| T O P I C R E V I E W |
| bernie426 |
Posted - 03/01/2013 : 6:29:10 PM
Origin Ver. and Service Release (Select Help-->About Origin):
Operating System:
I have an equation to fit with my experiment data points,but the curve fitting always shows magnitude lower at the y-axis. Could any expert help me out? F can be any constant larger than 0 and the only factor to be extract is n.
Here is my code:
x*=pi/180.0;// convert x to radian (x is independent variable)
t*=pi/180.0;// convert t to radian (t is know value, 20 degree)
double r;
r=z/(cos(t));//z set to 910
double u,v;
u=(r*cos(x))+((r*sin(x))*tan(x+t));
v=(r*cos(t))*(1/(cos(x+t)));
if(x>=0)
y=F*(n+1)*(1/(2*pi*u^2))*((cos(x))^n);
else if(x<=0)
y=F*(n+1)*(1/(2*pi*v^2))*((cos(x))^n)
The experiment data points are
x y
-12.54 770.5519
-8.18 768.7524
-3.92 787.47
0.19 774.24
4.13 726.31
7.78 675.0
11.41 637.15
14.71 569.86
17.8 512.13
20.68 465.64
23.34 420.52
25.82 355.25
28.11 314.7
30.22 274.03
32.19 247.61
34.00 202.18
35.69 187.7
37.13 170.9
Thanks
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| 2 L A T E S T R E P L I E S (Newest First) |
| cc261 |
Posted - 03/06/2013 : 11:07:00 PM
see the results below:
f 1715573154.26265
n 1.61268335005249 |
| Shirley_GZ |
Posted - 03/04/2013 : 02:44:04 AM
Hi Bernie,
I defined the fitting function as below:
double x1=x*pi/180;
double t=20;
double z=910;
t*=pi/180.0;// convert t to radian
double r;
r=z/(cos(t));
double u,v;
u=(r*cos(x1))+((r*sin(x1))*tan(x1+t));
v=(r*cos(t))*(1/(cos(x1+t)));
if(x1>=0)
y=F*(n+1)*(1/(2*pi*u^2))*((cos(x1))^n);
else if(x1<=0)
y=F*(n+1)*(1/(2*pi*v^2))*((cos(x1))^n)
And then, set the parameter n as 2 and fix it, set parameter F as 1E9.
Click Fit button to do fitting.
It is worthy to mention that in your code, after you simplify the equation, u is equal to v. so, for x>=0 and x<=0, the y equations are same.
Originlab Technical Service Team |
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