Discussion in 'General Electronics Chat' started by cmartinez, Jun 18, 2015.

1. ### cmartinez Thread Starter AAC Fanatic!

Jan 17, 2007
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Could anyone here (especially @joeyd999) tell me if the following reasoning is correct? Thanks in advance.

Say I'd like to measure a weight of up to 100Kg, with a 100Kg capacity load cell.
This load cell has an hysteresis of 0.02% of its rated output, and so it has a maximum theoretical accuracy of 20 gr, even with the best available ADC and perfect calibration.
Say now that I replaced said 100Kg load cell with five 20 Kg Load cells (never mind for the moment how I'd manage to distribute the weight evenly).
If each 20 Kg load cell also has an hysteresis of 0.02% of its rated output, then it would mean that the hysteresis value for this load cell is 4 gr. And that the maximum theoretical accuracy attainable is five times better than the single 100 Kg load cell.
Assuming meticulous calibration, would a system that simultaneously uses five 20Kg load cells (each with its own independent ADC) in parallel to measure a weight of up to 100Kg have a five fold improvement in accuracy compared to a single 100 Kg load cell?

2. ### OBW0549 Well-Known Member

Mar 2, 2015
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I'm no expert on load cells and strain gauges (I've only designed amplifiers for them), but I think there's a flaw in your logic: just as you will be adding the outputs of your five strain gauges (after amplification, of course), you will also be adding their five hysteresis errors, for little or no net improvement.

Actually, I suppose you might expect some slight improvement over the result from a single load cell, since the hysteresis errors from the five load cells probably won't correlate perfectly; but I wouldn't expect that improvement to be significant or for it to be consistent from unit to unit.

I defer to joeyd999 for the final word...

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3. ### cmartinez Thread Starter AAC Fanatic!

Jan 17, 2007
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Yes... I think I know what you mean. I'm a mechanical engineer and I'm beginning to see load cells under a more familiar light.

They're springs! With electronics attached, but springs nevertheless. And are therefore subject to not only hysteresis, but distortion, repeatability and fatigue as well.

But... what is keeping my mind busy is this:

A 20kg load cell is five times more accurate than a 100kg one in absolute terms.

If I were to measure five SEPARATE loads of 20kg EACH with a single 20kg load cell, the result would be more accurate than if I were to measure a single 100kg load with a single 100kg load cell.

Therefore, why can't I just measure ONE 100kg load with FIVE 20Kg load cells and still obtain an improved degree of accuracy?

4. ### OBW0549 Well-Known Member

Mar 2, 2015
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In absolute terms, yes; but not proportionately. They're all rated for 0.02% hysteresis error, right? When you distribute your 100 kg load over five load cells and then adding the outputs of the load cells, you're also adding whatever errors each load cell contributes.

Going to bed. TTY tomorrow.

Jan 15, 2015
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With an allowable hysteresis of .02% a 100 Kg cell with 10 volts excitation and a sensitivity of 2 mV/V will have an output of 20 mV under a full load. The hysteresis allowable would be 20 grams. Using 5 20 Kg cells with the same specifications the allowable hysteresis error for each cell would be 4 grams and the allowable errors would be 5 cells * 4 grams = 20 grams which is where I was with a single 100Kg cell. I may have this wrong but I see no improvement.The errors with 5 cells would be cumulative as I see it.

Ron

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6. ### cmartinez Thread Starter AAC Fanatic!

Jan 17, 2007
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Yes, but that's in a worst case scenario. Statistically maybe the errors would tend to cancel each other. Truth is, I'm not sure about that reasoning...

Let's look at the problem from a different angle. New question: Is it possible to measure 100 Kg with an error margin of ±4 grams using any number of conventional load cells of any capacity, but all with a standard hysteresis of 0.02% ?

Sadly, I'm beginning to think that both you and OBW may be right... I'm going to try to dig deeper and see if there's a way to solve this.

7. ### OBW0549 Well-Known Member

Mar 2, 2015
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The errors will ONLY tend to cancel one another if they are randomly distributed (and then, only partially). And hysteresis will most certainly NOT be randomly distributed, at least not in sign.

I'm where I was last night on this: you might achieve a slight improvement by using multiple load cells, but it won't be much-- certainly not enough to be worth the extra expense and complication.

