I'm scripting in python for starters.
Making this example simple, I have one edge, with uv Coordinates of ([0,0],[1,1]), so its a 45 degree angle. I have another edge that is ([0,0],[0,1]) so its angle is 0/360 degrees. My goal is to compare the angles of those two edges in order to get the difference so I can modify the angle of the second edge to match the angle of the first edge. Is there a way to do this via vector math?
Easiest to reconstruct and thus constructively remember is IMO the complex picture. To compute the angle from a=a.x+i*a.y to b=b.x+i*b.y rotate b back by multiplying with the conjugate of a to get an angle from the zero angle resp. the positive real axis,
arg((a.x-i*a.y)*(b.x+i*b.y))
=arg((a.x*b.x+a.y*b.y)+i*(a.x*b.y-a.y*b.x))
=atan2( a.x*b.y-a.y*b.x , a.x*b.x+a.y*b.y )
Note that screen coordinates use the opposite orientation to the Cartesian/complex plane, thus change use a sign switch as from atan2(y,x) to atan2(-y,x) to get an angle in the usual direction.
To produce a vector b rotated angle (in radians) w from a, multiply by cos(w)+i*sin(w) to obtain
b.x = cos(w)*a.x - sin(w)*a.y
b.y = cos(w)*a.y + sin(w)*a.x
You will have to rescale to get a specified length of b.
Related
When the Rodrigues function is called with a rotation matrix as argument it provides 2 results.
I understand that the first item returned is the vector around which the rotation occurs and that the magnitude of the vector provides the angle of rotation. It seems that it provides a number (in radians) in the range (0,180) degrees for rotations covering (0,360) degrees, therefore there must be a way to determine the sign of the rotation. How do you do that.
As a supplimentary question, I understand that the second result is a Jacobian matrix. How do you use that?
The rotation is always positive, and when it "needs" to be negative (equivalently, closer to 360 than 0 degrees), the vector is simply flipped to the other side, so now it can be positive.
There is the "right hand rule". Right hand grabs vector, thumb pointing along the vector. Fingers indicate positive rotation around vector.
Example: Place your (right) fist on the desk, thumb up. Going +90 degrees is a quarter turn counterclockwise (inward). Going -90 degrees is a quarter turn clockwise (outward)... or +90 degrees with your thumb pointing into the desk.
The Jacobian is a bunch of derivatives, a vector in output-space for each component of the input. It tells you how stable the calculation is, i.e. how easily perturbed the result is, were any of the elements of your input vector to fluctuate by a bit.
Jacobians also show up in robotics. You can use them for inverse kinematics, combined with a solver. Given the Jacobian of your "robot arm", a tool center point, and a target, some math involving a Jacobian tells you what joints to move (a little bit) in which way to get closer to the target. The Jacobian depends on the current pose (i.e. it's not a constant matrix), so you'd recalculate it all the time.
I would like to write a code in python that would solve for the smallest angle (θ) that would include all the points in a 2D plane given any number of points. The vertex of the angle is always centered at the origin (0,0). The points are defined using the Cartesian coordinate system with an (x,y) value. A figure below is shown for visualization. Any thoughts on how I should approach this problem?
Convert each of the Cartesian representations into polar coordinates.
Sort by reference angle.
Subtract adjacent reference angles to get the angles between adjacent vectors. Make sure to include the last and first points as one more angle to compute.
Identify the largest angle between adjacent vectors. The opposite side of this angle is the smallest angle that include all of the points.
For instance, using the canonical representation -- counter-clockwise from the x-positive ray -- you would find the reference angle for each polar vector. Sorted, you would have the list [d, c, b, a, e, f].
Next, you compute the angles dOc, cOb, bOa, ... and fOd.
You note that aOe is the largest angle of them all being somewhat in excess of a whole radian.
Therefore, eOa is the desired angle.
I have the following image where the four corners of the cattle housing are -[(x1,y1), (x2,y2), (x3,y3), (x4,y4)]. The camera to capture the image was positioned in the middle of the (x1,y1) and (x4,y4). As the (x3,y3) and (x2,y2) are far from the camera, in the image, x1-x4 not equal x2-x3.
I need to reproject the housing into a 2d rectangular plane with corners of [(x1,y1), (x2,y2), (x3,y3), (x4,y4)] and unlikely the original image, this new plane will have x1-x4 = x2-x3. Is there any viable option to do that? OpenCV comes with a perspective transformation function that can only be applied to an image. But, in this case, I will have some x,y locations of cattle on the original plane which need to be converted and drawn into the rectangular 2d plane to show the cattle position.
This problem is linear algebra more than programming. You have a linear transformation on a simple quadrilateral. The math is simpler, because you have two edges parallel to edges of the image.
First of all, we need to redefine some notation: for instance, you've used (x2, y2) to refer to both points on the posted image, and to the desired position of the upper-left corner of the transformed image. I will simplify this by declaring the transformed points to be A = (x1, y2) and B = (x4, y3): we're horizontally stretching the top of the trapezoid to form a rectangle.
Also note that y1=y4 and y2=y3 from the start; this simplifies the calculations. Visualize the new and old images overlaid, with a point of question Q in the interior, its coordinates marked on the boundary. We need to find the general equation for Q's transformed point, R, after the "stretching".
I have also marked the median of the original image, MN. Points on this line will not move during the stretching. As a side note, points along the bottom edge 1-4 will not move. Points on the outer fringes of 2-3 will move most. Let C be the point on edge 1-2 with the same y-coordinate as Q (and, later, R); let D be the corresponding point on MN.
A-----2-----M-----3---B
| |
Qy CR Q D |
| |
| |
1----Rx-Qx-N----------4
We merely need to pro-rate the amount that a chosen point moves. Finding the equations of MN and 1-2 are well-known (two-point formula). Substitute Qy into each of those equations to obtain Cx and Dx.
The "stretch" factor, in transforming CD to (x1, Qy) D is the ratio of their lengths: (Dx-x1) / (Dx-Cx). Q will move left by a proportion of that stretch factor, according to its distance left of D: (Dx-Qx) / (Dx-x1). Multiply those to get the distance Q moves. Subtract that amount from Qx to get Rx.
Yes, you now have several constants in the final, combined equation: x1, x2, x3, x4, y1, y2. You also have variables Qx and Qy. This is as it should be. This leaves you with a general equation to convert Qx => Rx for any point in the image.
If you plan to stretch vertically as well, the same proportioning will apply in the vertical direction. I suggest that you do one stretch at a time; this will keep the math modular: easier to check and debug in separate stages.
Does that get you moving?
D will not move;
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.
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.