Is it possible to express any planar transformation in terms of rotation?

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Discussion Overview

The discussion centers on the possibility of expressing any planar transformation solely in terms of rotation. Participants explore the implications of mixed transformations, which include both rotation and translation, on the radius of curvature of trajectories in a plane. The conversation involves theoretical considerations and practical calculations related to these transformations.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Priyanshu describes a discrepancy in the radius of curvature calculated using different methods for trajectories generated by mixed transformations compared to those generated solely by rotation.
  • Some participants propose that in 2 dimensions, any transformation can be expressed as a rotation, except for pure translation, which results in an infinite radius of curvature.
  • Others argue that in 3 dimensions, a combination of rotation and translation does not equate to a simple rotation, citing counterexamples.
  • Priyanshu questions why the radius of curvature varies significantly with increasing complexity in the curve when using mixed transformations, particularly when the translation is a function of rotation.
  • One participant requests to see the calculations used to derive the radius of curvature.
  • Priyanshu shares MATLAB code used for calculating the radius of curvature for transformed points of a square.
  • A later reply draws an analogy to navigational problems on the Earth's surface, discussing the implications of changing bearings versus maintaining a single direction.
  • Another participant clarifies that the discussion is focused on transformations in a 2D XY plane, not on navigational routes.

Areas of Agreement / Disagreement

Participants express differing views on the nature of planar transformations and the implications for radius of curvature calculations. There is no consensus on whether all planar transformations can be expressed solely in terms of rotation, particularly when considering mixed transformations.

Contextual Notes

Participants note that the calculations and conclusions may depend on specific assumptions about the nature of the transformations and the definitions used in the analysis. The discussion also highlights the complexity introduced by varying translation in conjunction with rotation.

priyansh
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Hi,

I'm am trying to calculate the radius of curvature of a trajectory in a plane. I am using Frenet-Serret formulas and perpendicular bisector approach to calculate the radius. It comes out that both the approaches lead to same answer when I calculate the radius for a trajectory generated using just rotation transformation. However, when I calculate the radius for a trajectory generated using mixed transformation (involving both rotation and translation) the answers are significantly different. This seems to me as strange as the trajectory must have only one radius at each point though it might vary along the arc length. So, the question boils down to whether it is possible to express any general planar transformation by just rotation?

Thanks.

Regards,
Priyanshu
 
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Any comments guys!
 
What do you mean by "a trajectory generated using mixed transformation (involving both rotation and translation)"?
 
I mean the homogeneous transformation matrix contains entry for both rotation and translation. Also, the rotation and translation vary as the trajectory is generated.
 
In 2 dimensions, the answer is yes except for the case of translation without rotation (infinite radius of curvature).

In 3 dimensions, a transformation that consists of a nontrivial rotation and a translation is not necessarily equivalent to a simple rotation. A counterexample is a rotation about some axis combined with translation along the same axis.
 
Thanks hamster. But I'm not clear about why is the radius obtained using both the approaches are significantly different when I increase the complexity of the curve in 2 dimensions. Say for t degrees of rotation, I translate by t^n and I vary t from 0 to 100. As I increase the value of n, the answer from both the approaches vary significantly. However, for a rotation+constant translation the radius by two approaches match exactly.
 
can you show your calculations?
 
Here is the MATLAB Code. Basically I'm applying the calculations to four corners of a square which is being transformed.

clear all
close all
clc

dt = 1;
Scale = 3; % Just defined to care of the plotting stuff
Sx = [0 0 1 1 0];
Sy = [0 1 1 0 0];
Sz = [0 0 0 0 0];
P1 = [];
P2 = [];
P3 = [];
P4 = [];
K = [];
X = [];
Y = [];
Z = [];
Radius = [];
Radius_PB = [];
Tprev = 0;

figure
plot(Sx,Sy)
axis(2*[-10 10 -10 10]*Scale);
grid on;
P = 2*Scale*[Sx; Sy; Sz;ones(size(Sx))];
Pprev = P;

for t = 1:dt:80
% R = makehgtform('zrotate',deg2rad(t),'translate',[(t/40) (t/40) 0]);
R = makehgtform('zrotate',deg2rad(t),'translate',[(t/50)^3 (t/50) 0]); % Homogeneous Transformation matrix
% R = makehgtform('zrotate',deg2rad(t),'translate',[2 4 0]);
% R = makehgtform('zrotate',deg2rad(t));
Pnew = R*P; % Evaluation of transformed points
P1 = [P1 Pnew(1:2,1)];
P2 = [P2 Pnew(1:2,2)];
P3 = [P3 Pnew(1:2,3)];
P4 = [P4 Pnew(1:2,4)];
PC = [P1 P2 P3 P4];

