size of the galaxy? In this exaggerated view it is obvious that the arc is appreciably longer than the line. As angle L gets smaller so the length of the arc more closely approaches the length of the line. At the very small angles that are involved for distant galaxies the difference becomes almost vanishingly small but nevertheless it is still present.

What do you think is the better calculation of the object's size, the line A-B or the arc A-C-B? Would it make a difference if what you are measuring is the distance between two different objects located at the two points A and B assuming that they are both the same distance away from us?
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#2
24-09-2016, 10:57 AM
 julianh72 (Julian) Registered User Join Date: Jan 2014 Location: Kelvin Grove Posts: 1,264
Two things to consider:

Firstly, when determining the size of a circular or spherical object, we would usually use the diameter, so the straight "chord length" is what you are trying to measure, not a slightly curved arc which deflects behind the straight chord.

Secondly, for small angles, such as astronomical targets, the straight chord length and curved arc length are almost identical anyway. The curved arc length is equal to the angle (in radians) times the radius, while the straight chord length is equal to 2 times the sine of half the angle (almost exactly equal to the sine of the angle, for all practical purposes).

Consider the Moon and the Sun, which are bigger than most targets, apart from the biggest galaxies and clusters. They are half a degree across; the curved arc length works out to be 0.00872665 times the radius, while the straight arc length is 0.00872662 times the radius, roughly 3 parts in a million difference. For smaller targets, the difference is even smaller.
#3
24-09-2016, 11:22 AM
 Robh (Rob) Registered User Join Date: Feb 2009 Location: Blue Mountains, Australia Posts: 1,311
Hi David,

You should note a couple of things ...

1. Most of the objects (e.g. globular clusters, nebulae, galaxies) do not have a definite boundary. They fade outwards gradually. In binoculars, Omega Centauri looks barely 15' across, in a telescope it might look 36', in images maybe 50'. The bigger the aperture or the longer the exposure, the bigger it will look. So any angular size is bound to be an approximation.

2. The distance to these objects is approximate. In fact, it is more likely to be a range of distances. Example, galaxy M66 in the NED database gives its distance as anywhere between 3.86 and 15.6 megaparsecs with a mean of approximately 9.8 megaparsec. One parsec = 3.26 light years.

The errors in these two will far outweigh the small difference in using the arc as against the chord.

Regards, Rob
#4
24-09-2016, 11:23 AM
 billdan (Bill) Registered User Join Date: Mar 2012 Location: Narangba, SE QLD Posts: 1,539
Gooday David,

In astronomy we usually only speak of angular size, which is the angle of an arc of a circle and then absolute size can be worked out from this formula.

absolute size = distance * tan(angular size)

Or you can work out the length of the circumference by using 2*pi*distance and then divide that length by 21,600 to give you the length of 1 arcmin. So if your angular size of the object was 5 arcmin then multiply your result by 5 to get absolute size.

Clear as mud

Bill

Last edited by billdan; 24-09-2016 at 11:41 AM.
#5
24-09-2016, 10:08 PM
 DavidLJ (David) Registered User Join Date: Jun 2011 Location: Sydney Posts: 120
Julian, Rob, Bill.

Firstly, thank you for your replies. They are much appreciated and have given much food for thought.

For an object spanning 60 degrees of arc I calculate that the length of the arc A-C-B would be 4.72% longer than the length of the chord A-B. 4% is not an insignificant amount. But of course an angle of 60 degrees is way beyond anything that will concern us in the night sky. Apart from the Milky Way, the largest object in our skies is possibly Barnard's Loop which, according to Wikipedia, spans about 10 degrees of arc. For such an angle the arc is only 0.13% longer than the chord. And by the time we get to more usual objects such as NGC 253 which spans 27.5 arc minutes the difference is only 0.001%. Add to that the uncertainty factors mentioned by Rob and it is clear that the question raised has little practical merit in the context of astronomical objects.

But it was an interesting mental exercise that kept the aging little grey cells ticking over and again I thank you for replying.

Last edited by DavidLJ; 24-09-2016 at 11:21 PM.

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