I have two 4x4 transformation matrices that define two coordinate spaces. I am trying to express a translate transformation in one coordinate space in terms of the other. Can anyone point me in the right direction?
Let A be the matrix from space a to global space G.
Let B be the matrix from space b to global space G.
Then to get from a to b we first transform with A to get into G and then with B^-1 (the inverse of B) to get from G to b. Hence
v_b = B^-1 · A ·v_a
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I am trying to workout the interception of two objects and try to code it into an application
A cannon is in Position A and a plane is in position B
The plane is moving with vector b (without affected by gravity)
A cannonball is shot (affected by gravity) with unit vector a and magnitude of m
They intercept at position C at T seconds
Knowns: A, m, B, b
Unknowns: a, C, T
The only thing I can think of to solve it in terms of code is split the equations into X, Y and Z component and substitute T as a value and increment it.
It will be nice if someone can kindly tell how to find one of the unknowns
Thanks
The answer is in here:
https://blog.forrestthewoods.com/solving-ballistic-trajectories-b0165523348c
from meowgoesthedog
I am not sure if a question like this was asked before but i searched and didn't found what i am looking for.
I know how to determine if a point is to the left or right of a 2D line. but suppose we have a vector in 3D. of course a 3D vector passes through infinite planes, but suppose we chose one plane of them in which we are interested, and we have a specific point on this plane which we want to know if it lies to the left or right or on our vector (with respect to the chosen plane). how to do this ?
You should explicitly define orientation of that plane - for example, define main (forward) normal N - like OZ axis is normal for OXY plane.
If you have A,B,C triangle and claim that it is oriented counterclockwise, you can calculate forward plane normal as N = AB x BC
For points A, B, D in given plane calculate mixed product (vector product of AB and AD, then scalar product of result and N)
mp = (AB x AD) . dot. N
Sign of this value is positive, if vectors AB, AD, N form right-handed triplet and D lies left to AB direction
An intuitive solution is to define a coordinate system for the plane as follows. Let's normalize the 3d vector in your question and call the resulting unit vector v, and let x be a point on your plane, whose unit normal we will denote as n. You can now chose a coordinate system centered at x, that is made by the three 3*1 unit vectors v, n and b=v.crossProduct(n).
The idea is that if you express a point in this coordinate system, then if its b coordinate is negative, you can says that it is, say, on the left. So, if its b coordinate is positive, it will be on the right.
Obviously, if you have a point q expressed in this coordinates system, you can write its expression q_w in world coordinates using
q_w=R*q+x
where the rotation matrix R is the matrix whose columns are the unit axes of the plane coordinate system:
R=[v n b]
So, if you have a point Q in world coordinates, using the inverse of the relation above, you compute transpose(R)*(Q-x), and look at whether the b coordinate is positive or negative.
Say I have 2 3D vectors, each describing a direction in 3D space (or a rotation, but I'm not sure if that terminology is correct). How would I calculate the difference between the two vectors as an Euler angle? That is, if I applied the angle to the first vector, it would rotate to equal the other? I understand how Euler angles have issues and are implementation-dependant, but I don't know what the implications of this are on a question such as mine.
To clarify a little, when I say "3D vectors", I'm picturing the "translation" gizmo you get in most 3D modelling packages or in Unity (which is what I'm using).
EDIT: Actually I just reviewed the "vectors" that I'm using, and what I said is not quite correct. I actually have 6 vectors, 3 for each rotation. Each vector is a position in 3D space offset from the centre of rotation. This probably makes an already-difficult question near-impossible, right?
ADDITIONAL INFORMATION: Ok, so I've worked out what I actually want to ask (because this question is really badly done), and it applies more to Unity, so I've asked a more Unity-specific question over on Unity Answers.
I'm don't normally understand mathematical formulae that are posted online, so C++-styled pseudo-code would be by far the most helpful for me.
Any help would be much appreciated, and if my question lacks certain information, please just ask for more :)
If you are going to do 3d you need to understand linear algebra and matrix notation.[1] Affine 4x4 matrices are the basis of space transformations in all 3d applications I've ever seen, (Euler angles just give alternate means to describe that matrix). Even unity uses matrices, although to be able to efficiently do the Euler-Lagrange particle motion equation they prefer to have the decomposed form. A matrix encodes a entire space with 4 vectors. This is conceptually easy in this case (not the only use for matrices), the matrix encodes the directions x y z and offset vector w.
The reason matrix notation is useful is: It becomes possible to manipulate the things like normal math symbols. If you remember from school solving x from:
a * x = b
Divide both sides by a and you get
a/a * x = b /a ->
x = b / a
Now if you have 2 spaces with 3 vectors each you essentially have 2 fully formed spaces at origin. Assuming the vectors span a 3D space (in other words dont point all in one plane, its even better if they are orthogonal to each other in which case you can just use transformation functions directly). That means you have 3 spaces. So your problem is given you know 2 spaces. You need to know the space transform form space A -> space B (its customary to give matrices big letters to denote they are more complex). This is mathematically:
A * X = B
Where * is a matrix multiplications and A, X and B are transformation matrices. So then divide by A, but alas there's no matrix division, fortunately there is inverse and division is multiplication by inverse so that's what we do instead. Also one other note rotations are not commutative so we need to specify on which side we multiply so because A is before X we multiply on the left hand side with inverse of A. So we get:
A^-1 * A * X = A^-1 * B
Where ^-1 denotes matrix inverse. This simplifies to :
I * X = A^-1 * B ->
X = A^-1 * B
X is the rotation space. In unity code this looks like:
X = A.inverse * B
Where X, A and B are Matrix4x4 elements The language you use may have other conventions I'm using the java script reference here. You can covert this matrix to a quaternion and from there to Euler angles an example of this can be found here.
