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# What is going wrong with my mesh command?

Asked by Gian Lorenzo on 6 May 2013

I am trying to visualize this function and eventually its contour lines:

```clear all
[x,y]=meshgrid(-10:0.1:10)
```
```d=max([sqrt(x.^2+y.^2);sqrt((x-5).^2+(y+1).^2);sqrt((x-4).^2+(y-6).^2);...
sqrt((x-1).^2+(y-3).^2)])+eps;
```
```mesh(d)
```

I get an error message saying Z should be a matrix. I looked at the example here http://www.mathworks.com/help/matlab/ref/mesh.html, and I don't understand why their function generates a matrix and mine don't. Altough mine is a 'weird' function with a max it shouldn't be conceptually too different, what am I missing here?

## Products

Answer by Iman Ansari on 6 May 2013
Edited by Iman Ansari on 6 May 2013

Hi.

```clear
[x,y]=meshgrid(-10:0.1:10)
```
```d=max(sqrt(x.^2+y.^2),sqrt((x-5).^2+(y+1).^2));
d=max(d,sqrt((x-4).^2+(y-6).^2));
d=max(d,sqrt((x-1).^2+(y-3).^2))+eps;
```
```mesh(x,y,d)
```

OR

```meshc(x,y,d)
```

Gian Lorenzo on 6 May 2013

Thanks! Can you explain me what's the wrong reasoning I am doing here? Does the max function handles at most two arguments? Is eps strictly necessary?

Iman Ansari on 6 May 2013

From Help: If A is a matrix, max(A) treats the columns of A as vectors, returning a row vector containing the maximum element from each column.

Your input in max function is 804*201, so output become 1*201.

No.

Answer by Walter Roberson on 6 May 2013
```d=max( cat(3, sqrt(x.^2+y.^2), sqrt((x-5).^2+(y+1).^2), sqrt((x-4).^2+(y-6).^2), sqrt((x-1).^2+(y-3).^2)), 3) + eps;
```

## 1 Comment

Gian Lorenzo on 6 May 2013

Never seen this cat function, very interesting, thank you