Altitude-azimuth calculation question?

Question

Somebody e-mailed me this question. I did simple subtraction of values, but I know there is more to it. Altitude is also affected by longitude and azimuth is affected by latitude. I also recognized that distance from the equator would affect the outcome, as well. Can someone give me a link where this kind of calculation is done, or maybe do it here, if it's not too difficult?

<<Message: One day when Aura went home to a certain location(10° 37' 58" N, 122° 58' 01" E)

she discovered a new asteroid in the sky.She called up her friend Gem and asked him to take a look.

Aura did not know the constellations well, so she could not tell which part

of the sky it can be found, so instead she told him that the object is at 60°

altitude and 180° azimuth. He computed the position seen from his location in

(14° 39' 10" N, 121° 04' 14" E) and was able to find it. What is

the azimuth and altitude from Gem's location?>>

Cuz geez, this isn't even a RA/dec to alt/az conversion problem and it's still too difficult for me.

Answer 1

I always get Starry Night to do calculations like this for me. I learned how to do these calculations back in second year university mathematics, but promptly forgot.

Answer 2

I wanted to do this a few years back.  You need spherical geometry to make it happen.  My own calculus text book doesn&#039;t have it.  So i went the library and checked out one that did.  Yes, that&#039;s right.  I read text books for fun.

I have a 1996 Atronomical Almanac.  While i&#039;m sure it has it, i&#039;m also sure i haven&#039;t seen it in there.  It would have the advantage of being astronomically oriented.  It has the disadvantage that it gives you the corrections for aberration, nutation, and a zillion other details.

I&#039;ve never seen so many sin&#039;s and cos&#039;s in my life.  These days, i leave such calculations to a computer.  Many planetarium programs let you plug in RA and Dec and will happily spit out alt/az.  You have to know where you are on Earth, and what time it is.

All i really wanted to know was the angular distance between two objects - to get a really quick field of view experiment for a scope.

There are tons of free planetarium programs.  See the links.

Answer 3

At that spot (10° 37&#039; 58&quot; N, 122° 58&#039; 01&quot; E), North Pole (celestial) has Al  = 10° 37&#039; 58&quot; N

Az = 0° (obvious)

Arc distance between the places is &#039;a&#039;, &amp; A = difference in longitude

cos a =  sin b sin c + cos b cos c cos A

a = 4° 25&#039; 30.06&quot;

Bearing from Aura&#039;s location to Gem&#039;s = 24° 30&#039; 32.73&quot;

From Aura&#039;s location Zenith is pointing to a Decl. 10° 37&#039; 58&quot; N

Asteroid&#039;s Clestial coordinates are

180° (Az) is South;

Al = 60°

=&gt; (90° - [60° + 10° 37&#039; 58&quot;]=19° 22&#039; 2&quot;); Decl.: -19° 22&#039; 2&quot;

&#039;RA&#039; does not matter as the observations were simultaneous.

From Gem&#039;s location (local) Zenith is at Decl. 14° 39&#039; 10&quot;

Az = 180° + (121°04&#039;14&quot;- 122°58&#039;01&quot;) = 181° 53&#039; 47&quot;

Al = 60°+ (14° 39&#039; 10&quot; -10° 37&#039; 58&quot;) = 64° 01&#039; 12&quot;

Answer 4

You&#039;re right, it looks as if it ought to be a simple subtraction, but it actually involves spherical trigonometry. This same question was posted here earlier today and I did half the calculation, but then decided it was getting too long-winded. I&#039;ll pick up the calculation and get back to you.

Later:

Most of these coordinate transformations boil down to visualising the relevant spherical triangle. The solution goes as follows.

From Aura&#039;s location, because the object is at azimuth 180°, the star is on Aura&#039;s local meridian and therefore has an hour angle (HA) of zero. Aura is at latitude 10°37&#039;58&quot; N, so the declination of an object which is overhead will also be +10°37&#039;58&quot;. This asteroid is at 60° altitude, therefore 30° from the zenith. It&#039;s declination (Dec) must therefore be 10°37&#039;58&quot; minus 30° which is -19°22&#039;02&quot;. So, for Aura the asteroid is at HA = 0° and Dec = -19°22&#039;02&quot;.

Moving to Gem&#039;s location, the difference in longitude is 122°58&#039;01&quot; - 121°04&#039;14&quot; which is 1°53&#039;47&quot;. Gem is west of Aura, so the HA of the asteroid is -1°53&#039;47&quot; (negative because the object is east of Gem&#039;s meridian and hour angle is measured positively to the west).

You now have to form a spherical triangle in Gem&#039;s sky. The three corners of the triangle are the overhead point (zenith), the celestial pole and the asteroid. Let&#039;s call these points A, B and C respectively. The angle at B is the hour angle. The arc BC is 90° - Dec and is therefore 109°22&#039;02&quot;. The arc AB is Gem&#039;s co-latitude which is 90° - 14°39&#039;10&quot; = 75°20&#039;50&quot;.

We have a spherical triangle where we know two sides and one of the angles. If the sides opposite to the angles A, B &amp; C are a, b &amp; c repectively, then

cos(b) = cos(a)cos(c) + sin(a)sin(c)cos(B) and

sin(A) = sin(a)sin(B)/sin(b)

Converting to decimal degrees, in the first equation, cos(b) = cos(109.3672°) cos(75.3389°) + sin(109.3672°) sin(75.3389°) cos(1.8964°)

= 0.82826

Therefore b = 34.0795°

In the second equation (ignoring the sign of B for the moment), sin(A) = sin(109.3672°) sin(1.8964°)/sin(34.0795°)

= 0.05572

Therefore A = 3.1939°

b is the zenith distance of the asteroid, so the altitude is 90° - 34.0795° = 55.9205° or 55°55&#039;14&quot;

A is the difference between the azimuth of the meridian and the azimuth of the asteroid. The azimuth of the meridian (to the south) is 180° so the azimuth of the asteroid is 180° - 3.1939° = 176.8061° or 176°48&#039;22&quot;. Gem is west of Aura, so the asteroid is east of Gem&#039;s meridian and therefore less than 180°.

