I am trying to precisely weigh objects (100 lbs/45 kgs max., +/- 0.1
gram accuracy) on a wheeled platform. I do not have a single scale

large enough to weigh the entire platform. Are there algorithms
available to combine the weights obtained by weighing each wheel of the
platform simultaneously (4 scales total)? Is it as simple as adding
the scale readings together and accounting for each scale's inaccuracy?
TIA

Yes and no.
You can do this with a car or a truck, because the wheels are sprung,
and any differences in the flatness of the surface that the thing sees
is accommodated by the springs.
If you have a platform with more than three wheels that doesn't have
nice compliant springs then you'll get into problems with having the
balance of weight between the various wheels upset.
With three wheels, you could measure the weight at each wheel, being
careful to keep the thing level for each measurement. You'll introduce
some error, but not much.
With four wheels and no springs you could measure the weight on each
pair of wheels (either front/rear or left/right), once again keeping the
thing level.
Beyond that you'll need to get a bigger scale, or extend the one you
have with clever use of lever arms.

I'm not sure I quite understand your question but I'll make a stab at
it. If you have a 4 wheel vehicle and you place 4 scales simultaneously
under those wheels then the sum of those values is the weight of the
vehicle plus the object. You would have to add the errors of each of
the scales to the final tally. However; here's where I don't seem to
get it. Are you going to remove the object so that you can subtract the
weight of the platform? If so, why not just weigh the object when it's
off the platform? If not, how were you going to tare off the platform
weight? Additionally, in my experience scales that are suitable for
weighing vehicles (assuming that the platform is "car like") are not
nearly accurate enough for you. I'm not even sure that you can weigh 45
kgs with +/- 0.1 gram accuracy without special equipment anyway. That
kind of accuracy is in the realm of scientific weigh scales and I'll bet
you'll have a hard time finding one for 45kgs let alone (45kgs) +
(vehicle platform weight).
Please update us if you find an answer.
Good Luck
S Penzes

I don't think that's strictly correct. If you know that your four scales
are all +/- 1 gram (for example), then the absolute maximum error would be
four grams, correct, but that's not a *reasonable* estimate of the accuracy
unless you have some reason to believe that all four scales would
consistently show a bias in the same direction.
I just did a quick Monte Carlo test as an example. Say we know that our
scales will always be within +/- 1 gram, and that the error is uniformly
distributed (so that any error between -1 and 1 is equally likely to occur)
and that the errors of the four scales are independent of one another.
Yes, the maximum error is 4 grams either way. But about 60% of the time the
error will be less than 1 gram, over 90% of the time, the error will be less
than 2 grams, and over 99% of the time, the error will be less than 3 grams.
Using 4 grams as the estimate of error is wildly over conservative.
-Paul

That is STILL a super precise scale-- 0.002%
I'd start by assuming reasonable precision, like 0.1%, which is 45 grams,
or around 10 grams if you split between 4 scales.

You're understanding is correct. Let me see if I can clarify things
further. Think of the object in terms of a payload. I don't have
direct access to either the platform/vehicle or the payload (remote
operations). A payload (payload weight << platform weight) will be
deposited on the platform, the platform/vehicle will move to a new
location, the payload will be removed/dumped, and the platform will
return to its original location. I need to weigh the platform at the
new location both before and after the payload is removed/dumped, given
the constraints in my original post. Believe it or not, the accuracy
and precision requirements of the scale are achievable, its just we've
never had to deal with this size platform before

OK
I'm suitably impressed regarding the weighscale. Could you tell me the
name of the scale vendor because we've been trying to accurately
(although not to the same extent as you) do this for a while but aren't
aware of suitable scales?
As to your answer regarding total error. I talked to one of the guys on
staff, a physicist and hence his statistical skills are very good. The
question I posed was "how to properly add multiple uniformly distributed
error bands"? I'm sure I'll get his answer wrong and I'm equally sure
that somebody with better statistical skills than mine will correct me.
But I'll give it a try anyway.
If you add distributions they will ultimately approach the normal
distribution due to the central limit theorem. If the individual
distributions are normal then the simple answer is that the final
distribution is normal with a variance equal to the sum of the variances
of the individual instances. However the trick comes from the fact that
your individual distributions are uniform. If you have a +/- 1 uniform
distribution then the variance is +/- 1/3. This can be shown by
integrating from -1 to 1 on (x-u)^2 with a factor of 1/2 thrown in
because the total area must equal 1. To get statistically significant
data you must have at least 6 samples (I missed this part but it had to
do with 6 additions being "near enough" to the variance of a normal
distribution).
So what does all this mean .... I'm not really sure. My guess would be
that you'll have to take multiple measurements (to get statistical
significance), then you'll compute the variance of the normal
distribution by adding up the individual variances, then convert the
variance to a standard deviation, then you'll compute the total weight
and make a statement something like:
"the final weight is X with a normally distributed error with Y standard
deviation".
If I recall properly (and it's been about 25 years), 1 standard
deviation covers about 65% of the area under the normal curve. This
would allow you to further state that;
"the final weight is X and 65% of the time the weight is within X +/- Y"
I think 3 sigma is about 99% therefore
"the final weight is X and 99% of the time the weight is within X +/- 3Y".
I'm sure I've got lots of this wrong so you'd better double check.
Hopefully somebody with a better grasp of statistics can help you out.
Good Luck
S Penzes

Allow me to express my skepticism. I have no doubt that you can achieve
the sensitivity you claim, and perhaps also repeatability. As for the
absolute accuracy you seem to claim, how would you know? Suppose that a
customer believed that the indicated weight of a 40-Kg load was half a
gram too high. How could you show that he's wrong? In terms of distance,
that's about .8" in a mile.
The errors are of two kinds, systematic and random, both of which may
depend on the weight being measured. The random errors have a particular
distribution and the sum of all of them is a new distribution. The
systematic errors simply add algebraically.
Jerry

--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Yes, you can add four scales together simultaneously to get the total
weight, as long as the entire weight is supported by the four scales.
Finding a scale with a resolution of 0.0002% is another story.
Don
Kansas City

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