I want to draw a plane which is given by the equation: Ax+By+Cz+D=0.
I first tried to draw him by setting x,y and then get z from the equation. this did not work fine because there are some planes like 0x+0y+z + 0 = 0 and etc...
my current solution is this:
- draw the plane on ZY plane by giving 4 coordinates which goes to infinity.
- find out to rotation that should be done in order to bring the normal of the given plane(a,b,c) to lay
on z axis.
- find the translation that should be done in order for that plane be on x axis.
- make the exactly opposite transformation to those rotation and to this translation hence i will get the
plane in his place.
ok
this is a great thing but I can make the proper math calculations(tried a lot of times...) with the dot product and etc....
can someone help me in understanding the exact way it should be done OR give me some formula in which I will put ABCD and get the right transformation?
You will want the following transformation matrix:
[ x0_x y0_x z0_x o_x ]
M = [ x0_y y0_y z0_y o_y ]
[ x0_z y0_z z0_z o_z ]
[ 0 0 0 1 ]
Here, z0 is the normal of your plane, and o is the origin of your plane, and x0 and y0 are two vectors within your plane orthogonal to z0 that define the rotation and skew of your projection.
Then any point (x,y) on your XY plane can be projected to a point (p_x, p_y, p_z) your new plane with the following:
(p_x, p_y, p_z, w) = M * (x, y, 0, 1)
now, z0 in your transformation matrix is easy, that's the normal of your plane and that is simply n = normalize(a,b,c).
In choosing the rest however you have distinctly more freedom. For the origin you could take the point that the plane intersects the Z axis, unless of course the plane is parallel to the Z axis, in which case you need something else.
So e.g.
if (c != 0) { //plane intersects Z axis
o_x = 0;
o_y = 0;
o_z = -d/c;
}
else if (b != 0) { // plane intersects Y axis
o_x = 0;
o_y = -d/b;
o_z = 0;
}
else { // plane must intersect the X axis
o_x = -d/a;
o_y = 0;
o_z = 0;
}
In practice you may want to prefer a different test than (c != 0), because with that test it will succeed even is c is very very small but just different from zero, leading your origin to be at say, x=0, y=0, z=10e100 which would probably not be desirable. So some test like (abs(c) > threshold) is probably preferable. However you could of course take an entirely different point in the plane to put the origin, perhaps the point closest to the origin of your original coordinate system, which would be:
o = n * (d / sqrt(a^2 + b^2 + c^2))
Then finally we need to figure out an x0 and y0. Which could be any two linearly independent vectors that are orthogonal to z0.
So first, let's choose a vector in the XY plane for our x0 vector:
x0 = normalize(z0_y, -z0_x, 0)
Now, this fails if your z0 happens to be of the form (0, 0, z0_z) so we need a special case for that:
if (z0_x == 0 && z0_y == 0) {
x0 = (1, 0, 0)
}
else {
x0 = normalize(z0_y, -z0_x, 0)
}
Finally let's say we do not want skew and choose y0 to be orthogonal to both x0, and y0, then, using the crossproduct
y0 = normalize(x0_y*y0_z-x0_z*y0_y, x0_z*y0_x-x0_z*y0_z, x0_x*y0_y-x0_y*y0_x)
Now you have all to fill your transformation matrix.
Disclaimer: Appropriate care should be taken when using floating point representations for your numbers, simple (foo == 0) tests are not sufficient in those cases. Read up on floating point math before you start implementing stuff.
Edit: renamed some variables for clarity
Is this what you're asking?
Transforming a simple plane like the xy plane to your plane is fairly simple:
your plane is Ax+By+Cz+D=0
the xy plane is simply z=0. i.e. A=B=D=0, while C=whatever you want. We'll say 1 for simplicity's sake.
When you have a plane in this form, the normal of the plane is defined by the vector (A,B,C)
so you want a rotation that will take you from (0,0,1) to (A,B,C)*
*Note that this will only work if {A,B,C} is unitary. so you may have to divide A B and C each by sqrt(A^2+B^2+C^2).
rotating around just two of the axes can get your from any direction to any other direction, so we'll pick x and y;
here are the rotation matrices for rotations by a about the x axis, and b about the y axis.
Rx := {{1, 0, 0}, {0, Cos[a], Sin[a]}, {0, -Sin[a], Cos[a]}}
Ry := {{Cos[b], 0, -Sin[b]}, {0, 1, 0}, {Sin[b], 0, Cos[b]}}
if we do a rotation about x, followed by a rotation about y, of the vector normal to the xy plane, (0,0,1), we get:
Ry.Rx.{0,0,1} = {-Cos[a] Sin[b], Sin[a], Cos[a] Cos[b]}
which are your A B C values.
i.e.
