Rotation of curve question

I am trying to rotate a curve about a point (yellow asterisk) and do not obtain a symmetric result.
Xr = (x-XXc)*cos(delta) + (yp-YYc)*sin(delta) + XXc;
Yr = - (x-XXc)*sin(delta) + (yp-YYc)*cos(delta) + YYc;
Where is the mistake ?
clear all; close all;clc;
ah=10; bh=1; ch = sqrt(ah^2+bh^2); % hyperbola
delta =10*pi/180; % rotation angle
XXc = 30; YYc = 0; % rotaion about point XXc, YYc
x = linspace(20, 40, 10000);
yp = bh*sqrt((x-10).^2/ah^2-1);
ym = - bh*sqrt((x-10).^2/ah^2-1);
Xr = (x-XXc)*cos(delta) + (yp-YYc)*sin(delta) + XXc;
Yr = - (x-XXc)*sin(delta) + (yp-YYc)*cos(delta) + YYc;
Xrm = (x-XXc)*cos(delta) + (ym-YYc)*sin(delta) + XXc;
Yrm = - (x-XXc)*sin(delta) + (ym-YYc)*cos(delta) + YYc;
xxcf0 = 10+ah; yycf0 = 0; %vertex
Xrm0 = (xxcf0-XXc)*cos(delta) + (yycf0-YYc)*sin(delta) + XXc;
Yrm0 = - (xxcf0-XXc)*sin(delta) + (yycf0-YYc)*cos(delta) + YYc;
figure(1)
plot(Xr,Yr,'r',Xrm,Yrm,'r', Xrm0,Yrm0,'*', [Xrm0 XXc],[Yrm0 YYc])
hold on
plot(x,yp,'b',x,ym,'b', XXc,YYc,'*',10+ah,0,'*')
hold off

4 Comments

Why do you think there's a mistake? Your parabola looks symmetric to me.
Ole
Ole on 8 Aug 2019
Edited: Ole on 8 Aug 2019
The purple asterisk is the vertex of the non rotated curve. The blue asterisk should be the vertex of the new rotated curve.
The line between the yellow asterisk and the blue asterisk should be the new axis X'.
The curve is supposed to be symmetric around X'. It looks the rotated vertex (the blue asterisk) is not the vertex of the new rotated curve.
Did not look (yet) your code but often if you do not run "axis equal" in the graphics you might be missleaded by the scaling.
Ole
Ole on 8 Aug 2019
Edited: Ole on 8 Aug 2019
Thank you!
The scaling was the issue.

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Answers (1)

Jon
Jon on 8 Aug 2019
Edited: Jon on 8 Aug 2019
I think you also have a sign error in your formula.
Xr = (x-XXc)*cos(delta) + (yp-YYc)*sin(delta) + XXc;
% should be
Xr = (x-XXc)*cos(delta) - (yp-YYc)*sin(delta) + XXc;
and
Xrm = (x-XXc)*cos(delta) + (ym-YYc)*sin(delta) + XXc;
% should be
Xrm = (x-XXc)*cos(delta) - (ym-YYc)*sin(delta) + XXc;
Also I would suggest taking advantage of MATLAB's ability to work directly with matrices and vectors and first define a rotation matrix R as
R = [cos(theta) -sin(theta);sin(theta) cos(theta)]
and then you can just use matrix vector multiplies for example defining your rotated vector Vr as
Vr = R*[x-XXc;yp-YYc] + [XXc;YYc]

1 Comment

Oh, sorry, never mind about the sign error. I see now that you are just rotating clockwise and have the negative term on the sin that is applied to x. I would still recommend using the rotation matrices though as it is more compact and makes the code much more readable.

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R2016b

Asked:

Ole
on 8 Aug 2019

Commented:

Jon
on 8 Aug 2019

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