How to plot 3D plot from polar plot

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 :)

Answer 1

[William Rose]

[moved this from Comment to Answer as I had originally intended]

@David Harra,

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.

[David Harra]

Hi William,

Apologies for the clustered post. Maybe I am misunderstanding what exactly I am needing.

So, if I plot my slownes curve using

figure
polar(0:pi/180:2*pi,1./phase_vel3)

I get something that looks like this

WHat I am trying to achieve , but not sure how to go about it should look like the image below. Velocity varying in z-direction

Thanks

Dave :)

[William Rose]

@David Harra,

I know you already know this, but anyway, let me say it anyway, to motivate the question I still have:

The polar plot you showed is a plot of r versus theta, where r=1/v. The second plot is a sphere that satisfies x^2+y^2+z^2=360, and the color of the sphere is assigned based on the z-coordinate of the surface.

I assume you want x and y to be assigned as the polar coordinates of theta and r, where r=1/v. What value do you want to assign to z, in your 3D plot?

[David Harra]

Hi William

I think that is my issue/ lack of understanding.

What you siad makes perfect sense, but I actually thought 1/v would be plotted in the z-direction as this shows the variation in velocity. I might need to revisit some of the maths etc.

Thanks for the input- its massively appreciated.

Thanks

Dave :)

[William Rose]

@David Harra, you're welcome.