I have two vector pairs (before and after rotation).
before rotation:
[x1,y1,z1]
[x2,y2,z2]
after rotation:
[x1',y1',z1']
[x2',y2',z2']
How to create a quaternion representing this rotation?
In most cases there is no rotation which transforms 2 vectors into 2 other vectors. Here is a simple way to visualize why: a rotation does not change the angle between vectors. If the angle between the 2 vectors before the rotation is different from the angle between the 2 vectors after the rotation, then there is no rotation which meets your criteria.
This said there may be an optimal quaternion with an acceptable error which "almost" rotates your 2 vector pairs. There are a number of algorithms which vary in speed and precision to find such a quaternion. I wrote a fast C++ algorithm for an Arduino application where the speed is critical but the precision is less important.
http://robokitchen.tumblr.com/post/67060392720/finding-a-rotation-quaternion-from-two-pairs-of-vectors
Before rotation: u0, v0. After rotation: u2, v2.
Quaternion q2 = Quaternion::fromTwoVectors(u0, u2);
Vector v1 = v2.rotate(q2.conjugate());
Vector v0_proj = v0.projectPlane(u0);
Vector v1_proj = v1.projectPlane(u0);
Quaternion q1 = Quaternion::fromTwoVectors(v0_proj, v1_proj);
return (q2 * q1).normalized();
If this does not meet the requirements of your own application try to google Wabha's problem.
Well, first you can find the rotation axis using vector-multiplication (cross-multiplication):
axis = v1 x v2;
Then you can compute the rotation angle:
sinA = |axis| / |v1|*|v2|
cosA = v1 . v2 / |v1|*|v2|
Here | | - is vector length operation, and . - is dot-multiplication
And finally, your quaternion is:
Q(w,x,y,z) = (cosA, axis.x * sinA, axis.y * sinA, axis.z * sinA)
I translated marcv81's very helpful blog post into Three.js:
const rotateVectorsSimultaneously = (u0, v0, u2, v2) => {
const q2 = new THREE.Quaternion().setFromUnitVectors(u0, u2);
const v1 = v2.clone().applyQuaternion(q2.clone().conjugate());
const v0_proj = v0.projectOnPlane(u0);
const v1_proj = v1.projectOnPlane(u0);
let angleInPlane = v0_proj.angleTo(v1_proj);
if (v1_proj.dot(new THREE.Vector3().crossVectors(u0, v0)) < 0) {
angleInPlane *= -1;
}
const q1 = new THREE.Quaternion().setFromAxisAngle(u0, angleInPlane);
const q = new THREE.Quaternion().multiplyQuaternions(q2, q1);
return q;
};
Because angleTo always returns a positive value, I manually flip the sign of the angle depending on which side of the u0-v0 plane v1 is on.
A mature solution to this problem is called Triad. Triad is one of the earliest and simplest solutions to the spacecraft attitude determination problem and is extremely efficient computationally.
With Triad, the idea is to replace your paired set of two vectors, with a paired set of three vectors, where the extra vector is generated with a cross-product. By normalizing the vectors, you can solve for a rotation matrix without a matrix inverse or an SVD (as is needed in more general instances of the problem -- see Wahba's Problem)
For full algorithm, see: https://en.wikipedia.org/wiki/Triad_method
You can then convert the solved rotation matrix from Triad to a rotation quaternion:
qw = √(1 + m00 + m11 + m22) /2
qx = (m21 - m12)/( 4 *qw)
qy = (m02 - m20)/( 4 *qw)
qz = (m10 - m01)/( 4 *qw)
In general to make the conversion to quaternion robust, you should consider looking at the matrix trace as discussed here: http://www.euclideanspace.com/maths/geometry/rotations/conversions/matrixToQuaternion/
Finally, consider an alternative to Triad that directly computes the optimal quaternion called QUEST.
It is fine to find the quaternion from v1 to v2.
The final q = (cos A/2, sin A/2 * axis), where the A is the angle between v1 and v2, axis is the normed axis.
Multiply both side by 2 * cos A/2,
Then we have
2 * cos A/2 *q = (1+cos A, sin A * axis)
(where cos A = dot(v1, v2)/|v1|/|v2| and
axis = cross(v1, v2).normalize() = cross(v1, v2)/|v1|/|v2|/sin A.)
