将单位向量转换为四元数

4.)四元数共轭是A = {a0，- a } 5.)四元数产物的缀合物是相反顺序的缀合物的产物. (A⊗ B) = B ⊗ A 6.)向量四元数的共轭是负数. X = {0，- x } = -X 7.)四元数范数为| A |. =√(A⊗ A )=√(a0² + a . a ) 8.)一个单元四元数是范数为1的单元. 9.)单位三向量 x = {x 1 ，x 2 ，x 3 }和 x . x = 1可表示为单位矢量四元数 X = {0， x }，| X | = 1. 10.)四元数矢量X的球面旋转，角度为θ.关于单位矢量轴 n 的值是Q⊗ X⊗ Q ，其中Q是四元数{cos(θ/2)，sin(θ/2) n }.注意| Q | = 1.注意四元数向量乘积的形式.给定向量四元数X 1 = {0， x 1 )和X 2 = {0， x 2 }，四元数乘积为X 2 ⊗ X 1 = { x 1 . x 2 ， x 1 和时间; x 2 }.四元数离婚了一百多年前，将点积重新组合为标量部分，将叉积重新组合为矢量部分.这些产品都不是可逆的，但是四元数的描述如下. 反转找到球面变换四元数Q 12 以旋转向量X 1 以与向量X 2 对齐.从上方 X 2 = Q 12 ⊗ X 1 ⊗ Q 12将两边都乘以X 1 ， X 2 ⊗ X 1 = Q 12 ⊗ X 1 ⊗(Q 12 ⊗ X 1 )请记住，旋转轴 n 来自叉积 x 1 × x 2 ，因此 n . x 1 =0.并且Q ⊗ X =(X⊗ Q) = X ⊗ Q，离开 X 2 ⊗ X 1 = Q 12 ⊗ X 1 ⊗ X 1 ⊗ Q 12 = Q 12 ⊗ Q 12因此四元数变换可以直接求解为 Q 12 =√(X 2 ⊗ X 1 )您自己依靠四元数平方根.有很多方法可以做到，最好的方法取决于您的应用程序，同时考虑速度和稳定性.-hth，弗雷德·克林格纳(Fred Klingener)So I'm very new to quaternions, but I understand the basics of how to manipulate stuff with them. What I'm currently trying to do is compare a known quaternion to two absolute points in space. I'm hoping what I can do is simply convert the points into a second quaternion, giving me an easy way to compare the two.What I've done so far is to turn the two points into a unit vector. From there I was hoping I could directly plug in the i j k into the imaginary portion of the quaternion with a scalar of zero. From there I could multiply one quaternion by the other's conjugate, resulting in a third quaternion. This third quaternion could be converted into an axis angle, giving me the degree by which the original two quaternions differ by.Is this thought process correct? So it should just be [ 0 i j k ]. I may need to normalize the quaternion afterwards, but I'm not sure about that.I have a bad feeling that it's not a direct mapping from a vector to a quaternion. I tried looking at converting the unit vector to an axis angle, but I'm not sure this would work, since I don't know what angle to give as an input. 解决方案 NotationQuaternions are defined in a four-space with bases {1, i, j, k}. Hamilton famously carved the fundamental relationship into the stone of the Brougham Bridge in Dublin:i = j = k = i j k = -1.There are many equivalent quaternion parameterizations, but here I'll use a {scalar, vector} form.1.) A = {a0, a} and B = {b0, b}, where A and B are quaternions, a0 and b0 are scalars, and a and b are three-vectors.2.) X = { 0, x } is a vector quaternion.3.) The (non-commutative) quaternion product derives directly from the properties of i, j and k above, A⊗B = {a0 b0 - a.b, a0 b + b0 a + a x b}4.) The quaternion conjugate is A = {a0, - a}5.) The conjugate of a quaternion product is the product of the conjugates in reverse order. (A⊗B) = B⊗A6.) The conjugate of a vector quaternion is its negative. X = {0, -x } = -X7.) The quaternion norm is |A| = √(A⊗A) = √( a0² + a.a )8.) A unit quaternion is one that has a norm of 1.9.) A unit three-vector x = {x1, x2, x3} with x . x = 1 is expressible as a unit vector quaternion X = { 0, x }, |X| = 1.10.) The spherical rotation of a quaternion vector X by an angle θ about a unit vector axis n is Q⊗X⊗Q,where Q is the quaternion {cos(θ/2), sin(θ/2) n }. Note that |Q| = 1.Notice the form of the quaternion vector product. Given vector quaternions X1 = { 0, x1 ) and X2 = { 0, x2 }, the quaternion product is X2⊗X1 = { x1.x2, x1 × x2 }. The quaternion reunites the dot product as the scalar part and cross product as the vector part, divorced over a hundred years ago. Neither of these products is invertible, but the quaternion is in the way described below.InversionFind the spherical transform quaternion Q12 to rotate vector X1 to align with vector X2.From aboveX2 = Q12⊗X1⊗Q12Multiplying both sides by X1,X2⊗X1 = Q12⊗X1⊗(Q12⊗X1)Remember that the rotation axis n derives from the cross product x1×x2, so n . x1 = 0. and Q⊗X = (X⊗Q) = X⊗Q, leavingX2⊗X1 = Q12⊗X1⊗X1⊗Q12 = Q12⊗Q12So the quaternion transform can be solved directly asQ12 = √(X2⊗X1)You're on your own for the quaternion square root. There are lots of ways to do it, and the best will depend on your application, considering speed and stability.--hth,Fred Klingener 这篇关于将单位向量转换为四元数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！