Find displacement in 3D

[eLg]

Hi,

Can someone tell me the steps on how to find the displacement in 3D?
I am trying to solve the problem with given:
seperate angles of two different vehicles from north, clockwise direction.
magnitude of their distances from the origin.
and the magnitude for elevation.
there is an attached image.

Thanks

[md65536]

Can someone tell me the steps on how to find the displacement in 3D?

Just find the locations of the vehicles as vectors, subtract one from the other, and find the magnitude of that vector.

 

Eg. Step

1) I'd convert to typical angles (counter clockwise from x-axis) just to match your diagram, but you don't have to do this.

angle = -theta - pi/2

x = cos( angle ) * magnitude_from_origin

y = sin( angle ) * magnitude_from_origin

z = elevation

 

2) displacement_vector = Loc2 - Loc1

or dx = x2 - x1, dy = y2 - y1, dz = z2 - z1

 

3) displacement = magnitude( displacement_vector )

= sqrt( dx^2 + dy^2 + dz^2 )

 

I've been sloppy with variable names, I hope the context is clear enough?

Edited October 5, 2013 by md65536

[studiot]

 

angle = -theta - pi/2

 

 

I think this should be

 

angle = - (theta - pi/2)

 

Otherwise you will get the wrong sign (and answer).

 

 

You can also use angle = (450 - theta) if you do not wish to use negative angles.

Edited October 5, 2013 by studiot

[md65536]

I think this should be

 

angle = - (theta - pi/2)

 

Otherwise you will get the wrong sign (and answer).

Oops, you're right.

 

Won't the answer come out the same though?

In fact couldn't you rotate both by any same angle, and the displacement will come out the same? In my example I've mistakenly rotated the vectors 180 degrees, making them backwards with respect to x and y.

If you name the coordinates (North, East, Height) instead of (x,y,z) then you could leave the angles as they are, and just do

North = cos(theta)*magnitude

East = sin(theta)*magnitude

Edited October 5, 2013 by md65536

[studiot]

 

Won't the answer come out the same though?

 

 

Not necessarily, but your formulae or mine will always work.

eLg presented a problem in the (mathematical) 4th quadrant.

 

Don't forget the sign rules for different quadrants.

 

Your formula is quite correct, but the sin and cos are reversed for normal surveying or navigation calculations, since they are measured from the North line, not the east line.

 

The correct terms are northing (y) and easting (x) or latidudes (y) and departures (x)

 

http://www.cfr.washington.edu/classes.esrm.304/Spring2011/Documents/Hurvitz_Schiess/procedures/latitudes_and_departures.html

Edited October 5, 2013 by studiot