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Adding Beltweigher Tolerances/Uncertainty

  1. May 9, 2012 #1

    I am hoping someone can set me straight on the following.

    At one of our power stations, we have a single conveyor which transfers coal from the mine to the power station. This single conveyor travels across two belt weighers mounted approx 5 metres apart. Both are the same model and made and installed by same OEM. Except one is a class 1 ±0.5% and the other is a class 0.5 ±0.25%.

    In an effort to try and keep track of what they are doing, I constructed a graph in excel to plot their daily measurements. As I do not know the actual weight of the coal passed across them, except that the same amount goes across both each day. All I have at the end of each day is a measurement from each weigher representing what each weigher claims has passed over it during the day. I determine the difference between the two measurements and covert it to a percentage against one of them. In the ideal world, if both weighers were measuring accurately, both readings would be the same. But given the allowable tolerance of each weigher, in my mind the maximum allowable difference would be ±0.75%, if one was on its upper limit and the other on its lower.

    But the thermal performance engineer I work with is telling me that the maximum allowable tolerance is ±0.56%, based on the following √(0.5^2+0.25^2)

    Based on what I have read regarding the adding of tolerance/uncertainty, this situation does not fit any example given.

    Am I missing something?

    Any thoughts most appreciated.

    Thank you
  2. jcsd
  3. May 9, 2012 #2

    Stephen Tashi

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    It won't be clear to the average reader what "class 1" means. It won't be clear to anyone who hasn't read the fine print of the equipment specifications what the [itex] \pm 0.5\% [/itex] means. Is that a specification for weighing accuracy that involves time, such as total weight per 8 hours of operation or is it a static type of specification that involves only per belt load ?

    After we settle those questions there is still the problem of what [itex] \pm 0.5\% [/itex] means. You are taking it as a guarantee that the measured abount will be within those limits. The other engineer is taking it as the standard deviation (or something proportional to it) of a distribution of measurements - in which case there will be some probablity that the measurements fall outside those limits. Only the manufacturer of the equipment or its documentation can tell you the meaning of [itex] \pm 0.5\%[/itex].

    If we assume 0.5% and 0.25% are standard deviations of independent random variables then his calculation is correct for he standard deviation of their difference. Whether this solves the problem correctly will depend on the precise definition of "tolerance" that is used.
  4. May 9, 2012 #3
    Class one is the term used for a weigher that has an accuracy of ±0.5%, and the class 0.5 is for a weigher that has an accuracy of ±0.25%

    It does not refer to time in any way, it just means that at any point in time when coal has, or is, going over the weigher the actual weight falls within that accuracy. Each year we have the weighers calibrated and the weighed, by law, must fall within the limits.

    It means that when the calibrator puts a known 1000Kg of coal across each weigher, it is acceptable for the class weigher to state anything between 1005 - 995, and the class 0.5 to state anything between 1002.5 - 997.5

    I hope this helps.

    Thank you
  5. May 10, 2012 #4

    Stephen Tashi

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    Taking your interpretation of the plus-or-minus percentages as an absolute gurantee of where the mesurements will be, my understanding is that you are computing

    [tex] F(e_1,e_2) = \frac{ W(1 + e_1) - W(1 + e_2)}{W(1 + e_2)} [/tex]
    [tex] = \frac{ W(e_1 - e_2)}{W(1 + e_2)} [/tex]
    [tex] = \frac{ e_1 - e_2}{1 + e_2} [/tex]

    [itex] W [/itex] is the unknown correct weight.
    [itex] e_1 [/itex] is the error of the class 1 machine, as a fraction of [itex] W [/itex].
    [itex] e_2 [/itex] is the error of the the other machine, as a fraction of [itex] W [/itex].

    We know
    [itex] -0.0050 \le e_1 \le +0.0050 [/itex]
    [itex] -0.0025 \le e_2 \le +0.0025 [/itex]

    We want to know the minumu and maximum values of [itex] F [/itex] subject to those constraints.

    I think the extrema of F occur at the boundary points (the corners) of the rectangle that bounds the possible values of [itex] (e_1,e_2) [/itex] So you have to evaluate the 4 possibilities:

    [itex] e_1 = -0.0050 , e_2 = -0.0025 [/itex]
    [itex] e_1 = -0.0050, e_2 = + 0.0025 [/itex]
    [itex] e_1 = +0.0050, e_2 = -0.0025 [/itex]
    [itex] e_1 = +0.0050, e_2 = +0.0025 [/itex]

    to see the extremes.