A = -Cos[a]Sin[b]
B = Sin[a]
C = Cos[a]Cos[b]
From here, it's simple.
a = aSin[B]
so now A = -Cos[aSin[B]]Sin[b]
Cos[aSin[x]] = sqrt(1-x^2)
so:
A = -Sqrt[1-B^2] * Sin[b]
b = aSin[-A/sqrt[1-B^2]]
a = aSin[B] (rotation about x axis)
b = aSin[-A/sqrt[1-B^2]] (rotation about y axis)
So we now have the angles about the x and y axes we need to rotate by.
After this, you just need to shift your plane up or down until it matches the one you already have.
The plane you have at the moment, (after those two rotations) is going to be Ax+By+Cz=0.
the plane you want is Ax+Bx+Cz+D=0. To find out d, we will see where the z axis crosses your plane.
i.e. Cz+D=0 -> z = -D/C
So we transform your z in Ax+By+Cz=0 by -D/C to give:
Ax+By+C(z+D/C) = Ax+By+Cz+D=0. Oh would you look at that!
It turns out you don't have to do any extra maths once you have the angles to rotate by!
The two angles will give you A,B, and C. To get D you just copy it from what you had.
Hope that's of some help, I'm not entirely sure how you plan on actually drawing the plane though...
Edited to fix some horrible formatting. hopefully it's better now.
Related
I have two 3D direction (normalized) vector A and B. I am looking for the Euler angles to rotate A into B. I know it has many solution because it is possible to rotate a normal vector anywhere with just using two axis like X and Y or roll and pitch. I have to find the solution where the Z rotation is Zero.
I would like to create a function like this:
Vector3 dir1 (0, 1, 0);
Vector3 someRotation(Pi / 4, Pi / 4, 0);
Vector3 dir2 = dir1.rotateXYZ(someRotation);
Vector3 xyRotation = dir1.eulerToDirection(dir2);
// now I expect that the eulerToDirection fv calculated the X, Y rotation from the vectors at Z = 0
// so xyRotation.x == Pi / 4 && xyRotation.y == Pi / 4 && xyRotation.z == 0 is true
// aside from the floating point error
Of corse the some rotation not always 0 at the Z. It is just for the example
First use atan2(z2-z1, y2-y1) to find the angle to rotate around the X-axis that aligns the y's and the z's. Then, use acosof the dot product between the just rotated vector and the final vector. This will be the angle needed for the rotation around Y. Depending on how you implemented the rotations, you might need to flip some signs.
I have some points (3D) all on the same (known) plane. Now I want to scale these points within the plane as opposed to the whole 3D space.
Is there some quick solution for this e.g. a modified scaling matrix?
Can someone help me?
Thanks.
EDIT: I'm more looking for an idea/pseudocode how to do this. If you want use MatLab or some convenient language
Your plane can be known by three non-collinear points P0, P1, P2, or by its implicit equation,
A.x + B.y + C.z + D = 0
In the first case, consider the vector P0P1 and normalize it (U = P0P1/|P0P1|). Then compute a second vector orthogonal with the first, V = P0P2 - (P0P2.U).U and normalize it.
In the second case you can take the three intersection points with the axes, (-D/A, 0, 0), (0, -D/B, 0), (0, 0, -D/C) and you are back in the first case (but mind degenerate cases).
Use the two vectors to compute the desired 2D coordinates of any point P = (X, Y, Z) by the dot products
(x, y) = (P.U, P.V)
(This transform is a rotation that makes P0P1 parallel to the x axis and brings P0P1P2 in the plane xy.)
I have a point a point (x, y, z) that is on a plane defined by ax+by+cz+d=0. I'm trying to figure out what the (x', y') relative to the plane, where it has a starting point of (x0, y0, z0) and the x'-axis is defined by (1,0) and the y'-axis is defined by (0,1).
My major goal is to have the mouse click on a surface, and know the 2D co-ordinates on a particular surface. I've managed to intersect the ray onto a plane quite trivially.
As a side-note, I'm using DirectX 9 - my familiarity with matrix/vector math is limited by the APIs provided to me through the D3DX libraries.
One thought I had was to use the angle of between one of the axis vectors and find the distance from origin, and figure out the x/y using simple trig. But I'm not sure if that's really an ideal solution or not - or if it can actually solve the issue at hand.
