[Robotics: Aerial Robotics][quiz]2.1

Question1：

旋轉矩陣是酉矩陣， 滿足 R*R^(T) = I

 r = [0.212 0.7743 0.5963; 0.212 -0.6321 0.7454;0.9540 -0.0316 -0.2981];

r*r' =  

1.0001   -0.0000    0.0000

   -0.0000    1.0001    0.0000

    0.0000    0.0000    1.0000

是以第一個是對的

 r = [2^(0.5)/2, 0, 2^(0.5)/2;0,1,0;-2^(0.5)/2, 0, 2^(0.5)/2];

 r*r'

ans =

    1.0000         0         0

         0    1.0000         0

         0         0    1.0000

第二個也是對的

第三個是二維的旋轉矩陣，滿足要求。

第四個元素中出現絕對值大于1的元素不符合要求。

Question2

建立函數：

function [ mat ] = getRzyz( thetaZ1, thetaY, thetaZ2 )
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
mat1 = [cos(thetaZ1), -sin(thetaZ1), 0;
    sin(thetaZ1), cos(thetaZ1), 0;
    0, 0, 1];

mat2 = [cos(thetaY), 0, -sin(thetaY);
    0,1,0;
    sin(thetaY), 0, cos(thetaY)];
mat3 = [cos(thetaZ2), -sin(thetaZ2), 0;
    sin(thetaZ2), cos(thetaZ2), 0;
    0, 0, 1];
mat = mat1*mat2*mat3;
end
           

getRzyz(0.2,0.1,0.6)符合

Question3

根據：

計算

 omega_b = [0,-1,0.9689;1,0,-0.2474;-0.9689,0.2474,0]

 r = [0.675,-0.1724,0.7174;0.2474,0.9689,0;-0.6951,0.1775,0.6967]

r_dot = r*omega_b

omega_s = r_dot*r'

          0   -0.6967    1.0000

    0.6967         0   -0.7174

   -1.0000    0.7174    0.0000

Question4

角度固定在了[0,pi]之間，那麼除了特例的情況下，是有解的

r = [0.2919,0.643,-0.7081;-0.643, -0.4161, -0.643;-0.7081, 0.643, 0.2919]

trace(r) = 0.1677;

acos((trace(r)-1)*0.5) 

2.0

0.5/sin(2)*(r-r')

ans =

  0    0.7071         0

   -0.7071         0   -0.7071

   0    0.7071         0

u = [0.7071, 0, -0.7071]

Question5

trace(r) = -1;

cos(theta) = (-1-1)/2 = -1

theta = pi;

轉軸 為u或是-u

沒有唯一解

Question6