| T O P I C R E V I E W |
| Breadfan |
Posted - 09/12/2018 : 09:28:37 AM
Origin Ver. and Service Release (Select Help-->About Origin): 8.5.1
Operating System:win10
Hello there, I have a problem solving a complex equation. I know that OriginLab can solve this equation, because I had success doning fittings. But now I want to calculate the parameter for a explicit value.
I have the following equations and want so solve for K. For the fittings of data before I got K without any issue.
There I used
L - independent variable / I - dependent variable / R - constant / K - parameter
But now I want to know K at a constant L and vary in I. In my opinion it should work, because at constant L, K just depends on I
A = (ln(2) + ln(2) * (1 - (2 / pi) * acos(1 / R)) - ln(2) * (1 - ((2 / pi) * acos(1 / R))^2));
B = (1 + 0.639 * (1 - (2 / pi) * acos(1 / R)) - 0.186 * (1 - ((2 / pi) * acos(1 / R))^2));
C = (A+(pi/(4*B*atan(L+(1/K))))+(1-A-(1/(2*B)))*(2/pi)*atan(L+(1/K)));
D = ((2.08/R^0.358)*(L-0.145/R)+1.585)/((2.08/R^0.358)*(L+0.0023*R)+1.57+ln(R)/L+(2/(pi*R))*ln(1+(pi*R)/(2*L)));
I = C+(D-1)/((1+2.47*L*K*R^0.31)*(1+(L^(0.006*R+0.113))*K^(-0.0236*R+0.91)));
Thanks in advance!
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| 7 L A T E S T R E P L I E S (Newest First) |
| Breadfan |
Posted - 10/02/2018 : 03:39:10 AM
Hello there,
I now managed to get Origin 2018, but even with the "equation solver" fromhttps://www.originlab.com/fileExchange/details.aspx?fid=393 I did not find a way to solve my problem. The equation solver also doesnt work for this complex equatin, if you want to solve for K. But it worked out when I did a control calculation with the same equation to solve for I.
A = (ln(2) + ln(2) * (1 - (2 / pi) * acos(1 / R)) - ln(2) * (1 - ((2 / pi) * acos(1 / R))^2));
B = (1 + 0.639 * (1 - (2 / pi) * acos(1 / R)) - 0.186 * (1 - ((2 / pi) * acos(1 / R))^2));
C = (A+(pi/(4*B*atan(L+(1/K))))+(1-A-(1/(2*B)))*(2/pi)*atan(L+(1/K)));
D = ((2.08/R^0.358)*(L-0.145/R)+1.585)/((2.08/R^0.358)*(L+0.0023*R)+1.57+ln(R)/L+(2/(pi*R))*ln(1+(pi*R)/(2*L)));
I = C+(D-1)/((1+2.47*L*K*R^0.31)*(1+(L^(0.006*R+0.113))*K^(-0.0236*R+0.91)));
with:
L=0.4
R=8 |
| Breadfan |
Posted - 09/17/2018 : 03:29:10 AM
Thanks
Aviel. I already did this on 12th september. I added a link to this topic. |
| arstern |
Posted - 09/14/2018 : 10:21:24 AM
Hi,
Would you mind sending your opj/data file to tech@originlab.com? I would like to test one of our apps with your data to make sure it works. The App is only for Origin 2018 and newer, however if it works, it may be worth it for you to upgrade from 8.5 to Origin 2018.
Thanks
Aviel
OriginLab |
| Breadfan |
Posted - 09/14/2018 : 03:11:50 AM
quote:
Originally posted by arstern
Hello,
If I understand correctly you need to find K as I changes. If you have the dataset for I and constant value for L, then you can calculate the equation to find K.
Aviel
Now you got it.
I indeed have the dataset for I and the value for L. But I dont know how to get K out of the complex equations, without converting it to a form K = ... (this would be to complex for me).
So, is there a easy way Originlab calculates K out of this equations?
Thanks in advance! |
| arstern |
Posted - 09/13/2018 : 11:26:34 AM
Hello,
If I understand correctly you need to find K as I changes. If you have the dataset for I and constant value for L, then you can calculate the equation to find K.
Aviel |
| Breadfan |
Posted - 09/12/2018 : 12:04:27 PM
Thanks
Aviel, thats exactly what I did to get "K" in the first experiment. But maybe I have to explain this more in detail. Imagine the following experiment:
You have an instrument above a sample, at a height of "L" and detecting some sort of output "I". If you now approching the surface of your sample ("L" decreases), "I" will change. How "I" will change depents on "K", a parameter of the surface activity wich is constant at a specific sample location. In this case the location you are you are approaching is constant and you are just vary the height of the instrument "L".
In the second experiment, you will keep the height "L" constant, but will scan your sample by change the location of the instrument. This time you will get different "I" depending on the location and activity parameters "K".
My probelm is now, that I have no idea to get "K" out of the data that is similar to the first experiment, just with a constant "L". I cant do a fitting for just one datapoint (there is just one height). I dont know how to set an command like "solve for K" |
| arstern |
Posted - 09/12/2018 : 10:22:22 AM
Hello,
Yes this should work. In the Fitting Function Builder did you select Function Type as Equations? When I input all of the information that you provided and select Quick Check in the Fitting Function Builder, a value for I is provided. What exactly is not working?
Also if you are willing to provide your XY data (I,L) that would be useful! Please send your data or opju file at tech@originlab.com.
Below is how I set it up:
Thanks
Aviel
OriginLab |
