CODEWITHSUNDEEP
c++mathgeometrytrigonometry
Shawn Buckley
Shawn Buckley

Reputation: 516

Calculating a line from a starting point and angle in 3d

I have a point in 3D space and two angles, I want to calculate the resulting line from this information. I have found how to do this with 2D lines, but not 3D. How can this be calculated?

If it helps: I'm using C++ & OpenGL and have the location of the user's mouse click and the angle of the camera, I want to trace this line for intersections.

Upvotes: 1

Views: 4510

Answers (2)

vdbuilder
vdbuilder

Reputation: 12994

In trig terms two angles and a point are required to define a line in 3d space. Converting that to (x,y,z) is just polar coordinates to cartesian coordinates the equations are:

x = r sin(q) cos(f)

y = r sin(q) sin(f)

z = r cos(q)

Where r is the distance from the point P to the origin; the angle q (zenith) between the line OP and the positive polar axis (can be thought of as the z-axis); and the angle f (azimuth) between the initial ray and the projection of OP onto the equatorial plane(usually measured from the x-axis).

Edit:

Okay that was the first part of what you ask. The rest of it, the real question after the updates to the question, is much more complicated than just creating a line from 2 angles and a point in 3d space. This involves using a camera-to-world transformation matrix and was covered in other SO questions. For convenience here's one: How does one convert world coordinates to camera coordinates? The answers cover converting from world-to-camera and camera-to-world.

Upvotes: 7

Motomotes
Motomotes

Reputation: 4237

The line can be fathomed as a point in "time". The equation must be vectorized, or have a direction to make sense, so time is a natural way to think of it. So an equation of a line in 3 dimensions could really be three two dimensional equations of x,y,z related to time, such as:

x = ax*t + cx
y = ay*t + cy
z = az*t + cz

To find that set of equations, assuming the camera is at origin, (0,0,0), and your point is (x1,y1,z1) then 

ax = x1 - 0
ay = y1 - 0
az = z1 - 0

cx = cy = cz = 0

so 

x = x1*t 
y = y1*t 
z = z1*t

Note: this also assumes that the "speed" of the line or vector is such that it is at your point (x1,y1,z1) after 1 second.

So to draw that line just fill in the points as fine as you like for as long as required, such as every 1/1000 of a second for 10 seconds or something, might draw a "line", really a series of points that when seen from a distance appear as a line, over 10 seconds worth of distance, determined by the "speed" you choose.

Upvotes: 3

Related Questions