Question about coordinate systems' differences

[Railly]

Hi,

Newbie to the forum. Hope somebody can help me with this question because it's been bugging me. It was recommended that I tried asking here.

If I measure an angular distance between two celestial objects at a specific time, using azimuth and altitude coordinates, why do the results differ somewhat from the results I would get if I measured using the equatorial RA and Declination coordinates? What accounts for the difference? Is it that equatorial coordinates are based on the sky (for practical purposes) being a perfect sphere and the Earth isn't, so that any angular distances derived from horizon coordinates will reflect the Earth's 'imperfect' shape and not match the perfectly spherical sky's RA and Dec distances? ... Or am I completely barking up the wrong tree here?

Thanks for any insights :-)

[wxsatuser]

Try this.

http://spiff.rit.edu/classes/phys445/lectures/radec/radec.html

[themos]

Are you talking about your own measurements or calculations based on published data, computer programs and the like?

[dph1nm]

Normally there shouldn't be any difference. I guess near the horizon the measured altitude is affect by refraction, but you would correct for that in the calculation, wouldn't you!?

NigelM

[Railly]

Thanks for the link, wxsatuser. While it contains this information,

It is possible to convert from (RA, Dec) to (alt, az), or vice versa. One needs to know two factors:

the location of the observer on Earth

the time of the observation

The calculations involve some spherical trigonometry. One can find the details in any good book on celestial mathematics, such as

Textbook on Spherical Astronomy by W. M. Smart

Computational Spherical Astronomy by L. G. Taff

Spherical Astronomy by R. M. Green

Practical Astronomy with your Calculator by P. J. Duffet-Smith

Astronomical Formulae for Calculators by J. Meeus

In these modern times, it's usually easiest to use one of the many fine planetarium programs on a computer to do this work.

... and then goes into some 'simple' spherical trigonometric calculations for finding angular distance (at which point my eyes glazed over LOL), it doesn't explain WHY an angular distance measured using horizon coordinates should be significantly different from the results gained by measuring using the equatorial system.

I'll give you an example to show you what I mean:

On the 10th Sept. 2013, at a UK location, at 20:00 BST, according to Sky View Cafe, the Moon was in Libra.

Using the equatorial system, I find that the Moon was 1.5 degrees east of beta Lib. and about 8.75 degrees south of it.

However, using the horizon system, I find that the Moon was about 5 degrees east of beta Lib. and about 7.3 degrees south of it.

The two systems yield different angular distances. All I need is a basic, K.I.S.S. explanation to account for those differences. I'm sure it will be something face-smackingly obvious in the end, but I can't work it out (other than maybe the idea in the OP).

[themos]

The most likely explanation is that the software is getting it wrong. Try to replicate its predictions in another software.

[Northern Soul man]

Hi Ya Railly, Your getting into real "BrainBox" Territory here mate. I'm no mathematician, but from what I can see is that I could understand that the distances for Solar System objects should be pretty accurate - so it seems that each calculation uses different methods for the outcome, you're really looking at a fixed body, in a fixed orbit (for the Moon anyway) and relatively close to us - what is its mean distance around a quarter of a million miles - when I say a fixed body - the orbit has been calculated for many years - with all perturbations (not sure of the spelling - told you I'm no good at this!!!) taken into account - as I say for the Solar System objects - in relative terms - in our back yard - I would expect calculations over interstellar space to be a lot more hit and miss - as these stars are, maybe in orbits of their own, all working against gravity, with the Earth in its orbit, against the Solar System in its orbit, against the Galaxy (Milky Way) in its orbit, against the local group in its orbit, against the group which the local group orbits - you tend to get a picture of the farther out you go - the more "anomalies" get added to the equations, Parallax, perturbations and precession - even thinking of the distances involved you could understand the differences here - but for the Moon - its beyond me - literally. Paul

[brantuk]

If you run up Stellarium and switch on the Alt/Az grid and EQ grid simultaneously, the differences will become apparent. I'm no mathematician but sometimes a practical demonstration helps understand a concept.

[BinocularSky]

Using the equatorial system, I find that the Moon was 1.5 degrees east of beta Lib. and about 8.75 degrees south of it.

However, using the horizon system, I find that the Moon was about 5 degrees east of beta Lib. and about 7.3 degrees south of it.

The equatorial data render an angular separation of 8.88 deg

The horizon data render an angular separation of 8.85 deg.

The difference is due to imprecision of the supplied data.

Later; this dawned on me just after I hit "Post":

Just in case you are doing something else, use the pairs of data as the two short sides of a right angled triangle and calculate the hypotenuse length to get the angular separation.

[Northern Soul man]

See I told Ya - but surely a 3 1/2 degree "shift" in RA and a 1 1/2 degree "shift" in Dec over this distance ain't right - oh and another thing - How big is the Moon in the sky again - 1/2 a degree ?????

[wxsatuser]

If you run up Stellarium and switch on the Alt/Az grid and EQ grid simultaneously, the differences will become apparent. I'm no mathematician but sometimes a practical demonstration helps understand a concept.

Exactly right.

Alt/Azi is a grid vertical and horizontal to your position.

The RA/DEC is inclined so the Moon and Beta Lib would be very near each other, but alt/azi they would appear further apart.

Try explain with this

Azimuth

RA

[Railly]

Ahhh. Thanks brantuk and wxsatuser for the screenshots. Very helpful. It *was* sort of face-smackingly obvious. I was barking up the wrong tree in the OP! It's purely the orientation of the two grid systems which will make the angles different. I assume, then, that the ecliptic system would yield *another* set of results which would vary from the other two. I'll need to get my head around it some more :-)

[brantuk]

The ecliptic (as I understand it) is just the closest path followed by the planets. I've not seen a grid system based on it.

[BinocularSky]

The ecliptic (as I understand it) is just the closest path followed by the planets.

The ecliptic is the apparent path followed by the Sun.

I've not seen a grid system based on it.

Ecliptic Co-ordinates

Most decent astronomical software has an option to display ecliptic co-ordinates.

[wxsatuser]

Ecliptic co-ordinates

[Railly]

Very interesting. It looks like the Azimuth and Ecliptic longitude distances are closer in value than the Ecliptic long. and RA ones. Hmm. Something to chew over. Thanks again.