Jan 15, 2015
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I believe that is what it will come down to. If I have a +/- 5% resistor value of 100 ohms and the true value is 97 Ohms that is fine. Then I have another with a true value of 103 Ohms that is fine also. So if I now parallel them I have the perfect 50 Ohm resistor.

Ron

9. ### WBahn Moderator

Mar 31, 2012
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As I think the others are homing in on, the question isn't just the magnitude, but also whether the error is random or systematic.

If the errors were random, then some would be positive and others would be negative and your total expected error would be the Pythagorean sum of the five, which would sqrt(5*(4g)^2) which would be sqrt(5)*4g or 9g. So you could expect about half the error as with a single load cell.

However, the fact that you are talking about hysteresis effects means that you are talking about predominantly systematic errors and you don't get the kind of cancellation that you do with random and independent errors. You probably will not see much, if any, improvement while picking up more error due to more measurement channels (though some of these errors will be random and be improved by the sqrt(n) effect) and also the complexity of distributing the weight evening. You will have some errors that go as the size of the reading and so having one cell carrying more of the weight may result in a larger error contribution than the sqrt(n) effect can overcome (even if the errors were random, which they aren't).

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10. ### RichardO Well-Known Member

May 4, 2013
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One more comment to go with the others' good advice...

All of this assumes that the load is evenly distributed between the load cells. Now, I will let the others tell you (and me) about the effect of mismatched loads.

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11. ### cmartinez Thread Starter AAC Fanatic!

Jan 17, 2007
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Ok, I think I see the picture now... so my second question still stands. Is there any way to measure 100Kg with a maximum error of ±4 gr using conventional class 2 load cells with a hysteresis of 0.02%?

12. ### WBahn Moderator

Mar 31, 2012
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You seem to be claiming that the hysteresis is the ONLY error.

Also, 1 gr is 1/7000 of a lb -- there are a bit over 15 gr in 1 g. You mean ±4 g.

13. ### cmartinez Thread Starter AAC Fanatic!

Jan 17, 2007
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Not my intention. I'm rather focusing in that error because it's the largest one in a normal arrangement.

I'm confused. I don't recall mixing units in my previous explanation. Mind elaborating?

14. ### WBahn Moderator

Mar 31, 2012
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But you did. You stated that the 100 Kg (which should be 100 kg) has a theoretical accuracy of 20 grains.

You have been asking if there is any way to measure 100 Kg with a maximum error of ±4 grains.

15. ### cmartinez Thread Starter AAC Fanatic!

Jan 17, 2007
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Well I'll be... I'm soooooo glad I wasn't one of the engineers working at Nasa's Mars Climate Orbiter mission, otherwise I'd be pulling my hair in despair and dodging the blame of its failure.

Yes W, you're absolutely right. One should mind the units carefully when posting any kind of question.... so, for the record, allow me to re-post my last question:
Is there any way to measure 100 kgf with a maximum error of ±0.004 kgf using conventional class 2 load cells that have a hysteresis of 0.02%?

Jan 15, 2015
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Grains and Grams? Very good mention on that note. I hand load my ammunition and here in the US the unit of measurement for a powder charge is weighed out in grains. When I see a load cell specified out as Kg I see Kilo Grams. Yep, a 1 Lb canister of powder is indeed 7,000 grains and if I am loading 50 grains to a cartridge case I can figure a pound of powder will yield about 140 loaded cartridges.

Something else I have noticed when looking at digital powder scales is they almost always readout in Grains or Grams so we choose Grains. However, if we open one up and observe the guts:

And research the actual strain gauge part numbers their data sheets reflect Grams and the electronics do the conversion.

Ron

17. ### strantor AAC Fanatic!

Oct 3, 2010
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I think in the theory your idea makes sense Martinez. Just like parallelling (5) 20% resistors with random accuracy error would result in a resistance that is much tighter accuracy, doing the same with load cells should yield a similar result. But in practice I think you would be disappointed. Many facets of the real-world would conspire against you. Consider the following:

Uneven loading- your load would need to be precisely centered in the middle of a circle of 5 load cells. The sum of the 5 should add to the proper value even if one is higher than the others, but what if that one higher one is >100% (or 150% typ.) and no longer is contributing to the sum?

...sorry I had more to say but gotta go.

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