How to plot an orientated rectangle having coordinates and angle?
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Hello everybody,
I have a simple problem that I cannot manage to solve in Matlab.
I am representing the behavior of a vehicle traveling on a given car track and I would like to plot orientated rectangles instead of the standard markers present in the plot environment. I am able to reproduce the orientation through the quiver function, but I am not able to use rectangles instead of the arrows.
Note that I would like to have a rectangle for every instant k (i.e.: if I have 10 instants, ten different rectangles for every position).
This is the code I use for the moment with quiver:
scale_factor = 10;
r = 1; % magnitude (length) of arrow to plot
x = x_c; y = y_c;
angle = alpha+beta;
u = r * cos(angle); % convert polar (theta,r) to cartesian
v = r * sin(angle);
h = quiver(x,y,u*scale_factor,v*scale_factor,'--','LineWidth',3,'AutoScale','off');
I have also tried this one translating manually each point but it does not work:
function[]= draw_rectangle1(x_c,y_c,theta)
L = 20;
H = 10;
x_low_left = x_c - L/2;
x_low_right = x_c + L/2;
x_up_right = x_c + L/2;
x_up_left = x_c - L/2;
y_low_left = y_c - H/2;
y_low_right = y_c - H/2;
y_up_right = y_c + H/2;
y_up_left = y_c + H/2;
x_low_left_new = cos(theta)*(x_low_left-x_c)-sin(theta)*(y_low_left-x_c)+x_c;
y_low_left_new = sin(theta)*(x_low_left-x_c)+cos(theta)*(y_low_left-x_c)+y_c;
x_low_right_new = cos(theta)*(x_low_right-x_c)-sin(theta)*(y_low_right-x_c)+x_c;
y_low_right_new = sin(theta)*(x_low_right-x_c)+cos(theta)*(y_low_right-x_c)+y_c;
x_up_right_new = cos(theta)*(x_up_right-x_c)-sin(theta)*(y_up_right-x_c)+x_c;
y_up_right_new = sin(theta)*(x_up_right-x_c)+cos(theta)*(y_up_right-x_c)+y_c;
x_up_left_new = cos(theta)*(x_up_left-x_c)-sin(theta)*(y_up_left-x_c)+x_c;
y_up_left_new = sin(theta)*(x_up_left-x_c)+cos(theta)*(y_up_left-x_c)+y_c;
x_coor=[x_low_left_new x_low_right_new x_up_right_new x_up_left_new];
y_coor=[y_low_left_new y_low_right_new y_up_right_new y_up_left_new];
fill(x_coor, y_coor,'r');
end
I have also given a chance to patch but no way.
If anyone can help me with this simple but tricky code it would be awesome.
Thank you!
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采纳的回答
Jim Riggs
2019-1-17
编辑:Jim Riggs
2019-1-17
You have to perform the rotation in an object-centered coordinate frame.
Define the rectangle:
X = [-L/2 L/2 L/2 -L/2];
Y = [-H/2 -H/2 H/2 H/2];
Now perform the rotation in the object-centered frame:
theta = % The rotation angle
cth = cos(theta) ;
sth = sin(theta);
Xrot = X*cth + Y*sth;
Yrot = -X*sth + Y*cth;
Now when you generate the graphic, add the true position:
fill(Xrot + Xc, Yrot + Yc,'r')
2 个评论
Jim Riggs
2019-1-22
I thought about that detail after I posted my answer. I defined theta to be positive for a clockwise rotation. If you define theta to be positive for as counter-clockwise rotation, then you can either use the minus sign (as you suggest) or you can use the inverse rotation:
Xrot = X*cth - Y*sth;
Yrot = X*sth + Y*cth;
This will perform the rotation in the oposite direction.
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