Is the Azimuth on the equator equal to the one not on the equator?

Question

I am trying to fully understand the concept of Azimuth and I am encountering some inconsistencies (or maybe it is my error).

I show you some examples that do not match, hoping somebody can explain me how this really works.

I show the coordinates in EPSG:900913, in PostGIS and using my own JavaScript function.

MY FUNCTION

/* Difference between the two longitudes */
var dLon = lon2 - lon1;
/* Y value */
var y = Math.sin(dLon) * Math.cos(lat2);
/* X value */
var x = Math.cos(lat1) * Math.sin(lat2) - Math.sin(lat1) * Math.cos(lat2) * Math.cos(dLon);
/* Calculates the azimuth between the two points and converts it to degrees */
var angle = Math.atan2(y, x) / Math.PI * 180;

EXAMPLES

/* Same Y, not on the equator */
Point A: (-81328.998084106, 7474929.8690234)
Point B: (4125765.0381464, 7474929.8690234)
Result in PostGIS: 90 degrees
Result in my JS function: 74.232 degrees

/* Same Y, on the equator */
Point A: (-81328.998084106, 0)
Point B: (4125765.0381464, 0)
Result in PostGIS: 90 degrees
Result in my JS function: 90 degrees

I understand, that on the equator, the Azimuth is 90 (or 270) for a horizontal line. think that if you draw a horizontal line a bit North (or South) of the equator, then the Azimuth is not 90 degrees anymore. But... PostGIS tells me that it is always 90 degrees when we have the same Y.

Additionally, this calculator also show that the Azimuths of horizontal lines are not 90 degrees when Y != 0 (not on the equator).

How is it correct?

Thanks

possible duplicate of [How to calculate the Azimuth between two points in PostGIS?](http://stackoverflow.com/questions/25526684/how-to-calculate-the-azimuth-between-two-points-in-postgis) — Peter O., Aug 28 '14 at 17:19
This question appears to be off-topic because it is not about programming (though it uses programming). See [What topics can I ask about here](http://stackoverflow.com/help/on-topic) in the Help Center. Perhaps the [Geography Stack Exchange](https://gis.stackexchange.com/) would be a better place to ask. — jww, Aug 28 '14 at 17:53
@jww, thank you. I know GIS.StackExchange is better. I tried both and the aswer here was better. So, a big thank you to the person who answered! :) I'll be more careful in the future. — joaorodr84, Aug 28 '14 at 18:06
@joaorodr84 - awesome; glad you got a good answer. (And please don't take the close personally. I've had to close my own questions (did I really ask that question years ago?)). — jww, Aug 28 '14 at 18:12
No problem, @jww. :) I don't take it personally. You're absolutely right. ;) — joaorodr84, Aug 28 '14 at 19:17

score 2 · Accepted Answer · edited Aug 28 '14 at 17:17

In your example you have used EPSG:900913, which is planar, projected and in meters. This means that the formula used will be the atan2, which will always be 90 when the latitudes are the same, as the formula is:

azimuth = atan2(y1-y2, x1-x2)

And the 2nd part will always be 0, giving an azimuth of 90. So, using planar coordinates, yes, the azimuth will always be the same for pairs of coordinates with identical latitudes and this is why Postgis always gives the same answer when using EPS:900913.

If you switch to the geography datatype, and therefore use geodesic coordinates, this is no longer the case.

For example:

select degrees(   
  st_azimuth(
   st_makepoint(0, 10)::geography, 
   st_makepoint(90, 10)::geography));

Gives 80.1318065 in Postgis and gives 80.139 on the calculator page you linked.

As the x/longitudes get closer together, the closer the values get to 90, for a given latitude. For example,

select degrees(   
  st_azimuth(
   st_makepoint(0, 10)::geography, 
   st_makepoint(1, 10)::geography));

Now gives 89.9131737 in Postgis and and 89.333 in the online calculator (slightly more discrepancy).

All of this is due to the fact that the formula now accounts for curvature, so the angle between the projections of the two vectors with equal latitude will no longer be 90, except on the equator.

Look at the equation in the Wikipedia azimuth article for the spheroid version. That should be easy enough to code up in JavaScript and that ought to give comparable answers to Postgis with the geography type.

Hi @JohnBarça. I was using PostGIS the wrong way. Thank you. As you can see, I am already using the [function from Wikipedia](http://upload.wikimedia.org/math/2/3/8/238805bcce98ded92525289f49b3d6f9.png), slightly modified. Or do you mean the [second one](http://upload.wikimedia.org/math/e/5/e/e5ecad955f1b1f7e2f84c3dd36eb4296.png)? — joaorodr84, Aug 27 '14 at 16:40
I meant the 2nd one, that accounts for the oblate spheroid, starting with http://upload.wikimedia.org/math/7/6/e/76e90f1c786b852709ce30d9504cb19a.png. If you use that you should get closer to the values you get using geography in Postgis. — John Powell, Aug 27 '14 at 16:43

Is the Azimuth on the equator equal to the one not on the equator?

1 Answers1

Linked