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I am new here so I hope I'm posting it at the correct place.

I am having some troubles with some of my force sensors. My case is like this: I have a body which have a weight. However, the surface is not even. I wanted to measure the weight of the whole body, but with a small sensor (circular force sensing resistors with diameter of 2 cm). Hence, I know that the reading that I am getting on that specific region of my sensors are then as pressure (am I correct on this??)

The figure of the intended system

I tried to make a figure out of it to help to ease things out (hopefully). So, imagine the 2 orange circles are the 2 cm diameter circular FSR sensors. The blocks over it is the weight, but it is not evenly distributed. Hence the front side have a heavier weight compared to the back side.

If so, is it possible for me to add up the "weights" that I got from the two sensors (shown as the orange circles), and then calculate the pressure by:

Pressure = (Weight1 + Weight 2)/total Area of 2 sensors

By this sense, can I assume that I am able to get the total weight, by using this formula:

Total Weight= Pressure (from eq. on top) * Total area of the "grey" box

Any help would be really great!

Thank you in advance for your help!

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  • $\begingroup$ I'm not familiar with FSR sensors, but what does it give you, force or pressure? Both? And if it gives pressure, can it do so if the force per unit area of the sensor surface is not uniform (the orange disc to the left)? $\endgroup$
    – Bob D
    Commented Aug 20, 2019 at 22:12
  • $\begingroup$ Hi, for now I get the weight from each of the sensor. And what I get from the readings are that the weights from the sensors are different. Hence I assume that I actually got a pressure instead of weight or force here. Do you get what I mean? I am quite confused how to explain this aswell.. but in any case, can I sum up the weights that I got from the 2 sensors and get the "overall pressure"? Does it work that way? $\endgroup$
    – Stanley Setiawan
    Commented Aug 21, 2019 at 9:03
  • $\begingroup$ Is the yellow sensor to the left not completely covered by the weight? It looks that way in the picture. $\endgroup$
    – Bob D
    Commented Aug 21, 2019 at 10:56

2 Answers 2

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As I am not familiar with force sensing resistors (FRS) but after watching a couple of Utube videos I do wonder about their accuracy and the proper way to use them. In particular, the effect of not uniformly pressing the entire surface of the disc and the non linear relationship of resistance versus force concern me.. In any event, my answer is based on the assumption they are accurate.

If so, is it possible for me to add up the "weights" that I got from the two sensors (shown as the orange circles), and then calculate the pressure by:

Pressure = (Weight1 + Weight 2)/total Area of 2 sensors

This equation is incorrect. There is no total pressure as implied by the left side of your equation. There is only a pressure on each sensor, which will be different unless the center of gravity of the combined weights is exactly between the sensors and the contact areas on the two sensors are the same. Otherwise, you can't add pressures like you do forces.

The total weight is given by

$$W=P_{A}A_{A}+P_{B}A_{B}$$

where $P_A$ and $P_B$ are the pressures on sensors A and B and $A_A$ and $A_B$ are the areas of sensors A and B in contact with the grey weight, and $W$ is the total weight (Weight 1 + Weight 2)

Based on your first equation, you have

$$P_{tot}= \frac{W}{A_{A}+A_{B}}$$

Or

$$W=P_{tot}A_{A}+P_{tot}A_{B}$$

You can see the two equations are not the same because $P_{A}$ does not equal $P_{B}$ unless as already stated the center of gravity of the total mass is exactly between the sensors and the contact areas are the same.

By this sense, can I assume that I am able to get the total weight, by using this formula:

Total Weight= Pressure (from eq. on top) * Total area of the "grey" box

It's not clear to me what you mean by the "total area of the grey"? However, as already shown the first equation is incorrect, so any conclusions you reach based on it would also be incorrect.

Hope this helps.

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    $\begingroup$ Great! thank you for your answer. It gives me much more of a clearer view of my current issue. Thanks for all of your troubles! have a great day! $\endgroup$
    – Stanley Setiawan
    Commented Aug 21, 2019 at 20:19
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    $\begingroup$ @StanleySetiawan You are very welcome. $\endgroup$
    – Bob D
    Commented Aug 21, 2019 at 20:21
  • $\begingroup$ @BobD I not like the 0 score with the green check mark; the answer is very nice. $\endgroup$
    – Sebastiano
    Commented Aug 25, 2019 at 11:36
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If the weights are not moving, the total weight is the weight measured by sensor 1 plus the weight measured by sensor 2. Assuming your drawing is correct and the grey weight is totally supported by the yellow sensors.

If your sensors are measuring weight, why use your pressure analogy? Just read the weight on each sensor.

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  • $\begingroup$ Hi, thanks for answering! It seemed my drawing is a little misleading then. The weights are kinda moving every time. Maybe for easier visualization, imagine a human foot. And several sensors of 2cm in diameter are placed below the foot. Hence, the sensor is not representing all of the weight, bcs some of the weights are placed outside of the sensors. In this case, is it available on what I was suggesting? $\endgroup$
    – Stanley Setiawan
    Commented Aug 20, 2019 at 20:53
  • $\begingroup$ By not moving I meant there was no acceleration of the weights so that F = ma = 0. Which means the weights are in equilibrium so Force pushing up on weights equals force of gravity pulling down. Then F1 from sensor 1 + F2 from sensor 2 = total weight of objects. $\endgroup$
    – Natsfan
    Commented Aug 20, 2019 at 21:26
  • $\begingroup$ @jmh Are you familiar with FSR sensors? $\endgroup$
    – Bob D
    Commented Aug 20, 2019 at 22:13
  • $\begingroup$ no I'm not. sorry! $\endgroup$
    – Natsfan
    Commented Aug 20, 2019 at 23:54
  • $\begingroup$ Ah i see, okay thanks @jmh. But however, the orange circles are actually so thin, hence the other surfaces are on the surface of the floor aswell. I am trying to say that the grey surface is also being "supported" by the floor. can it then be F1 + F2 = total force? $\endgroup$
    – Stanley Setiawan
    Commented Aug 21, 2019 at 9:06

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