# Verification of weighing scale 150kg capacity 0.05g accuracy

#### Dongzkie

##### Involved In Discussions
Just for example I want to verify a weighing scale 150kg capacity 0.05g accuracy. If this are my test wts calibration result (20kg each) : 19.996, 19.996, 19.993,19.994 and 19.990, If I am going to verify the scale to 40kg, so I am going to add the results of two test weights as reference 19.996 + 19.996=39.992kg? or 19.99 + 19.99=39.98kg? which of this reference were correct 39.992kg or 39.98kg ? should i include adding the last digit?

#### Hershal

##### Metrologist-Auditor
Staff member
Super Moderator
I would say that the first point is that you need to verify the scales at several points, if a 150kg scale. I would select zero, then something close to the 150kg, perhaps 140kg, then three other points.

Given you state it as a verification and not calibration, you get a bit more leeway.

You mention 20kg, so it should be a point.

Then trend your results. Excel can do this over time.

Oh, the mass standards you use need to be calibrated by an accredited calibration lab with MASS in the scope - NOT scales and balance. Also, use gloves to handle the mass standards.

Hope this helps.

N

When you add the weights what is the scale displaying? I would say if it is displaying what the weight is then it's verified and as long as the weight has been cal'd then the scale is cal'd. Counting scales are a little different. They can be out of cal and still count correctly. So be cautious.

#### Dongzkie

##### Involved In Discussions
need some more explanations.

#### dgriffith

##### Quite Involved in Discussions
need some more explanations.
In order to help better, so do we.
Type of scale, stated accuracy--not the same thing as readability/resolution, and is 0.05g the internal or display resolution or the actual accuracy statement?
The ratio of 0.05g to 150kg FS is 3,000,000:1, better than anything I've found from the scale house mfr's (2,000,000:1, and that was internal resolution for a counting scale, and not for 150kg). Are you sure 0.05g is correct?

As for the test masses, I would add the full individual values and then round the result if needed. But this all seems too simplified to me. Where's the mass uncertainty or tolerance, and what will you do with it?

#### Dongzkie

##### Involved In Discussions
sorry my mistake its 0.05kg
1.)as for example ( 2 test weights ) : calibration result 19.998 and 19.996 unc of meas is +/-0.002kg, if im going to add the two wts it would be 39.994kg as my reference, if my scale's actual reading is 40.00kg, the error is 0.01kg?
2.) what if ill just add excluding the last digit 0.008 and 0.006, 19.99+19.99=39.98kg as my reference, is this correct?
In order to help better, so do we.
Type of scale, stated accuracy--not the same thing as readability/resolution, and is 0.05g the internal or display resolution or the actual accuracy statement?
The ratio of 0.05g to 150kg FS is 3,000,000:1, better than anything I've found from the scale house mfr's (2,000,000:1, and that was internal resolution for a counting scale, and not for 150kg). Are you sure 0.05g is correct?

As for the test masses, I would add the full individual values and then round the result if needed. But this all seems too simplified to me. Where's the mass uncertainty or tolerance, and what will you do with it?

#### dgriffith

##### Quite Involved in Discussions
I will assume the test weights were calibrated at the same time and place by the same reference masses. Therefor, since their uncertainties have the same source directly, they are correlated.
I would not round or truncate the test wt. values, and the uncertainties add linearly, not RSS.
(19.998 ?0.002) kg +
(19.996 ?0.002) kg =
-------------------
(39.994 ?0.004) kg; (39.990 to 39.998) kg.

If the reading is 40.00 kg, then the indicated error could be from 0.00 to 0.01. Remember, the 0.05kg is likely the readability/resolution, not the accuracy (perhaps 0.1% FS?). The scale should be ok even though it reads 0.01kg high.
This is all superficial and not at all rigorous, but I hope it helps.

A

#### Automatic

Just for example I want to verify a weighing scale 150kg capacity 0.05g accuracy. If this are my test wts calibration result (20kg each) : 19.996, 19.996, 19.993,19.994 and 19.990, If I am going to verify the scale to 40kg, so I am going to add the results of two test weights as reference 19.996 + 19.996=39.992kg? or 19.99 + 19.99=39.98kg? which of this reference were correct 39.992kg or 39.98kg ? should i include adding the last digit?
If the two masses are KNOWN and you're doing a span calibration at 40.000kg, you'll use the sum of the value. I would focus on corner load and repeatability.

This is one of the most common misunderstanding. The smallest change that can be read by a scale is called resolution. This may or may not be the same as the "e" value. Usually, it is not.

A 100kg x 0.05 resolution will be rated d=0.05, but the e value will most likely be 0.1 to 0.5. The "e" value is the smallest unit that can be used for pricing.

You shouldn't be using precision weights except for the span verification and calibration adjustment. Something like plastic coated gym dumbells or buckets of sand are adequate for linearity and repeatability. You really just need to write down the value for one of them.

Say dumbell one reads 20.755kg. Ok, well write it down. We'll call it "x".
Weigh it repeatedly ten times and take the standard deviation of all the values. Check that the STANDARD DEVIATION is within specs.
20.760, 20.750, 20.755 . . . .

Start in the middle, then measure half way to each direction and document. if the spec is +/- 10g at about 20kg, it means if the middle reads 20.755kg, left reads 20.735, right reads 20.775, it's out of specs.

Linearity is another.
so forth until you get near 150kg

Hysteresis:
Repeat in increasing and decreasing direction and compare the difference.
Preload the scale to about 40kg. zero. add the 20.755kg dumbell. When you read 20.765 going up, that's a 10g linearity error. Zero it, then remove the dumbell. If you read minus 20.750 this time, your hysteresis in this span range is 15g.

A scale could have a passing span calibration using one precision weight right in the dead center and only in the ascending direction. This is the "accuracy check" as often mistakenly called. Scales can pass this and fail the other tests miserably. Normally failure of corner load and hysteresis is caused by cell damage.

#### Dongzkie

##### Involved In Discussions
So do you mean that it is recommended or required that adding 2 test weights should be calibrated at the same time in a accredited labs?
Thanks..
I will assume the test weights were calibrated at the same time and place by the same reference masses. Therefor, since their uncertainties have the same source directly, they are correlated.
I would not round or truncate the test wt. values, and the uncertainties add linearly, not RSS.
(19.998 ?0.002) kg +
(19.996 ?0.002) kg =
-------------------
(39.994 ?0.004) kg; (39.990 to 39.998) kg.

If the reading is 40.00 kg, then the indicated error could be from 0.00 to 0.01. Remember, the 0.05kg is likely the readability/resolution, not the accuracy (perhaps 0.1% FS?). The scale should be ok even though it reads 0.01kg high.
This is all superficial and not at all rigorous, but I hope it helps.

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