Curve fitting equation does not give the input z back

I have made a 3D plot of experimental data to use the curve fitter in order to get an equation. When I use the equation and insert an x and y value, the value of z is way higher than in de 3D plot.

Curve fit data:

Linear model Poly33:

f(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p30*x^3 + p21*x^2*y

+ p12*x*y^2 + p03*y^3

where x is normalized by mean 8.327e+06 and std 6.911e+06

and where y is normalized by mean 330 and std 22.62

Coefficients (with 95% confidence bounds):

p00 = 2.653e-05 (2.273e-05, 3.033e-05)

p10 = 2.195e-05 (1.646e-05, 2.744e-05)

p01 = -1.158e-05 (-1.83e-05, -4.858e-06)

p20 = 5.127e-06 (1.941e-06, 8.312e-06)

p11 = -1.045e-05 (-1.238e-05, -8.524e-06)

p02 = 4.16e-06 (1.917e-06, 6.402e-06)

p30 = -4.23e-06 (-7.447e-06, -1.013e-06)

p21 = 3.459e-06 (1.103e-06, 5.814e-06)

p12 = 2.015e-06 (-2.528e-07, 4.284e-06)

p03 = -1.098e-06 (-4.878e-06, 2.683e-06)

Goodness of fit:

SSE: 1.113e-09

R-square: 0.9555

Adjusted R-square: 0.9438

RMSE: 5.722e-06

Validation of equation:

pA = 8300000;

T = 323;

mu = 2.653e-05 + 2.195e-05*pA + -1.158e-05*T + 5.127e-06*pA^2 + -1.045e-05*pA*T + 4.16e-06*T^2 + -4.23e-06*pA^3 + 3.459e-06*pA^2*T + 2.015e-06*pA*T^2 + -1.098e-06*T^3

mu =

-2.4186e+15

The input z valeus are in order of 10^-5 which is logical but I don't understand why the formula gives a value of -2.4186e+15. Does anyone have experience with this kind of deviations?

I would like to thank you in advance!

[Star Strider]

What were the original ‘x’, ‘y’ and ‘z’ values used to estimate the parameters?

[Olaf Schipper]

y x z

300 100000 0.00001502

300 500000 0.00001505

300 1000000 0.00001511

300 2500000 0.00001539

300 5000000 0.00001667

300 7500000 0.000061

300 10000000 0.00007244

300 12500000 0.00007974

300 15000000 0.00008555

300 17500000 0.00009053

300 20000000 0.00009496

320 100000 0.00001598

320 500000 0.00001601

320 1000000 0.00001606

320 2500000 0.0000163

320 5000000 0.00001719

320 7500000 0.00001969

320 10000000 0.0000324

320 12500000 0.00005096

320 15000000 0.00006004

320 17500000 0.00006654

320 20000000 0.00007188

340 100000 0.00001693

340 500000 0.00001696

340 1000000 0.000017

340 2500000 0.00001721

340 5000000 0.00001791

340 7500000 0.00001941

340 10000000 0.0000227

340 12500000 0.00002995

340 15000000 0.00004006

340 17500000 0.00004833

340 20000000 0.00005477

360 100000 0.00001787

360 500000 0.00001789

360 1000000 0.00001793

360 2500000 0.00001812

360 5000000 0.0000187

360 7500000 0.0000198

360 10000000 0.00002175

360 12500000 0.00002507

360 15000000 0.00003014

360 17500000 0.00003631

360 20000000 0.00004233

Answer 1

[Torsten]

@Olaf Schipper

Did you read that you have to use the normalized values for pA and T to evaluate mu(pA,T) ?

f(x,y) = p00 + p10*x + p01*y + p20*x^2 + p11*x*y + p02*y^2 + p30*x^3 + p21*x^2*y

+ p12*x*y^2 + p03*y^3

where x is normalized by mean 8.327e+06 and std 6.911e+06

and where y is normalized by mean 330 and std 22.62

So my guess is you have to evaluate the function with x' = (x-8.327e+06)/6.911e+06 and y' = (y-330)/22.62.

But you should use the parameters in full length. Use f(x,y) directly to evaluate the fit result for given values of x and y.

pA = (8300000-8.327e+06)/6.911e+06;
T = (323-330)/22.62;
mu = 2.653e-05 + 2.195e-05*pA + -1.158e-05*T + 5.127e-06*pA^2 + -1.045e-05*pA*T + 4.16e-06*T^2 + -4.23e-06*pA^3 + 3.459e-06*pA^2*T + 2.015e-06*pA*T^2 + -1.098e-06*T^3

mu = 3.0445e-05

[Olaf Schipper]

Thank you very much Torsten. I totally read over that information and thought x was directly usable. I should read more carefully. But thank you very much! This helped a lot!