Help to implement an erosion coefficient in morphotectonics

[AllanLopez]

Dear Wolfgang
Although erosion and the erosion coefficient are vital elements of the stream power incision model and its equation and it is also implicit in the Steepness, Chi and Hypsometric integral, there is a need for a relative reference value to correlate with the previous cited metrics. Meanwhile absolute values determined from cosmogenic nucleides are expensive.
The authors of the following paper proposed an erosion coefficient K= E/ Ksn, directly derived from TTT, but I unable to implement it due to my Matlab deep limitations, so I ask you please Wolfgang, to do it along with a Shp export capability.
My deep appreciation in advance
Allan
Impact of fault damage on eastern Tibet topography Heather M. Kirkpatrick, Seulgi Moon*, An Yin and T. Mark Harrison (Geology, 2020; open access at https://pubs.geoscienceworld.org/gsa/geology/article/49/1/30/590535/Impact-of-fault-damage-on-eastern-Tibet-topography)
They mention that:
You can extract channel steepness (ksn) from "chiplot" function in topo toolbox.

Then, you can use fitlm or fitnlm to fit ksn and erosion rate and examine which fit explain the overall relationship.
Once you decide your functional form, you can calculate erosion coefficient using the best-fit parameters (e.g., power-law exponents)

for example,
xda1 = ksn; yda1 = erosion;
beta0 = [1];
nonlm0 = fitnlm(xda1, yda1 , @(b,xda1)b(1).*xda1.^1,beta0); %linear without intercept

beta0 = [1e-2 1];
nonlm1 = fitnlm(xda1, yda1 , @(b,xda1)exp(xda1.*b(1) +b(2)),beta0) ;%exponential

beta0 = [1e-2 1];
nonlm2 =fitnlm(xda1, yda1 , @(b,xda1)b(1).*xda1.^b(2),beta0) ; %power law

beta0 = [1 0];
nonlm3 = fitnlm(xda1, yda1 , @(b,xda1)b(1).*xda1.^1+b(2),beta0); %linear with intercept

b_n2 = nonlm2.Coefficients{:,1}
erocoef_n2 = erosion./(ksn.^b_n2(2)); %n = 0.49, coefficient for nonlinear powerlaw fit
erocoef = erosion./(ksn); %n = 0.49, coefficient for linear fit without intercept

Hope this helps!
Seulgi

[AllanLopez]

Impact of fault damage on eastern Tibet topography.pdf

[wschwanghart]

Dear Allan,
you have been following my blog for years, and you are still no MATLAB expert? What have I done wrong? ;-) (just joking)
Please give me some sample code, so that I can help. What have you done so far to solve the problem.
Cheers, Wolfgang

[AllanLopez]

Dear Wolfgang, Thank you for your response. After all these years I have learned to detect my mistakes in the execution, I hope. What is obvious for you is martian language to me (also joking). I guess a bridge is needed to fill the gap between Ksn extracted from chipplot and the following example given by the author below. I mean I have not worked anything; I do not know what to do. Later I shall mention and idea to visit you for a couple of weeks next year. All the best, Allan for example, Then, you can use fitlm or fitnlm to fit ksn and erosion rate and examine which fit explain the overall relationship. Once you decide your functional form, you can calculate erosion coefficient using the best-fit parameters (e.g., power-law exponents) xda1 = ksn; yda1 = erosion; beta0 = [1]; nonlm0 = fitnlm(xda1, yda1 , @(b,xda1)b(1).*xda1.^1,beta0); %linear without intercept beta0 = [1e-2 1]; nonlm1 = fitnlm(xda1, yda1 , @(b,xda1)*exp*(xda1.*b(1) +b(2)),beta0) ;%exponential beta0 = [1e-2 1]; nonlm2 =fitnlm(xda1, yda1 , @(b,xda1)b(1).*xda1.^b(2),beta0) ; %power law beta0 = [1 0]; nonlm3 = fitnlm(xda1, yda1 , @(b,xda1)b(1).*xda1.^1+b(2),beta0); %linear with intercept b_n2 = nonlm2.Coefficients{:,1} erocoef_n2 = erosion./(ksn.^b_n2(2)); %n = 0.49, coefficient for nonlinear powerlaw fit erocoef = erosion./(ksn); %n = 0.49, coefficient for linear fit without intercept El lun, 5 jun 2023 a la(s) 13:47, Wolfgang Schwanghart ( ***@***.***) escribió:

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Dear Allan, you have been following my blog for years, and you are still no MATLAB expert? What have I done wrong? ;-) (just joking) Please give me some sample code, so that I can help. What have you done so far to solve the problem. Cheers, Wolfgang — Reply to this email directly, view it on GitHub <#37 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AB3VH4RJEZDLBXT3ASVZONTXJYZUZANCNFSM6AAAAAAY2FKEOQ> . You are receiving this because you authored the thread.Message ID: ***@***.***>