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Any idea on how to determine the discrimination of the measure,ment system?What is the increment of time?
thanks and regards
I know Mini and JMP provide a measure of discrimination. It usually is in the form of how many increments your measurement system can detect. If you get a discrimination of 2 it means you can tell good and bad but no further scale is possible. The rule of thumb is for the value to be 4 or more so at least you can make some distinction between items.
Thank You.Please elaborate more to clarify the concept.when to use it ?At which control chart (XmR ?).Appreciate the enlightenment?
To get the discrimination of your measurement system you use the MSA process. This has nothing to do with the type of control chart you are using.
You perform the MSA, and you divide the sigma (parts) by the sigma (gage) and multiply by 1.14, then truncate the answer to an integer.
This tells you what is called the number of distinct categories. That is within the variation of your parts, how many groups can be discriminared by your gage.
A number of distinct categories of 4 is considered adequate for some applications.
If you need good discrimination for fine control, a number of distinct categoriest of 10 is considered excellent.
Jim Shelor
As Jim explained, this has to do with MSA and not control charts. It is probably a good idea to do an MSA before collecting data and plotting on a control chart. If your measurement system has poor discrimination then the signals that might come from a control chart will have you needlessly chasing variation that can’t even be measured by your measurement system. If your measurement system can only discriminate whole integers, reacting to points which might vary by the second decimal point, is just noise. A nice conceptual analogy I like to use is time….one measurement system with really poor discrimination can only measure daytime and nighttime. One slightly better might be able to discriminate by hour and an even better one down to minutes. If you want to be able to identify time by seconds, none of the above measurement systems have sufficient discrimination. If you want to be able to determine morning, afternoon and evening, one of the above will be sufficient.
Discrimination is the resolution of the measurement system to detect small changes in parts.
Recommended discrimination : 1/10 of 6 times of the measurement standard deviation.
” Number of distinct categories ” will give you this measurement. Generally it has to be more than four or five to have adequate discrimination.
regards
joe
You can perofrm this in Minitab or JMP and see the number of distinct categories (ndc). The idela is that it should be >4.
Hi Jim
You multiply by 1.41, not 1.14. Typo?
Yep, typo. I hate it when I do that!
Hi Jim,
I tried to reply to your e-mail but it bounced back. Do you have an alternate?
OOPS! I meant the TO line, not the subject line.
No alternate.
That happens to most people because they spell my last name wrong.
Copy and paste j.shelor@verizon.net into the subject line of the message.
You can use a technique called an ISOPLOT which is a special type of regression analysis (Scatter Diagram). This is a technique developed by the late Dorian Shainin and is an excellent tool.
The Discrimination is figured by determining the Product Variation and dividing it by the Measurement Variation. One must acheive 6 or greater. 10 is best.
ISOPLOT is a crappy tool. Use real math.
The plot is a good way to visualize a problem, but not to analyze a problem.
People using Shainin tools are stuck in the 1970’s.
Heay maaaan,
that ISOPLOT… It’s really groovy maaaaan.
maaaan, I’ve got the munchies
peace.
hbgb b^2,
If you get some time could you give me a call tomorrow?
Regards
The Isoplot works fine for me and our processes despite your ignorant comments.
and your solution is? I noticed you criticize often but have no solutions. Please entertain us with your answer to the discrimination issue.
Hi Jim. I understand your response on discrimination but am wondering why multiply by 1.14? Thanks, bob
Bob,
There are two errors in my original post. I said:
You perform the MSA, and you divide the sigma (parts) by the sigma (gage) and multiply by 1.14, then truncate the answer to an integer.
That sentence should have said:
You perform the MSA, and you divide the sigma (parts) by the sigma (measurement) and multiply by 1.41, then truncate the answer to an integer.
I do not know where the 1.41 comes from, but now that you asked, I am going to have to do a little study to see if I can find out. It seems like we are picking up a sqrt(2) from somewhere.
Regards,
Jim Shelor
It should be 6
Bob,
The link below will take you to the derivation of the NDC equation.
http://www.minitab.com/support/docs/Answers/Root2.pdf
It shows where the 1.41 comes from.
I hope this helps.
Jim Shelor
Go read Wheeler’s evaluating the measurement system – it will explain it all.
Thanks much, your response was very helpful. bob
Thanks, I will check it out. I got another posting with a web site also. bob
Excellent Elaboration
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