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I am looking to measure the force applied (in this case by a foot to a pedal).

I have a working prototype that allows up to 450N of force, but wish to double this.

The load cell I have been using is available in a 900N rating however there is no stock of this nor is it likely to be available any time soon. The output of the loadcell is as a value transmitted in the I2C protocol

As a late night brainwave purchase I bought a second identical load cell, with the thought process that if I divided the load between both and then summed the values I would have the force applied, while increasing the load measuring capacity to the desired range.

I am however getting myself very confused over whether I am just measuring the same force twice.

My current plan / arrangement is to have the load cells at the end of a rod, all held within a tube, so:

Fixed end -- load cell -- rod -- load cell -- pedal

While the basic physics is current, my brain is getting older and would like a bit of help working this out (other than through experiment) before I start machining metal. Its over 20 years since I needed to consider anything like this.

Thanks in advance

James Hounslow
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1 Answers1

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"Fixed end -- load cell -- rod -- load cell -- pedal"

You are indeed measuring the same force twice. You need to build a little mechanism that shares the force between the two load cells, basically a seesaw.

load cell seesaw

Or for half the cost just use a 2:1 lever

lever

Greg Locock
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  • I think the problem here (in my head) is that I am thinking of the load cell as a spring, which it is not. This is the same reason why my head is struggling with your half the cost solution. I'm probably going to have to do a lot more reading and relearning, but thats not always a bad thing – James Hounslow Feb 04 '23 at 10:55
  • @JamesHounslow Load cells do act like springs you got that right. The main point is that they are imperceptibly compressible. So if you put them into series they both experience the same force. If you put them in parallel (like above) they split the force. – NMech Feb 04 '23 at 15:35
  • @NMech - it is this aspect that is confusing me. In electronics exactly the opposite is true - in parallel a resistor sees the whole voltage and in series it is divided between them. This is what is catching me out in my head. – James Hounslow Feb 05 '23 at 17:12
  • @JamesHounslow you are right they are the opposite, so it can be confusing at first. – NMech Feb 05 '23 at 19:01
  • That's why when univerity profs are confusing students by using Firestone electrical analogies to mechanical systems they use current to represent force, so they can use the same circuit diagram. – Greg Locock Feb 05 '23 at 20:52