I am projecting a 3D matrix of density values into 3 2D planes (ZX,ZY,XY). I then rotate each projection by 3 different angles: Pzx, Pzy, Pxy using the rotation matrix below:
How do I convert these 3 separate angles so that I can apply them to a 3D transformation matrix which will rotate the 3D object about X,Y,Z (or Z,Y,X) such as the rotation matrix below:
To be clear, I do not wish to apply angles Pzx, Pzy, Pxy to the 3D object, but instead calculate what those individual rotations in 2D would translate to in 3D.
This problem yields a system of equations. Let R_3d be the rotation in 3d space, R_xy the rotation in the xy plane, and [*]_xy be the projection of * onto the xy plane. Then for any point v:
I: [R_3d v]_zx = R_zx [v]_zx
II: [R_3d v]_zy = R_zy [v]_zy
III: [R_3d v]_xy = R_xy [v]_xy
We see that every coordinate is present in two equations. Let's check the relevant equations for the x-coordinate:
a := alpha, b := beta, c := gamma
I: cos b cos c x - cos b sin c y + sin b z = sin Pzx z + cos Pzx x
III: cos b cos c x - cos b sin c y + sin b z = cos Pxy x - sin Pxy y
We see that the following relation mus hold for any v (right hand side of both equations):
sin Pzx z + cos Pzx x = cos Pxy x - sin Pxy y
Similar equations exist for the other two coordinates. Only if these conditions are met, an exact 3d rotation can exist. If I'm not mistaken, that's only the case if Pzx=Pzy=Pxy=0. In general, an approximate solution can be calculated. I would suggest a least-squares solution based on the following energy:
E(a, b, c) = Sum { for all v in data set } ( || [R_3d v]_zx - R_zx [v]_zx ||^2
+ || [R_3d v]_zy - R_zy [v]_zy ||^2
+ || [R_3d v]_xy - R_xy [v]_xy ||^2 )
And the optimal rotation parameters are:
{a, b, c}* = arg min {a, b, c} E(a,b,c)
This solution will minimize the distance of the two projections of corresponding points.
Unfortunately, the problem is not a linear least-squares problem which would be easy to solve. Instead, iterative methods can solve this problem (e.g. Levenberg–Marquardt). Look for an implementation of that algorithm in your programming language, plug in the energy and solve for the optimal rotation parameters.
Related
I have been trying to understand Jacobian Determinant.
I hope someone is able to give me a pointer.
Most material that I found on Internet didn't provide
derivation of Jacobian Determinant.
One such web site is:
http://tutorial.math.lamar.edu
(Which I find quite good, otherwise.)
I spent a lot of time trying to deepen my understanding of
Jacobian Determinant.
I played with Transformations that define uv-axes and
how integration of a function over a Region/area would work
with the Transformations.
For example, when I started with simple Transformations of:
u = ( x - y )/√2
v = ( x + y )/2√2
which is uv-axes rotated -45° from Cartesian xy-axes,
and with v-axis at 2 times the scale,
that is, v = 1 maps to 2 units length in xy-coords.
So, I say that uscale = 1, vscale = 2,
for the above transformations.
With this uv-axes, I can simplify a 10x20 rectangle Region
which is rotated at 45° from x-axis,
such that the longer dimension points at 45° from x-axis.
With such examples, I begin to develop intuition
how Jacobian Determinant works.
I understand Jacobian Determinant to be a Scaling Factor
to convert area measurement in uv-axes to xy-dimensions.
Area measurement in uv-axes is given simply by formula
Δu x Δv, where Δu = 10, Δv = 10, because vscale = 2).
Jacobian Determinant Scaling Factor = uscale x vscale
(quite intuitively).
Area in xy-dimensions = Δu x Δv x (uscale x vscale)
= 10 x 10 x 1 x 2 = 200.
Integration of volume over such a simpler uv Square,
could be easier than over the same xy Region,
appearing at an angle.
With the above initial understanding,
I am trying to work out how Jacobian Determinant is derived.
