Teory of Applied Robotics (Successive rotation about global axes – Penyelesaian soal)

1. Exercise/Example 3

 

Figure 2.5. Point P on slab

Tema: Successive rotation about global axes.

The final position of the corner P (5, 30, 10) of the slab shown in Figure 2.5 after 30deg rotation about the Z-axis, followed by 30deg about the X-axis, and then 90deg about the Y-axis can be found by first multiplying QZ, 30 by [5, 30, 10]T to get the new global position after first rotation

% Skrip MATLAB 2011b

% diketahui

% definisi variabel untuk sudut P dari slab/papan

p1 = 5;

p2 = 30;

p3 = 10;

% matrik P =

P = [p1;p2;p3];

%——————————–

 

% Z_axis = 30 derajat pada sumbu z

% X_axis = 30 derajat pada sumbu x

% Y_axis = 90 derajat pada sumbu y

 

Z_axis = pi/6; %30 derajat pada sumbu z

X_axis = pi/6; %30 derajat pada sumbu x

Y_axis = pi/2; %90 derajat pada sumbu y

 

% pertanyaan: Posisi P setelah rotasi pada

% sumbu z=30 derajat, dilanjutkan sumbu x=30 derajat, dan

% sumbu y=90 derajat.

 

%—Penyelesaian————————————–

% Maka global position setelah rotasi pertama sumbu z

% |X2|  |cos(Z_axis) -sin(Z_axis) 0|

% |Y2| =|sin(Z_axis)  cos(Z_axis) 0| * [P] = ?

% |Z2|  |0            0           1|

 

%Pembentukan matrik Pz

Pz = [cos(Z_axis) -sin(Z_axis) 0;

sin(Z_axis)  cos(Z_axis) 0;

0            0           1];

 

Qz_30 = Pz * P;

 

%———————————————

% kemudian dirotasi pada sumbu x=30 derajat

% Maka global position setelah rotasi kedua sumbu x

% |X3|  |1      0            0     |

% |Y3| =|0 cos(X_axis) -sin(X_axis)| * [Qz_30]= ?

% |Z3|  |0 sin(X_axis)  cos(X_axis)|

 

 

 

 

%Pembentukan matrik Px

Px = [1 0 0;

0 cos(X_axis) -sin(X_axis);

0 sin(X_axis)  cos(X_axis)];

 

Qx_30 = Px * Qz_30;

 

%———————————————

% kemudian dirotasi pada sumbu y=90 derajat

% Maka global position setelah rotasi kedua sumbu x

% |X4|  |cos(Y_axis)   0   sin(Y_axis)|

% |Y4| =|    0         1        0     | * [Qx_30] = ?

% |Z4|  |-sin(Y_axis)  0   cos(Y_axis)|

 

%Pembentukan matrik Py

Py = [cos(Y_axis)  0 sin(Y_axis);

0            1 0;

-sin(Y_axis) 0 cos(Y_axis)];

 

Qy_90 = Py * Qx_30;

 

%———–selesai——————–

 

Hasil:

Posisi titip P setelah dirotasi pada sumbu z sebesar 30’ adalah:

 

 

 

Kemudian posisi titip P setelah dirotasi pada sumbu x sebesar 30’ adalah:

 

 

 

 

 

Kemudian posisi titip P setelah dirotasi pada sumbu y sebesar 90’ adalah:

 

 

2. Exercise/example 198

Rotation of a body point about a global axis. The slab shown in Figure 2.5 is turning about the Z-axis with 10 deg /s. The global velocity of the corner point P(5, 30, 10), when the slab is at = 30 deg, is:

 

 

% Skrip MATLAB 2011b

% definisi variabel untuk konstanta GRB

a1=sin(pi/6);  a2=-cos(pi/6);  a3=0;

a4=cos(pi/6);  a5=-sin(pi/6);  a6=0;

a7=0;          a8=0;           a9=0;

 

%matrik GRB nya

GRB = [a4 a5 a3;

a1 a4 a6;

a7 a8 a9];

 

GRBt = [a5 a2 a3;

a4 a5 a6;

a7 a8 a9];

 

 

%definisi variabel untuk BrP

b1=5;

b2=30;

b3=10;

BrP = [b1;b2;b3];

 

%nilai sudut putaran = alpha

a=10;

alpha=(a*pi)/180;

 

% global velocity of the corner point

GvP = alpha * GRBt *BrP;

 

% global positionnya

GrP = GRB * BrP;

 

%———–selesai——————————

Hasil:

à global velocity of the corner P point

à global position

 

 

 

 

 

 

% Example 72 Moving body coordinate frame.

% Figure 4.2 shows a point P at BrP = 0.1i + 0.3j + 0.3k in a body frame B, which is rotated 50deg about the Z-axis, and translated 1 along X, 0.5 along Y, and 0.2 along the Z axes.

% The position of P in global coordinate frame is:

% Gr = (GRB * BrP) + Gd

 

%Pengujian Example 72

%definisi variabel untuk konstanta GRB

i = 0.1;

j = 0.3;

k = 0.3;

%BrP = poin P pada body frame

BrP = [i;j;k];

%——————————–

az = 50; %50 derajat pada sumbu z

a1=cos(az);  a2=-sin(az); a3=0;

a4=sin(az);  a5=cos(az);  a6=0;

a7=0;        a8=0;        a9=1;

 

%matrik GRB nya

GRB = [a1 a2 a3;

a4 a5 a6;

a7 a8 a9];

%——————————–

tx = -1;

ty = 0.5;

tz = 0.2;

 

%matrik Gd = nilai translasi (translation value)

Gd = [tx;ty;tz];

 

% The position of P in global coordinate frame is

Gr1 = GRB;

Gr2 = BrP;

Gr = GRB * (BrP + Gd);

 

 

 

 

 

 

 

 

 

% Example 198 Rotation of a body point about a global axis.

% The slab shown in Figure 2.5 is turning about the Z-axis with % a = 10 deg /s. The global velocity of the corner point P(5, 30, 10), when the slab is at ? = 30 deg, is:

 

%Pengujian Example 198

%definisi variabel untuk konstanta GRB

a1=sin(pi/6); a2=-cos(pi/6);  a3=0;

a4=cos(pi/6);  a5=-sin(pi/6);  a6=0;

a7=0;          a8=0;           a9=0;

 

%matrik GRB nya

GRB = [a4 a5 a3;

a1 a4 a6;

a7 a8 a9];

 

GRBt = [a5 a2 a3;

a4 a5 a6;

a7 a8 a9];

 

%definisi variabel untuk BrP

b1=5;

b2=30;

b3=10;

BrP = [b1;b2;b3];

 

%nilai sudut putaran = alpha

a=10;

alpha=(a*pi)/180;

 

%Global positionnya

GrP = GRB * BrP;

 

% global velocity of the corner point

GvP = alpha * GRBt *BrP;

%——————————————————–

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3. Exercise No.2 halaman 81 (Body point and global rotations.)

The point P is at BrP = [1, 2, 1]T in a body coordinate B(Oxyz).
Find the final global position of P after

1. A rotation of 30 deg about the X-axis, followed by a 45 deg rotation about the Z-axis
2. A rotation of 30 deg about the Z-axis, followed by a 45 deg rotation about the X-axis.

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