trasformation of a unit vector quiver3

Hi everyone , i would like to know if it is possibile to obtain a trasformation of quiver3 object . I have a normal quiver3 object and i want to trasform with my homogenous traformation matrix hM(4*4). i don't want modify my q_w but get another one. thank you very much.

q_w=quiver3(zeros(3,1),zeros(3,1),zeros(3,1),[1;0;0],[0;1;0],[0;0;1]);
q_w.LineWidth=3;
q_w.AutoScaleFactor=8;

rz=[ cos(psi) -sin(psi)  0 ;
        sin(psi) cos(psi)   0  ;
        0      0       1] ;
    
    ry=[ cos(theta)   0   sin(theta) ;
        0        1      0     ;
        -sin(theta)    0  cos(theta)];
    
    rx=[  1      0         0    ;
        0    cos(fi)  -sin(fi);
        0    sin(fi)  cos(fi)];
    
    rM=ry*rx*rz; % giusta
    % rMf=matlabFunction(rM);
    %creaiamo la nostra matrice omogenea

    transition=[x y z]';
    % transitionF=matlabFunction(transition);
    
    one=ones(1);
    
    hM= [ rM  transition ;
        zeros   one    ];

Answer 1

[Ameer Hamza]

Try this

vec1 = [1;0;0];
vec2 = [0;1;0];
vec3 = [0;0;1];
q_w=quiver3(zeros(3,1),zeros(3,1),zeros(3,1),vec1,vec2,vec3);
q_w.LineWidth=3;
q_w.AutoScaleFactor=8;

psi = pi/4;
theta = pi/3;
fi = pi/6;

rz=[ cos(psi) -sin(psi)  0 ;
    sin(psi) cos(psi)   0  ;
    0      0       1] ;

ry=[ cos(theta)   0   sin(theta) ;
    0        1      0     ;
    -sin(theta)    0  cos(theta)];

rx=[  1      0         0    ;
    0    cos(fi)  -sin(fi);
    0    sin(fi)  cos(fi)];

rM=ry*rx*rz; % giusta
figure;
q_w=quiver3(zeros(3,1),zeros(3,1),zeros(3,1),rM*vec1,rM*vec2,rM*vec3);
q_w.LineWidth=3;
q_w.AutoScaleFactor=8;

Also see eul2rotm(): https://www.mathworks.com/help/releases/R2020a/robotics/ref/eul2rotm.html to generate the rotation matrix.

[Andrea Gusmara]

i understan and how can i translate quiver3 object.

[Ameer Hamza]

Add a constant value to the x, y, and z coordinates. Following translate the vectors to (1 2 3)

q_w=quiver3(zeros(3,1)+1,zeros(3,1)+2,zeros(3,1)+3,rM*vec1,rM*vec2,rM*vec3);

[Andrea Gusmara]

ok i can follow you but when i try to apply this changes into my matlab program i encotered the following error.

Index in position 1 exceeds array bounds (must not exceed 1).

Error in prova (line 103)

q_w=quiver3(zeros(3,1)+x,zeros(3,1)+y,zeros(3,1)+z,rM*vec1,rM*vec2,rM*vec3);

This is my full program. i've tried for hours to understand why the error occursbut i can't fix the problem .

clear all 
close all


surf=gca;
xlim(surf,[-15 +15]);
xlabel('x');
zlim(surf,[-15 +15]);
zlabel('z');
ylim(surf,[-15 +15]);
ylabel('y');
daspect([1 1 1]);




dim=input('enter the dimensions of your cuboid [x,y,z]\n ');
dx=dim(1);
dy=dim(2);
dz=dim(3);


vertices=[-dx/2,-dy/2,0;-dx/2,+dy/2,0;dx/2,dy/2,0;dx/2,-dy/2,0; -dx/2,-dy/2,dz;-dx/2,+dy/2,dz;dx/2,dy/2,dz;dx/2,-dy/2,dz];
S.Vertices=vertices;
faces=[1,2,3,4;1,2,6,5;3,2,6,7;4,3,7,8;4,1,5,8;5,6,7,8];
S.Faces=faces;
color=[1 0 0; 0 1 0 ; 0 0 1 ; 0 1 1 ; 1 0 1 ; 1 1 0 ];
p=patch(S);
p.FaceVertexCData=color;
p.FaceColor="flat";
hold('on');
q_w=quiver3(zeros(3,1),zeros(3,1),zeros(3,1),[1;0;0],[0;1;0],[0;0;1]);
q_w.LineWidth=3;
q_w.AutoScaleFactor=8;


