Python 3D rotation of a point - python

I have a point (x,y,z) in 3d that I would like to rotate. First I would like to rotate the point around another point (0,0,0) 360 degrees. Then I would like to change the plane that the point rotates in by 1 degree and repeat. I have been looking at the rotation_matrix function in http://www.lfd.uci.edu/~gohlke/code/transformations.py.html , however it seems as if the rotation only goes around the x,y or z axis rather than an arbitrary angle. Does anyone know how to accomplish this?

Rotating around the x, y, and z pretty much is the only way to do it. Rotating a plane is exactly like rotating around an axis. Check out my scratch project for the math.

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Arranging the points in clockwise manner viewing from from surface

I have 3d points from the surface. So corresponding to one normal vector they becomes 2d point. That normal vector is anything , standard x,y or z or any other. How can I arrange the points in clockwise direction.
I cannot find out any logic. Sorry for that. Please help

Convert Eye Gaze (Pitch and yaw) into screen coordinates (Where the person is looking at?)

I am asking this questions as a trimmed version of my previous question. Now that I have a face looking some position on screen and also gaze coordinates (pitch and yaw) of both the eye. Let us say
Left_Eye = [-0.06222888 -0.06577308]
Right_Eye = [-0.04176027 -0.44416167]
I want to identify the screen coordinates where the person probably may be looking at? Is this possible? Please help!
What you need is:
3D position and direction for each eye
you claim you got it but pitch and yaw are just Euler angles and you need also some reference frame and order of transforms to convert them back into 3D vector. Its better to leave the direction in a vector form (which I suspect you got in the first place). Along with the direction you need th position in 3D in the same coordinate system too...
3D definition of your projection plane
so you need at least start position and 2 basis vectors defining your planar rectangle. Much better is to use 4x4 homogenous transform matrix for this because that allows very easy transform from and in to its local coordinate system...
So I see it like this:
So now its just matter of finding the intersection between rays and plane
P(s) = R0 + s*R
P(t) = L0 + t*L
P(u,v) = P0 + u*U +v*V
Solving this system will lead to acquiring u,v which is also the 2D coordinate inside your plane yo are looking at. Of course because of inaccuracies this will not be solvable algebraicaly. So its better to convert the rays into plane local coordinates and just computing the point on each ray with w=0.0 (making this a simple linear equation with single unknown) and computing average position between one for left eye and the other for right eye (in case they do not align perfectly).
so If R0',R',L0',L' are the converted values in UVW local coordinates then:
R0z' + s*Rz' = 0.0
s = -R0z'/Rz'
// so...
R1 = R0' - R'*R0z'/Rz'
L1 = L0' - L'*L0z'/Lz'
P = 0.5 * (R1 + L1)
Where P is the point you are looking at in the UVW coordinates...
The conversion is done easily according to your notations you either multiply the inverse or direct matrix representing the plane by (R,1),(L,1),(R0,0)(L0,0). The forth coordinate (0,1) just tells if you are transforming vector or point.
Without knowing more about your coordinate systems, data accuracy, and what knowns and unknowns you got is hard to be more specific than this.
If your plane is the camera projection plane than U,V are the x and y axis of the image taken from camera and W is normal to it (direction is just matter of notation).
As you are using camera input which uses a perspective projection I hope your positions and vectors are corrected for it.

How to recalculate the coordinates of a point after scaling and rotation?