Deriving from the above Transformations formula:
dx/du = √2 / 2
dx/dv = √2
dy/du = -√2 / 2
dy/dv = √2
I can also derive from Geometry that:
dx/du = uscale cos Θ
dy/du = uscale sin Θ
dx/dv = vscale cos (90° - Θ)
dy/dv = vscale sin (90° - Θ)
I could get:
areaInXY / areaInUV = uscale x vscale
which matches my understanding.
However, Jacobian Determinant formula is:
∂(x, y) / ∂(u, v) = ∂x/∂u ∂y/∂v - ∂x/∂v ∂y/∂u
= uscale * vscale * cos 2Θ
This leaves me quite puzzled why I have the extra cos 2Θ factor
which isn't making intuitive sense -- why would the
area Scaling Factor depends on how the rectangle is rotated
and thus how uv-axes are rotated?!
Anybody can see where my reasoning went wrong above?
Let me try to explain what basically the Jacobian determinant does. This is true in general for smooth functions mapping from R^n to R^n, but for the sake of simplicity, assume we are working on R^2. Let F(x,y) a smooth R^2 to R^2 function. Then we can say that F(x,y) sends the x coordinate to f1(x,y) and the y coordiate to f2(x,y) at point (x,y). Then think about an infinitesimal rectangular area, defined by the points (x,y),(x+dx,y),(x,y+dy) and (x+dx,y+dy). Now, the area of this infinitesimal rectangle is dxdy. What happens to this rectangle when it goes through the F(x,y) transformation? We apply F(x,y) to each of the four coordinates and obtain the following points:
A:(x,y)->(f1(x,y),f2(x,y))
B:(x+dx,y) -> (f1(x+dx,y),f2(x+dx,y)) (approx.)= (f1(x,y) + (∂f1/∂x)dx,f2(x,y) + (∂f2/∂x)dx)
C:(x,y+dy) -> (f1(x,y+dy),f2(x,y+dy)) (approx.)= (f1(x,y) + (∂f1/∂y)dy,f2(x,y) + (∂f2/∂y)dy)
D:(x+dx,y+dy) -> (f1(x+dx,y+dy),f2(x+dx,y+dy)) (approx.)=(f1(x,y) + (∂f1/∂x)dx + (∂f1/∂y)dy,f2(x,y) + (∂f2/∂x)dx + (∂f2/∂y)dy)
The equalities are approximately equal and exactly hold in the limit where dx and dy goes to 0, they are the best linear approximation to the function F at new points. (We obtain these from the first order parts of the Taylor approximation of the functions f1 and f2).
If we look to the new (approximated) area under the transformation F(x,y), we see the new distance vectors between the transformed points a:
B-A:((∂f1/∂x)dx,(∂f2/∂x)dx)
C-A:((∂f1/∂y)dy,(∂f2/∂y)dy)
D-C:((∂f1/∂x)dx,(∂f2/∂x)dx)
D-B:((∂f1/∂y)dy,(∂f2/∂y)dy)
As you can see, the newly transformed infinitesimal area is a parallelogram. Let:
u=((∂f1/∂x)dx,(∂f2/∂x)dx)
v=((∂f1/∂y)dy,(∂f2/∂y)dy)
These vectors constitute the edges of our parallelogram. It can be shown with the help of the cross product between u and v, that the area of the parallelogram is:
area^2 = (u1v2 - u2v1)^2 = ((∂f1/∂x)(∂f2/∂y)dxdy - (∂f2/∂x)(∂f1/∂y)dxdy)^2
area^2 = ((∂f1/∂x)(∂f2/∂y) - (∂f2/∂x)(∂f1/∂y))^2 (dxdy)^2
area = |(∂f1/∂x)(∂f2/∂y) - (∂f2/∂x)(∂f1/∂y)|dxdy (dx and dy are positive)
area = |det([∂f1/∂x, ∂f1/∂y],[∂f2/∂x, ∂f2/∂y])|dxdy
So, the matrix we are going to take the determinant of is simply the Jacobian matrix. Like I said in the beginning, this derivation can be extended to arbitrary dimensions of n,given the coordinate transformation function F is smooth and the Jacobian matrix is hence invertible, with non-zero determinant.
A good visual explanation of this is given at: http://mathinsight.org/double_integral_change_variables_introduction
I tried using a raycasting-style function to do it but can't get any maintainable results. I'm trying to calculate the intersection between two tangents on one circle. This picture should help explain:
I've googled + searched stackoverflow about this problem but can't find anything similar to this problem. Any help?
