Complex question: Want to find [x,y] when z > 0.5 from z of other surfaces.

For instance, I have three basic surface plots:

x = [-10:10]; y = [-10:10]; test1 = x + 0.5*y; test2 = x + y; test3 = x + 1.5*y;

Which equate to:

test1 =

Columns 1 through 9

-15.0000  -13.5000  -12.0000...

test2 =

Columns 1 through 9

   -20   -18   -16...

test3 =

 Columns 1 through 13

-25.0000  -22.5000  -20.0000...

Goal: I want to find out [x,y] when test2 deviates from test1 AND test3 by > 4.5.

So in the example above, I would want to know the [x,y] when test2 = -20, because clearly it is when test2 deviates from test1 AND test3 by > 4.5.

(i.e., column1 in all three equations is where absolutevalue(-20-(-15)) AND absolutevalue(-20-(-25) is > 4.5; so I want to know at what [x,y] this occurs)

Thank you!

Answer 1

[Star Strider]

This works:

x = [-10:10]; 
y = [-10:10]; 
test1 = @(x,y) x + 0.5*y; 
test2 = @(x,y) x + y; 
test3 = @(x,y) x + 1.5*y;

[X,Y] = meshgrid(x,y);

T1 = test1(X,Y);                    % Generate surfaces
T2 = test2(X,Y);
T3 = test3(X,Y);

D = (abs(T2-T1)+abs(T2-T3)) > 4.5;  % Differences

[Dr,Dc] = find(D);                  % Indices where D == 1

figure(1)
surf(X,Y,T1)
hold on
surf(X,Y,T2)
surf(X,Y,T3)
hold off
grid on

figure(2)
surf(X, Y, double(D))

[A]

Dear Star Strider,

That is PERFECT!! just what I wanted.

Just one question:

Do you think this:

D = (abs(T2-T1)+abs(T2-T3)) > 4.5; % Differences

Can somehow become this:

D = (abs(T2-T1) && abs(T2-T3)) > 4.5; % Differences

Because I want T2-T1 to be > 4.5 AND I want T2-T3 > 4.5; not the summation of them?

Would I need an if statement?

Thanks

[A]

x = [-10:10]; 
y = [-10:10]; 
test1 = @(x,y) x + 0.5*y; 
test2 = @(x,y) x + y; 
test3 = @(x,y) x + 1.5*y;
[X,Y] = meshgrid(x,y);
T1 = test1(X,Y);                    % Generate surfaces
T2 = test2(X,Y);
T3 = test3(X,Y);
*D = (abs(T2-T1) > 4.5) & (abs(T2-T3) > 4.5);  % Differences* 
[Dr,Dc] = find(D);                  % Indices where D == 1
figure(1)
surf(X,Y,T1)
hold on
surf(X,Y,T2)
surf(X,Y,T3)
hold off
grid on
figure(2)
surf(X, Y, double(D))

How about this? I think this works.

[Image Analyst]

Please mark his answer as Accepted then.

[Star Strider]

A — Your restatement for D will do what you want.

Image Analyst — Thank you!

[A]

Hi Star Strider,

I hope you are well. I had a follow-up question and request for help.

Presently, the graph I get is binary. I can find out areas where z is > 4.5 (denoted by 1) and areas where it isn't > 4.5 (denoted by 0).

Is it possible to find out the actual difference instead of getting a binary value of 1 or 0? So I want to only find values that are > 4.5, but it's find if they remain 0 where it is < 4.5?

Does this make sense what I'm asking for?

Thank you

[Star Strider]

II’m not certain I understand.

However if you want to find the actual differences rather than the logical test for the threshold, simply remove the threshold condition and other logical operators. So D becomes:

D = (abs(T2-T1)+abs(T2-T3));

It now returns a double result and will be plotted in figure(2) as a continuous surface. The rest of the code remains the same.

[A]

Thank you! I think that works. I used that in combination with the min() function to get it working.

[Star Strider]

My pleasure!