Then 2 * cos A/2 *q = (1+dot(v1, v2)/|v1|/|v2|, cross(v1, v2)/|v1|/|v2|)
Finally q = (1+dot(v1, v2)/|v1|/|v2|, cross(v1, v2)/|v1|/|v2|).normalize()
Related
I am trying to calculate an intercepting vector based on Velocity Location and time of two objects.
I found an post covering my problem but was left over with some technical questions i could not ask because my reputation is below 50.
Calculating Intercepting Vector
The answer marked as best goes over the process of how to solve my problem, however when i tried to calculate myself, i could not understand how the vectors of position and velocity are converted to a real number.
Using the data provided here for the positions and speeds of the target and the interceptor, the solving equation is the following:
plugging in the numbers, the coefficients of the quadratic equation in t are:
s_t = [120, 40]; v_t = [5,2]; s_i = [80, 80]; v_i = 10;
a = dot(v_t, v_t)-10^2
b = 2*dot((s_t - s_i),v_t)
c = dot(s_t - s_i, s_t - s_i)
Solving for t yields:
delta = sqrt(b^2-4*a*c)
t1 = (b + sqrt(b^2 - 4*a*c))/(2*a)
t2 = (b - sqrt(b^2 - 4*a*c))/(2*a)
With the data at hand, t1 turns out to be negative, and can be discarded.
I have been trying to work out how to calculate Poincaré sections for a system of non-linear ODEs, using a paper on the exact system as reference, and have been wrestling with numpy to try and make it run better. This is intended to run within a bounded domain.
Currently, I have the following code
import numpy as np
from scipy.integrate import odeint
X = 0
Y = 1
Z = 2
def generate_poincare_map(function, initial, plane, iterations, delta):
intersections = []
p_i = odeint(function, initial.flatten(), [0, delta])[-1]
for i in range(1, iterations):
p_f = odeint(function, p_i, [i * delta, (i+1) * delta])[-1]
if (p_f[Z] > plane) and (p_i[Z] < plane):
intersections.append(p_i[:2])
if (p_f[Z] > plane) and (p_i[Z] < plane):
intersections.append(p_i[:2])
p_i = p_f
return np.stack(intersections)
This is pretty wasteful due to the integration solely between successive time steps, and seems to produce incorrect results. The original reference includes sections along the lines of
whereas mine tend to result in something along the lines of
Do you have any advice on how to proceed to make this more correct, and perhaps a little faster?
To get a Pointcaré map of the ABC flow
def ABC_ode(u,t):
A, B, C = 0.75, 1, 1 # matlab parameters
x, y, z = u
return np.array([
A*np.sin(z)+C*np.cos(y),
B*np.sin(x)+A*np.cos(z),
C*np.sin(y)+B*np.cos(x)
])
def mysolver(u0, tspan): return odeint(ABC_ode, u0, tspan, atol=1e-10, rtol=1e-11)
you have first to understand that the dynamical system is really about the points (cos(x),sin(x)) etc. on the unit circle. So values different by multiples of 2*pi represent the same point. In the computation of the section one has to reflect this, either by computing it on the Cartesian product of the 3 circles. Let's stay with the second variant, and chose [-pi,pi] as the fundamental period to have the zero location well in the center. Keep in mind that jumps larger pi are from the angle reduction, not from a real crossing of that interval.
def find_crosssections(x0,y0):
u0 = [x0,y0,0]
px = []
py = []
u = mysolver(u0, np.arange(0, 4000, 0.5)); u0 = u[-1]
u = np.mod(u+pi,2*pi)-pi
x,y,z = u.T
for k in range(len(z)-1):
if z[k]<=0 and z[k+1]>=0 and z[k+1]-z[k]<pi:
# find a more exact intersection location by linear interpolation
s = -z[k]/(z[k+1]-z[k]) # 0 = z[k] + s*(z[k+1]-z[k])
rx, ry = (1-s)*x[k]+s*x[k+1], (1-s)*y[k]+s*y[k+1]
px.append(rx);
py.append(ry);
return px,py
To get a full picture of the Poincare cross-section and avoid duplicate work, use a grid of squares and mark if one of the intersections already fell in it. Only start new iterations from the centers of free squares.