Well, if your variables are:
C = (cx, cy) - Circle center
A = (x1, y1) - Tangent point 1
B = (x2, y2) - Tangent point 2
The lines from the circle center to the two points A and B are CA = A - C and CB = B - C respectively.
You know that a tangent is perpendicular to the line from the center. In 2D, to get a line perpendicular to a vector (x, y) you just take (y, -x) (or (-y, x))
So your two (parametric) tangent lines are:
L1(u) = A + u * (CA.y, -CA.x)
= (A.x + u * CA.y, A.y - u * CA.x)
L2(v) = B + v * (CB.y, -CB.x)
= (B.x + v * CB.y, B.x - v * CB.x)
Then to calculate the intersection of two lines you just need to use standard intersection tests.
The answer by Peter Alexander assumes that you know the center of the circle, which is not obvious from your figure http://oi54.tinypic.com/e6y62f.jpg.
Here is a solution without knowing the center:
The point C (in your figure) is the intersection of the tangent at A(x, y) with the line L perpendicular to AB, cutting AB into halves. A parametric equation for the line L can be derived as follows:
The middle point of AB is M = ((x+x2)/2, (y+y2)/2), where B(x2, y2). The vector perpendicular to AB is N = (y2-y, x-x2). The vector equation of the line L is hence
L(t) = M + t N, where t is a real number.
I have a set of points (x1, x2,..xn) that lie on the plane define by Ax+ By+Cz+d=0.
I would like to find the transformation matrix to translate and rotate to XY plane. So, the new point coordinate will be x1'=(xnew, ynew,0).
A lot of answer give quaternion, dot or cross product matrix. I am not sure which one is the correct way.
Thank you
First of all, unless in your plane equation, d=0, there is no linear transformation you can apply. You need to instead perform an affine transformation.
One way to do this is to determine an angle and vector about which to rotate to make your pointset lie in a plane parallel to the XY plane (ie. the Z component of your transformed pointset to all have the same values). Then you simply drop the Z component.
For this, let V be the normalized plane normal for the plane containing your points. For convenience define from your plane equation above Ax+By+Cz+d=0:
V = (A, B, C)
V' = V / ||V|| = (A', B', C')
Z = (0, 0, 1)
where
A' = A / ||V||
B' = B / ||V||
C' = C / ||V||
||V|| = (A2+B2+C2)1/2
The angle will simply be:
θ = cos-1(Z∙V / ||V||)
= cos-1(Z∙V')
= cos-1(C')
The axis R about which to rotate is just the cross product of the normalized plane normal V' and Z. That is
R = V'×Z
= (B', -A', 0)
You can now use this angle / axis pair to build the quaternion rotation needed to rotate all of the points in your dataset to a plane parallel to the XY plane. Then, a I said earlier, just drop the Z component to perform an orthogonal projection onto the XY plane.
Update: antonakos makes a good point about normalizing the R before using an API taking axis / angle pairs.
I have a 3D Plane defined by two 3D Vectors:
P = a Point which lies on the Plane
N = The Plane's surface Normal
And I want to calculate any vector that lies on the plane.
Take any vector, v, not parallel to N, its vector cross product with N ( w1 = v x N ) is a vector that is parallel to the plane.
You can also take w2 = v - N (v.N)/(N.N) which is the projection of v into plane.
A point in the plane can then be given by x = P + a w, In fact all points in the plane can be expressed as
x = P + a w2 + b ( w2 x N )
So long as the v from which w2 is "suitable".. cant remember the exact conditions and too lazy to work it out ;)
If you want to determine if a point lies in the plane rather than find a point in the plane, you can use
x.N = P.N
for all x in the plane.
If N = (xn, yn, zn) and P = (xp, yp, zp), then the plane's equation is given by:
(x-xp, y-yp, z-zp) * (xn, yn, zn) = 0
where (x, y, z) is any point of the plane and * denotes the inner product.
And I want to calculate any vector
that lies on the plane.
