Nonlinear equation numerical solution

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Tadej Cigler
Tadej Cigler 2020 年 7 月 21 日
編集済み: Matt J 2020 年 7 月 21 日
Hi,
I have a following relation
C = 1.2;
D = 9420;
E = -2.0;
B = 0.1973;
f = D*sin(C*atan(B*x-E*(B*x-atan(B*x))));
and I would like to X if Y = 9250.
When I use below code:
syms x0 y0
y0 = 9250;
eqn = y0 == StiffnessMagicFormula(D,C,B,E,x0);
slip_angle = vpasolve(eqn, x0)
I obtained following result:
slip_angle =
-4.1564971021120118282594483668007e191020734
But actually result should be around 6.8 .
I tryed also with functions: fsolve, roots and find but were all unsuccessful.
BR, Tadej

回答 (2 件)

Matt J
Matt J 2020 年 7 月 21 日
編集済み: Matt J 2020 年 7 月 21 日
I tryed also with functions: fsolve, roots and find but were all unsuccessful.
fsolve is overkill for such a simple 1D problem. Use fzero instead:
C = 1.2;
D = 9420;
E = -2.0;
B = 0.1973;
y0 = 9250;
f = @(x) D*sin(C*atan(B*x-E*(B*x-atan(B*x))))-y0;
x0=fzero(f,6.8)
For me, this results in
x0 =
6.9606

Fabio Freschi
Fabio Freschi 2020 年 7 月 21 日
% params
C = 1.2;
D = 9420;
E = -2.0;
B = 0.1973;
y0 = 9250;
% function
fun = @(x)D*sin(C*atan(B*x-E*(B*x-atan(B*x))))-y0;
% solution
x0 = fsolve(fun,0)

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