N=20
grid = np.zeros([N,N], dtype=int)
for i in range(N):
for j in range(N):
if grid[i,j]>0: continue;
x0, y0 = (2*i+1)*pi/N-pi, (2*j+1)*pi/N-pi
px, py = find_crosssections(x0,y0)
for rx,ry in zip(px,py):
m, n = int((rx+pi)*N/(2*pi)), int((ry+pi)*N/(2*pi))
grid[m,n]=1
plt.plot(px, py, '.', ms=2)
You can now play with the density of the grid and the length of the integration interval to get the plot a little more filled out, but all characteristic features are already here. But I'd recommend re-programming this in a compiled language, as the computation will take some time.
What I have tried already: d = |v||PQ|sin("Theta")
Now, I need to determine what theta is, so I set up a position on a makeshift graph, the graph I made was on the xy plane only as the z plane complicates things needlessly for finding theta. So, I ended up with an acute angle, and if the angle is acute, then I have to find theta which according to dot product facts is greater than 0.
I do not have access to theta, so I used the same princples from cross dots. u * v = |u||v|cos("theta") but in this case, u and v are PQ and v. A vector is a vector, right?
so now I have theta = acos((v*PQ)/(|v||PQ))
with that I get (4sqrt(10))/15 = 32.5125173162 in degrees, so the angle is 32.5125173162 degrees.
So, now that I have theta, I plug it into my distance formula |v||PQ|sin(32.5125173162)
3*sqrt(10)*sin(32.5125173162) = 5.0990195136
or for the sake of simplicity, 5.1
I however want to know if this question is correct.
If it is NOT correct, what can I do to correct it? At what points did I use incorrect information?
This is not a question with a definitive answer in the back of the book, its a question on the side of a page that said: "try this!"
There are a couple of problems with this question.
From the context it looks like you mean for both v and PQ to be vectors. The "distance" between two vectors is an awkward (not well defined) question because vectors are not position bound.
You are using the cross product formula and I have no idea why:
|AxB| = |A||B|Sin(theta)
I think what you are actually trying to do is calculate the distance between the terminal points of the vectors, (2, 1, 2) and (1, 0, 3). Just use the Pythagorean Theorem (extended to 3D) for this.
d = sqrt( (x1 - x2)^2 + (y1 - y2)^2 + (z1 - z2)^2 )
d = sqrt( (2 - 1)^2 + (1 - 2)^2 + (2 - 3)^2 )
d = sqrt( 1^2 + (-1)^2 + (-1)^2 )
d = sqrt(3)
Edit:
If what you need really is the magnitude of the cross product, |AxB| then just find the cross product (using the determinant) and then calculate the magnitude of the result. There is no need for the formula you were using.
I had attached also the schematic to depict my question.
I need to rotate the vector V with the base point P by an angle and find the new vector V'.
The rotation axis is say for is about a local y axis at point P (which is parallel to global Y axis)
Subsequently, I need to rotate the initial vector V about x axis which is parallel to global Y axis.
The main reason for the rotation is to find the new vector V' at point P. Both the rotations are independent and each of the rotation provides a new V'. I'm programming this in VB.net and output is a double() of new vector V'.
Just apply the two rotations independently (see Wikipedia). The base point does not play any role in this because it is just a constant offset that never changes. If I got your description right, you want the following:
//rotation about y-axis
iAfterRot1 = cos(phi1) * i + sin(phi1) * k
jAfterRot1 = j
kAfterRot1 = -sin(phi1) * i + cos(phi) * k
//rotation about x-axis
iAfterRot2 = iAfterRot1
jAfterRot2 = cos(phi2) * jAfterRot1 - sin(phi2) * kAfterRot1
kAfterRot2 = sin(phi2) * jAfterRot1 + cos(phi2) * kAfterRot1
I have a green vehicle which will shortly collide with a blue object (which is 200 away from the cube)
It has a Kinect depth camera D at [-100,0,200] which sees the corner of the cube (grey sphere)
The measured depth is 464 at 6.34° in the X plane and 12.53° in the Y plane.
