I have a point on a sphere that needs to be rotated. I have 3 different degrees of rotation (roll, pitch, yaw). Are there any formulas I could use to calculate where the point would end up after applying each rotation? For simplicity sake, the sphere can be centered on the origin if that helps.
I've tried looking at different ways of rotation, but nothing quite matches what I am looking for. If I needed to just rotate the sphere, I could do that, but I need to know the position of a point based on the rotation of the sphere.
Using Unity for an example, this is outside of unity in a separate project so using their library is not possible:
If the original point is at (1, 0, 0)
And the sphere then gets rotated by [45, 30, 15]:
What is the new (x, y, z) of the point?
If you have a given rotation as a Quaternion q, then you can rotate your point (Vector3) p like this:
Vector3 pRotated = q * p;
And if you have your rotation in Euler Angles then you can always convert it to a Quaternion like this (where x, y and z are the rotations in degrees around those axes):
Quaternion q = Quaternion.Euler(x,y,z);
Note that Unity's euler angles are defined so that first the object is rotated around the z axis, then around the x axis and finally around the y axis - and that these axes are all the in the space of the parent transform, if any (not the object's local axes, which will move with each rotation).
So I suppose that the z-axis would be roll, the x-axis would be pitch and the y axis would be yaw.You might have to switch the signs on some axes to match the expected result - for example, a positive x rotation will tilt the object downwards (assuming that the object's notion of forward is in its positive z direction and that up is in its positive y direction).
Related
I have three non-colinear 3D points, let's say pt1, pt2, pt3. I've computed the plane P using the sympy.Plane. How can I find the orientation of this plane(P) i.e. RPY(euler angles) or in quaternion?
I never used sympy, but you should be able to find a function to get the angle between 2 vectors (your normal vector and the world Y axis.)
theta = yaxis.angle_between(P.normal_vector)
then get the rotation axis, which is the normalized cross product of those same vectors.
axis = yaxis.cross(P.normal_vector).normal()
Then construct a quaternion from the axis and angle
q = Quaternion.from_axis_angle(axis, theta)
I have two normalized vectors:
A) 0,0,-1
B) 0.559055,0.503937,0.653543
I want to know, what rotations about the axes would it take to take the vector at 0,0,-1 to 0.559055,0.503937,0.653543?
How would I calculate this? Something like, rotate over X axis 40 degrees and Y axis 220 (that's just example, but I don't know how to do it).
Check this out. (google is a good thing)
This calculates the angle between two vectors.
If Vector A is (ax, ay, az) and
Vector B is (bx, by, bz), then
The cos of angle between them is:
(ax*bx + ay*by + az*bz)
--------------------------------------------------------
sqrt(ax*ax + ay*ay + az*az) * sqrt(bx*bx + by*by + bz*bz)
To calculate the angle between the two vectors as projected onto the x-y plane, just ignore the z-coordinates.
Cosine of Angle in x-y plane =
(ax*bx + ay*by)
--------------------------------------
sqrt(ax*ax + ay*ay) * sqrt(bx*bx + by*by
Similarly, to calculate the angle between the projections of the two vectors in the x-z plane, ignore the y-coordinates.
It sounds like you're trying convert from Cartesian coordinates (x,y,z) into spherical coordinates (rho,theta,psi).
Since they're both unit vectors, rho, the radius, will be 1. This means your magnitudes will also be 1 and you can skip the whole denominator and just use the dot-product.
Rotating in the X/Y plane (about the Z axis) will be very difficult with your first example (0,0,-1) because it has no extension in X or Y. So there's nothing to rotate.
(0,0,-1) is 90 degrees from (1,0,0) or (0,1,0). If you take the x-axis to be the 0-angle for theta, then you calculate the phi (rotation off of the X/Y plane) by applying the inverse cos upon (x,y,z) and (x,y,0), then you can skip dot-products and get theta (the x/y rotation) with atan2(y,x).
Beware of gimbal lock which may cause problems.
So I need to map a texture to a sphere from within a pixel/fragment shader in Cg.
