Accuracy Equation - Linear gage with a stated accuracy of (1.5+L/50) ?m L= mm

C

Curtis317

We have a linear gage with a stated accuracy of (1.5+L/50) ?m L= mm. How would you set up calibration tolerances over a range of 0-200 mm?

Would you insert the gage block dimension for L and use that for the tolerance of the measurement for the gage block?

Just wanting confirmation or maybe another way to look at the issue.

Thanks
 

dgriffith

Quite Involved in Discussions
Re: Accuracy Equation - Linear gage with a stated accuracy of (1.5+L/50) µm L= mm

We have a linear gage with a stated accuracy of (1.5+L/50) µm L= mm. How would you set up calibration tolerances over a range of 0-200 mm?

Would you insert the gage block dimension for L and use that for the tolerance of the measurement for the gage block?

Just wanting confirmation or maybe another way to look at the issue.

Thanks
Question: are you calibrating the linear gage with the gage block, or the gage block with the linear gage?
Sorry, don't do much in the dimensional areas.
 
P

Pezikon

Re: Accuracy Equation - Linear gage with a stated accuracy of (1.5+L/50) µm L= mm

Code:
    Test Point    Tolerance
10%        20 mm   ± 1.9 µm
20%        40 mm   ± 2.3 µm
30%        60 mm   ± 2.7 µm
40%        80 mm   ± 3.1 µm
50%       100 mm   ± 3.5 µm
60%       120 mm   ± 3.9 µm
70%       140 mm   ± 4.3 µm
80%       160 mm   ± 4.7 µm
90%       180 mm   ± 5.1 µm
100%      200 mm   ± 5.5 µm

In (1.5+L/50), L is the testpoint. It is nominal. It is the size of the standard gage block you are using to test the linear gage.
 

dgriffith

Quite Involved in Discussions
Re: Accuracy Equation - Linear gage with a stated accuracy of (1.5+L/50) µm L= mm

Is the formula you gave correct? By the order of operations, at 50% scale [(1.5+L/50) µm L= mm], 100mm / 50 = 2mm. 2mm = 2,000um. 2000 + 1.5 = 2001.5um, not 3.5um.
Any thoughts? Am I seeing this incorrectly?
Are you using high grade blocks?

In any event, if you are using blocks to calibrate the linear gage, then the tolerance of the block (or assigned value) should be between 4 and 10 times better than the tolerance of the linear gage at each test point.

Using the tolerances you listed above, for 100mm, a UUT tolerance of 3.5um should have gage blocks with a tolerance of 0.875um or better.
I think that's equivalent to GGG grade 3 or B89 grade AS1.
 
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P

Pezikon

Re: Accuracy Equation - Linear gage with a stated accuracy of (1.5+L/50) µm L= mm

Is the formula you gave correct? By the order of operations, at 50% scale [(1.5+L/50) µm L= mm], 100mm / 50 = 2mm. 2mm = 2,000um. 2000 + 1.5 = 2001.5um, not 3.5um.
Any thoughts? Am I seeing this incorrectly?

If you google "mitutoyo linear gage" you will inevitably find the spec sheet for their "linear gages". Accuracy is published as: (1.5+L/50)µm (L=mm). They sport a resolution of 0.5µm or 1µm and ranges such as 10 or 50 mm. It would be unreasonable to presume the accuracy is anywhere near 2mm with a resolution and range like that. With this accuracy equation, you forget about the units once the input value (L) has been properly scaled; proper scaling is denoted by: "L=mm". And whatever the result, put µm on the end; hence "(...)µm". I can understand the confusion since specifications are typically not formatted like that.

There's bound to be even more confusion when people realize that the calculated specification has greater precision than the resolution of the device/meter. That is to say, how is one supposed to discern 4.7µm with a meter resolution of 1µm? The meter will read 4 or 5µm. The mitutoyo specification also states that quantizing error is not included in the accuracy spec. If it were, this wouldn't be an issue. The answer is to simply round the calculated accuracy up to be consistent with the resolution of the meter.
 
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