I want to calculate the position of the corner as it would appear if there was a camera F at [150,0,0], which would see this:
in other words transform the red vector into the yellow vector. I know that this is achieved with a transformation matrix but I can't find out how to compute the matrix from the D-F vector [250,0,-200] or how to use it; my high-school maths dates back 40 years.
math.se has a similar question but it doesn't cover my problem and I can't find anything on robotices.se either.
I realise that I should show some code that I've tried, but I don't know where to start. I would be very grateful if somebody could help me to solve this.
ROS provides the tf library which allows you to transform between frames. You can simply set a static transform between the pose of your camera and the pose of your desired location. Then, you can get the pose of any point detected by your camera in the reference frame of your desired point on your robot. ROS tf will do everything you need and everything I explain below.
The longer answer is that you need to construct a transformation tree. First, compute the static transformation between your two poses. A pose is a 7-dimensional transformation including a translation and orientation. This is best represented as a quaternion and a 3D vector.
Now, for all poses in the reference frame of your kinect, you must transform them to your desired reference frame. Let's call this frame base_link and your camera frame camera_link.
I'm going to go ahead and decide that base_link is the parent of camera_link. Technically these transformations are bidirectional, but because you may need a transformation tree, and because ROS cares about this, you'll want to decide who is the parent.
To convert rotation from camera_link to base_link, you need to compute the rotational difference. This can be done by multiplying the quaternion of base_link's orientation by the conjugate of camera_link's orientation. Here's a super quick Python example:
def rotDiff(self,q1: Quaternion,q2: Quaternion) -> Quaternion:
"""Finds the quaternion that, when applied to q1, will rotate an element to q2"""
conjugate = Quaternion(q2.qx*-1,q2.qy*-1,q2.qz*-1,q2.qw)
return self.rotAdd(q1,conjugate)
def rotAdd(self, q1: Quaternion, q2: Quaternion) -> Quaternion:
"""Finds the quaternion that is the equivalent to the rotation caused by both input quaternions applied sequentially."""
w1 = q1.qw
w2 = q2.qw
x1 = q1.qx
x2 = q2.qx
y1 = q1.qy
y2 = q2.qy
z1 = q1.qz
z2 = q2.qz
w = w1 * w2 - x1 * x2 - y1 * y2 - z1 * z2
x = w1 * x2 + x1 * w2 + y1 * z2 - z1 * y2
y = w1 * y2 + y1 * w2 + z1 * x2 - x1 * z2
z = w1 * z2 + z1 * w2 + x1 * y2 - y1 * x2
return Quaternion(x,y,z,w)
Next, you need to add the vectors. The naive approach is to simply add the vectors, but you need to account for rotation when calculating these. What you really need is a coordinate transformation. The position of camera_link relative to base_link is some 3D vector. Based on your drawing, this is [-250, 0, 200]. Next, we need to reproject the vectors to your points of interest into the rotational frame of base_link. I.e., all the points your camera sees at 12.53 degrees that appear at the z = 0 plane to your camera are actually on a 12.53 degree plane relative to base_link and you need to find out what their coordinates are relative to your camera as if your camera was in the same orientation as base_link.
For details on the ensuing math, read this PDF (particularly starting at page 9).
To accomplish this, we need to find your vector's components in base_link's reference frame. I find that it's easiest to read if you convert the quaternion to a rotation matrix, but there is an equivalent direct approach.
To convert a quaternion to a rotation matrix:
def Quat2Mat(self, q: Quaternion) -> rotMat:
m00 = 1 - 2 * q.qy**2 - 2 * q.qz**2
m01 = 2 * q.qx * q.qy - 2 * q.qz * q.qw
m02 = 2 * q.qx * q.qz + 2 * q.qy * q.qw
m10 = 2 * q.qx * q.qy + 2 * q.qz * q.qw
m11 = 1 - 2 * q.qx**2 - 2 * q.qz**2
m12 = 2 * q.qy * q.qz - 2 * q.qx * q.qw
m20 = 2 * q.qx * q.qz - 2 * q.qy * q.qw
m21 = 2 * q.qy * q.qz + 2 * q.qx * q.qw
m22 = 1 - 2 * q.qx**2 - 2 * q.qy**2
result = [[m00,m01,m02],[m10,m11,m12],[m20,m21,m22]]
return result
Now that your rotation is represented as a rotation matrix, it's time to do the final calculation.
