Resistance of a sphere of resistivity rho and radius S

[Sadiq]

I calculated the resistance between two points (P and Q) on a sphere that are located on the two ends of a diameter by dividing the sphere into thin strips of thickness dz that are in series perpendicular to the line PQ (say the z-axis). I can get an expression for the indefinite integral in theta (the polar angle theta, z = S Cos (theta), and S= radius), however I run into problem when evaluating this definite integral for theta = 0 to pi. Apparently the fact that the strip element becomes a points at the ends is causing this problem. I use dR=rho * dl/A with dl = -S Sin(theta) d(theta) and A(theta) = pi {S Sin(theta)}^2. What is going wrong here? Because I know that there must be a finite resistance between two such points.

 

[ardcarlos]

I need help!

I'm a spanish coach from Barcelona and I have a problem.

I need some information about "curve fenomenon" in velodrom cycling. When a cyclist go into a curve power is less than in straight and velocity is high to a straight. Please do you explain this fenomenon?.

My e-mail is ardcarlos@hotmail.com

Thank you very much!

 

[Graeme Robinson]

It seems that you are on the right track with your approach to calculating the resistance between two points on a sphere. However, there are a few things to consider in your calculation.

Firstly, when dividing the sphere into thin strips, it is important to make sure that the strips are evenly spaced and have a consistent width. This will ensure that your calculation is accurate and that you are not missing any areas of the sphere.

Secondly, when evaluating the definite integral for theta = 0 to pi, you may encounter some issues due to the fact that the strip element becomes a point at the ends. This is because at the poles (theta = 0 and theta = pi), the strip element has a width of 0, resulting in a singularity in your expression. To solve this issue, you can use the limit definition of the integral to evaluate the integral at the poles separately.

Lastly, it is important to check your units in your calculations. The resistance should have units of ohms (Ω), so make sure that all of your variables and constants are in the correct units.

Overall, it seems that your approach is correct, but there may be some small errors or oversights in your calculations that are causing the issues you are encountering. I suggest double-checking your work and making sure that all of the variables and units are correct.