How to form A and B from the vectors? Just put the vector for starting space to A's columns 0-2 and destination spaces correspondingly to B columns[2].
[1] Yes its compulsory, its much simpler than it may seem at first. While you can live quite far without them they aren't any harder to use than saying rotate about x axis fro so and so. Also learn quats.
[2] I should check this, but unity seems to use column matrices so it should be right
PS: By the way if you have noisy data and more then 3 vectors per instance then you can use least squares to average the matrix t a 3 by 3 sub matrix.
I'm doing something where I have a plane in a coord sys A with a set of points already on it. I also have a normal vector in space N. How can I rotate the points on coord sys A so that the underlying plane will have the same normal direction as N?
Wondering if any one has a good idea on how to do this. Thanks
If you have, or can easily compute, the normal vector to the plane that your points are currently in, I think the easiest way to do this will be to rotate around the axis common to the two planes. Here's how I'd go about it:
Let M be the vector normal to your current plane, and N be the vector normal to the plane you want to rotate into. If M == N you can stop now and leave the original points unchanged.
Calculate the rotation angle as
costheta = dot(M,N)/(norm(M)*norm(N))
Calculate the rotation axis as
axis = unitcross(M, N)
where unitcross is a function that performs the cross product and normalizes it to a unit vector, i.e. unitcross(a, b) = cross(a, b) / norm(cross(a, b)). As user1318499 pointed out in a comment, this step can cause an error if M == N, unless your implementation of unitcross returns (0,0,0) when a == b.
Compute the rotation matrix from the axis and angle as
c = costheta
s = sqrt(1-c*c)
C = 1-c
rmat = matrix([ x*x*C+c x*y*C-z*s x*z*C+y*s ],
[ y*x*C+z*s y*y*C+c y*z*C-x*s ]
[ z*x*C-y*s z*y*C+x*s z*z*C+c ])
where x, y, and z are the components of axis. This formula is described on Wikipedia.
For each point, compute its corresponding point on the new plane as
newpoint = dot(rmat, point)
where the function dot performs matrix multiplication.
This is not unique, of course; as mentioned in peterk's answer, there are an infinite number of possible rotations you could make that would transform the plane normal to M into the plane normal to N. This corresponds to the fact that, after you take the steps described above, you can then rotate the plane around N, and your points will be in different places while staying in the same plane. (In other words, each rotation you can make that satisfies your conditions corresponds to doing the procedure described above followed by another rotation around N.) But if you don't care where in the plane your points wind up, I think this rotation around the common axis is the simplest way to just get the points into the plane you want them in.
If you don't have M, but you do have the coordinates of the points in your starting plane relative to an origin in that plane, you can compute the starting normal vector from two points' positions x1 and x2 as
M = cross(x1, x2)
(you can also use unitcross here but it doesn't make any difference). If you have the points' coordinates relative to an origin that is not in the plane, you can still do it, but you'll need three points' positions:
M = cross(x3-x1, x3-x2)
A single vector (your normal - N) will not be enough. You will need another two vectors for the other two dimensions. (Imagine that your 3D space could still rotate/spin around the normal vector, and you need another 2 vectors to nail it down). Once you have the normal and another one on the plane, the 3rd one should be easy to find (left- or right-handed depending on your system).
Make sure all three are normalized (length of 1) and put them in a matrix; use that matrix to transform any point in your 3D space (use matrix multiplication). This should give you the new coordinates.
I'm thinking make a unit vector [0,0,1] and use the dot-product along two planes to find the angle of difference, and shift all your points by those angles. This is assuming you want the z-axis to align with the normal vector, else just use [1,0,0] or [0,1,0] for x and y respectively.
As far as I know, Direct3D works with an LH coordinate system right?
So how would I get position and x/y/z axis (local orientation axis) out of a LH 4x4 (world) matrix?
Thanks.
In case you don't know: LH stands for left-handed
If the 4x4 matrix is what I think it is (a homogeneous rigid body transformation matrix, same as an element of SE(3)) then it should be fairly easy to get what you want. Any rigid body transformation can be represented by a 4x4 matrix of the form
g_ab = [ R, p;
0, 1]
in block matrix notation. The ab subscript denotes that the transformation will take the coordinates of a point represented in frame b and will tell you what the coordinates are as represented in frame a. R here is a 3x3 rotation matrix and p is a vector that, when the rotation matrix is unity (no rotation) tells you the coordinates of the origin of b in frame a. Usually, however, a rotation is present, so you have to do as below.
The position of the coordinate system described by the matrix will be given by applying the transformation to the point (0,0,0). This will well you what world coordinates the point is located at. The trick is that, when dealing with SE(3), you have to add a 1 at the end of points and a 0 at the end of vectors, which makes them vectors of length 4 instead of length 3, and hence operable on by the matrix! So, to transform point (0,0,0) in your local coordinate frame to the world frame, you'd right multiply your matrix (let's call it g_SA) by the vector (0,0,0,1). To get the world coordinates of a vector (x,y,z) you multiply the matrix by (x,y,z,0). You can think of that as being because vectors are differences of points, so the 1 in the last element goes the away. So, for example, to find the representation of your local x-axis in the world coordinates, you multiply g_SA*(1,0,0,0). To find the y-axis you do g_SA*(0,1,0,0), and so on.
The best place I've seen this discussed (and where I learned it from) is A Mathematical Introduction to Robotic Manipulation by Murray, Li and Sastry and the chapter you are interested in is 2.3.1.