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Acceptable calibration accuracy of a 60" linear measuring device

qcman

Registered Visitor
#1
With typical length tolerances of ± .015 / 0.39mm what would be an acceptable calibration accuracy of a 60" linear measuring device?
 

qcman

Registered Visitor
#3
That is what I thought as I do not see a need for 10x with the intended use. Do you spread that 4x over the entire 60"? In other words I am checking every 5" so would the 4x be every 5"? I have always outsourced the calibration of these units believing they were actually being calibrated :mad: Anyway I have purchased grade blocks and will bring the calibration in house hence the questions.
 
#4
I would need more information on the tolerance before I could give an answer.

First, if it is a "typical" tolerance of ±0.015" that is what it should be, not what it must be. You can't really call a "typical" device out of specification since it is not a warranted tolerance...

Also, assuming that this might be a warranted tolerance, is it over the full range of the device (60")? Or is there also a linearity specification (such as 0.005" per 12") in addition.
 

Al Rosen

Staff member
Super Moderator
#5
That is what I thought as I do not see a need for 10x with the intended use. Do you spread that 4x over the entire 60"? In other words I am checking every 5" so would the 4x be every 5"? I have always outsourced the calibration of these units believing they were actually being calibrated :mad: Anyway I have purchased grade blocks and will bring the calibration in house hence the questions.
I would Calibrate it where you are using it on the scale. Measuring +/-.015" over 60" with a steel rule is tight.
 

qcman

Registered Visitor
#6
Maybe I need to better clarify. We cut a variety parts to different lengths with most having a tolerance of ± .015. I checked the linear every 5" and it measures within .00047 or below up to 50". 55" measures at +.001 which is more than acceptable for our process.
 

Ninja

Looking for Reality
Trusted
#7
55" measures at +.001 which is more than acceptable for our process.
Pretty sure you just answered your own question...with the "more than acceptable for our process" statement....however...

Math says +0.001" = 0.0254mm (assumed inches since you didn't state units on 0.001)

Tolerance 0.015mm / 4 = 0.004mm

Math says it's not good enough...assuming I assumed units correctly.
 

Al Rosen

Staff member
Super Moderator
#8
Pretty sure you just answered your own question...with the "more than acceptable for our process" statement....however...

Math says +0.001" = 0.0254mm (assumed inches since you didn't state units on 0.001)

Tolerance 0.015mm / 4 = 0.004mm

Math says it's not good enough...assuming I assumed units correctly.
His stated tolerance in the op is +/-.015 inches, not mm. I think qcman is stating it in inches.
 

Ninja

Looking for Reality
Trusted
#9
Yup, looks so...

With typical length tolerances of ± .015 / 0.39mm
Didn't do the math on that one to realize it was inches/mm. Thanks.

Thus Math says +0.001"

Tolerance 0.015" / 4 = 0.004"
If it is indeed +/- 0.015"...the math looks better.

On the edge of 4x...but the statement that it more than meets needs rules (until a customer disagrees).
 

qcman

Registered Visitor
#10
Yes inches. I am trying to drop english all together here but it is incredible how much resistance there is to this :confused: Anyway that would be .015 / 4 = .0037 inches. This may sound like a dumb question but would that be .0037 for any given zone checked but not to exceed .0037 over 60"? Btw this is a Mitutoyo KA-200 counter with a AT715 scale which should have an accuracy of 7 micrometers / 0.00027 inches.
 
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