Propagation of Errors in Calculations - Thickness of a coating material inside a tube

[Brad Gover]

Hi All, I have a problem in calculating the thickness of a coating material inside a tube. The tubes inside diameter is measured before and after coating. The coating thickness is calculated by substracting the diameters and dividing by two to get the thickness of the coated material. The before coat diameter is 0.3949". The after coat diameter is 0.3893". The Total Gage R&R standard deviation is 0.000127". I need to calculate the coating thickness value and its associated error.

I am thinking the thickness without error = (0.3949 - 0.3893)/2 = 0.0028"
The associated error from the resulting calculation would be
± Error = [5.15 * (0.000127”2 + 0.000127” 2).5]/2 = ± 0.000462”
I would record the thickness as 0.0028 ± 0.000462”?

I saw some examples where they standard deviation was used in place of the 5.15 multiplier value i.e., 0.0028 ± 0.00018. I would think this would only capture some 66% of the uncertainty.
P.S. I used Big 3 MSA value of 5.15 instead of 6
Thanks, Brad

[Stijloor]

A Quick Bump!

Can someone help Brad?

Thank you very much!!

Stijloor.

[Stijloor]

Again A Quick Bump!

Can someone help?

Thank you very much!!

Stijloor.

[AdamP]

In Reply to Parent Post by [COLOR=black

I need to calculate the coating thickness value and its associated error.[/COLOR]

I am thinking the thickness without error = (0.3949 - 0.3893)/2 = 0.0028"
The associated error from the resulting calculation would be
± Error = [5.15 * (0.000127”2 + 0.000127” 2).5]/2 = ± 0.000462”
I would record the thickness as 0.0028 ± 0.000462”?

HI - A couple of questions. If you're stating the margin of error around the average coating thickness I'm not sure why you've included the GRR StDEV twice. Also, the CI formula (margin of error) has us taking the Z value (here 5.15 or 6) * StDEv/SQRT of sample size - I don't see the SQRT of n in your calculation - CAn you share the actual formula you're using to express the margin of error?

Also - yeah folks have moved to using 6 rather than 5.15 as it shows a confidence level of 99.73 rather than 99.

Cheers,

Adam

[Brad Gover]

Well, I am thinking the reason I am using the standard deviation twice is we measure the ID twice and calculate the difference between the two measurements to get the thickness of the coating. i.e.,
Machined ID = 0.3955"
Coated ID = 0.3925"
Coating thickness = (Machined ID - Coating ID)/2 = (0.3955" - 0.3925")/2 = 0.003".
Each measurement is made by the same instrument and each individual measurement would induce error. In this case we made two measurements to get the coating thickness. (If I only made one measurement I think I would only use the error Std Dev once.) The first measurement error is compounded with the second measurement error when both are used in a calculation. (See attachment if I am lucky enough to have it uploaded for an explanation.) The standard deviation comes from using Minitab's Gage RR via a two factor cross random ANOVA array. We performed this using three operators and three trials with 10 parts. Thanks for the help!
Brad

[Allattar]

In propagating errors you add variances, not standard deviations.

You would square both standard deviations, and then add them. Then square root to get the standard deviation of the system. In which case your error, if assuming the same standard deviation on measurements before and after becomes Std Dev * root 2.

The problem can be broken down into

Difference + Error(overall) = (Before + error(Measure) - (After + error(measure)

but my heads to frazzled today to work it further than that

[Bev D]

Allatar is on the right track. unfortunately AIAG has the wrong formulas for truly extracting measurement error from total observed variation. As Allater said - you do not use standard deviations you use variances.

But befroe commenting furhter, I need to ask why you want to know the measurement error? what will you do with the info? how will you use it?
the answers will provide us wiht the means for giving you a useful answer.

for example, if you aren't doing anything with the data, but are simply reporting it to your customer, then we woudl give one answer. If you were going to use it to guardband yoru measurements or develop a better measurement sytem, we would give a very different answer...