Apparent resistivity for homogeneous halfspace

[jiajiasun]

I calculated the electric potential and the apparent resistivity for a homogeneous halfspace of 500 ohm*m using the equation and model setup shown in the following slide (taken from the DISC course). But, my results do not match what is shown on the slide.

Here is how I calculated the potential at a location 10 meters away from the injection point which is located at 0 m, according to the above picture. V = rho * I/(2 * pi * r) = 500 * 1/(2 * pi * 10) = 7.9577 Volts (Here I assume the inject current is 1 A). But in the above picture, 6.9 Volts is shown.

I also calculated the apparent resistivity: rho = 2 * pi * r * V / I = 2 * pi * 10 * 6.9/1 = 433.5398 Ohm * m. Shouldn't it be 500 ohm * m (because we are talking about homogeneous halfspace)?

Would you please double check what is wrong here? I would like my students to do similar calculations tomorrow in class. Thanks!

[thast]

Hi @jiajiasun, you are right, good catch! I do not know at the moment how this particular figure was generated (maybe @sgkang knows). However there is a few things I can think of:

• most of these plots are probably taken from the DC apps. These apps for fast computation use a 2.5D approach instead of a full 3D. Maybe that can come from that.
• Also, the 2D mesh have the center top cell cell-centered on (0,0), however the DC code used here (you have the option in SimPEG) is Node centered. As the cells have a 0.5m size, the distance in this picture is between -0.25m for A and 10.25m for M. This lead to a voltage of V=7.57 V. So still a discrepancy here I agree.
• Using one of the apps (in this case the DC_Plate_2_5D app) I regenerated the same setup. It seems I get an answer much closer to the truth (7.6V for 7.57V expected). So the apps would be a reliable tool for you to generate the pictures you need:
  

I hope this is helpful.

[lheagy]

Hi @jiajiasun, this is a subtle point. I believe we created that with an earlier version of the 2.5D DC layered earth app (DC_Layer_Cylinder_2_5D.ipynb)

In this case, it looks like we are not necessarily satisfying the boundary conditions (the mesh was trimmed to speed up the computation), so a correction factor is used under the hood so that the halfspace apparent resistivity is correct (cc @thast).

With the image @thast showed above, it looks like we are doing a better job of mesh design in the plate app, so you might want to check out the 2.5D plate app if you would like the students to calculate it by hand.

[thast]

Seems also the panels are no longer aligning in some DC apps :/ I created an issue
geoscixyz/em-apps#54

[jiajiasun]

Hi, @thast @lheagy Thank you very much for your response and explanation! It now makes more sense! Following your explanation, I ran the two apps: DC_Plate2_5D and DC_Layer_Cylinder_2_5D, and was able to reproduce what @thast showed (i.e., 7.6e+00 V). Thanks again!

I also ran another app in DC_Cylinder_2D.ipynb with the same set-up, i.e., pole-pole survey with electrodes located at -0.25 and 10.25 m for a homogeneous halfspace of 500 ohm*m. And I got something interesting.

The potential at the M electrode is 360 Volts.

Is that because this is a 2D simulation (as compared to 2.5D or 3D)? How to make sense of 360 V?

Once again, thank you very much! All the materials and apps you have spent so much time on are super helpful. My students are benefiting from all your efforts. I posted the slides that I have used so far in class in my Azure library (https://notebooks.azure.com/jsunatuh/libraries/uhelectromagnetics). I will keep posting more slides and all the lab exericses as I move on.

[thast]

@jiajiasun , yes the 360V is because the DC_cylinder_2D is a pure 2D app. As you can see in the upper panel, solving the Poisson equation in 2D lead to a logarithmic decay of the potential (instead of the 1/r decay in 3D). This is because in a pure 2D problem, the source is an infinite line instead of a single point/electrode. This is why 2.5D schemes are necessary, to approximate the point source and 1/r decay on a 2D mesh.

The advantage of pure 2D apps is that they are really fast, which is great for interactivity. There are helpful for building understanding (secondary fields...) but not for calculation.

Same as before (cc @lheagy ), we compute a correction factor to return a proper apparent resistivity.

[jiajiasun]

Hi, @thast Thanks a lot! That makes total sense. That was what I suspected might be happening. Thanks for your informative explanation!

[lheagy]

@jiajiasun I am going to close this issue. Please feel free to re-open if you have further related questions