More on the erosion coefficient

[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

[wschwanghart]

Hi Allan,
well, you'll need basin-wide erosion rates which Seulgi named as vector erosion.
Then, you'll have to calculate basin-averaged ksn values. Richard Ott explains here how to do this. Also, these values should be stored as a vector and must have the same length as the vector erosion. Then, there are the options detailed by Seulgi. In the simplest case, there is a linear relation between both variables so that you can do it like she does:

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

In fact, this can also be done by

beta = xda1\yda1;

This gives you a scalar value, which is the erosion coefficient. So, no need to write anything as a shapefile. The result is just a single value which you retrieved from multiple basin-wide averaged denudation rates and basin-wide averaged ksn values.

Cheers, Wolfgang

[AllanLopez]

Dear Wolfgang, Thank you very much indeed for your kind assistance and guidance into the erosion affair. All the best, Allan El jue, 8 jun 2023 a la(s) 14:07, Wolfgang Schwanghart ( ***@***.***) escribió:

…

Hi Allan, well, you'll need basin-wide erosion rates which Seulgi named as vector erosion. Then, you'll have to calculate basin-averaged ksn values. Richard Ott <https://github.com/Richard-Ott> explains here <https://topotoolbox.wordpress.com/2021/07/14/mean-basin-ksn-and-smoothing-madness/> how to do this. Also, these values should be stored as a vector and must have the same length as the vector erosion. Then, there are the options detailed by Seulgi. In the simplest case, there is a linear relation between both variables so that you can do it like she does: xda1 = ksn; yda1 = erosion; beta0 = [1]; nonlm0 = fitnlm(xda1, yda1 , @(b,xda1)b(1).*xda1.^1,beta0); %linear without intercept In fact, this can also be done by beta = xda1\yda1; This gives you a scalar value, which is the erosion coefficient. So, no need to write anything as a shapefile. The result is just a single value which you retrieved from multiple basin-wide averaged denudation rates and basin-wide averaged ksn values. Cheers, Wolfgang — Reply to this email directly, view it on GitHub <#38 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AB3VH4T7WUIU3VX5GLJCN5TXKIWJZANCNFSM6AAAAAAY3UEVLU> . You are receiving this because you authored the thread.Message ID: ***@***.***>