Functions with multiple variables

2 次查看(过去 30 天)
Ben
Ben 2011-11-15
Hi
I am working on functions with multiple variables:
IFF1 = @(s22,s33) (s22+s33+sqrt((s22-s33)*(s22-s33)+4*s23*s23))/(2*St22)-1
I would like to gain all combinations of s22 and s33 so the function equals 0. Then i would like to plot these values, result should be an ellipsis.
Thank you for you help!
Greetings Ben
  1 个评论
Walter Roberson
Walter Roberson 2011-11-15
Is the St22 reference a typo, or a constant in the scope of the defining routine?
Likewise for the s23 and s33 references.

请先登录,再进行评论。

回答(4 个)

Jonathan
Jonathan 2011-11-15
Ben,
Here is a neat way of solving your problem. There are certainly more efficient methods. This was the least amount of code for me to right, though. Let me know how it goes for you.
s23 = rand;
St22 = rand;
IFF1 = @(s22,s33) (s22+s33+sqrt((s22-s33)*(s22-s33)+4*s23*s23))/(2*St22)-1;
[X, Y, Z] = meshgrid(linspace(-10, 2, 50), linspace(-10, 2, 50), [0 0]);
V = nan(size(X));
for i = 1:numel(X)
V(i) = IFF1(X(i), Y(i));
end
patch(isosurface(X,Y,Z,V,0))

Ben
Ben 2011-11-15
Re
If i understand correct, your code plots my function. Though it does not solve my optimization problem. Any hints?
Thank you for your support!

Walter Roberson
Walter Roberson 2011-11-15
There are an infinite number of solutions, not confined to any region of the graph. For any s33,
s22 = (s23^2+s33*St22-St22^2)/(s33-St22)
This is clearly an infinite curve with a discontinuity at s33 = St22

Ben
Ben 2011-11-15
Sorry I should have added more information:
St22 = 73.00 [MPA]
s23 = 30 [MPA]
(s22+s33+sqrt((s22-s33)*(s22-s33)+4*s23*s23))/(2*St22) = 1
I have been able to solve this problem with an optimization approach. But was hoping to find an easyer way to visualize the result.
  1 个评论
Walter Roberson
Walter Roberson 2011-11-15
Same issue as before:
s22 = (-4429+73*s33)/(s33-73)
which is an infinite curve with a discontinuity at s33 = 73

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Annotations 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by