how set the appropriate grids (SX DX)

[ThreeIcug]

Recently, I try to set a new model to do the study. My vertical fault geometry is about 2*2km patches. The total length of fault is 200km，and the depth is 30km. So I set the SX = 512， then set the Dx = 0.35 for a best result by trial and error. However the range of SX*DX is small than I want. When I increase the SX to 1024 ， the code don't output any result. Then i found the max value of SX is about 800. In addition, if I increase the DX, the reslut is not good as the Dx =0.35. So my question is how to set an appropriate grids values to get the best result and the most range both.
Thank you.

[sbarbot]

Hi, The sampling should be about a fifth of the patch size that describes the fault geometry for coseismic displacement. This allows the numerical model to capture the deformation in the near field of any slip patch. This requirement is to use a numerical solution (analytic in the Fourier domain) for the coseismic slip distribution and the resulting stress. However, the postseismic deformation may require less stringent conditions if the viscoelastic layer is far from the coseismic stress change. In that case, you can compute the coseismic displacement in closed form with the analytic solution of Okada (1992). Then, a numerical solution is used for the postseismic relaxation. I have seen issues with hardware/software limitation on the maximum model size. This has to do with the OpenMP parameters that limit the size of arrays used in reduction operation (in this case, an integration). Check your OpenMP configuration to maximize this parameter. You can also compile the code without OpenMP. This will allow larger models, but it will be slower to compute the solution. Best wishes, Sylvain

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________________________________________ From: ThreeIcug ***@***.***> Sent: Tuesday, March 15, 2022 2:23 AM To: geodynamics/relax Cc: Subscribed Subject: [geodynamics/relax] how set the appropriate grids (SX DX) (Issue #4) Recently, I try to set a new model to do the study. My vertical fault geometry is about 22 patches. The total length of fault is 200km，and the depth is 30km. So I set the SX = 512， then set the Dx = 0.35 for a best result by trial and error. However the range of SXDX is small than I want. When I increase the SX to 1024 ， the code don't output any result. Then i found the max value of SX is about 800. In addition, if I increase the DX, the reslut is not good as the Dx =0.35. So my question is how to set an appropriate grids values to get the best result and the most range both. Thank you. — Reply to this email directly, view it on GitHub<https://urldefense.com/v3/__https://github.com/geodynamics/relax/issues/4__;!!LIr3w8kk_Xxm!8jSmSyraYSQoLoUPNiao1fmmmK6627TuZazWwUmDKAFu91gseWa-3d57EepJlDg$>, or unsubscribe<https://urldefense.com/v3/__https://github.com/notifications/unsubscribe-auth/ABJHTWDLRWXO64L77BBO2DDVABJJLANCNFSM5QYBPZ7A__;!!LIr3w8kk_Xxm!8jSmSyraYSQoLoUPNiao1fmmmK6627TuZazWwUmDKAFu91gseWa-3d57VgjZoZc$>. Triage notifications on the go with GitHub Mobile for iOS<https://urldefense.com/v3/__https://apps.apple.com/app/apple-store/id1477376905?ct=notification-email&mt=8&pt=524675__;!!LIr3w8kk_Xxm!8jSmSyraYSQoLoUPNiao1fmmmK6627TuZazWwUmDKAFu91gseWa-3d57n8fF4ZQ$> or Android<https://urldefense.com/v3/__https://play.google.com/store/apps/details?id=com.github.android&referrer=utm_campaign*3Dnotification-email*26utm_medium*3Demail*26utm_source*3Dgithub__;JSUlJSU!!LIr3w8kk_Xxm!8jSmSyraYSQoLoUPNiao1fmmmK6627TuZazWwUmDKAFu91gseWa-3d57-roDQnY$>. You are receiving this because you are subscribed to this thread.Message ID: ***@***.***>

[ThreeIcug]

Dear Sylvain,
Thank you for your detail reply. That solves my question.
With the gradual deepening of learning, I encountered new problem about the setting of some parameters.

In generally, we always like using meter to indicate the fault slip and using Km to indicate the fault geometry( default in Relax is meter). the manual suggest the stress can be compensated for by multiplying the Lame parameters by a factor of 1E-3 and the buoyancy parameter (Gamma) by a factor of 1E3.

According to the formula of the buoyancy wavelength ,its unit is 1/m. So if we want to use Km to indicate the fault geometry, we need to multiply a factor of 1e3. But according to the rate-strengthening constitutive law, the unit of slip rate is related to the reference slip rate and didn't depend on the part of sinh(). So what is the reason of multiplying the Lame parameters and mu by a factor of 1E-3.

I tried to understand the question, but wasn't sure it was right. Showing in your 2009 JGR paper, the effective stress drop (delta τ) = C*(s/L)*G. Where s is fault slip in meter, So if we want to keep the unit of delta τ(MPa) and change L to Km, we need multiply G ( namely mu ) with a factor of 1E-3 (MPa to GPa).
In addition, in the JGR paper, a ~ 1E-3, a-b <1E-2, sigma = 100MPa . According to the above analysis，I set the parameters as below,
# lambda(GPa), mu(GPa), gamma(Km)
30 30 8.33E-4
# n depth(Km) gamma0(m/a) (a-b)sigma(MPa) friction cohesion
1 5 5 1 0.6 0
Is the unit in the setting right? The gamma0 in your 2009 JGR is about 5E-5 m/a, and other paper is in the magnitude of 1E-7.

Thank you.
Jingwei