Hi, John.
I've just had a quick skim through your worksheet. Wow, that's a lot of work, and it would take me a long time to get to grips with it.
However, I think I might have spotted the cause of your unexpected nested vector of 49x1 vectors:
Vectorization only applies to the vectors it sees in the variables/explicity defined matrices. It does not apply to the results of an expression that uses those vectorized variables. Hence, the behaviour you see in the clip above.
Stuart
PS. Edited to add that I used Mathcad Prime 10 to view your M15 worksheet.
Are you sure that L.xx(1,2) should be a vector but g.xy(1,2) is just a scalar?
I guess L.xx(1,2) should be a scalar, too. Otherwise it would be wrong to call an expression which contains L.xx using vectorization.
Its a good idea to test functions which expect scalar arguments if the result is as expected before you call them vectorized with vector arguments.
According the cross product I am a little bit confused what it should denote. As far as I know a cross product of two vectors is only defined for 3 and for 7 dimensions!
Can't comment on the vector n
You promised a working Prime 11 worksheet which you did not attach - guess you meant Mathcad 11 as Prime 11 does not yet exist in public use 😉
And if its a Mathcad 11 sheet, you should be able to open it in Mathcad 15 (or 14) and use it !?
Just a comment on the polynomial fit of your data. I don't understand why the data is rounded down to 0 decimals before doing the polynomial fit.
Furthermore the coefficients for the polynomial fit could also be calculated by using polyfitc, the function defined via polyfit.
There seems to be no need for the spline interpolation.
But then, I have not looked in full detail in your sheet, so I may be wrong, indeed.
Mcad ts supplier an off done in 11.
Sorry, I don't understand.
I’ll post the pdf.
I understood that you have a working Mathcad or Prime worksheet which would be more valuable than a pdf.
Still waiting for an explanation what this vector n should denote.
As already written, a cross product is only defined for vectors in exactly three or seven dimensions.
It would make sense if n should be a normal vector of the surface, but then it would not be constant but rather depends on x and y
Still confused and wondering why you don't post the pdf file !??
As I already had written here Re: Curvature of a surface MCAD 15 M050 - PTC Community, your normal vector should be a function in x and y and not a constant as you tried it to define.
So you should use what you have written here
but without the equal sign at the end and with n'(x0',y0):= at the left hand side. Its a function, not a constant.
I also wonder why you call the first function argument x.0' and the second only y.0.
Looks quite strange to me, but then the name of the formal arguments of a function do not matter at all.
Being lazy I would simply use
Keep in mind that x and y here have nothing to do with the vectors defined earlier - they are just the formal function arguments.
You may equally use a and b or any other names here.
The same applies to the other functions like
I would define them without vectorization and apply vectorization only when you call the function with vectors as arguments.
BTW, is there any reason why you named the first partial derivative of f wrt x x'.0 and not simply f.x ? I found the sheet much harder to read with names like that.
Furthermore I am confused about what L.xx etc, should be
You write as an explanation
which would mean that L.xx is a scalar, but when you define L.xx you additionally multiply it with vector n, which of course makes L.xx a 3x1 vector, which I guess it should not be. Especially as you later define a matrix L(x,y) with L.xx etc. as its arguments. What you define in Mathcad as L(x,y) seems to have nothng to do with the determinant of the expression shown to the left of it, nor with what is shown in the picture to the right of it as your definition uses the first derivatives and not the second ones.
I guess you misinterpreted this handwritten formula
Here X is a vector (x,y,f(x,y)^T and a second derivate (no matter wrt which variable) means you get (0, 0, f.xi,xj)). Scalar multiplication by unit vector n simply results in f.xi,xj / sqrt(....).
So it should be that way:
Similar
and then
Enjoy simple, not nested result vectors and real result for the quadratic equation
Conclusion: Once you have your function f(x,y) forget about the data vectors X, Y or x' and y' and set up your calculations as functions in simple scalar arguments x and y. Only when you need the results for your vectors call those function (vectorized) with the data vectors as their arguments.
Hope this helps!
