Calculate 3D point coordinates using horizontal and vertical angles and slope distance

Question

I am trying to learn how to calculate the XYZ coordinates of a point using the XYZ coordinates of an origin point, a horizontal and vertical angle, and 3d distance. I can make the calculations simply by projecting the points onto 2D planes, but is there a more straightforward way to do this in 3D?

I am trying to understand how a surveying total station calculates new point locations based on it's measured location, the 3d (slope) distance that it measures to a new point, and the measured horizontal and vertical angles that sight onto the new point location.

Thanks,

E

Answer 1

Just a note on conventions: 2D polar coordinates often use (radius, theta), where theta is the 'horizontal' or 'azimuth' angle. It ranges from: theta=0, the 2D point (x,y) = (radius,0) on the X-axis, to: theta=2*PI on the XY plane - an anti-clockwise direction as theta increases. Now to confuse matters...

3D spherical coordinates (maintaining a right-handed coordinate system) often use coordinates: (radius, theta, phi). In this case, theta is used for the 'vertical' or 'zenith' angle, ranging from theta=0 (the Z axis) to theta=PI (the -Z axis). phi is used for the azimuth angle.

Other texts will use different conventions - but this seems to be favoured by physicists and (some) mathematics texts. What matters is that you pick a convention and use it consistently.

Following this:

radius: distance to the point. given an point (x,y,z) in cartesian coordinates, we have the (pythagorean) radius: r = sqrt(x * x + y * y + z * z), e.g., 0 <= radius < +infinity

theta: the zenith angle, where theta=0 is directly above (the +Z axis), and theta=PI is directly below (the -Z axis), and theta=PI/2 is what you would consider an 'elevation' of 0 degrees, e.g.,
0 <= theta <= PI

phi: the azimuth angle, where phi=0 is to the 'right' (the +X axis), and as you turn 'anticlockwise', phi=PI/2 (the +Y axis), phi=PI (the -X axis), phi=3*PI/2 (the -Y axis), and phi=2*PI - equivalent to the phi=0 (back to the +X axis). e.g., 0 <= phi < 2*PI

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Pseudo-code: (standard math library trigonometric functions)

From (radius, theta, phi) you can find the point (x,y,z) :

x = radius * sin(theta) * cos(phi);
y = radius * sin(theta) * sin(phi);
z = radius * cos(theta);

Conversely, you can find a (radius, theta, phi) from (x,y,z) :

radius = sqrt(x * x + y * y + z * z);
theta = acos(z / radius);
phi = atan2(y, x);

Note: it is important to use atan2 in the final equation, not atan!