How to convert Matrix4x4 imported from .X to Unity? [duplicate] - unity3d

I would like to change a 4x4 matrix from a right handed system where:
x is left and right, y is front and back and z is up and down
to a left-handed system where:
x is left and right, z is front and back and y is up and down.
For a vector it's easy, just swap the y and z values, but how do you do it for a matrix?

Let me try to explain it a little better.
I need to export a model from Blender, in which the z axis faces up, into OpenGL, where the y axis faces up.
For every coordinate (x, y, z) it's simple; just swap the y and z values: (x, z, y).
Because I have swapped the all the y and z values, any matrix that I use also needs to be flipped so that it has the same effect.
After a lot of searching I've eventually found a solution at gamedev:
If your matrix looks like this:
{ rx, ry, rz, 0 }
{ ux, uy, uz, 0 }
{ lx, ly, lz, 0 }
{ px, py, pz, 1 }
To change it from left to right or right to left, flip it like this:
{ rx, rz, ry, 0 }
{ lx, lz, ly, 0 }
{ ux, uz, uy, 0 }
{ px, pz, py, 1 }

I think I understand your problem because I am currently facing a similar one.
You start with a world matrix which transforms a vector in a space where Z is up (e.g. a world matrix).
Now you have a space where Y is up and you want to know what to do with your old matrix.
Try this:
There is a given world matrix
Matrix world = ... //space where Z is up
This Matrix changes the Y and Z components of a Vector
Matrix mToggle_YZ = new Matrix(
{1, 0, 0, 0}
{0, 0, 1, 0}
{0, 1, 0, 0}
{0, 0, 0, 1})
You are searching for this:
//same world transformation in a space where Y is up
Matrix world2 = mToggle_YZ * world * mToggle_YZ;
The result is the same matrix cmann posted below. But I think this is more understandable as it combines the following calculation:
1) Switch Y and Z
2) Do the old transformation
3) Switch back Z and Y

It is often the case that you want to change a matrix from one set of forward/right/up conventions to another set of forward/right/up conventions. For example, ROS uses z-up, and Unreal uses y-up. The process works whether or not you need to do a handedness-flip.
Note that the phrase "switch from right-handed to left-handed" is ambiguous. There are many left-handed forward/right/up conventions. For example: forward=z, right=x, up=y; and forward=x, right=y, up=z. You should really think of it as "how do I convert ROS' notion of forward/right/up to Unreal's notion of forward/right/up".
So, it's a straightforward job to create a matrix that converts between conventions. Let's assume we've done that and we now have
mat4x4 unrealFromRos = /* construct this by hand */;
mat4x4 rosFromUnreal = unrealFromRos.inverse();
Let's say the OP has a matrix that comes from ROS, and she wants to use it in Unreal. Her original matrix takes a ROS-style vector, does some stuff to it, and emits a ROS-style vector. She needs a matrix that takes an Unreal-style vector, does the same stuff, and emits an Unreal-style vector. That looks like this:
mat4x4 turnLeft10Degrees_ROS = ...;
mat4x4 turnLeft10Degrees_Unreal = unrealFromRos * turnLeft10Degrees_ROS * rosFromUnreal;
It should be pretty clear why this works. You take a Unreal vector, convert it to ROS-style, and now you can use the ROS-style matrix on it. That gives you a ROS vector, which you convert back to Unreal style.
Gerrit's answer is not quite fully general, because in the general case, rosFromUnreal != unrealFromRos. It's true if you're just inverting a single axis, but not true if you're doing something like converting X→Y, Y→Z, Z→X. I've found that it's less error-prone to always use a matrix and its inverse to do these convention switches, rather than to try to write special functions that flip just the right members.
This kind of matrix operation M * X * inverse(M) comes up a lot. You can think of it as a "change of basis" operation; to learn more about it, see https://en.wikipedia.org/wiki/Matrix_similarity.