What I have as "input" in every pass are the Cartesian coordinates x, y, z for the point on the sphere where I want the texture to be sampled. I then transform those coordinates into Spherical coordinates and use the angles Phi and Theta as U and V coordinates, respectively, like this:
u = atan2(y, z)
v = acos(x/sqrt(x*x + y*y + z*z))
I know that this simple mapping will produce seams at the poles of the sphere but at the moment, my problem is that the texture repeats several times across the sphere. What I want and need is that the whole texture gets wrapped around the sphere exactly once.
I've fiddled with the shader and searched around for hours but I can't find a solution. I think I need to apply some sort of scaling somewhere but where? Or maybe I'm totally on the wrong track, I'm very new to Cg and shader programming in general... Thanks for any help!
Since the results of inverse trigonometric functions are angles, they will be in [-Pi, Pi] for u and [0, Pi] for v (though you can't have searched for hours with at least basic knowledge of trigonometrics, as acquired from school). So you just have to scale them appropriately. u /= 2*Pi and v /= Pi should do, assuming you have GL_REPEAT (or the D3D equivalent) as texture coordinate wrapping mode (which your description sounds like).
If i have a point (x,y,z) how to project it on to a sphere(x0,y0,z0,radius) (on its surface).
My input will be the coordinates of point and sphere.
The output should be the coordinates of the projected point on sphere.
Just convert from cartesian to spherical coordinates?
For the simplest projection (along the line connecting the point to the center of the sphere):
Write the point in a coordinate system centered at the center of the sphere (x0,y0,z0):
P = (x',y',z') = (x - x0, y - y0, z - z0)
Compute the length of this vector:
|P| = sqrt(x'^2 + y'^2 + z'^2)
Scale the vector so that it has length equal to the radius of the sphere:
Q = (radius/|P|)*P
And change back to your original coordinate system to get the projection:
R = Q + (x0,y0,z0)
Basically you want to construct a line going through the spheres centre and the point. Then you intersect this line with the sphere and you have your projection point.
In greater detail:
Let p be the point, s the sphere's centre and r the radius then x = s + r*(p-s)/(norm(p-s)) where x is the point you are looking for. The implementation is left to you.
I agree that the spherical coordinate approach will work as well but is computationally more demanding. In the above formula the only non-trivial operation is the square root for the norm.
It works if you set the coordinates of the center of the sphere as origin of the system (x0, y0, z0). So you will have the coordinates of the point referred to that origin (Xp', Yp', Zp'), and converting the coordinates to polar, you discard the radius (distance between the center of the sphere and the point) and the angles will define the projection.
I have gone through all available study resources in the internet as much as possible, which are in form of simple equations, vectors or trigonometric equations.
I couldn't find the way of doing following thing:
Assuming Y is up in a 3D world.
I need to draw two 2D trajectories orthogonally (not the projections) for a 3D trajectory, say XY-plane for side view of the trajectory w.r.t. the trajectory itself and XZ-plane for top view for the same.
I have all the 3D points of the 3D trajectory, initial velocity, both the angles can be calculated by vector mathematics.
How should I proceed further?
refer:
Below a curve in different angles, which can loose its significance if projected along XY-plane. All I want is to convert the red curve along itself, the green curve along green curve and so on. and further how would I map side view to a plane. Top view is comparatively easy and done just by taking X and Z ordinates of each points.
I mean this the requirement. :)
I don't think I understand the question, but I'll answer my interpretation anyway.
You have a 3D trajectory described by a sequence of points p0, ..., pN. We are given an angle v for a plane P parallel to the Y-axis, and wish to compute the 2D coordinates (di, hi) of the points pi projected onto that plane, where hi is the height coordinate in the direction Y and di is the distance coordinate in the direction v. Assume p0 = (0, 0, 0) or else subtract p0 from all vectors.
Let pi = (xi, yi, zi). The height coordinate is hi = yi. Assume the angle v is given relative to the Z-axis. The vector for the direction v is then r = (sin(v), 0, cos(v)), and the distance coordinates becomes di = dot(pi, r).