As a rough check, the change in the asteroid&#039;s altitude is very nearly the same as the difference in latitudes of the two observers (4°01&#039;12&quot;), which is what would be expected when the asteroid is close to both observers&#039; meridians.

I followed suitti and Geoff&#039;s idea and did a check of the result using Starry Night Enthusiast, choosing a suitable star close to the required declination and checking its azimuth and altitude as seen from the two geographical locations. The result checks out O.K.

I told you it was long-winded.

Even later:

In my answer to the other question, I asked the questioner whether she&#039;d studied spherical trigonometry. If not, then I made the same point as you, namely that the person who set the original question may have incorrectly assumed the answer to be obtainable by simple subtraction.

Answer 5

Geez, one of those messy questions -- but it&#039;s not that bad.

First, here’s the answer:

position for Gem: altitude = 55.9288 azimuth = 176.8056

The only complication here is that the usual formulas convert from (hour angle, delta) to (az, alt).  I could derive the inverted formula, but that&#039;s not necessary here because of the simplifying fact that Aura&#039;s azimuth is 180 degrees.  (I assume that 180 azimuth means due south.)

OK, first let&#039;s look at the basic facts.  I hate dealing with deg/min/sec, so let&#039;s convert everything to decimal.  Also, the formulas I give at the bottom of the answer assume that longitude in the western hemisphere is *positive*, so the longitudes for Aura and Gem are negative.

basic facts (all units in degrees):

Aura is at lat = 10.6328 long = -122.9669

Aura sees object at alt = 60 az = 180 (due south)

Gem is at lat = 14.6528 long = -121.0706

Because the object is on Aura&#039;s meridian, we know that the hour angle of the object at her location must be 0:

HA of object for Aura = 0

For observers at two different locations, the difference in HA is simply the difference in longitude.  Since Gem lives at a more westerly location, the HA is a little earlier (just as time zones get earlier to the west):

HA of object for Gem = (HA for Aura) + 121.0706 - 122.9669

= 0 - 1.8963

= -1.8963

= 358.1037

(In the last step, I added 360 degrees to keep quantities positive.  Normally, HA is given in hours and minutes, but here we use degrees.)

What is the declination of the object?  Since Aura is at latitude 10.6328, the declination of the zenith is 10.6328.  The object is on the meridian 30 degrees below the zenith, so its declination is

object dec = 10.6328 - 30

= -19.3672

Now we need to calculate the azimuth and altitude for Gem.  Let&#039;s review the current information:

dec of object = -19.3672

hour angle of object at Gem&#039;s location = 358.1037

Gem&#039;s lat = 14.6528

Gem&#039;s long = -121.0706

We&#039;ve now cast the problem in a more common way -- namely, given the declination and hour angle of an object, what is its azimuth and altitude?  The formulas for this calculation are given below, and they yield the following answer:

altitude for Gem = 55.9288

azimuth for Gem = 176.8056

There are different versions of the formulas to convert (hour angle, dec) to (az, alt) -- here&#039;s one:

convert (H,δ) to (A,h):

sin h = sin δ * sin φ + cos δ * cos φ * cos H

A = atan2 (- cos δ * sin H, sin δ * cos φ - cos δ * sin φ * cos H)

where

h = altitude

A = azimuth

H = hour angle

φ = latitude

δ = declination

atan2 (y,x) = atan (y/x)

(Finally, let me explain why the numbers in a previous answer are clearly wrong.  The object is southern [at declination -19.3672].  Aura and Gem are seeing the object on or near the meridian, but Gem lives at a higher latitude.  When you view a southern object and move northward in the northern hemisphere, the object sinks lower in the sky.  The other answer says that Gem sees the object *higher* in the sky, and that can&#039;t be right.)

-- edit

Simplified above explanation, and discarded irrelevant references to right ascension.

-- edit

(Changed the above formula for A to a different [but equivalent] one.)

This problem really has three steps:

1) Convert the (A,h) coordinates for Aura to (H,δ).

2) Calculate the H value for Gem, based on Aura&#039;s H value and the difference in longitudes.

3) Convert (H,δ) for Gem to (A,h).

The first part was easy because the object was on the meridian, but what if it was not?  We can do the (A,h) to (H,δ) conversion in general with the following formulas:

convert (A,h) to (H,δ):

sin δ = sin h * sin φ + cos h * cos φ * cos A

H = atan2 (- cos h * sin A, sin h * cos φ - cos h * sin φ * cos A)

If you compare this with the formulas presented earlier, you&#039;ll see that they&#039;re identical, except that I&#039;ve interchanged A with H, and also interchanged h with δ.

-- edit

What if there&#039;s a 15-minute (900 seconds) delay from the time Aura sees the object to the time Gem sees it?  That&#039;s easy to handle --  you just add the following number of degrees to Gem&#039;s hour angle:

360*900/D

where D=86164.1, the number of seconds in a sidereal day.  There will be some error because we&#039;re not accounting for the asteroid&#039;s motion relative to the background stars in this time, but that should be small.