I have the coordinates of 6 points in an image
(170.01954650878906, 216.98866271972656)
(201.3812255859375, 109.42137145996094)
(115.70114135742188, 210.4272918701172)
(45.42426300048828, 97.89037322998047)
(167.0367889404297, 208.9329833984375)
(70.13690185546875, 140.90538024902344)
I have a point as center [89.2458, 121.0896]. I am trying to re-calculate the position of points in python using 4 rotation degree (from 0,90,-90,180) and 6 scaling factor (0.5,0.75,1,1.10,1.25,1.35,1.5).
My question is how can I rotate and scale the abovementioned points relative to the center point and get the new coordinates of those 6 points?
Your help is really appreciated.
Mathematics
A mathematical approach would be to represent this data as vectors from the center to the image-points, translate these vectors to the origin, apply the transformation and relocate them around the center point. Let's look at how this works in detail.
Representation as vectors
We can show these vectors in a grid, this will produce following image
This image provides a nice way to look at these points, so we can see our actions happening in a visual way. The center point is marked with a dot at the beginning of all the arrows, and the end of each arrow is the location of one of the points supplied in the question.
A vector can be seen as a list of the values of the coordinates of the point so
my_vector = [point[0], point[1]]
could be a representation for a vector in python, it just holds the coordinates of a point, so the format in the question could be used as is! Notice that I will use the position 0 for the x-coordinate and 1 for the y-coordinate throughout my answer.
I have only added this representation as a visual aid, we can look at any set of two points as being a vector, no calculation is needed, this is only a different way of looking at those points.
Translation to origin
The first calculations happen here. We need to translate all these vectors to the origin. We can very easily do this by subtracting the location of the center point from all the other points, for example (can be done in a simple loop):
point_origin_x = point[0] - center_point[0] # Xvalue point - Xvalue center
point_origin_y = point[1] - center_point[1] # Yvalue point - Yvalue center
The resulting points can now be rotated around the origin and scaled with respect to the origin. The new points (as vectors) look like this:
In this image, I deliberately left the scale untouched, so that it is clear that these are exactly the same vectors (arrows), in size and orientation, only shifted to be around (0, 0).
Why the origin
So why translate these points to the origin? Well, rotations and scaling actions are easy to do (mathematically) around the origin and not as easy around other points.
Also, from now on, I will only include the 1st, 2nd and 4th point in these images to save some space.
Scaling around the origin
A scaling operation is very easy around the origin. Just multiply the coordinates of the point with the factor of the scaling:
scaled_point_x = point[0] * scaling_factor
scaled_point_y = point[1] * scaling_factor
In a visual way, that looks like this (scaling all by 1.5):
Where the blue arrows are the original vectors and the red ones are the scaled vectors.
Rotating
Now for rotating. This is a little bit harder, because a rotation is most generally described by a matrix multiplication with this vector.
The matrix to multiply with is the following
(from wikipedia: Rotation Matrix)
So if V is the vector than we need to perform V_r = R(t) * V to get the rotated vector V_r. This rotation will always be counterclockwise! In order to rotate clockwise, we simply need to use R(-t).
Because only multiples of 90° are needed in the question, the matrix becomes a almost trivial. For a rotation of 90° counterclockwise, the matrix is:
Which is basically in code:
rotated_point_x = -point[1] # new x is negative of old y
rotated_point_y = point[0] # new y is old x
Again, this can be nicely shown in a visual way:
Where I have matched the colors of the vectors.
A rotation 90° clockwise will than be
rotated_counter_point_x = point[1] # x is old y
rotated_counter_point_y = -point[0] # y is negative of old x
A rotation of 180° will just be taking the negative coordinates or, you could just scale by a factor of -1, which is essentially the same.
As last point of these operations, might I add that you can scale and/or rotated as much as you want in a sequence to get the desired result.
Translating back to the center point
After the scaling actions and/or rotations the only thing left is te retranslate the vectors to the center point.
retranslated_point_x = new_point[0] + center_point_x
retranslated_point_y = new_point[1] + center_point_y
And all is done.
Just a recap
So to recap this long post:
Subtract the coordinates of the center point from the coordinates of the image-point
Scale by a factor with a simply multiplication of the coordinates
Use the idea of the matrix multiplication to think about the rotation (you can easily find these things on Google or Wikipedia).
Add the coordinates of the center point to the new coordinates of the image-point
I realize now that I could have just given this recap, but now there is at least some visual aid and a slight mathematical background in this post, which is also nice. I really believe that such problems should be looked at from a mathematical angle, the mathematical description can help a lot.

Get a point on a plane at specific distance from initial point

I got this plane
from sympy.geometry.plane import Plane
p=Plane((0,5,0),(0,0,0),(5,7,0)) #2d first to make it easier
now I want to get a point on that plane 45deg from plane point one and lenght=sqrt(2). In this case that point will be (1,6,0)
I've tried this:
a=np.sqrt(2)*p.arbitrary_point(pi/4)
but it does not work as a coordinates return (1.0,8.07106781187,0.0)
the problem is that arbitrary_point returns a point in a circle of radius 1 about p1 of the Plane. I want to be able to change that radius.
Multiplying all of a point's coordinates by sqrt(2) moves it away from the origin (0,0,0). You want to move it away from p1 of the plane. This is what the scale method is for.
p.arbitrary_point(np.pi/4).scale(np.sqrt(2), np.sqrt(2), np.sqrt(2), p.p1)
returns Point3D(1, 6, 0).
(I'm assuming import numpy as np here, to match the setting of the question; the standard math module could be used instead of numpy.)

How to plot two linear lines on a graph in python with radius 1 and center at origin and measure the angle difference?

I have an issue that I can't seem to solve. I have already acquired data from another source and created 2 polynomials that are identical in shape but not in orientation, that is one is rotated x degrees compared to the other, and if you rotate the graph x degrees back they will match.
I have already taken the derivative of both of the graphs at a certain point.
I would like to graph these slopes onto a unit circle on a polar graph, and somehow find the angle difference between these two line segments of slope i and j that extend from the origin.
I'm fairly new to python so I so not know how to begin plotting these in polar or finding a way to determine the angle difference. I know that by hand, you can take the inverse tangent but that will only give you a range from +90 to -90. I would like my number to fall in the range from 0 to 360 for rotation.
Any help is appreciated. If this isn't enough info or if it isn't clear enough I can provide more.

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