I have been working on converting the Unity SteamVR_Utils.RigidTransform to ROS geometry_msgs/Pose and needed to convert Unity left handed coordinate system to the ROS right handed coordinate system.
This was the code I ended up writing to convert coordinate systems.
var device = SteamVR_Controller.Input(index);
// Modify the unity controller to be in the same coordinate system as ROS.
Vector3 ros_position = new Vector3(
device.transform.pos.z,
-1 * device.transform.pos.x,
device.transform.pos.y);
Quaternion ros_orientation = new Quaternion(
-1 * device.transform.rot.z,
device.transform.rot.x,
-1 * device.transform.rot.y,
device.transform.rot.w);
Originally I tried using the matrix example from #bleater, but I couldn't seem to get it to work. Would love to know if I made a mistake somewhere.
HmdMatrix44_t m = device.transform.ToHmdMatrix44();
HmdMatrix44_t m2 = new HmdMatrix44_t();
m2.m = new float[16];
// left -> right
m2.m[0] = m.m[0]; m2.m[1] = m.m[2]; m2.m[2] = m.m[1]; m2.m[3] = m.m[3];
m2.m[4] = m.m[8]; m2.m[5] = m.m[10]; m2.m[6] = m.m[9]; m2.m[7] = m.m[7];
m2.m[8] = m.m[4]; m2.m[9] = m.m[6]; m2.m[10] = m.m[5]; m2.m[11] = m.m[11];
m2.m[12] = m.m[12]; m2.m[13] = m.m[14]; m2.m[14] = m.m[13]; m2.m[15] = m.m[15];
SteamVR_Utils.RigidTransform rt = new SteamVR_Utils.RigidTransform(m2);
Vector3 ros_position = new Vector3(
rt.pos.x,
rt.pos.y,
rt.pos.z);
Quaternion ros_orientation = new Quaternion(
rt.rot.x,
rt.rot.y,
rt.rot.z,
rt.rot.w);

After 12 years, the question is still misleading because of the lack of description of axis direction.
What question asked for should probably be how to convert to .
The answer by #cmann is correct for the above question and #Gerrit explains the reason. And I will explain how to graphically get that conversion on the transform matrix.
We should be clear that orthogonal matrix contains both rotation matrix and point reflection(only point reflection will change the coordinate system between left-handed and right-handed). Thus they can be expressed as a 4x4 matrix and obey to transform matrix multiplying order. "The matrix of a composite transformation is obtained by multiplying the matrices of individual transformations."
to contains both rotation matrix and point reflection. But we can get the composite transformation graphically.
According to above image, after transformation, in RhC(Right-handedCorrdinate) will be in LfC as below
where is a transform bring points expressed in above RhC to points expressed in LhC.
Now We are able to convert () to () accroding to transform matrix multiplying order as below image.
The result is the same as #cmann's.
Result:

It depends if you transform your points by multiplying the matrix from the left or from the right.
If you multiply from the left (e.g: Ax = x', where A is a matrix and x' the transformed point), you just need to swap the second and third column.
If you multiply from the right (e.g: xA = x'), you need to swap the second and third row.
If your points are column vectors then you're in the first scenario.

Change sin factor to -sin for swaping coordinate spaces between right and left handed

Since this seems like a homework answer; i'll give you a start at a hint: What can you do to make the determinant of the matrix negative?
Further (better hint): Since you already know how to do that transformation with individual vectors, don't you think you'd be able to do it with the basis vectors that span the transformation the matrix represents? (Remember that a matrix can be viewed as a linear transormation performed on a tuple of unit vectors)

