Calculating bearing?

[thesage1014]

Okay, I need to figure out how to get an angle from 2 gps detectors. I have made a diagram >.<
Right, so how would you find that angle? My brother said he only knows how to do it with inverse tan. Any help at all would be great!

[kymagic]

Well why dont you do it with trig?

[Norseman]

Inverse tangent in a great way to do it.

Angle = atan (X,Y)

What&#39;s so hard about that?

[pogo]

Can be done completely with the expression tool.

In: AX AY AZ BX BY BZ
Out: Dist X Y Z Bearing Elevation

X = BX - AX
Y = BY - AY
Z = BZ - AZ
Dist = sqrt( sqrt( X^2 + Y^2 ) + Z^2 )

Bearing = atan( Y/X )
Elevation = atan( Z/sqrt(X^2 + Y^2) )[/b]

[Psawhn]

That Distance formula isn&#39;t quite right. It should be

Dist = sqrt( sqrt(X^2+Y^2)^2 + Z^2)

or [code]Dist = sqrt(X^2+Y^2 + Z^2)

[thesage1014]

Can be done completely with the expression tool.

In: AX AY AZ BX BY BZ
Out: Dist X Y Z Bearing Elevation

X = BX - AX
Y = BY - AY
Z = BZ - AZ
Dist = sqrt( sqrt( X^2 + Y^2 ) + Z^2 )

Bearing = atan( Y/X )
Elevation = atan( Z/sqrt(X^2 + Y^2) )[/b]

But that isn&#39;t the bearing angle, it&#39;s just the offset. I need to know what it&#39;s angle is.

[Psawhn]

Well, you&#39;re right that it&#39;s not the bearing - it&#39;s actually the angle as in that diagram in your first post.
The better formula would be:

Bearing = atan2(X, Y)

Well, actually that formula gives the azimuth, which are degrees clockwise of &#39;North,&#39; or +y referenced to the world.

[thesage1014]

Thanks! That&#39;s what I wanted. Sorry for not exactly explaining.