    It's 1:08 AM here, so I'm not going to attempt that now!
  6. May 10, 2012 #5
    If you know and you're positive that the same amount go across both each day you should consider as the maximum allowable tolerance 0.25, that is the one from class 0.5.

    Let me explain, imaging you have a bucket with water and I give you two thermometers to check the temperature, one let's say class 10 with an accuracy of ±10 degrees and one class 1 with an accuracy of ±1 degree.

    So imaging class 10 tells you 20 degrees and class 1 tells you 30 degrees, would you even consider that the actual temperature is below 29? Can you see how it makes no sense to estimate the accuracy that way when both thermometer are measuring the same temperature?

    As a matter of fact you can further reduce the maximum allowable tolerance from 0.25 by using the information given by the second weighter using techniques like Kalman filtering and others, but I think that would be an overkill in your case, just stick with the most accurate measurement of class 0.5 with ±0.25.

    Good Luck!
    Last edited: May 10, 2012
  7. May 10, 2012 #6

    Stephen Tashi

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    Doing the calculations this morning, I agree with you. [itex] F [/itex] should be between -0.00749 and +0.00752. Computing [itex] F [/itex] makes sense as a rough "sanity check".

    It isn't clear from your original post whether your main goal is merely to check that the weights from the two machines are sane or whether your main goal is to estimate the actual weight of coal transported per day. If you want the "best" estimate of the actual weight of coal transported then we have to formulate some probability model for the errors in weighing and say what "best" means in mathematical terms.

    Contemplating how a belt weigher works is interesting. It's trying to estimate the weight of coal it transports, correct? I suppose it could measure its own speed and weight at closely spaced intervals of time and do a calculation to determine how much has moved across it. Off hand, I don't see how it could make a more direct measurement of the amount transported.
  8. May 10, 2012 #7
    Thank you all for you responses.

    If I may add: I agree that what I am trying to achieve is not a definitive check as I never know the the correct weight of coal that goes across both, only that it is the same quantity. So all I can do is monitor the difference between the measurement of each weigher each day. Then if this amount is greater as a percentage than the 0.75% I believe is the maximum, then all I know is that one of the weighers is outside its designed tolerance. If both weighers were to stay within the 0.75% of each other and both drift high or low, then this check will not indicate this. I am just trying to use what data I do get each day to keep an eye on them.
  9. May 10, 2012 #8
    Ah OK, then your concern is about one of the weighter going outside its tolerance? I thought you wanted to know how accurate were the measurements.

    Trying to find out if a weighter is going outside its tolerance by checking the difference between measurements might be not the best idea since they might both lose its calibration at the same time (especially if both are the same model) and then you would not notice it. You say that by law it must fall within the limits, yet the kind of measurements you're doing cannot guarantee that.

    Anyway, if the only data you have is those measurements, and since you are collecting the data, I would recommend you to work with the empirical distribution of the differences between both weighters. Once you have that distribution you can better determine what range of values are acceptable to you. For example, you might find out that differences of 0.2 might be beyond 99% of the differences and then that 0.2 would be already a sign that something is wrong without having to wait for differences of 0.75
  10. May 11, 2012 #9
    I agree it is not a definitive check and is not meant to be legal. Each year they are checked, calibrated and re-certified. Normally no problems are experienced year to year, but as I am collecting the data on a day to day bases, I thought it would be a simple, non-legal, sanity check that all is as expected. If the occasion did occur whereby the difference in readings became greater than 0.75%, then I could suggest that someone go out and see if there was anything obviously wrong, such as build up around one or more load cells of the cradle of one of the weighers.
  11. May 11, 2012 #10
    Right, then definitely I would suggest the empirical distribution for a more sensitive method to detect anomalies, there are more elaborate options but this one is very simple and it suffice for your problem. I am not very familiar with the statistical capabilities of Excel but this is pretty simple and I am sure Excel can handle it.

    Anyway, If you can post the data I can give you some examples with R.
  12. May 14, 2012 #11
    Thank you all for your comments, has given me a bit more to think about.
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