Rotating a coordinate with a rotation matrix

So I'm working with a rotation matrix, basically trying to simulate

Where and are coordinates. H is a transformation matrix such as rotation

rot = [cosd(5),sind(5);-sind(5),cosd(5)];

Now, according to the equation, multiplying the transformation matrix with a coordinate would result in a coordinate but if is [9,1] for example, if i multiply with the rotation matrix.

test_coor = [9;1];
h = [cosd(5),sind(5);-sind(5),cosd(5)];
out = test_coor * h;

I would get:

out =

    9.0529
    0.2118

But it's in decimals. So do i have to round it up the values after? Or am i supposed to do something with it to get ?

[Bruno Luong]

Why you want to round up? The coordinates are real numbers, that might contain fractional part.

[Stewart Tan]

@Bruno Luong because the equation is saying, the transformation applied to would become . I have with me, say [9,0] hence the equation would allow me to check if after being rotated would be equal to [9,0] for example. Hence i wonder if the decimals should be kept.

[Bruno Luong]

So you want to find out the (xi,yi) such that

H*[xi,yi] = [9;1] % 1 or 0, make your own mind

?

In that case

H = [cosd(5),sind(5);-sind(5),cosd(5)];
xy_j = [9;1];
xy_i = H'xy_j

you can then check

H*xy_i

Answer 1

[Jim Riggs]

There seems to be some confusion regarding what the coordinate rotation transform is doing.

One way to think about it is that it expresses the coordinates of a point given in one reference frame in terms of some other frame. For a pure rotation, the frames are co-located, i.e. the point [0,0] is the same point in both frames.

Consider the point (in blue) in the figure, below. I have superimposed two different, co-located reference frames in the figure, one in black with the i subscript, and another one in red with a j subscript. The "j" frame is shown rotated through a positive angle "a" relative to the "i" frame.

You can think of the point as existing on it's own, appart from any reference frame. Then when we impose a reference frame, we can now describe the position of that point in that frame. Changing from the black frame to the red frame does not changethe position of the blue point, it only changes the way I observe it.

In the above figure, the coordinates of the point in the "i" frame (on the black grid) are [4; 2]. The figure implies a rotation of +13.6441 degrees, such that the coordinates of the point in the "j" frame (on the red grid) are [4.3589, 1.0000].

The rotation of a point in the i frame, expressed in the j frame is computed by

h = [cosd(a), sind(a) ; -sind(a), cosd(a)]

[ xj ; yj ] = h * [ xi ; yi ]

so, for [ xi; yi ] = [ 4; 2 ] (or [ 4.0000; 2.0000 ] ) and a = +13.6441 degrees,

[ xj; yj ] = [4.3589, 1.0000]

We are simply describing the point using 2 different sets of axes, which are related by the angle a.

Bellow is an illustration of your values;

[ xi, yi ] = [ 9; 1 ] and a = 5 degrees.

From the figure you can see that in the j frame, the point is just a bit more than 9, and a little above zero.

[Stewart Tan]

Thank you for the detailed explanation. If i were to have and initially instead of the angle a, how do i find the angle a in this case?

[Jim Riggs]

First, compute the sine and cosine of the angle. Then compute the angle using the atan2 function:

sa = (xi*yj - xj*yi) / (-xi^2 - yi^2);  % sine of a
ca = (yi*yj + xi*xj) / (xi^2 + yi^2);   % cosine of a
a = atan2(sa,ca);   % Angle a

DERIVATION

[Jim Riggs]

...and I forgot to mention; both sets of points should be the same distance from the origin; i.e. xi^2 + yi^2 = xj^2 + yj^2. If this is not true, then the calculated angle is not valid, because these two points are not on the same circle. So, beforce calculating the angle, make sure that the point coordinates are a valid case.

[Stewart Tan]

@Jim Riggs thank you so much! Very clear explanation. Cheers!

[Bruno Luong]

Remarks: no need to divide by denominator

a = atan2(xi*yj - yi*xj, xi*xj + yi*yj)

This 2D formula is the "z-oriented version" (counter-clockwise convention) given many time in 3D for angle between two 3D vectors v1 and v2:

atan2(norm(cross(v1,v2)), dot(v1,v2))

in 2D it's also given here

Answer 2

[Timothy Simon Thomas]

Rotation Matrix acting on a Vector

Parameters

Theta holds the angle to be rotated by, vector is the initial vector.

vector=[1;6]
theta=270;

Plot Initial Vector

plot([0 vector(1)],[0 vector(2)],'r--^','LineWidth',2);
title("Vector Rotation","BackgroundColor",'y')
hold on;

Rotation Matrix

2D-rotation matrix on the X-Y plane.

Rot_by_theta=[cosd(theta) -sind(theta) ; sind(theta) cosd(theta)]

Rotate

Multiply the initial vector with the rotation matrix to get the rotated vector.

rotated_vector=(Rot_by_theta*vector);

Plot Rotated Vector

Insert 0 at the begining to visualize the vector

plot([0 rotated_vector(1)],[0 rotated_vector(2)],'m-o','LineWidth',2);
legend("Ini_vector","Rotated")

Aesthetics

Draws X-Y axes, sets limts between -10 and 10 on both axes

grid on
xlim([-10 10])
ylim([-10 10])
line([xlim],[0 0]);
line([0 0],[ylim]);
hold off