How to plot 3D plot from polar plot
18 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
David Harra
le 9 Août 2022
Commenté : William Rose
le 17 Août 2022
I am looking at plotting the the slowness profiles of Titinium (Inverser of velocity) by solving the christoffell equation.
I have managed to get a polar plot of this but would like to visualise this in 3d. For my polar plot I have done.
c11 = 162000000000;
c12=92000000000;
c44=47000000000;
c13= 69000000000;
c33= 181000000000;
c66= 35000000000
rho = 4430;
C=[c11,c12,c13,0,0,0;
c12,c11,c13,0,0,0;
c13,c13,c33,0,0,0;
0,0,0,c44,0,0;
0,0,0,0,c44,0;
0,0,0,0,0,c66];
phase_vel3=zeros(1,181);
r=0;
for theta=-pi:pi/180:pi
r=r+1;
n=[cos(theta),0,sin(theta)];
L_11=(C(1,1)*n(1)^2+C(6,6)*n(2)^2+C(5,5)*n(3)^2+2*C(1,6)*n(1)*n(2)+2*C(1,5)*n(1)*n(3)+2*C(5,6)*n(2)*n(3));
L_12=(C(1,6)*n(1)^2+C(2,6)*n(2)^2+C(4,5)*n(3)^2+(C(1,2)+C(6,6))*n(1)*n(2)+(C(1,4)+C(5,6))*n(1)*n(3)+(C(4,6)+C(2,5))*n(2)*n(3));
L_13=(C(1,5)*n(1)^2+C(4,6)*n(2)^2+C(3,5)*n(3)^2+(C(1,4)+C(5,6))*n(1)*n(2)+(C(1,3)+C(5,5))*n(1)*n(3)+(C(3,6)+C(4,5))*n(2)*n(3));
L_22=(C(6,6)*n(1)^2+C(2,2)*n(2)^2+C(4,4)*n(3)^2+2*C(2,6)*n(1)*n(2)+2*C(4,6)*n(1)*n(3)+2*C(2,4)*n(2)*n(3));
L_23=(C(5,6)*n(1)^2+C(2,4)*n(2)^2+C(3,4)*n(3)^2+(C(4,6)+C(2,5))*n(1)*n(2)+(C(3,6)+C(4,5))*n(1)*n(3)+(C(2,3)+C(4,4))*n(2)*n(3));
L_33=(C(5,5)*n(1)^2+C(4,4)*n(2)^2+C(3,3)*n(3)^2+2*C(4,5)*n(1)*n(2)+2*C(3,5)*n(1)*n(3)+2*C(3,4)*n(2)*n(3));
Christoffel_mat=[L_11, L_12, L_13;
L_12,L_22,L_23;
L_13,L_23,L_33];
[ev,d]=eig(Christoffel_mat);
% eigvecs = polarisation vectors
pl=ev(:,3);
% eigvals ~ slowness (phase)
phase_vel3(r)=sqrt(rho./d(3,3)); % quasi-longitudinal
r=0;
for theta=-pi:pi/180:pi
r=r+1;
n=[cos(theta),0,sin(theta)];
L_11=(C(1,1)*n(1)^2+C(6,6)*n(2)^2+C(5,5)*n(3)^2+2*C(1,6)*n(1)*n(2)+2*C(1,5)*n(1)*n(3)+2*C(5,6)*n(2)*n(3));
L_12=(C(1,6)*n(1)^2+C(2,6)*n(2)^2+C(4,5)*n(3)^2+(C(1,2)+C(6,6))*n(1)*n(2)+(C(1,4)+C(5,6))*n(1)*n(3)+(C(4,6)+C(2,5))*n(2)*n(3));
L_13=(C(1,5)*n(1)^2+C(4,6)*n(2)^2+C(3,5)*n(3)^2+(C(1,4)+C(5,6))*n(1)*n(2)+(C(1,3)+C(5,5))*n(1)*n(3)+(C(3,6)+C(4,5))*n(2)*n(3));
L_22=(C(6,6)*n(1)^2+C(2,2)*n(2)^2+C(4,4)*n(3)^2+2*C(2,6)*n(1)*n(2)+2*C(4,6)*n(1)*n(3)+2*C(2,4)*n(2)*n(3));
L_23=(C(5,6)*n(1)^2+C(2,4)*n(2)^2+C(3,4)*n(3)^2+(C(4,6)+C(2,5))*n(1)*n(2)+(C(3,6)+C(4,5))*n(1)*n(3)+(C(2,3)+C(4,4))*n(2)*n(3));
L_33=(C(5,5)*n(1)^2+C(4,4)*n(2)^2+C(3,3)*n(3)^2+2*C(4,5)*n(1)*n(2)+2*C(3,5)*n(1)*n(3)+2*C(3,4)*n(2)*n(3));
Christoffel_mat=[L_11, L_12, L_13;
L_12,L_22,L_23;
L_13,L_23,L_33];
[ev,d]=eig(Christoffel_mat);
% eigvecs = polarisation vectors
pl=ev(:,3);
% eigvals ~ slowness (phase)
phase_vel3(r)=sqrt(rho./d(3,3)); % quasi-longitudinal
figure
polar(0:pi/180:2*pi,1./phase_vel3)
Any assistance how to visualise this in 3d would be appreciated.
Thanks
Dave :)
0 commentaires
Réponse acceptée
William Rose
le 9 Août 2022
[moved this from Comment to Answer as I had originally intended]
Please post the simplest possible code fragment that clearly illustrates what you want to do. This is always a good idea, in order to raise the odss of getting a useful answer.
I cannot tell what you want to plot on the third dimension.
Matlab cannot directly plot in cylindrical or spherical coordinates, as far as I know. But it does have a built-in function to convert from cylindrical to Cartesian, which I use below.
z=0:200; theta=z*pi/25; r=1-z/400;
[x,y,z]=pol2cart(theta,r,z); %convert cylindrical to Cartesian coords
figure;
subplot(121); polar(theta,r); %polar plot
subplot(122); plot3(x,y,z); %3D plot
I hope that helps.
4 commentaires
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Polar Plots dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!