Related

Flip a matrix for scatter plot

I want to flip an "arc" to the right side. I have tried imrotate but it gives me "arc" facing down instead of right side as shown in figure below. Please have a look at the code I am using. Thank you in advance.
R = 8; x_c = 5; y_c = 8;
thetas = 0:pi/499:pi;
xs = x_c + R*cos(thetas);
ys = y_c + R*sin(thetas);
% Now add some random noise to make the problem a bit more challenging:
mult = 0.5;
xs = xs+mult*randn(size(xs));
ys = ys+mult*randn(size(ys));
c = linspace(1,50,500);
D = [xs' ys'];
Dx = imrotate(D, 180, 'crop');
Dy=Dx;
Dy = imrotate(Dy, 180, 'crop') ;
subplot(211)
scatter(Dx(:,1), Dx(:,2), 140, c, 'filled', 'LineWidth',1.5)
subplot(212)
scatter(Dy(:,1), Dy(:,2),[],c, 'd','LineWidth',1.5)
You want to reflect the distribution for Dx. The operation of reflection means take x and make it -x.
Here the x-axis for Dx is Dx(:,1), so at first we need to write Dx(:,1) = - Dx(:,1).
When you do so and plot you'll see that the values of your x-axis has shifted to negative values. Maybe that is good enough for your purpose and that's it.
If I understand you correctly, this is not enough. Now, in order to bring this distribution back to positive x-axis value it needs to be translated.
This means:
Dx(:,1) = - Dx(:,1) + some number that translated x-axis to positive values.
you can choose that number by first taking the average (or center of mass) of the distribution, i.e. mean(Dx(:,1)), this is the value around which your value are distributed. If you only subtract the mean from the distribution you end up with values around zero, to bring it to the same distance on the positive side you need to subtract another time that mean.
Dx(:,1) = - Dx(:,1) + 2 * mean(Dx(:,1))
Reading this line means that the mean(Dx) is the calculated before the reflection so it has a positive value...
Another solution would be to set the 'XDir' property of the axes to 'reverse':
set(gca,'XDir','reverse')
This flips the axes, so it increases to the left instead of to the right. The data plotted is still identical, it is just shown differently. This might or might not be what you're after.
fliplr can work, but not with your Dx array, as it is. fliplr operates on the data, not the image, so with an array of size 500 x 2, the function flips x-values for y-values, and vice versa, which is why you get a rotation instead of a reflection.
One way to get fliplr to work is to capture the image data after you plot it, then flip it:
scatter(Dx(:,1), Dx(:,2), 140, c, 'filled', 'LineWidth',1.5)
f = getframe;
flipped = fliplr(f.cdata);
imshow(flipped)

Convert an Array from Spherical to Cartesian Coordinates in MATLAB

I am working in MATLAB with a formula which indexes points on a unit sphere in spherical coordinates.
[i, j] = ndgrid(1:N_theta, 1:N_phi);
theta = (i-1)*2*pi/N_theta;
phi = (j)*pi/(N_phi+1);
r = 1;
b = [theta(:).'; phi(:).'; r * ones(1,numel(theta))];
Let's assume I choose particular values for N_theta and N_phi and that each point has a position vector in spherical coordinates, where the first component is theta, the second component is phi and the third component is r. Running the formula then creates an array (I've called it b) which takes the position vector for each of the N points and slots them all next to each other to make a 3xN matrix.
I essentially just need to take that array and convert it so it's the same array with the vectors all next to each other but now the position vectors are in Cartesian coordinates (we could call the new array B).
I have looked up the sph2cart function in MATLAB which is designed for that purpose but I'm not sure if I am using it correctly and am hoping someone could point it what I am doing wrong. I have tried this, for example
B=sph2cart(b(1,:),b(2,:),b(3,:));
and
B = sph2cart(theta,phi,r);
but they both create matrices which are too small, so something is obviously going wrong.

