Weighing objects on multiple scales

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

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

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

"eromlignod" wrote in news:1156881494.586319.51650 @b28g2000cwb.googlegroups.com:
I was wondering about that myself.

"Paul Skoczylas" wrote in news:vL1Jg.20320 $Ch.14839@clgrps13:
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

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