将向量的差值转换为0到1之间 [英] converting the difference in vectors to between 0 and 1

问题描述

我正在处理两个单位向量，但不确定如何计算.我需要这样，如果它们指向同一方向，则答案为1，相反方向的答案为0，垂直(向上或向下)的答案为0.5，等等.

示例:对于两个向量(1,0)和(-1,0)(所以，相反的向量)，我得到的答案是0.对于两个向量(1,0)和(1/sqrt(2)，1/sqrt(2))(因此，指向45度角的单位向量)我得到0.25.对于两个向量(0,1)和(-1,0)(所以，垂直向量)，我得到0.5

谢谢您的帮助！

解决方案

了解

如果2个归一化向量指向相同方向，则点积为1；如果点朝相反方向，则点积为-1；如果向量垂直，则点积为0.

在pygame中，点积可以通过 math.Vector2.dot() .如果 A 和 B 是 pygame.math.Vector2 对象:

  uA = A.normalize()uB = B.normalize()AdotB = uA.dot(uB) 

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在上面的示例中， AdotB 的范围为[-1.0，1.0]. AdotB * 0.5 + 0.5 的范围为[0.0，1.0]，并且 math.acos(AdotB)/math.pi + 1 映射 A 和 B 线性地排列在[0.0，1.0]范围内.

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此外， pygame.math.Vector2.angle_to() 计算与给定矢量的角度(以度为单位).可以根据以下公式计算范围[0.0，2.0]范围内的值:

  w = 1-A.angle_to(B)/180 

I'm working with two unit vectors but not sure how to calculate this. I need it so that if they point in the same direction the answer is 1, opposite directions the answer is 0, perpendicular (either up or down) the answer is 0.5, etc.

Examples: For two vectors (1,0) and (-1,0) (so, opposite vectors), the answer I get is 0. For two vectors (1,0) and (1/sqrt(2),1/sqrt(2)) (so, the unit vector pointing at a 45 degree angle) I get 0.25. For two vectors (0,1) and (-1,0) (so, perpendicular vectors) I get 0.5

Thank you for any help with this!

解决方案

Read about the Dot product In general The dot product of 2 vectors is equal the cosine of the angle between the 2 vectors multiplied by the magnitude (length) of both vectors.

dot( A, B ) == | A | * | B | * cos( angle_A_B ) 

This follows, that the dot product of 2 unit vectors is equal the cosine of the angle between the 2 vectors, because the length of a unit vector is 1.

uA = normalize( A )
uB = normalize( B )
cos( angle_A_B ) == dot( uA, uB )

If 2 normalized vectors point in the same direction, then the dot product is 1, if the point in the opposite direction, the dot product is -1 and if the vectors are perpendicular then the dot product is 0.

In pygame the dot product can be computed by math.Vector2.dot(). If A and B are pygame.math.Vector2 objects:

uA = A.normalize()
uB = B.normalize()
AdotB = uA.dot(uB)

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In the example above, AdotB is in range [-1.0, 1.0]. AdotB * 0.5 + 0.5 is in range [0.0, 1.0] and math.acos(AdotB) / math.pi + 1 maps the angle between A and B linearly to the range [0.0, 1.0].

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Furthermore, pygame.math.Vector2.angle_to() calculates the angle to a given vector in degrees. A value in range [0.0, 2.0] dependent on the angle can be computed by

w = 1 - A.angle_to(B) / 180

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