Decomposing rotation matrix (x,y',z'') - Cartesian angles

Decomposing rotation matrix (x,y',z'') - Cartesian angles
Im currently working with rotation matrices and I have the following problem:
Given three coordinate systems (O0,x0,y0,z0; O1,x1,y1,z1; O2,x2,y2,z2) which coincide. We rotate first the frame #1 with the respect to frame #0, then the frame #2 with respect to frame #1.
The order of the rotations: R = Rx_alpha * Ry_beta * Rz_gamma, so first about x, then y', then z'', which are also known as the Cartesian angles.
If R1 stands for the 1st and R2 for the 2nd rotation, we are looking for the angles of the 2nd frame with respect to initial frame (#0) after both of the rotations. This can be done by decomposing the rotation matrix R (where:R = R1*R2 ). There are many literature available, how it can be done by Euler- and RPY-angles, but I don't find any, how to solve this problem in case of Cartesian angles.
I have a matlab function which works only by simple rotations. If all the angles have values different than 0 (example below), then the result becomes really unstable.
Orientation of the 1st frame with respect to the frame #0:
alpha1 = 30*pi/180;
beta1 = 10*pi/180;
gamma1 = 0*pi/180;
Orientation of the 2nd frame with respect to the frame #1
alpha2 = 10*pi/180;
beta2 = 10*pi/180;
gamma2 = 0*pi/180;
The matlab function I was using for solving the problem:
function [q] = cartesian_angles(R)
beta = asin(R(1,3));
*% Catching the numerical singularty*
if abs(abs(beta)-pi/2) > eps;
*% singulartiy of acos*
gamma1 = acos(R(1,1) / cos(beta));
gamma2 = asin(-R(1,2) / cos(beta));
if gamma2<0
gamma=2*pi-gamma1;
else
gamma=gamma1;
end
alpha1 = acos(R(3,3) / cos(beta));
alpha2 = asin(-R(2,3) / cos(beta));
if alpha2<0
alpha = 2*pi-alpha1;
else
alpha = alpha1;
end
else
fprintf('beta=pi/2 \n')
gamma = 0;
alpha = 0;
beta = 0;
end;
alpha = alpha*180/pi;
beta = beta*180/pi;
gamma = gamma*180/pi;
q = [alpha; beta; gamma];
Thank you for any help! If you have some questions don't hesitate to ask!
Marci
First, I'm going to assume you are passing into your function a well conditioned, right-handed rotation matrix. I'm going to use the same rotation sequence as you listed above, X Y' Z''
If you know the symbolic construction of the rotation matrix you are trying to extract angles from, the math is pretty straight forward. Below is an example of matlab code to determine the construction of the rotation matrix of order X-Y'-Z''
a = sym('a');%x
b = sym('b');%y
g = sym('g');%z
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)];
Rz = [cos(g) -sin(g) 0;sin(g) cos(g) 0;0 0 1];
R = Rz*Ry*Rx
The output looks like this:
R =
[ cos(b)*cos(g), cos(g)*sin(a)*sin(b) - cos(a)*sin(g), sin(a)*sin(g) + cos(a)*cos(g)*sin(b)]
[ cos(b)*sin(g), cos(a)*cos(g) + sin(a)*sin(b)*sin(g), cos(a)*sin(b)*sin(g) - cos(g)*sin(a)]
[ -sin(b), cos(b)*sin(a), cos(a)*cos(b)]
Here's the same result in a nicer looking format:
Now let's go over the math to extract the angles from this matrix. Now would be a good time to become comfortable with the atan2() function.
First solve for the beta angle (by the way, alpha is the rotation about the X axis, beta is the rotation about Y' axis, and gamma is the angle about the Z'' axis):
beta = atan2(-1*R(3,1),sqrt(R(1,1)^2+R(2,1)^2))
Written more formally,
Now that we have solved for the beta angle we can solve more simply for the other two angles:
alpha = atan2(R(3,2)/cos(beta),R(3,3)/cos(beta))
gamma = atan2(R(2,1)/cos(beta),R(1,1)/cos(beta))
Simplified and in a nicer format,
The above method is a pretty robust way of getting the Euler angles out of your rotation matrix. The atan2 function really makes it much simpler.
Finally I will answer how to solve for the rotation angles after a series of rotations. First consider the following notation. A vector or rotation matrix will be notated in the following way:
