| T O P I C R E V I E W |
| computon |
Posted - 08/12/2009 : 11:18:43 PM
HI
i used boltzmann function to nonlinear fit the data of Charpy impact data(temparature ,Charpy Energy),as follows:
E=(A1-A2)/(1+exp(T-Tt)/¦È1)
E-Charpy Energy;T-temperature; Tt-an index temperature at the mid-energy:0.5(A1+A2);¦È1-parameter.
The experiment data is as follows:
T E
25 75
25 70
25 77.5
10 73
10 67
10 72
0 54.5
0 57
0 56.5
-10 57
-10 48
-10 54.5
-20 38.5
-20 40
-20 44
-30 36.5
-30 35.5
-30 36
-40 39
-40 33
-40 34
-50 30
-50 27
-50 27.5
-60 26
-60 33
-60 30
The fitting result is as follows:
Parameters A1 A2 Tt ¦È1
Unit. J J ¡æ
value 28.42¡À2.17 79.03¡À4.26 -6.72¡À2.90 13.65¡À2.85
Now i want to know the uncertainty of the energy?
can i use the equation to calculate the uncertainty of the energy?
¦Ò(E)=(¦²£¨Ei-E£¨i-fit£©£©^2/(n-1))
Ei ¡ªthe i-th charpy energy at the i-th temparature;
E(i-fit)¡ªthe calculating energy using the fitting function, Boltzmann function ,by inputing the i-th temperatue corresponding to the i-th charpy energy;
If i calculate an index temperature at the energy 41J,T-41J:
T-41J=¦È1*ln((A2-A1)/(A2-41)-1)+Tt
how to calculate the uncertainty of the index tempaterature T-41J,¦Ò(T-41J)?
can i use the equation as following:
¦Ò(T-41J)=¦È1*ln((A2-A1)/(A2-41)-1)-¦È1*ln((A2-A1)/(A2-41+¦Ò(E))-1)
what's the relationship between the parameter uncertainty in the fitting and the uncertainty of the curve /the index temprature T-4J?
thank you for you help and your reply?
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| 7 L A T E S T R E P L I E S (Newest First) |
| computon |
Posted - 08/15/2009 : 12:20:13 AM
hi
i have read your link careful.but i still can not find the answer to to evaluate the uncertainty of T-41J? because the T-41J is not the fitting parameter. so it must be deduced from the uncertainty of Energy or parameter?
T-41J--the indext temperatue when the energy is 41J in the boltzmann fitting curve.
you can deduce the T-41J using the equation:
T-41J=θ1*ln((A2-A1)/(A2-41)-1)+Tt
how to evaluate the uncertainty of T-41J? |
| computon |
Posted - 08/15/2009 : 12:15:44 AM
We can be told from the parameter standard errors that how precision of the fitted values. These value give you a primary impression on goodness of the fitted parameters. Usually, we would like to see the magnitude of the standard error values are less than the fitted values. If these values much larger than the fitted values, your fitting model may be overparameterized.
for example,the parameter A2=79.03±4.26 ,can we think that the uncertainty at the upper self energy (A2) is 4.26J.or we must use the uncertainty of Reduced Chi Square ?
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| larry_lan |
Posted - 08/14/2009 : 11:23:45 PM
Please read this link carefully. We have provided all of the equations we used in nonlinear curve fitting.
Thanks
Larry |
| computon |
Posted - 08/14/2009 : 11:15:32 PM
T-41J--the indext temperatue when the energy is 41J in the boltzmann fitting curve.
you can deduce the T-41J using the equation:
T-41J=θ1*ln((A2-A1)/(A2-41)-1)+Tt
how to evaluate the uncertainty of T-41J?
you mentioned that the uncertainty of the energy in the whole temperature range can be deduced by the Reduced Chi Square ,what is the equation? is it as follows?
σ(E)=(Σ((Ei-E(i-fit)^2)/(n-1))
or
σ(E)=(Σ((Ei-E(i-fit)^2)/(n-4))?
or
σ(E)=(Σ((Ei-E(i-fit)^2)/(n))?
what the relationship between the uncertainty of energy and prediction interval/predition interval?
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| larry_lan |
Posted - 08/14/2009 : 07:08:06 AM
Hi:
First of all, we already output the "uncertainty of the energy" you mentioned, it's Reduced Chi Square in our report.
Second, you'd better explain the "index temperature" clearly. what's 41J stands for? When you calculate (T-41J), are you going to minus 41 for all T? And how the equation for (T-41J) comes from?
Thanks
Larry |
| computon |
Posted - 08/13/2009 : 8:59:43 PM
i think you don't understant my question.i used the pridiction band and confidence band in the nonlinear fitting process and gained the parameters' value and standard daviation,as you can see below.i want to know the uncertainty of the index temperature T41-J,σ(T41-J),it is can not be gained immediately after fitting and it must deduce.
i copy my question and revise the gibberish in my post ,as follows?
thank you for your patience and looking forward your detailed reply.
HI
i used boltzmann function to nonlinear fit the data of Charpy impact data(temparature ,Charpy Energy),as follows:
E=(A1-A2)/(1+exp(T-Tt)/θ1)
E-Charpy Energy;T-temperature; Tt-an index temperature at the mid-energy:0.5(A1+A2);θ1-parameter.
The experiment data is as follows:
T E
25 75
25 70
25 77.5
10 73
10 67
10 72
0 54.5
0 57
0 56.5
-10 57
-10 48
-10 54.5
-20 38.5
-20 40
-20 44
-30 36.5
-30 35.5
-30 36
-40 39
-40 33
-40 34
-50 30
-50 27
-50 27.5
-60 26
-60 33
-60 30
The fitting result is as follows:
Parameters A1 A2 Tt θ1
Unit. J J ℃
value 28.42±2.17 79.03±4.26 -6.72±2.90 13.65±2.85
Now i want to know the uncertainty of the energy?
can i use the equation to calculate the uncertainty of the energy?
σ(E)=(Σ((Ei-E(i-fit)^2)/(n-1))
Ei -the i-th charpy energy at the i-th temparature;
E(i-fit)-the calculating energy using the fitting function, Boltzmann function ,by inputing the i-th temperatue corresponding to the i-th charpy energy;
If i calculate an index temperature at the energy 41J,T-41J:
T-41J=θ1*ln((A2-A1)/(A2-41)-1)+Tt
how to calculate the uncertainty of the index tempaterature T-41J,θ(T-41J)?
can i use the equation as following:
σ(T-41J)=θ1*ln((A2-A1)/(A2-41)-1)-θ1*ln((A2-A1)/(A2-41+σ(E))-1)
what's the relationship between the parameter uncertainty in the fitting and the uncertainty of the curve /the index temprature T-4J?
thank you for you help and your reply?
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| larry_lan |
Posted - 08/13/2009 : 02:29:06 AM
Hi:
I think you need the confidence band, right? Check the following checkbox to output these values.
And I think this link should be helpful for you.
P.S: Could you please review your post after submitting? You know, this forum is in English, so please use Western or Unicode Character Coding in your browser, or we may see gibberish in your post. (For IE, select View : Encoding; For Firefox, select View : Character Encoding)
Thanks
Larry |
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