Using Matlab, how can I calculate a two points in Equilateral triangle if there are known one point and the Center of Gravity in a 3D ?
I know there is a infinite solutions but i need just a random one.
Thank you.
Take the vector pointing from the center of gravity to the point.
Create an orthogonal vector (this can be done in a few ways, I usually take the first vector, add 1.0 to each component until it is not parallel, then take the cross product with the original vector).
Rotate your vector 120 degrees about the orthogonal vector. (look up the rotation matrix about an arbitrary vector)
Create your second point by adding that vector to your center of gravity.
Create your third point by rotating it again or in the opposite direction.
Related
I've got a 3d sphere which I've been able to plot a point on using longitude and latitude thanks to some work of another developer I've found online. I think I understand what its doing.
What I need to do now is rotate my planet so the point is always at the top most point (ie the north pole) but I'm not sure how to do this. I'm probably missing some important fundamentals here so I'm hoping the answer can assist in my future learning.
Here's an image showing what I have - The blue line is a line coming from the longitude and latitude I have plotted and I need to rotate the planet so that line is basically pointing directly upwards.
https://ibb.co/2y24FxS
If anyone is able to advise it'd be very much appreciated.
If I'm not mistaken, Unity uses a coordinate system where the y-axis points up.
If the point on your sphere was in the xy-plane, you'd just have to determine the angle between the radius-vector (starts in the center of the sphere, ends on the point in question) and the y-axis, and than rotate by that amount around the z-axis, so that the radius vector becomes vertical. But your point is at an arbitrary location in 3D space - see the image below. So one way to go about it is to first bring the point to the xy-plane, then continue from there.
Calculate the radius vector, which is just r = x-sphereCenter. Make a copy of it, set y to zero, so that you have (x, 0, z) - which is just the projection of the vector r on the horizontal xz-plane - let's call the copy rXZ.
Determine the signed angle between the x-axis and rXZ (use Vector3.SignedAngle(xAxis, rXZ, yAxis), see docs), and create a rotation matrix M1 that rotates the sphere in the opposite direction around the vertical (negate the angle). This should place your point in the xy-plane.
Now determine the angle between r and the y-axis (Vector3.SignedAngle(r, yAxis, zAxis)), and create a new rotation matrix M2 that rotates by that angle around the zAxis. (I think for this second one, the simpler Vector3.Angle will work as well.)
So, what you want now is to combine the two matrices (by multiplying them) into a single transform (I'm assuming this is a transformation in the local coordinate system of the sphere, where (0, 0, 0) is the sphere's center). If I'm not mistaken, Unity uses column-major matrices, so the multiplication order should be M = M2 * M1 (the rightmost matrix is applied first).
Reorient your globe using M as a local transform, and it should bring your point to the top. You can also create M3 = M1.inverse, and then do M = M3 * M2 * M1, to preserve the original angular offset from the xy-plane.
Check for edge cases, such as r already being vertical (pointing straight up, or straight down).
I have two independent 3D shapes; one is a square and another is a cone.
Let's assume the cone lies inside the square. How can I find out that the surface of the cone touches the square's surface when I move the cone in any direction?
It will be helpful if anyone can suggest an algorithm to check whether the surface touches another shape.
I am working with MATLAB, but the underlying logic will be appreciated in any language will be appriciated.
https://in.mathworks.com/matlabcentral/answers/367565-findout-surface-to-surface-intersection-between-two-3d-shapes
There is a relatively easy solution, thanks to the fact that the truncated cone is a convex shape, and finding its AABB is not so hard.
First rotate space so that the cube become axis aligned (and the cone in arbitrary position). Then to find the AABB of the base, is suffices to get the maxima an minima of the coordinates, using the parametric equation, C + R cos t + R' sin t, where C is the position vector of the center, and R, R' two orthogonal radii. You find the limit angles by canceling the derivative.
After finding the extrema on the three coordinates, the global bounding box is the one that surrounds these six points plus the apex.
By comparing the AABB to the cube, you can tell what distance remains before a collision, in any direction.
I have a convex polygon in 3D. For simplicity, let it be a square with vertices, (0,0,0),(1,1,0),(1,1,1),(0,0,1).. I need to arrange these vertices in counter clockwise order. I found a solution here. It is suggested to determine the angle at the center of the polygon and sort them. I am not clear how is that going to work. Does anyone have a solution? I need a solution which is robust and even works when the vertices get very close.
A sample MATLAB code would be much appreciated!
This is actually quite a tedious problem so instead of actually doing it I am just going to explain how I would do it. First find the equation of the plane (you only need to use 3 points for this) and then find your rotation matrix. Then find your vectors in your new rotated space. After that is all said and done find which quadrant your point is in and if n > 1 in a particular quadrant then you must find the angle of each point (theta = arctan(y/x)). Then simply sort each quadrant by their angle (arguably you can just do separation by pi instead of quadrants (sort the points into when the y-component (post-rotation) is greater than zero).
Sorry I don't have time to actually test this but give it a go and feel free to post your code and I can help debug it if you like.
Luckily you have a convex polygon, so you can use the angle trick: find a point in the interior (e.g., find the midpoint of two non-adjacent points), and draw vectors to all the vertices. Choose one vector as a base, calculate the angles to the other vectors and order them. You can calculate the angles using the dot product: A · B = A B cos θ = |A||B| cos θ.
Below are the steps I followed.
