The X angle between two 3D vectors?

问题:

I have two 3D vectors called A and B that both only have a 3D position. I know how to find the angle along the unit circle ranging from 0-360 degrees with the atan2 function by doing:

EDIT: (my atan2 function made no sense, now it should find the "y-angle" between 2 vectors):

toDegrees(atan2(A.x-B.x,A.z-B.z))+180

But that gives me the Y angle between the 2 vectors. I need to find the X angle between them. It has to do with using the x, y and z position values. Not the x and z only, because that gives the Y angle between the two vectors. I need the X angle, I know it sounds vague but I don't know how to explain. Maybe for example you have a camera in 3D space, if you look up or down than you rotate the x-axis. But now I need to get the "up/down" angle between the 2 vectors. If I rotate that 3D camera along the y-axis, the x-axis doens't change. So with the 2 vectors, no matter what the "y-angle" is between them, the x-angle between the 2 vectors wil stay the same if y-angle changes because it's the "up/down" angle, like in the camara.

Please help? I just need a line of math/pseudocode, or explanation. :)

回答1:

Computing the altitude angle

OK, it might be I finally understood your comment below about the result being independent of the y angle, and about how it relates to the two vectors. It seems you are not really interested in two vectors and the angle between these two, but instead you're interested in the difference vector and the angle that one forms against the horizontal plane. In a horizontal coordinate system (often used in astronomy), that angle would be called “altitude” or “elevation”, as opposed to the “azimuth” you compute with the formula in your (edited) question. “altitude” closely relates to the “tilt” of your camera, whereas “azimuth” relates to “panning”.

We still have a 2D problem. One coordinate of the 2D vector is the y coordinate of the difference vector. The other coordinate is the length of the vector after projecting it on the horizontal plane, i.e. sqrt(x*x + z*z). The final solution would be

x = A.x - B.x
y = A.y - B.y
z = A.z - B.z
alt = toDegrees(atan2(y, sqrt(x*x + z*z)))
az = toDegrees(atan2(-x, -z))

The order (A - B as opposed to B - A) was chosen such that “A above B” yields a positive y and therefore a positive altitude, in accordance with your comment below. The minus signs in the azimuth computation above should replace the + 180 in the code from your question, except that the range now is [-180, 180] instead of your [0, 360]. Just to give you an alternative, choose whichever you prefer. In effect you compute the azimuth of B - A either way. The fact that you use a different order for these two angles might be somewhat confusing, so think about whether this really is what you want, or whether you want to reverse the sign of the altitude or change the azimuth by 180°.

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Orthogonal projection

For reference, I'll include my original answer below, for those who are actually looking for the angle of rotation around some fixed x axis, the way the original question suggested.

If this x angle you mention in your question is indeed the angle of rotation around the x axis, as the camera example suggests, then you might want to think about it this way: set the x coordinate to zero, and you will end up with 2D vectors in the y-z plane. You can think of this as an orthogonal projection onto said plain. Now you are back to a 2D problem and can tackle it there.

Personally I'd simply call atan2 twice, once for each vector, and subtract the resulting angles:

toDegrees(atan2(A.z, A.y) - atan2(B.z, B.y))

The x=0 is implicit in the above formula simply because I only operate on y and z.

I haven't fully understood the logic behind your single atan2 call yet, but the fact that I have to think about it this long indicates that I wouldn't want to maintain it, at least not without a good explanatory comment.

I hope I understood your question correctly, and this is the thing you're looking for.

回答2:

atan2(crossproduct.length,scalarproduct)

The reason for using atan2 instead of arccos or arcsin is accuracy. arccos behaves very badly close to 0 degrees. Small computation errors in argument will lead to disproportionally big errors in result. arcsin has same problem close to 90 degrees.

回答3:

Just like 2D Vectors , you calculate their angle by solving cos of their Dot Product

You don't need atan, you always go for the dot product since its a fundamental operation of vectors and then use acos to get the angle.

double angleInDegrees = acos ( cos(theta) ) * 180.0 / PI;