# Possible to derive Lat and Lon or Cartesian X,Y,Z from Azimuth and Elevation?

k80sg

## Homework Statement

Is there a univeral conversion formulae to convert Azimuth and Elevation to Lat Lon or Cartesian X,Y,Z?

## Homework Equations

So far, I have only been able to source out and successfully convert Cartesian X,Y,Z to Lat Long Alt.

No findings yet.

Thanks.

AC130Nav

## Homework Statement

Is there a univeral conversion formulae to convert Azimuth and Elevation to Lat Lon or Cartesian X,Y,Z?

## Homework Equations

So far, I have only been able to source out and successfully convert Cartesian X,Y,Z to Lat Long Alt.

## The Attempt at a Solution

No findings yet.

Thanks.

I am unsure why you would want to convert xyz to Lat/Long; and am, therefore, not sure how to frame a reply. Are you trying to deal with star coordinates?

p1ayaone1
azimuth and elevation are angles.
latitude and longitude are angles.

there isn't a formula per se, just check the conventions for each.

And for cartesian XYZ, you'll need to know the raduis of your sphere. from either lat/lon or az/elev angles, get the equivalent representation in sphereical coordinates (r-theta-phi) and convert that to rectangular. I don't know those formluae off the top of my head, but it's not too hard and readily availible. Wiki and google are your friends.

AC130Nav

## Homework Statement

Is there a univeral conversion formulae to convert Azimuth and Elevation to Lat Lon or Cartesian X,Y,Z?

## Homework Equations

So far, I have only been able to source out and successfully convert Cartesian X,Y,Z to Lat Long Alt.

## The Attempt at a Solution

No findings yet.

Thanks.

It turns out wiki has the formula: http://en.wikipedia.org/wiki/Polar_coordinate_system" [Broken] . (It's about a page down.)

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Staff Emeritus
It turns out wiki has the formula: http://en.wikipedia.org/wiki/Polar_coordinate_system" [Broken] . (It's about a page down.)

That's for converting polar coordinates to cartesian, AC130Nav. Latitude and longitude are not polar coordinates for one thing (the Earth is not spherical). Moreover, and much more importantly, there is no way to convert azimuth and elevation to either latitude and longitude or cartesian x/y/z. There are two reasons. First off, you need to know distance. Azimuth and elevation specify a direction, not a coordinate. Secondly, you need to know the latitude, longitude, and altitude of the observer. Think of it this way: The azimuth and elevation of the Moon are quite different for observers in New York and in California who see the Moon at the same time. Azimuth and elevation are local coordinates.

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AC130Nav
That's for converting polar coordinates to cartesian, AC130Nav. Latitude and longitude are not polar coordinates for one thing (the Earth is not spherical). Moreover, and much more importantly, there is no way to convert azimuth and elevation to either latitude and longitude or cartesian x/y/z. There are two reasons. First off, you need to know distance. Azimuth and elevation specify a direction, not a coordinate. Secondly, you need to know the latitude, longitude, and altitude of the observer. Think of it this way: The azimuth and elevation of the Moon are quite different for observers in New York and in California who see the Moon at the same time. Azimuth and elevation are local coordinates.

My apologies to all that the Wiki article didn't go far enough and I didn't look closely enough.

Actually, DH, it happens latitude and longitude are SPHERICAL Polar Coordinates, and it doesn't matter that the Earth isn't sperical (not that the k80sg specified that he was talking about the Earth, there's also the celestial spere), because the coordinates are essentially angles. (Did you know that the length of a nautical mile is variable?)

Navigation is not yet the subject of this thread, since K80sg hasn't specified; and I wouldn't
use the moon by preference for a navigation example because it's apparent motion is more
irregular. Let's say, however, you happen to know where the subpoint of the moon was at the time you measured its bearing and height; you would then know roughly where you are. The assumed position of the observed is used for plotting (because the great circle of equidistance from the subpoint is too large to plot and generally too far away for the bearing to make any sense) and knowing where approximately to look for a star among the myriads. I'm not sure this has anything to do with the question, though.

The distance from the center of the Earth (if there can be considered to be such a point) is
significant in calculating xyz coordinates in that it is the radius in the trig equations. It is
hard to determine to any degree of accuracy (although the GPS people are continually bettering their capability); however, we still don't know if that's what k80sg had in mind.

So playaone1 is quite right in so far as he went. Maybe some illustration will help with the
homework without giving all the math away:

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Staff Emeritus
Actually, DH, it happens latitude and longitude are SPHERICAL Polar Coordinates, and it doesn't matter that the Earth isn't sperical (not that the k80sg specified that he was talking about the Earth, there's also the celestial spere), because the coordinates are essentially angles.
It most certainly does matter that the Earth is not spherical. You are talking about geocentric latitude, which is something that is not used for the Earth. The latitude you find on maps, and on globes, is geodetic latitude. A better model for the shape of the Earth is an oblate spheroid. If the Earth truly was an oblate spheroid, the latitude at some location on the surface of the Earth would be defined by the angle between the local vertical (extended downward to the Earth's equatorial plane) and the equatorial plane. The Earth is not quite an oblate spheroid, either; latitude is defined with respect to a particular oblate spheroid model, the WGS84 model.
(Did you know that the length of a nautical mile is variable?)
No, it isn't. A nautical mile is 1852 meters, exactly.