The 3D planar polygon can be rotated to 2D plane using the known formulas. Use the one under the section Rotation matrix from axis and angle.
Then as indicated by #Glenn, an internal points needs to be calculated to find the angles. I take that internal point as the mean of the vertex locations.
Using the x-axis as the reference axis, the angle, on a 0 to 2pi scale, for each vertex can be calculated using atan2 function as explained here.
The non-negative angle measured counterclockwise from vector a to vector b, in the range [0,2pi], if a = [x1,y1] and b = [x2,y2], is given by:
angle = mod(atan2(y2-y1,x2-x1),2*pi);
Finally, sort the angles, [~,XI] = sort(angle);.
It's a long time since I used this, so I might be wrong, but I believe the command convhull does what you need - it returns the convex hull of a set of points (which, since you say your points are a convex set, should be the set of points themselves), arranged in counter-clockwise order.
Note that MathWorks recently delivered a new class DelaunayTri which is intended to superseded the functionality of convhull and other older computational geometry stuff. I believe it's more accurate, especially when the points get very close together. However I haven't tried it.
Hope that helps!
So here's another answer if you want to use convhull. Easily project your polygon into an axes plane by setting one coordinate zero. For example, in (0,0,0),(1,1,0),(1,1,1),(0,0,1) set y=0 to get (0,0),(1,0),(1,1),(0,1). Now your problem is 2D.
You might have to do some work to pick the right coordinate if your polygon's plane is orthogonal to some axis, if it is, pick that axis. The criterion is to make sure that your projected points don't end up on a line.
As shown in image, there is a binary polygonal image. I want to find the principal direction in the image with respect to X-axis. I have shown the principal direction and X-axis with blue line. This can be done using PCA but my problem is such a small rectangle will have around 1000 pixels and I have to find Principal directions for around 100 polygons (polygon can be of arbitrary shape).
One approach that I have thought is:
Project that rectangle onto a line which is oriented at degrees at an interval (say) 5 degrees. The projection which has the maximum variance is the desired projection axis, and thus that is the desired angle. But this also falls under a greedy approach and thus will take time. Is there a smarter approach?
Also, if anybody could explain the exact procedure to do this using PCA, it would be helpful. I know the steps:
1. Take the covariance matrix.
2. Get the top eigenvector corresponding to largest eigenvalue -> that will be the principal direction.
But I am confused in the following statement which I often read everywhere:
A column vector: [0.5 0.5] is the first principal component and it gives the direction of the maximum variance. So can do I exactly calculate the angle by which I should rotate the data so that it will become parallel to X-axis.
Compute the eigenvector associated with the highest eigen value. Call that v. Normalize v. v = v/norm(v);
Compute angle between that and the horizontal direction: angle=acos(sum(v.*[1,0]))
Rotate by -angle, transformation matrix T = [cos(-angle) -sin(-angle); sin(-angle) cos(-angle)], multiply all points by that matrix. Do that for all polygons.
I have a 3D matrix of data in matlab, but I want to extract an arbitrarily rotated slice of data from that matrix and store it as a 2D matrix, which I can access. Similar to how the slice() function displays data sliced at any angle, except I would also like to be able to view and modify the data as if it were an array.
I have the coordinates of the pivot-point of the plane as well as the angles of rotation (in x, y and z axis), I have also calculated the equation of the plane in the form:
Ax + By + Cz = D
and can extract a 3D matrix containing only the data that fall on that plane, but I don't know how to then convert that into a simple 2D array.
Another way of doing it would be to somehow rotate the source matrix in the opposite direction of the angle of the plane, so as to line up the plane of data with the XY axis, and simply extract that portion of the matrix, but I do not know if rotating a matrix like that is possible.
I hope this hasn't been answered elsewhere, I've been googling it all day, but none of the problems seem to exactly match mine.
Thanks
You can take a look at the code here. I think the function is similar to what you are trying to solve.
The function extracts an arbitrary plane from a volume given the size of the plane, the center point of the plane, and the plane normal, i.e. [A,B,C]. It also outputs the volumetric index and coordinate of each pixel on the plane.
Aha! May have just solved it myself.
To produce the plane equation I rotate a normal vector of (0,0,1) using rotation matrices and then find D. If I also rotate the following vectors:
(1,0,0) //step in the x direction of our 2D array
and
(0,1,0) //step in the y direction of our 2D array
I'll have the gradients that denote how much my coordinates in x,y,z have to change before I step to the next column in my array, or to the next row.
I'll mock this up ASAP and mark it as the answer if it works
EDIT: Ok slight alteration, when I'm rotating my vectors I should also rotate the point in 3D space that represents the xyz coordinates of x=0,y=0,z=0 (although I'm rotating around the centre of the structure, so it's actually -sizex/2,-sizey/2,-sizez/2, where size is the size of the data, and then I simply add size/2 to each coordinate after the rotations to translate it back to where it should be).
Now that I have the gradient change in 3D as I increase the x coordinate of my 2D array and the gradient change as I increase the y coordinate, I can simply loop through all possible x and y coordinates (the resulting array will be 50x50 for a 50x50x50 array, I'm not sure what it will be for irregular sizes, which I'll need to work out eventually) in my 2D array and calculate the resulting 3D coordinates on my plane in the data. My rotated corner value serves as the starting point. Hooray!
Just got to work out a good test for this encompassing all angles and then I'll approve this as an answer