wrt=eye(4);
rsb=eye(4);
zeros=zeros(1,3);
continua=true;
count=0;
while continua
    
    
    rs=input('insert body if you want to transform with respect to the body reference system or insert world if you want with respect to the world  \n ');
    
    dof=input('enter the transformation values [fi,theta,psi,x,y,z] \n ');

    
    
    
    fi=deg2rad(dof(1));
    theta=deg2rad(dof(2));
    psi=deg2rad(dof(3));
    x=dof(4);
    y=dof(5);
    z=dof(6);
    
    rz=[ cos(psi) -sin(psi)  0 ;
        sin(psi) cos(psi)   0  ;
        0      0       1] ;
    
    ry=[ cos(theta)   0   sin(theta) ;
        0        1      0     ;
        -sin(theta)    0  cos(theta)];
    
    rx=[  1      0         0    ;
        0    cos(fi)  -sin(fi);
        0    sin(fi)  cos(fi)];
    
    rM=ry*rx*rz; % giusta
    % rMf=matlabFunction(rM);
    %creaiamo la nostra matrice omogenea

    transition=[x y z]';
    % transitionF=matlabFunction(transition);
    
    one=ones(1);
    
    hM= [ rM  transition ;
        zeros   one    ];
    
    if strcmp(rs,'body')
        rsb=rsb*hM;
        newVertices=rsb*[vertices ones(1,8)']';
    else
        rsb=wrt*hM;
        newVertices=rsb*[vertices ones(1,8)']';
    end
    
    count=count+1;
    newVertices=newVertices';
    S.Vertices=[S.Vertices ; newVertices(:,1:3)];
    S.Faces=[S.Faces ; faces+(8*count)];
    p=patch(S);
    for j=0:count
        p.FaceVertexCData=[p.FaceVertexCData ; color];
    end
    p.FaceColor="flat";
     hold('on');
    q_=bquiver3(zeros(3,1)+x,zeros(3,1)+y,zeros(3,1)+z,rM*vec1,rM*vec2,rM*vec3);
    q_b.LineWidth=3;
    q_b.AutoScaleFactor=8;
    
    
    risp=input('do you wish to continue? (yes,no) \n ');
    if strcmp(risp,'yes')
        continua=true;
    else
        continua=false;
    end
end

[Ameer Hamza]

What inputs did you use for input() statements?

[Andrea Gusmara]

[0,45,0,5,5,5]

[Ameer Hamza]

You haven't defined vec1, vec2, and vec3, as I have defined in the answer.

[Andrea Gusmara]

yes you are right , i forgot it . but now , the error persists.

[Ameer Hamza]

You have used a line

zeros=zeros(1,3);

But zeros is also the name of the MATLAB function. Change the name of the variable here and everywhere else in the code.

[Andrea Gusmara]

q_w=quiver3(zeros(3,1),zeros(3,1),zeros(3,1),[0.7071;0;-0.7071],[0;1;0],[0.7071;0;0.7071]);

Index in position 1 exceeds array bounds (must not exceed 1).

this is the istruction with the unit_vector rotated.

But the strange thing is also without rotation of the unit_vectors i encontered the error

Index in position 1 exceeds array bounds (must not exceed 1).

Error in roto_traslazione_spazio (line 106)

q_w=quiver3(zeros(3,1),zeros(3,1),zeros(3,1),vec1,vec2,vec3);

[Ameer Hamza]

Have you modified the name of variable zeros? Also run

clear zeros

to remove the variable from workspace.

[Andrea Gusmara]

you are right , the problem was the variable zeros , thank you so much

for your patience and attention.

[Ameer Hamza]

I am glad to be of help.