Since you have a 2D image on that plane, you apparently want to match its coordinate system. To do so, determine the unit vectors of the picture. That is, take the 3d coordinates B for the picture position (x,0) for any x>0, and subtract from that the 3d coordinates A for the origin (0,0) of the picture. The resulting vector B − A will describe the positive x direction of your image. Do the same for the y direction. Then normalize both these vectors. This means dividing them by their length, sqrt(x²+y²+z²), but D3Dx has a function D3DXVec3Normalize for this. Let's call the resulting 3d vectors X and Y. To compute the x and y coordinate of any 3D point p, simply subtract the origin A from p, i.e. compute the vector p − A. Then compute the dot product between the result and the unit vectors X and Y. This will give you two numbers: the desired coordinates. This is because the dot product can be used to compute an orthogonal projection.
Translating this into D3Dx, it should look somewhat like the following. As I have never used it, this might have mistakes.
D3DXVECTOR3 *p; // input point
D3DXVECTOR3 a, b, c, ab, ac, ap; // helper vectors
FLOAT x, y; // output coordinates
imagePosTo3D(&a, 0, 0); // a = origin of image
imagePosTo3D(&b, 1, 0); // b = anywhere on positive x axis, perhaps a corner
imagePosTo3D(&c, 0, 1); // c = anywhere on positive y axis, perhaps a corner
D3DXVec3Subtract(&ab, &b, &a); // ab = b - a
D3DXVec3Subtract(&ac, &c, &a); // ac = c - a
D3DXVec3Normalize(&ab, &ab); // ab = ab / |ab|
D3DXVec3Normalize(&ac, &ac); // ac = ac / |ac|
// the above has to be done once for the image, the code below for every p
D3DXVec3Subtract(&ap, p, &a); // ap = p - a
x = D3DXVec3Dot(&ab, &ap); // x = ab∙ap
y = D3DXVec3Dot(&ac, &ap); // y = ac∙ap
How do you find the 3 euler angles between 2 3D vectors?
When I have one Vector and I want to get its rotation, this link can be usually used: Calculate rotations to look at a 3D point?
But how do I do it when calculating them according to one another?
As others have already pointed out, your question should be revised. Let's call your vectors a and b. I assume that length(a)==length(b) > 0 otherwise I cannot answer the question.
Calculate the cross product of your vectors v = a x b; v gives the axis of rotation. By computing the dot product, you can get the cosine of the angle you should rotate with cos(angle)=dot(a,b)/(length(a)length(b)), and with acos you can uniquely determine the angle (#Archie thanks for pointing out my earlier mistake). At this point you have the axis angle representation of your rotation.
The remaining work is to convert this representation to the representation you are looking for: Euler angles. Conversion Axis-Angle to Euler is a way to do it, as you have found it. You have to handle the degenerate case when v = [ 0, 0, 0], that is, when the angle is either 0 or 180 degrees.
I personally don't like Euler angles, they screw up the stability of your app and they are not appropriate for interpolation, see also
Strange behavior with android orientation sensor
Interpolating between rotation matrices
At first you would have to subtract vector one from vector two in order to get vector two relative to vector one. With these values you can calculate Euler angles.
To understand the calculation from vector to Euler intuitively, lets imagine a sphere with the radius of 1 and the origin at its center. A vector represents a point on its surface in 3D coordinates. This point can also be defined by spherical 2D coordinates: latitude and longitude, pitch and yaw respectively.
In order "roll <- pitch <- yaw" calculation can be done as follows:
To calculate the yaw you calculate the tangent of the two planar axes (x and z) considering the quadrant.
yaw = atan2(x, z) *180.0/PI;
Pitch is quite the same but as its plane is rotated along with yaw the 'adjacent' is on two axis. In order to find its length we will have to use the Pythagorean theorem.
float padj = sqrt(pow(x, 2) + pow(z, 2));
pitch = atan2(padj, y) *180.0/PI;
Notes:
Roll can not be calculated as a vector has no rotation around its own axis. I usually set it to 0.
The length of your vector is lost and can not be converted back.
In Euler the order of your axes matters, mix them up and you will get different results.
It took me a lot of time to find this answer so I would like to share it with you now.
first, you need to find the rotation matrix, and then with scipy you can easily find the angles you want.
