Triangles can never be non-planar; anything with more than 3 points can be non-planar and thus un-renderable unless converted to triangles. For example: A square is two triangles that are on the same plane, if all the points that make up the square are co-planar. It takes a lot of calculations to make sure all … Read more
This is just my two cents. Is your texture follow the isgl3D requirement? from isgl3D Tutorial 3 – Texture mapping http://isgl3d.com/tutorials/4/tutorial_3_texture_mapping For standard image files, the image size must be factor of two compliant: eg 64×128, 256×32, etc. For pvr textures, the image file must also be square: eg 64×64, 256×256, etc. Hope this help.
The solution I’m currently using seems to be missing here. Assuming that the plane normal is normalized (|Vn| == 1), the signed angle is simply: For the right-handed rotation from Va to Vb: atan2((Va x Vb) . Vn, Va . Vb) For the left-handed rotation from Va to Vb: atan2((Vb x Va) . Vn, Va … Read more
Here’s a project with a rotating camera. Looking through the source it seems to just move the camera position in a circle. function onDocumentMouseMove( event ) { event.preventDefault(); if ( isMouseDown ) { theta = – ( ( event.clientX – onMouseDownPosition.x ) * 0.5 ) + onMouseDownTheta; phi = ( ( event.clientY – onMouseDownPosition.y ) … Read more
I have been facing this dilemma for some months now, and made exhaustive research. Algorithms Mainly there are 2 categories of algorithms: computation geometry, and implicit surfaces. Computation Geometry They fit the mesh on the existing points. Probably the most famous algorithm of this group is powercrust, because it is theoretically well-established – it guarantees … Read more
If you want to rotate a vector you should construct what is known as a rotation matrix. Rotation in 2D Say you want to rotate a vector or a point by θ, then trigonometry states that the new coordinates are x’ = x cos θ − y sin θ y’ = x sin θ + … Read more
[*] Alright, I came here looking for an answer and didn’t find something simple and straightforward, so I went ahead and did the dumb but effective (and relatively simple) thing: Monte Carlo optimisation. Very simply put, the algorithm is as follows: Randomly perturb your projection matrix until it projects your known 3D coordinates to your … Read more
There is no clear answer to your question. It depends entirely how your data is organized. Something to keep in mind: Quadtrees work best for data that is mostly two dimensional like map-rendering in navigation systems. In this case it’s faster than octrees because it adapts better to the geometry and keeps the node-structures small. … Read more
It is a little complicated, but you can draw all the objects by the following code: from mpl_toolkits.mplot3d import Axes3D import matplotlib.pyplot as plt import numpy as np from itertools import product, combinations fig = plt.figure() ax = fig.gca(projection=’3d’) ax.set_aspect(“equal”) # draw cube r = [-1, 1] for s, e in combinations(np.array(list(product(r, r, r))), 2): … Read more
try: %matplotlib notebook see jakevdp reply here EDIT for JupyterLab users: Follow the instructions to install jupyter-matplotlib Then the magic command above is no longer needed, as in the example: # Enabling the `widget` backend. # This requires jupyter-matplotlib a.k.a. ipympl. # ipympl can be install via pip or conda. %matplotlib widget # aka import … Read more