Here "U" represents the universal frame, or global coordinate system. "Fn" represents the nth local coordinate system that is different from U. R means rotation matrix (this notation could also be used for homogeneous transformations). The left side superscript will always represent the parent frame of reference of the rotation matrix or vector. The left side subscript indicates the child frame of reference. For example, if I have a vector in F1 and I want to know what it is equivalently in the universal frame of reference I would perform the following operation:
To get the vector resolved in the universal frame I simply multiplied it by the rotation matrix that transforms things from F1 to U. Notice how the subscripts are "cancelled" out by the superscript of the next item in the equation. This is a clever notation to help someone from getting things mixed up. If you recall, a special property of well conditioned rotation matrices is that the inverse matrix is the transpose of the matrix, which is will also be the inverse transformation like this:
Now that the notation details are out of the way, we can start to consider solving for complicated series of rotations. Lets say I have "n" number of coordinate frames (another way of saying "n" distinct rotations). To figure out a vector in the "nth" frame in the universal frame I would do the following:
To determine the Cardan/Euler angles that result from "n" rotations, you already know how to decompose the matrix to get the correct angles (also known as inverse kinematics in some fields), you simply need the correct matrix. In this example I am interested in the rotation matrix that takes things in the "nth" coordinate frame and resolves them into the Universal frame U:
There is it, I combined all the rotations into the one of interest simply by multiplying in the correct order. This example was easy. More complicated cases come when someone wants to find the reference frame of one rigid body resolved in the frame of another and the only thing the two rigid bodies have in common is their measurement in a universal frame.
I want to also note that this notation and method can also be used with homogeneous transformations but with some key differences. The inverse of a rotation matrix is its transpose, this is not true for homogeneous transformations.
Thank you for you answer willpower2727, your answer was really helpful!
But I would like to mention, that the code you have shown is useful to decompose rotational matrices, which are built in the following way:
R = Rz*Ry*Rx
What I'm looking for:
R = Rx*Ry*Rz
Which results into the following rotational matrix:
However, it's not a problem, as following the method how you calculate the angles alpha, beta and gamma, it was easy to modify the code so it decomposes the matrix shown above.
The angles:
beta = atan2( R(1,3), sqrt(R(1,1)^2+(-R(1,2))^2) )
alpha = atan2( -(R(2,3)/cos(beta)),R(3,3)/cos(beta) )
gamma = atan2( -(R(1,2)/cos(beta)),R(1,1)/cos(beta) )
One thing is still not clear though. The method is perfectly useful, bunt only if I calculate the angles after one rotation. As there are more rotations linked after each other, the results are false. However, it's still solvable, I guess, considering the following way: Let's say, we have two rotations linked after each other (R1 and R2). q1 shows the angles of R1, q2 of R2. after decomposing the single matrices. The total angle of rotation of the matrix R=R1*R2can be easily calculated through summing up the rangles before: q=q1+q2
Is there no way, how to calculate the angles of the total rotation, not by summing the partial angles, but decomposing the matrix R=R1*R2?
UPDATE:
Considering the following basic example. The are to rotations linked after each other:
a1 = 10*pi/180
b1 = 20*pi/180
g1 = 40*pi/180
R1 = Rx_a1*Ry_b1_Rz_g1
a2 = 20*pi/180
b2 = 30*pi/180
g2 = 30*pi/180
R2 = Rx_a2*Ry_b2*Rz_g2
Decomposing the individual matrices R1 and R2 results in the rights angles. The problem occures, when I link the rotations after each other and I try to determinate the angles of the last frame in the inertial frame. Theoretically this could be done by decomposing the product of all rotational matrices of the chain of transformations.
R = R1*R2
Decomposing this matrix gives the following false result shown in degrees:
a = 0.5645
b = 54.8024
g = 61.4240
Marci

