将2 3D点转换为方向向量再转换为欧拉角 [英] Convert 2 3D Points to Directional Vectors to Euler Angles
问题描述
本质上是我的问题。同样，也许我对欧拉角还不够熟悉，而且我尝试做的事情也不可能。
Here's essentially my problem. Also maybe I am not familiar enough with Euler angles and what I'm attempting to do is not possible.
我在3d空间中有2个点。
I have 2 points in 3d space.
p1（1,2,3）
p2（4,5,6）
p1 (1,2,3)
p2 (4,5,6)
在为了获得这两点的单位矢量，我基本上是在做。
In order to get the unit vectors for these two points I'm doing this basically.
var productX = (position.X2 - position.X1);
var productY = (position.Y2 - position.Y1);
var productZ = (position.Z2 - position.Z1);
var normalizedTotal = Math.sqrt(productX * productX + productY * productY + productZ * productZ);
var unitVectorX, unitVectorY, unitVectorZ;
if(normalizedTotal == 0)
{
unitVectorX = productX;
unitVectorY = productY;
unitVectorZ = productZ;
}
else
{
unitVectorX = productX / normalizedTotal;
unitVectorY = productY / normalizedTotal;
unitVectorZ = productZ / normalizedTotal;
}
所以现在我对这2个3d点有一个单位向量xyz。
So now I have a unit vector x y z for these 2 3d points.
我现在正尝试从方向矢量转换为欧拉角。这可能吗。我在这里缺少什么，因为我找不到如何执行此操作的好资源。
I'm attempting now to convert from directional vector to euler angle. Is this possible. What am I missing here as I can't find any good resource on how to do this.
感谢您的帮助。
有时图片会有所帮助。
也许这将为我要解决的问题提供一个更好的例子。
maybe this will give a better example of what i'm trying to solve for.
给出2个点，我确定了一个中点，长度，现在我试图找出要设置的角度，以便圆柱体正确地围绕x定向，y，z轴。我想我需要找出所有3个角度，而不仅仅是1和2是正确的吗？我认为来自方向矢量的欧拉角会刺穿您。
Given 2 points, I have determined a midpoint, length, and now i'm trying to figure out hte angles to set so that the cylinder is correctly oriented around the x,y,z axis. I think I need to figure out all 3 angles not just 1 and 2 is that correct? I think the euler angles from a directional vector bit through you off.
推荐答案
您想要的是矢量笛卡尔坐标的转换
What you want is a transformation from Cartesian coordinates of the vector
v = (v_x, v_y, v_z)
到球形坐标 r ，ψ和θ其中
v = ( r*COS(ψ)*COS(θ), r*SIN(θ), r*SIN(ψ)*COS(θ) )
可通过以下公式完成
r = SQRT(v_x^2+v_y^2+v_z^2)
TAN(ψ) = (v_z)/(v_x)
TAN(θ) = (v_y)/(v_x^2+v_z^2)
到获取角度ψ和θ，请使用 ATAN2（dy，dx） 函数类似于
To get the angles ψ and θ, use the ATAN2(dy,dx) function as in
ψ = ATAN2(v_z, v_x)
θ = ATAN2(v_y, SQRT(v_x^2+v_z^2))
现在您有了沿方向向量
j = ( COS(ψ)*COS(θ), SIN(θ), SIN(ψ)*COS(θ) )
您可以从中获得两个垂直向量
you can get the two perpendicular vectors from
i = ( SIN(ψ), 0, -COS(ψ) )
k = ( COS(ψ)*SIN(θ), -COS(θ), SIN(ψ)*SIN(θ) )
这三个向量组成了3×3旋转矩阵的列
These three vectors make up the columns of the 3×3 rotation matrix
| SIN(ψ) COS(ψ)*COS(θ) COS(ψ)*SIN(θ) |
E =[i j k] = | 0 SIN(θ) -COS(θ) |
| -COS(ψ) SIN(ψ)*COS(θ) SIN(ψ)*SIN(θ) |
就欧拉角而言，以上等同于
In terms of Euler angles the above is equivalent to
E = RY(π/2-ψ)*RX(π/2-θ)
示例
两点 p_1 =（3,2,3）和 p_2 =（5,6,4）定义向量
v = (5,6,4) - (3,2,3) = (2,4,1)
注意：我将 v [i] 表示为第个第向量的元素，如上面的 v [1] = 2 所示。这既不像从零开始的 C ， Python ，也不像 VB ， FORTRAN 或 MATLAB ，它们使用括号（）
NOTE: I am using the notation of v[i] for the i-th element of the vector, as in v[1]=2 above. This is neither like C, Python which is zero based, nor like VB, FORTRAN or MATLAB which uses parens () for the index.
使用上面的表达式
r = √(2^2+4^2+1^2) = √21
TAN(ψ) = 1/2
TAN(θ) = 4/√(2^2+1^2) = 4/√5
ψ = ATAN2(1,2) = 0.463647
θ = ATAN2(4,√5) = 1.061057
现在可以找到方向矢量
j = ( COS(ψ)*COS(θ), SIN(θ), SIN(ψ)*COS(θ) ) = (0.4364, 0.87287, 0.21822 )
i = ( SIN(ψ), 0, -COS(ψ) ) = (0.44721, 0, -0.89443 )
k = ( COS(ψ)*SIN(θ), -COS(θ), SIN(ψ)*SIN(θ) ) = (0.78072, -0.48795, 0.39036)
将方向向量作为本地坐标到世界坐标的列变换（旋转）
Put the direction vectors as columns of the local to world coordinate transformation (rotation)
E[1,1] = i[1] E[1,2] = j[1] E[1,3] = k[1]
E[2,1] = i[2] E[2,2] = j[2] E[2,3] = k[2]
E[3,1] = i[3] E[3,2] = j[3] E[3,3] = k[3]
| 0.447213595499957 0.436435780471984 0.780720058358826 |
| |
E = | 0 0.872871560943969 -0.487950036474266 |
| |
| -0.894427190999915 0.218217890235992 0.390360029179413 |
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