See pages 13-15 in "Electrical Contacts: Principles and Applications" for reference.

Imagine the following system:

  • Two flat, but rough, surfaces of the same finite area $A$ being brought together.
  • The roughness of the surfaces is much less in magnitude than the total thickness of the two bodies hosting these surfaces.
  • These bodies have the same electrical resistivity $\rho$.
  • Body 1/Surface 1 is fixed in space.
  • Surface 2 is pressed against Surface 1 with force $F$.
  • Each body has the same Meyer hardness $H$.
  • The surfaces are perfectly clean.

The true mechanical contact area $A_m$ is much smaller than the nominal contact area $A_n=A$. Given this fact, the actual pressure at the contacts is much higher than you might initially expect, typically of the order of the strength of the material that the bodies are made of. It is suggested that the contact pressure is equal to the flow pressure of the bodies and plastic flow occurs.

Taking the above, the we can make the assertion:


It seems quite counter-intuitive at first that there is no presence of $A$ in the equation. In other words, it doesn't matter how large the contacting surfaces are, for the same force they will have the same $A_m$. I'll leave the book (link above) to explain this further.

Following some additional working in the book, we obtain the following important equation, that appears to be generally accepted:

$$R_c=\sqrt{\frac{\rho^2\eta\pi H}{4F}}$$

Here $\eta$ is related to how clean the surfaces are. We will make $\eta=1$, based on our earlier assumption that the surfaces are clean. We will now say:


Where $k=\sqrt{\frac{\rho^2\pi H}{4}}$.

So, $R_c$ is proportional to the inverse square root of $F$. I find this relationship a little odd. Why I find it odd, should make more sense based on a parallel problem below.

For a fixed $F$ let's model $R_c$ as just a simple resistor. Say I split $A$ into 4 smaller but equal areas, $A_s$, such that $A=4A_s$. I should be able to treat each $A_s$ as a resistor with resistance $R_s$. Now we have 4 resistors of resistance $R_s$ that are in parallel with each other, one for each area $A_s$. Given that these resistors are in parallel with each other, it should be pretty straightforward, via Ohm's Law, to relate $R_c$ to $R_s$:


The resistance of $A_s$ is 4 times larger than the resistance of $A$. This makes sense to me, 1/4 of the cross-sectional area means 4 times the resistance. Sum them all together in parallel and you get $R_c$. It all adds up.

But think about it in relation to our $\sqrt{F}$ relationship above. For a fixed $F$, evenly distributed, isn't the force applied to $A_s$ equal to $\frac{F}{4}$? If so, our equation for $R_s$ becomes:


Relating this to equation for $R_c$ above gives us the relationship:


I appear to have two contradicting statements for how to relate $R_s$ to $R_c$, with a factor of $\sqrt{N}$ difference, where $N$ is the number of smaller areas I choose to split the surface into. If $R_c$ was proportional to $\frac{1}{F}$ instead, this would fix the issue.

Given that I am trying to do finite element analysis, I have a significant issue here. Any ideas where I am going wrong?

Some additional thoughts

I've been doing some more thinking and went down the dimensional analysis route (thanks Ben Crowell) to see what my, possibly incorrect, assumption about evenly distributed force and resistance might lead to as an equation. Strangely enough, it seems to give a more sensible contact resistance for realistic values of the parameters in the equation, for my scenario. Clearly, this could just be coincidence, but here goes.

If we introduce a new parameter $\alpha$ to describe the characteristic "height" of the roughness of the surface (units of length), we can obtain the following relationship from dimensional analysis:

$$R_c\propto\frac{\rho H \alpha}{F}$$

This seems to make some logical sense:

  • Increase in resistivity, hardness and roughness gives an increase in contact resistance.
  • Increase in force gives a decrease in contact resistance.

Do you agree that the dimensions stack up? Have you seen something of this form anywhere else? Am I just doing bad physics now?

  • 2
    $\begingroup$ Good reasoning. I also find the formula suspicious. It does not contain A, and yet when we need a good electrical connection from a practical standpoint we definitely do use connectors with larger area. I would expect both area and force to be important. $\endgroup$
    – Dale
    Oct 17, 2019 at 11:40
  • $\begingroup$ Thanks @Dale. I completely agree. Yes, it is a little strange, yet the logic in the book doesn't seem too unreasonable and the equation seems to be generally accepted. However, I just can't see the flaw in my thinking with the later parallel problem. $\endgroup$
    – Tim W
    Oct 17, 2019 at 12:55

1 Answer 1


Interesting question! These are just some thoughts based on general knowledge of physics. I don't have any detailed knowledge of this topic. I could be wrong.

If the resistance is going to depend on the variables $\rho$, $H$, and $F$, then the exponents follow from dimensional analysis. Given the units associated with the Meyer hardness, plus this assumption, we have to have a proportionality to $F^{-1/2}$.

In your argument about splitting the surface up into four surfaces in parallel, you're assuming that the force gets equally distributed, i.e., that the pressure is uniform. Actually this is unlikely to be true. The surface is rough, so the pressure is likely to be nonuniform. This model can probably be interpreted as implying some statistical (and possibly fractal) model of the surface. In your four surfaces, probably one of them is sustaining almost all of the force.

  • $\begingroup$ Thanks for the helpful thoughts here. They definitely help me to head in the right direction. You are absolutely right about the dimensional analysis. The book states $H$ as the Meyer Hardness which somewhat unhelpfully few people seem to use. I wonder if it's all about this fractal nature of the surface. Perhaps the $\sqrt{N}$ term appears from the statistics of that. I'll keep thinking, but thanks for the insight. $\endgroup$
    – Tim W
    Oct 17, 2019 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.