There is no short way to do this.
so let's first declare some functions...
import numpy as np
from scipy.spatial.transform import Rotation
def normalize(v):
return v / np.linalg.norm(v)
def find_additional_vertical_vector(vector):
ez = np.array([0, 0, 1])
look_at_vector = normalize(vector)
up_vector = normalize(ez - np.dot(look_at_vector, ez) * look_at_vector)
return up_vector
def calc_rotation_matrix(v1_start, v2_start, v1_target, v2_target):
"""
calculating M the rotation matrix from base U to base V
M # U = V
M = V # U^-1
"""
def get_base_matrices():
u1_start = normalize(v1_start)
u2_start = normalize(v2_start)
u3_start = normalize(np.cross(u1_start, u2_start))
u1_target = normalize(v1_target)
u2_target = normalize(v2_target)
u3_target = normalize(np.cross(u1_target, u2_target))
U = np.hstack([u1_start.reshape(3, 1), u2_start.reshape(3, 1), u3_start.reshape(3, 1)])
V = np.hstack([u1_target.reshape(3, 1), u2_target.reshape(3, 1), u3_target.reshape(3, 1)])
return U, V
def calc_base_transition_matrix():
return np.dot(V, np.linalg.inv(U))
if not np.isclose(np.dot(v1_target, v2_target), 0, atol=1e-03):
raise ValueError("v1_target and v2_target must be vertical")
U, V = get_base_matrices()
return calc_base_transition_matrix()
def get_euler_rotation_angles(start_look_at_vector, target_look_at_vector, start_up_vector=None, target_up_vector=None):
if start_up_vector is None:
start_up_vector = find_additional_vertical_vector(start_look_at_vector)
if target_up_vector is None:
target_up_vector = find_additional_vertical_vector(target_look_at_vector)
rot_mat = calc_rotation_matrix(start_look_at_vector, start_up_vector, target_look_at_vector, target_up_vector)
is_equal = np.allclose(rot_mat # start_look_at_vector, target_look_at_vector, atol=1e-03)
print(f"rot_mat # start_look_at_vector1 == target_look_at_vector1 is {is_equal}")
rotation = Rotation.from_matrix(rot_mat)
return rotation.as_euler(seq="xyz", degrees=True)
Finding the XYZ Euler rotation angles from 1 vector to another might give you more than one answer.
Assuming what you are rotation is the look_at_vector of some kind of shape and you want this shape to stay not upside down and still look at the target_look_at_vector
if __name__ == "__main__":
# Example 1
start_look_at_vector = normalize(np.random.random(3))
target_look_at_vector = normalize(np.array([-0.70710688829422, 0.4156269133090973, -0.5720613598823547]))
phi, theta, psi = get_euler_rotation_angles(start_look_at_vector, target_look_at_vector)
print(f"phi_x_rotation={phi}, theta_y_rotation={theta}, psi_z_rotation={psi}")
Now if you want to have a specific role rotation to your shape, my code also supports that!
you just need to give the target_up_vector as a parameter as well.
just make sure it is vertical to the target_look_at_vector that you are giving.
if __name__ == "__main__":
# Example 2
# look and up must be vertical
start_look_at_vector = normalize(np.array([1, 2, 3]))
start_up_vector = normalize(np.array([1, -3, 2]))
target_look_at_vector = np.array([0.19283590755300162, 0.6597510192626469, -0.7263217228739983])
target_up_vector = np.array([-0.13225754322703182, 0.7509361508721898, 0.6469955018014842])
phi, theta, psi = get_euler_rotation_angles(
start_look_at_vector, target_look_at_vector, start_up_vector, target_up_vector
)
print(f"phi_x_rotation={phi}, theta_y_rotation={theta}, psi_z_rotation={psi}")
Getting Rotation Matrix in MATLAB is very easy
e.g.
A = [1.353553385, 0.200000003, 0.35]
B = [1 2 3]
[q] = vrrotvec(A,B)
Rot_mat = vrrotvec2mat(q)
I have a quaternion rotation, as usually described by 4 values: a b c d.
Lets say it transforms the x axis so that i look at some object from the front. Now i want to change this rotation so i look at the object from the back.
So basicly i want to change the viewpoint from front to back, but do that using this rotation.
How can the opposite rotation be computed?
Learning from the wikipedia page, it seems that if you want to perform a 180° rotation around the z axis, then the corresponding Quaternion rotation would simply be:
0 0 0 1
The key here is the formula , where (w,x,y,z) = (a,b,c,d).
Indeed, since cos(90°) = 0 and sin(90°) = 1, then replacing alpha with 180° and u with (0, 0, 1), gives you (0, 0, 0, 1).
Edit: As Christian has pointed out, the up direction need not be z, but may be any unit vector u = (x,y,z) (otherwise normalize it by dividing by its norm). In that case, the corresponding 180° quaterion rotation would be
0 x y z
Now to apply this rotation in order to move around the object, say you have the position an the direction vetors of your camera c_pos and c_dir, then simply (left) conjugate it by q = (0 x y z), and move the camera position accordingly. Something like
c_dir = q * c_dir * q^-1
c_pos = 2 * o_pos - c_pos
where o_pos is the position of the object, and c_dir should be converted to a quaternion with 0 real part.
In my case, hel me this..
original quat (x y z w)
opposite oriented quat (y -x w -z)