%Radius calculation using Frenet-Serret formulas
dr = (Pnew-Pprev); %Calculating dr
Tnew = [dr(:,1)/(sum(dr(:,1).^2))^0.5,...
dr(:,2)/(sum(dr(:,2).^2))^0.5,...
dr(:,3)/(sum(dr(:,3).^2))^0.5,...
dr(:,4)/(sum(dr(:,4).^2))^0.5,...
dr(:,5)/(sum(dr(:,5).^2))^0.5]; %Calculating dr/ds
dT = (Tnew-Tprev);
dTds = [dT(:,1)/(sum(dr(:,1).^2))^0.5,...
dT(:,2)/(sum(dr(:,2).^2))^0.5,...
dT(:,3)/(sum(dr(:,3).^2))^0.5,...
dT(:,4)/(sum(dr(:,4).^2))^0.5,...
dT(:,5)/(sum(dr(:,5).^2))^0.5];

k = [(sum(dTds(:,1).^2))^0.5,...
(sum(dTds(:,2).^2))^0.5,...
(sum(dTds(:,3).^2))^0.5,...
(sum(dTds(:,4).^2))^0.5]
radius = [1/k(1),1/k(2),1/k(3),1/k(4)]
K = [K; k];
Radius = [Radius; radius];

%Radius calculation using Perperndicular bisector method
drxyz = [dr(1:3,1), dr(1:3,2), dr(1:3,3), dr(1:3,4)];
Z = repmat([0; 0; 1],1,4);
Nr = cross(drxyz,Z);
Nr = [Nr(1:2,1)/sum(Nr(1:2,1).^2)^0.5, -Nr(1:2,2)/sum(Nr(1:2,2).^2)^0.5,...
Nr(1:2,3)/sum(Nr(1:2,3).^2)^0.5, -Nr(1:2,4)/sum(Nr(1:2,4).^2)^0.5];
MP = (Pnew+Pprev)/2;
DMP12 = [MP(1:2,2)-MP(1:2,1)];
DMP34 = [MP(1:2,4)-MP(1:2,3)];
lambda = [inv(Nr(:,1:2))*DMP12, inv(Nr(:,3:4))*DMP34]
r1 = MP(1:2,1)+lambda(1)*Nr(:,1)-Pnew(1:2,1); %Calculating the radius for 1st corner
r2 = MP(1:2,2)+lambda(2)*Nr(:,2)-Pnew(1:2,2);
r3 = MP(1:2,3)+lambda(3)*Nr(:,3)-Pnew(1:2,3);
r4 = MP(1:2,4)+lambda(4)*Nr(:,4)-Pnew(1:2,4);
Radius_PB = [Radius_PB; sum(r1.^2)^0.5, sum(r2.^2)^0.5,sum(r3.^2)^0.5, sum(r4.^2)^0.5 ]

%Plotting and annotating
plot(Pnew(1,:),Pnew(2,:),'-k',P1(1,:),P1(2,:),'-r',P2(1,:),P2(2,:),'-b',P3(1,:),...
P3(2,:),'-g',P4(1,:),P4(2,:),'-y');
text(Pnew(1,1),Pnew(2,1),'1');
text(Pnew(1,2),Pnew(2,2),'2');
text(Pnew(1,3),Pnew(2,3),'3');
text(Pnew(1,4),Pnew(2,4),'4');
axis(2*[-10 10 -10 10]*Scale);
grid on;
pause(0.01);
Pprev = Pnew;
Tprev = Tnew;
% pause;
end

figure
s = length(K)
h = plot(1:s,K(1:s,1),'-b',1:s,K(1:s,2),'-r',1:s,K(1:s,3),'-g',1:s,K(1:s,4),'-y');
axis tight
xlabel('Sample No \rightarrow','fontsize',12)
ylabel('\kappa \rightarrow','fontsize',12);
legend(h,'Vertex 1','Vertex 2','Vertex 3','Vertex 4');
grid on

figure
h = plot(1:s,Radius(1:s,1),'-b',1:s,Radius(1:s,2),'-r',1:s,Radius(1:s,3),'-g',1:s,Radius(1:s,4),'-y');
axis tight
xlabel('Sample No \rightarrow','fontsize',12)
ylabel('Radius of Curvature \rightarrow','fontsize',12);
legend(h,'Vertex 1','Vertex 2','Vertex 3','Vertex 4');
grid on

figure
h = plot(1:s,Radius_PB(1:s,1),'-b',1:s,Radius_PB(1:s,2),'-r',1:s,Radius_PB(1:s,3),'-g',1:s,Radius_PB(1:s,4),'-y');
axis tight
Title(' Radius of Curvature from perpendicular bisector','fontsize',12);
xlabel('Sample No \rightarrow','fontsize',12)
ylabel('Radius of Curvature \rightarrow','fontsize',12);
legend(h,'Vertex 1','Vertex 2','Vertex 3','Vertex 4');
grid on
 
Is this related to the navigational problem of moving about the surface of the earth, at sea, say? There are many routes from A to B. One of them requires the navigator to set a single bearing direction, but, in general this is not the shortest route, which would be part of a great circle. The shortest route requires the navigator to keep changing the bearing.
 
  • #10
No, this is just a square in an XY plane which is being transformed in plane.
 

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