If I understand correctly You need to check if point belongs to the plane?
http://en.wikipedia.org/wiki/Plane_%28geometry%29
You mast check if this equation: nx(x − x0) + ny(y − y0) + nz(z − z0) = 0 is true for your point.
where: [nx,ny,nz] is normal vector,[x0,y0,z0] is given point, [x,y,z] is point you are checking.
//edit
Now I'm understand Your question. You need two linearly independent vectors that are the planes base. Sow You need to fallow Michael Anderson answerer but you must add second vector and use combination of that vectors. More: http://en.wikipedia.org/wiki/Basis_%28linear_algebra%29
Let's say I have two points in 3D space (a and b) and a fixed axis/unit vector called n.
I want to create a rotation matrix that minimizes the euclidan distance between point a (unrotated) and the rotated point b.
E.g:
Q := matrix_from_axis_and_angle (n, alpha);
find the unknown alpha that minimizes sqrt(|a - b*Q|)
Btw - If a solution/algorithm can be easier expressed with unit-quaternions go ahead and use them. I just used matrices to formulate my question because they're more widely used.
Oh - I know there are some degenerated cases ( a or b lying exactly in line with n ect.) These can be ignored. I'm just looking for the case where a single solution can be calculated.
sounds fairly easy. Assume unit vector n implies rotation around a line parallel to n through point x0. If x0 != the origin, translate the coordinate system by -x0 to get points a' and b' relative to new coordinate system origin 0, and use those 2 points instead of a and b.
1) calculate vector ry = n x a
2) calculate unit vector uy = unit vector in direction ry
3) calculate unit vector ux = uy x n
You now have a triplet of mutually perpendicular unit vectors ux, uy, and n, which form a right-handed coordinate system. It can be shown that:
a = dot(a,n) * n + dot(a,ux) * ux
This is because unit vector uy is parallel to ry which is perpendicular to both a and n. (from step 1)
4) Calculate components of b along unit vectors ux, uy. a's components are (ax,0) where ax = dot(a,ux). b's components are (bx,by) where bx = dot(b,ux), by = dot(b,uy). Because of the right-handed coordinate system, ax is always positive so you don't actually need to calculate it.
5) Calculate theta = atan2(by, bx).
Your rotation matrix is the one which rotates by angle -theta relative to coordinate system (ux,uy,n) around the n-axis.
This yields degenerate answers if a is parallel to n (steps 1 and 2) or if b is parallel to n (steps 4, 5).
I think you can rephrase the question to:
what is the distance from a point to a 2d circle in 3d space.
the answer can be found here
so the steps needed are as following:
rotating the point b around a vector n gives you a 2d circle in 3d space
using the above, find the distance to that circle (and the point on the circle)
the point on the circle is the rotated point b you are looking for.
deduce the rotated angle
...or something ;^)
The distance will be minimized when the vector from a to the line along n lines up with the vector from b to the line along n.
Project a and b into the plane perpendicular to n and solve the problem in 2 dimensions. The rotation you get there is the rotation you need to minimize the distance.
Let P be the plane that is perpendicular to n.
We can find the projection of a into the P-plane, (and similarly for b):
a' = a - (dot(a,n)) n
b' = b - (dot(b,n)) n
where dot(a,n) is the dot-product of a and n
a' and b' lie in the P-plane.
We've now reduced the problem to 2 dimensions. Yay!
The angle (of rotation) between a' and b' equals the angle (of rotation) needed to swing b around the n-axis so as to be closest to a. (Think about the shadows b would cast on the P-plane).
The angle between a' and b' is easy to find:
dot(a',b') = |a'| * |b'| * cos(theta)
Solve for theta.
Now you can find the rotation matrix given theta and n here:
http://en.wikipedia.org/wiki/Rotation_matrix
Jason S rightly points out that once you know theta, you must still decide to rotate b clockwise or counterclockwise about the n-axis.
The quantity, dot((a x b),n), will be a positive quantity if (a x b) lies in the same direction as n, and negative if (a x b) lies in the opposite direction. (It is never zero as long as neither a nor b is collinear with n.)
If (a x b) lies in the same direction as n, then b has to be rotated clockwise by the angle theta about the n-axis.
If (a x b) lies in the opposite direction, then b has to be rotated clockwise by the angle -theta about the n-axis.