geometry

问题 While looking at various methods for point-in-triangle testing (2D case), I found that the method which uses barycentric coordinates is the most used one. Here is a StackOverflow answer which explains it. Why is this method the most preferred one? It probably has to do with doing less calculations, but what about numerical stability? Is this algorithm better suited than say, the "same side" technique, for cases in which the point is particularly near the border? 回答1: If you solve it: p = p0 +

问题 While looking at various methods for point-in-triangle testing (2D case), I found that the method which uses barycentric coordinates is the most used one. Here is a StackOverflow answer which explains it. Why is this method the most preferred one? It probably has to do with doing less calculations, but what about numerical stability? Is this algorithm better suited than say, the "same side" technique, for cases in which the point is particularly near the border? 回答1: If you solve it: p = p0 +

问题 While looking at various methods for point-in-triangle testing (2D case), I found that the method which uses barycentric coordinates is the most used one. Here is a StackOverflow answer which explains it. Why is this method the most preferred one? It probably has to do with doing less calculations, but what about numerical stability? Is this algorithm better suited than say, the "same side" technique, for cases in which the point is particularly near the border? 回答1: If you solve it: p = p0 +

问题 While looking at various methods for point-in-triangle testing (2D case), I found that the method which uses barycentric coordinates is the most used one. Here is a StackOverflow answer which explains it. Why is this method the most preferred one? It probably has to do with doing less calculations, but what about numerical stability? Is this algorithm better suited than say, the "same side" technique, for cases in which the point is particularly near the border? 回答1: If you solve it: p = p0 +

问题 I have been playing around with the example of a rotating cube here. I have generated 2 cubes that should rotate around the Y-axis. However, it doesn't seem to work as expected and I can't figure out what the problem of it is. Here is a working code example: import sys import math import pygame from pygame.math import Vector3 from enum import Enum class Color(Enum): BLACK = (0, 0, 0) SILVER = (192,192,192) class Cube(): def __init__(self, vectors, screen_width, screen_height, initial_angle=25

问题 I have been playing around with the example of a rotating cube here. I have generated 2 cubes that should rotate around the Y-axis. However, it doesn't seem to work as expected and I can't figure out what the problem of it is. Here is a working code example: import sys import math import pygame from pygame.math import Vector3 from enum import Enum class Color(Enum): BLACK = (0, 0, 0) SILVER = (192,192,192) class Cube(): def __init__(self, vectors, screen_width, screen_height, initial_angle=25

问题 I have been playing around with the example of a rotating cube here. I have generated 2 cubes that should rotate around the Y-axis. However, it doesn't seem to work as expected and I can't figure out what the problem of it is. Here is a working code example: import sys import math import pygame from pygame.math import Vector3 from enum import Enum class Color(Enum): BLACK = (0, 0, 0) SILVER = (192,192,192) class Cube(): def __init__(self, vectors, screen_width, screen_height, initial_angle=25

问题 Equation of first ellipse=> (((x*cos(A)+y*sin(A)-H1)^2)/(a1^2))+(((x*sin(A)-y*cos(A)-K1)^2)/(b1^2))=1 Equation of the second ellipse=> (((x*cos(B)+y*sin(B)-H2)^2)/(a2^2))+(((x*sin(B)-y*cos(B)-K2)^2)/(b2^2))=1 I know that the ellipse will intersect at One Point Two Point Three Point Four Point No intersection at all Is there a general set of equation to solve the same. 回答1: You can transform these equations to general form of conic section: A*x^2+2*B*x*y+C*y^2+D*x+E*y+F=0 and solve system of

问题 I wanted to ask you if anyone knew how to check if a point was inside a given triangle, on a reference system in space. I am aware that speaking of 2d systems I can obtain this point with the following procedures: To determine if a given point v lies within a given triangle, consider a single vertex, denoted v0, v1 and v2 are the vectors from the other two vertices v0. Expressing the vector from v0 to v in terms of v1 and v2 then gives v = v0 + av1 + bv2 where a, b are constant. Solve for a,

问题 I'm nuilding a graphical editor where you can select elements, scale them, rotate, resize etc. One thing I've been stuck on is positioning an element after it's rotated. I have the following javascript class that handles resizing: const rotateCoords = (x, y, rotation) => { const rad = (rotation * Math.PI) / 180 return { y: y * Math.cos(rad) - x * Math.sin(rad), x: x * Math.cos(rad) + y * Math.sin(rad), } } export default ( e, ratio, resizeDirection, resizeMouseX, resizeMouseY, zoomScale,