Staff Emeritus
So playaone1 is quite right in so far as he went. Maybe some illustration will help with the homework without giving all the math away.
Suppose you are a surveyor and you take sightings toward a mountain. You find that the azimuth and elevation of the peak are 245° degrees and 5° degrees, respectively. What are the latitude and longitude of the peak?

The question is meaningless because not enough information has been specified. You need to know your own latitude, longitude, and elevation, plus the distance between you and the peak, before that question can be answered.

p1ayaone1
D H, I understand exactly what you mean, but I still have a question about surveying. If you survey a peak as described, you determine those values for azimuth and elevation. If the observer is at some known position (alt/lat/lon), he can determine the summit altitude by measuring the elevation angle (and then doing some trig). That is measured as the "elevation" above his local horizontal (Earth surface tangent in direction of summit) so no problem. With respect to what is the azimuth measured? Is there some convention like ccw from the north side of the local north-south meridian?

Anyways, the basis for my comparison of azimuth/elevation and latitude and longitude is that if you assume the Earth is spherical and you are standing at the center along the north-south axis, the azimuth is analogous to specifying a line of longitude. Likewise, an "elevation" is analogous to specifying a line of latitude. If your elevation is say 0 degrees, then that specifies the equator. If you azimuth is zero degrees, then you are looking at the Greenwich Meridian; and if it's 180 degrees it;s the International Dateline (more or less).

Coming back to the OP's question, maybe he has observed the moon at a particular UTC time, and based on those observations he wants to derive the lat/lon of where the observation was made. (That's pure speculation since we haven't heard back...) I'm not sure if that's enough information to solve the problem, there is a nice wiki on why determining longitude is difficult.

http://en.wikipedia.org/wiki/History_of_longitude

Staff Emeritus
Suppose you see some object and you want to know its azimuth and elevation. Imagine projecting the line to the object onto the local horizontal plane. The angle between the line to the object and the projection onto the plane is elevation. A positive elevation value means the object is above the local horizontal. The angle between true north and that projection onto the horizontal plane is the azimuth angle, measured positive to the east (or clockwise when viewed from above).

Azimuth and elevation are a local coordinate system. The concept was invented for use on the surface of the Earth, not at the center of the Earth.

AC130Nav
Suppose you see some object and you want to know its azimuth and elevation. Imagine projecting the line to the object onto the local horizontal plane. The angle between the line to the object and the projection onto the plane is elevation. A positive elevation value means the object is above the local horizontal. The angle between true north and that projection onto the horizontal plane is the azimuth angle, measured positive to the east (or clockwise when viewed from above).

Azimuth and elevation are a local coordinate system. The concept was invented for use on the surface of the Earth, not at the center of the Earth.

So now we're doing spheroids, not that I condede aything that's been said. But I think you're bit adrift.

AC130Nav
And D H was just wrong.

AC130Nav
It most certainly does matter that the Earth is not spherical. You are talking about geocentric latitude, which is something that is not used for the Earth. The latitude you find on maps, and on globes, is geodetic latitude. A better model for the shape of the Earth is an oblate spheroid. If the Earth truly was an oblate spheroid, the latitude at some location on the surface of the Earth would be defined by the angle between the local vertical (extended downward to the Earth's equatorial plane) and the equatorial plane. The Earth is not quite an oblate spheroid, either; latitude is defined with respect to a particular oblate spheroid model, the WGS84 model.

No, it isn't. A nautical mile is 1852 meters, exactly.

Do you work for ARPA?

AC130Nav
The nautical mile is defined by latitude, due to the very deformation of the globe heretofore stated.

Staff Emeritus
And D H was just wrong.
Regarding what, may I ask? Certainly not the nautical mile. It is exactly 1852 meters, by definition. That has been the definition since 1929. Not exactly new.

That common or geodetic latitude is the angle between the equatorial plane and local vertical rather than angle between the equatorial plane and a line from the center of the Earth is not new. People have been using geodetic latitude since the 1700s.

That azimuth and elevation are a local coordinate system is not new, either. People have needed to precisely aim things like surveying equipment and telescopes for a long time. Azimuth and elevation specify the direction in which those equipment are oriented.

Do you work for ARPA?
No. ARPA? ARPA doesn't really do much in this regard. The US agency that is responsible for defining things like the nautical mile is the National Institute of Standards and Technology. NASA obviously has to worry about things like azimuth and elevation so it can aim its telescopes and ground-based antenna in the right direction.

BTW, I work for NASA.

k80sg
Hi folks, thanks for all the information provided. As much as I tried to understand some of you, its just beyond me. Btw, I did bump into this site recently which provides a formulae to convert Azimuth, Elevation and Range to XYZ coordinates.

http://geostarslib.sourceforge.net/main.html [Broken]

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p1ayaone1
Hi folks, thanks for all the information provided. As much as I tried to understand some of you, its just beyond me. Btw, I did bump into this site recently which provides a formulae to convert Azimuth, Elevation and Range to XYZ coordinates.

http://geostarslib.sourceforge.net/main.html [Broken]

cool nice find. looks like a well-made library

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Staff Emeritus