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4th order polynomial coefficients for pressure at temperature readings, Help!!!

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Nominal Animal:
If you provide a table with three entries – temperature count, pressure count, normalized pressure – with as many samples as you can muster, I'd be happy to provide you with a reasonable bivariate approximation, that takes in the temperature and pressure counts, and spits out normalized pressure.

Or, if you want, four entries – temperature count, pressure count, actual temperature, normalized pressure – and I'll provide you with two bivariate polynomials: one spits out the actual temperature, and the other normalized pressure, when given just the temperature and pressure counts.

I'll even show how I do that with Gnuplot.  (Why gnuplot?  It's free, takes in simple tabulated data, and can produce publication-quality graphs, making it quite popular among computational scientists.)

ETITsynthesizer:
I think Nominal Animal had the best response in this whole thread. I don't know if this is helpful but I would have tried the polynomial regression tool in excel. this gives the answer in the standard polynomial form. let the computer do the math. I have never worked with a bivariate polynomial in math before but I can dig it. I will trust that he/she knows what they are talking about since they have experience with this kind of regression.

itsbiodiversity:
I believe you've got the answer here.  My apologies for any explanation errors - I am working "backwards" from available knowledge in many aspects. Did you happen to look at the spreadsheet I attached on the last post?  The only data that they used for this product was taking PSI readings from the Low to High Range at 3 different Temperatures. Looking at the spreadsheet it appears they have a Zero: Regression Output with x coefficients.  These x coefficients are used in the formula +$A:$N$11*C7+$A:$O$11*D7+$A:$P$11*E7+$A:$Q$11*F7 
where N11 is an x coefficient, C7 is the temperature counts, O11 is another x coefficient, D7 is the UUT PSI reading, P11 is the third x coefficient, E7 is Temperature Counts (C7) x UUT PSI reading (D7), Q11 is the fourth x coefficient, and F7 is temperature counts squared (C7 x C7) multiplied by UUT PSI reading (D7). 

Applied line by line the formula does a great job of correcting for temperature area. The author of the instrument stated the following:
Enter Temperature Coefficients
This command provides for the entry of temperature coefficients that will compensate the sensor for ambient temperature conditions. The coefficients are determined during the factory calibration process.
tcomp <a0> <a1> <a2> <a3> <a4>  (THESE ARE WHAT I AM TRYING TO IDENTIFY) :)
The operator specifies five coefficients, which are used in a fourth order polynomial that corrects temperature readings for the ambient temperature at the sensor.
The currently effective temperature compensation coefficients can be viewed using the coef command, which is described in the user’s manual.
default is 0 0 1 0 0.

Does this make sense?  I feel like this is close to what you were describing.  I'm attaching another spreadsheet here, and I'm also downloading the program you mentioned.  Thank you so much.

itsbiodiversity:
Pretty sure this is the first time I've encountered anyone rude like this on here.  I usually hang in the Metrology forum, which is what my career is (Chemist/Metrologist).  I'm not sure if him saying "good luck with your homework" is supposed to make me feel inferior, but an answer surely would've been respected.  I'm positing that if you knew the answer you'd have provided it, and if you do have the answer I'd still be appreciative. 

I truly appreciate those of you that have spent time and thought towards helping with this question.

DrG:

--- Quote from: itsbiodiversity on August 19, 2020, 01:42:28 pm ---Pretty sure this is the first time I've encountered anyone rude like this on here.  I usually hang in the Metrology forum, which is what my career is (Chemist/Metrologist).  I'm not sure if him saying "good luck with your homework" is supposed to make me feel inferior, but an answer surely would've been respected.  I'm positing that if you knew the answer you'd have provided it, and if you do have the answer I'd still be appreciative. 

I truly appreciate those of you that have spent time and thought towards helping with this question.

--- End quote ---

Read me carefully...I spent plenty of time with your problem last night...repeatedly presenting graphs and coefficients and trying to help you. Trying, in fact, to understand what it is that you were trying to do. It was clearly a mistake for me to have even tried to help you and I regret spending anytime on it at all. Obviously you did not appreciate the effort and, equally obvious, is that it did not help you. I did, however. learn from the experience.

I don't think you explained yourself very well and I believe you when you said how confused you were about some things about the task. You can certainly understand that it is difficult to ascertain what someone else knows or does not know on a forum.

I am glad that you seem to have found your answer. I do think it is some kind of homework problem and I don't think that you explained what you were trying to do very well. I am saying good luck with your homework (or whatever it is you are trying to do) in just exactly as it says. Beyond that, I don't much care what you think or how you feel.


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