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Understanding and Measuring Flatness Spec

Understanding and Measuring Flatness Spec

Understanding and Measuring Flatness Spec

Hello Everyone,

I recently started at a new plant and we seem to be having flatness issues with two of our parts. The two parts have a flatness spec that reads .0002/IN, while our other products are all .0005 for the entire surface. The quality department says the .0002/IN can just be multiplied by the width of the part for your entire surface spec (ie. a 3 inch part would have an entire flatness spec of .0006). The engineering department believes it is .0002 inches within a circle with a radius of .5 inches.

Two questions:
1. Which is the correct meaning of the spec (.0002/IN)
2. What is the best method of measuring this? We currently use and inverted gauge installed under a granite block. How would you measure a radius of .5 inches?


RE: Understanding and Measuring Flatness Spec

I take it to mean that any 1" portion of the surface cannot exceed 0.0002" out of flat.
Let's assume that you have a 3" square surface, and that you only have any deviation in one corner of it.
If that corner is out of flat more than 0.0002" within 1" then it is out of spec, even if the entire surface is only out by 0.0003" across the entire thing.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

RE: Understanding and Measuring Flatness Spec

I'd agree with Ed. Typically, such a specification means that ANY one inch circle must meet the spec. Your quality department's recommendation means that you can potentially have a 0.0006 in deviation within a one inch circle and still meet their interpretation.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
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RE: Understanding and Measuring Flatness Spec

IRstuff points out the flaw in their logic. They are over-simplifying the requirement which invites error.

You could end up with a perfectly flat part with a peak/valley w/in 1/2" of 0.0006" and it still pass, despite not being flat w/in .0002" in that sq-in or in-dia area. ____/\____

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