Finding exact point on the surface

Question

M 2023-9-12

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2020161-finding-exact-point-on-the-surface

评论： M 2023-9-14

Hi all,

I have a 3D surface obtained from the equation y = f(x, z). I'm wondering how I can determine the exact value of z when I have the values of x and y. It's challenging to find the equation for z = g(x, y), which is why I created the surface based on x and z.

Essentially, I have a rectangular region on the x-y plane, and I need to project it onto the surface. To achieve this, I need to calculate the exact z values for each corner of the rectangle based on the given x and y values.

I appreciate any assistance with this.

Thank you!

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M 2023-9-12

It could be two, I mean surface has two sheets, so I want the value on the lower sheets.

Torsten 2023-9-12

Use "fsolve" to solve y - f(x,z) = 0 with known x and y and unknown z.

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Answer 1

Matt J 2023-9-13

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2020161-finding-exact-point-on-the-surface#answer_1308536

在 MATLAB Online 中打开

z = fzero(@(z)y-f(x,z), [z1,z2])

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M 2023-9-14

编辑：M 2023-9-14

在 MATLAB Online 中打开

I changed it a little bit and again gives another error:

gamma=5.5;
T=1/(gamma*40);
kh=0.1;
p=0.09;
delta=0.1;
ktau=0.04;
Kc=0.2; 
Khat=0.000015;
Kp=0.3;
kb=0.4;
Vs=T*0.9;
v_pmm=T*0.07;
alph0=T*0.003;
alph1=T*0.01;
Ke=14;
ks=0.2;
Kf=T*40;
kplc=0.11;
ki=2;
tmax=200/T;
e=0.0016;
vss=Vs/e;
K=(Khat)/ktau^4;
alpha0=delta.*(alph0)/ktau^4;
alpha1=delta.*(alph1)/ktau^4;
v_pm=delta.*(v_pmm)/ktau^4;
tmaxhat=tmax*ktau^4;
%[c,ct]=meshgrid(0:0.01:10);
A=@(ct,c) (-(vss.*c.^2)./(ks.^2))+((Vs.*K.*gamma.^2.*ct.^2)./(ks.^2))+alpha0+alpha1.*((Ke.^4)./(Ke.^4+(gamma.*ct).^4));
h=@(ct,c) (-0.4.*A.*((Kc.^4).*(Kp.^2))./((p.^2.*c.^4.*gamma.*ct.*Kf)));
% Given values of ct and h
ct = 0.32;
h = 0.18;
% Define a function for the equation to solve
equation_to_solve = @(c) h(ct, c) - h;
% Initial guess for c
c0 = 0.010663;
% Use fzero 
z_optimized = fzero(equation_to_solve, c0);
Error using fzero>localFirstFcnEvalError
FZERO cannot continue because user-supplied function_handle ==> @(c)h(ct,c)-h failed with the error below.

 Index in position 1 is invalid. Array indices must be positive integers or logical values.

Error in fzero (line 302)
        localFirstFcnEvalError(FunFcn,FunFcnIn,ME);
disp(['The optimized z value is approximately z = ', num2str(z_optimized)]);

Dyuman Joshi 2023-9-14

在 MATLAB Online 中打开

You are over-writing the function handle h with the numerical value 0.18 and then trying to use it as a function handle.

Also, it's a better idea to define constants before defining function handles.

gamma=5.5;
T=1/(gamma*40);
kh=0.1;
p=0.09;
delta=0.1;
ktau=0.04;
Kc=0.2; 
Khat=0.000015;
Kp=0.3;
kb=0.4;
Vs=T*0.9;
v_pmm=T*0.07;
alph0=T*0.003;
alph1=T*0.01;
Ke=14;
ks=0.2;
Kf=T*40;
kplc=0.11;
ki=2;
tmax=200/T;
e=0.0016;
vss=Vs/e;
K=(Khat)/ktau^4;
alpha0=delta.*(alph0)/ktau^4;
alpha1=delta.*(alph1)/ktau^4;
v_pm=delta.*(v_pmm)/ktau^4;
tmaxhat=tmax*ktau^4;
%[c,ct]=meshgrid(0:0.01:10);
%Renamed to constant to h0
% Given values of ct and h0
ct = 0.32;
h0 = 0.18;
A=@(c) (-(vss.*c.^2)./(ks.^2))+((Vs.*K.*gamma.^2.*ct.^2)./(ks.^2))+alpha0+alpha1.*((Ke.^4)./(Ke.^4+(gamma.*ct).^4));
h=@(c) (-0.4.*A(c).*((Kc.^4).*(Kp.^2))./((p.^2.*c.^4.*gamma.*ct.*Kf)));
% Define a function for the equation to solve
equation_to_solve = @(c) h(c) - h0;
% Initial guess for c
c0 = 0.010663;
% Use fzero 
z_optimized = fzero(equation_to_solve, c0);
disp(['The optimized z value is approximately z = ', num2str(z_optimized)]);
The optimized z value is approximately z = -0.2563

Dyuman Joshi 2023-9-14

在 MATLAB Online 中打开

Change the initial guess

gamma=5.5;
T=1/(gamma*40);
kh=0.1;
p=0.09;
delta=0.1;
ktau=0.04;
Kc=0.2; 
Khat=0.000015;
Kp=0.3;
kb=0.4;
Vs=T*0.9;
v_pmm=T*0.07;
alph0=T*0.003;
alph1=T*0.01;
Ke=14;
ks=0.2;
Kf=T*40;
kplc=0.11;
ki=2;
tmax=200/T;
e=0.0016;
vss=Vs/e;
K=(Khat)/ktau^4;
alpha0=delta.*(alph0)/ktau^4;
alpha1=delta.*(alph1)/ktau^4;
v_pm=delta.*(v_pmm)/ktau^4;
tmaxhat=tmax*ktau^4;
%[c,ct]=meshgrid(0:0.01:10);
%Renamed to constant to h0
% Given values of ct and h0
ct = 0.32;
h0 = 0.18;
A=@(c) (-(vss.*c.^2)./(ks.^2))+((Vs.*K.*gamma.^2.*ct.^2)./(ks.^2))+alpha0+alpha1.*((Ke.^4)./(Ke.^4+(gamma.*ct).^4));
h=@(c) (-0.4.*A(c).*((Kc.^4).*(Kp.^2))./((p.^2.*c.^4.*gamma.*ct.*Kf)));
% Define a function for the equation to solve
equation_to_solve = @(c) h(c) - h0;
% Initial guess for c
c0 = 0.010663;
% Use fzero 
z_optimized = fzero(equation_to_solve, 0.5);
disp(['The optimized z value is approximately z = ', num2str(z_optimized)]);
The optimized z value is approximately z = 0.2563