Following the MIT lecture notes from my link above, I'll arbitrarily name the vector to your point of interest from the camera A.
Find the rotation matrix that corresponds with the quaternion that represents the rotation between base_link and camera_link and simply perform a matrix multiplication. If you're in Python, you can use numpy to do this, but in the interest of being explicit, here is the long form of the multiplication:
def coordTransform(self, M: RotMat, A: Vector) -> Vector:
"""
M is my rotation matrix that represents the rotation between my frames
A is the vector of interest in the frame I'm rotating from
APrime is A, but in the frame I'm rotating to.
"""
APrime = []
i = 0
for component in A:
APrime.append(component * M[i][0] + component * M[i][1] + component * m[i][2])
i += 1
return APrime
Now, the vectors from camera_link are represented as if camera_link and base_link share an orientation.
Now you may simply add the static translation between camera_link and base_link (or subtract base_link -> camera_link) and the resulting vector will be your point's new translation.
Putting it all together, you can now gather the translation and orientation of every point your camera detects relative to any arbitrary reference frame to gather pose data relevant to your application.
You can put all of this together into a function simply called tf() and stack these transformations up and down a complex transformation tree. Simply add all the transformations up to a common ancestor and subtract all the transformations down to your target node in order to find the transformation of your data between any two arbitrary related frames.
Edit: Hendy pointed out that it's unclear what Quaternion() class I refer to here.
For the purposes of this answer, this is all that's necessary:
class Quaternion():
def __init__(self, qx: float, qy: float, qz: float, qw: float):
self.qx = qx
self.qy = qy
self.xz = qz
self.qw = qw
But if you want to make this class super handy, you can define __mul__(self, other: Quaternion and __rmul__(self, other: Quaternion) to perform quaternion multiplication (order matters, so make sure to do both!). conjugate(self), toEuler(self), toRotMat(self), normalize(self) may also be handy additions.
Note that due to quirks in Python's typing, the above other: Quaternion is only for clarity. You'll need a longer-form if type(other) != Quaternion: raise TypeError('You can only multiply quaternions with other quaternions) error handling block to make that into valid python :)
The following definitions are not necessary for this answer, but they may prove useful to the reader.
import numpy as np
def __mul__(self, other):
if type(other) != Quaternion:
print("Quaternion multiplication only works with other quats")
raise TypeError
r1 = self.qw
r2 = other.qw
v1 = [self.qx,self.qy,self.qz]
v2 = [other.qx,other.qy,other.qz]
rPrime = r1*r2 - np.dot(v1,v2)
vPrimeA = np.multiply(r1,v2)
vPrimeB = np.multiply(r2,v1)
vPrimeC = np.cross(v1,v2)
vPrimeD = np.add(vPrimeA, vPrimeB)
vPrime = np.add(vPrimeD,vPrimeC)
x = vPrime[0]
y = vPrime[1]
z = vPrime[2]
w = rPrime
return Quaternion(x,y,z,w)
def __rmul__(self, other):
if type(other) != Quaternion:
print("Quaternion multiplication only works with other quats")
raise TypeError
r1 = other.qw
r2 = self.qw
v1 = [other.qx,other.qy,other.qz]
v2 = [self.qx,self.qy,self.qz]
rPrime = r1*r2 - np.dot(v1,v2)
vPrimeA = np.multiply(r1,v2)
vPrimeB = np.multiply(r2,v1)
vPrimeC = np.cross(v1,v2)
vPrimeD = np.add(vPrimeA, vPrimeB)
vPrime = np.add(vPrimeD,vPrimeC)
x = vPrime[0]
y = vPrime[1]
z = vPrime[2]
w = rPrime
return Quaternion(x,y,z,w)
def conjugate(self):
return Quaternion(self.qx*-1,self.qy*-1,self.qz*-1,self.qw)