Change from one cartesian 3D co-ordinate system to another by translation and rotation

There are two reasons for me to ask this question:
I want to know if my understanding on this issue is correct.
To clarify a doubt I have.
I want to change the co-ordinate system of a set of points (Old cartesian coordinates system to New cartesian co-ordinate system). This transformation will involve Translation as well as Rotation. This is what I plan to do:
With respect to this image I have a set of points which are in the XYZ coordinate system (Red). I want to change it with respect to the axes UVW (Purple). In order to do so, I have understood that there are two steps involved: Translation and Rotation.
When I translate, I only change the origin. (say, I want the UVW origin at (5,6,7). Then, for all points in my data, the x co-ordinates will be subtracted by 5, y by 6 and z by 7. By doing so. I get a set of Translated data.)
Now I have to apply a rotation transform (on the Translated data). The Rotation matrix is shown in the image. The values Ux, Uy and Uz are the co-ordinates of a point on the U axis which has unit distance from origin. Similarly, the values Vx, Vy and Vz are the coordinates of a point on the V axis which has a unit distance from origin. (I want to know if I am right here.) Wx, Wy, Wz is calculated as ((normalized u) X (normalised v))
(Also, if it serves any purpose, I would like to let you know that I am using MATLAB.)
edit:
I have a set of 42 points in 3D (42 X 3 matrix A) I want the first point to be considered as origin of UVW plane. So the values of the first point will be my translation vector. Correct?
Next, to calculate the Rotation vector: According to my requirement, the 6th row of matrix A has to be the U axis while 37th row has to be V axis. Consequently, vector u will be (1st row minus 6th row) of matrix A. While vector v will be (1st row minus 37th row) of matrix A.
The first row of Rotation Matrix will be vector u/|u| (normalized). Second row will be vector v/|v| (v normalized). The third row will be (u X v) . Am I right here?
Given this information, how can I calculate the value of Wx, Wy and Wx. How can I calculate the 3rd row of rotation matrix R?
Since you already have U and V, the two basis vectors of the orthonormal UVW system, the W basis vector would be the cross product of U and V. The cross product gives out the vector that is perpendicular to its operands; hence W = U × V. The components of W would fill in the third row of the rotation matrix.
Is my approach correct?
The order of the transforms matter; changing the order would lead to different results. When doing transformations of systems, usually scaling and rotation are tackled first and translation is dealt with lastly. The reason for this is that rotation would always be with respect to the origin. If the new system isn't on the old one's origin then applying rotation would rotate the new system not around its own origin but around the old system's origin. See the rightside case of figure 3-4 on this page to understand the difference what would happen if it's not on the origin; imagine the pot as the UVW coordinate system.
Think of both the coordinate systems being super-imposed (laid one atop the other). Now when you rotate UVW system with respect to the origin of XYZ, you will end up with the effect of rotating UVW w.r.t its own origin. Once rightly oriented, you can apply translation to it. However, if you'd already translated, then rotating would lead to translated rotation.
If you're using column-vector convention then TR would be the order i.e. rotation followed by translation. If you're using the row-vector convention then RT would be the order, again the order is rotation followed by translation.
You can apply the cross product of the Vectors OU and OV.
I think it's easier to perform it in steps. 1) Translation. 2) Rotation about x-axis. 3) Rotation about y-axis. 4) Rotation about z-axis.
% Assuming this is your coordinates before any operation
x0 = 5; y0 = 5; z0 = 5;
% This is the new origin
u = 5; v = 6, w = 7;
% If you wish to rotate pi/4 about x-axis, pi/3 about y-axis, pi/2 about z- axis, the three representative rotation matrix will be:
rx = [1 0 0; 0 cos(pi/4) -sin(pi/4); 0 sin(pi/4) cos(pi/4)];
ry = [cos(pi/3) 0 sin(pi/3); 0 1 0; -sin(pi/3) 0 cos(pi/3)];
rz = [cos(pi/2) -sin(pi/2) 0; sin(pi/2) cos(pi/2) 0; 0 0 1];
% First perform translation
xT = x0-u; yT = y0-v; zT = z0-w;
% Then perform rotation about x
rotated_x = mtimes( rx,[xT;yT;zT]);
% Then perform rotation about y
rotated_xy = mtimes( ry, rotated_x);
% Then perform rotation about z
rotated_xyz = mtimes( rz, rotated_xy);

obtaining the rotation and size of a UIImageView based on its transformation matrices

If I have the original transform matrix of a rectangular UIImageView and this image is scaled and rotated and by the end I can read the final transform matrix of this same view, how can I calculate how much the image scaled and rotated?
I suppose that somehow these matrix contain these two informations. The problem is how to extract it...
any clues?
thanks for any help.
A little bit of matrix algebra and trigonometric identities can help you solve this.
We'll work forward to generate a matrix that scales and rotates, and then use that to figure out how to extract the scale factors and rotations analytically.
A scaling matrix to scale by Sx (in the X axis) and Sy (in the Y axis) looks like this:
⎡Sx 0 ⎤
⎣0 Sy⎦
A matrix to rotate clockwise by R radians looks like this:
⎡cos(R) sin(R)⎤
⎣-sin(R) cos(R)⎦
Using standard matrix multiplication, the combined scaling and rotation matrix will look like this:
⎡Sx.cos(R) Sx.sin(R)⎤
⎣-Sy.sin(R) Sy.cos(R)⎦
Note that linear transformations could also include shearing or other transformations, but I'll assume for this question that only rotation and scaling have occurred (if a shear transform is in the matrix, you will get inconsistent results from following the algebra here; but the same approach can be used to determine an analytical solution).
A CGAffineTransform has four members a, b, c, d, corresponding to the 2-dimensional matrix:
⎡a b⎤
⎣c d⎦
Now we want to extract from this matrix the values of Sx, Sy, and R. We can use a simple trigonometric identity here:
tan(A) = sin(A) / cos(A)
We can use this with the first row of the matrix to conclude that:
tan(R) = Sx.sin(R) / Sx.cos(R) = b / a and therefore R = atan(b / a)
And now we know R, we can extract the scale factors by using the main diagonal:
a = Sx.cos(R) and therefore Sx = a / cos(R)
d = Sy.cos(R) and therefore Sy = d / cos(R)
So you now know Sx, Sy, and R.