niemi24s
Aluminum
- Joined
- Sep 26, 2010
- Location
- South Of Lake Superior, USA
My understanding of GD&T as practiced on Army Ordnance blueprints for the M1911A1 pistol during the 1940 to 1980 period was that it was based on Maximum Material Condition (MMC) guidelines of...
• Nominal value was of the largest pin with a negative tolerance,
• Nominal value was of the smallest hole with a positive tolerance and
• Nominal value for the location of a hole was the average value with a ± tolerance.
...as seen for most dimensions on this section of an Army Ordnance blueprint:
But many I come into contact with feel the nominal values on this and other Ordnance blueprints are something special and represent the optimum/best/ideal/preferred/optimal/most desired value - and not merely the MMC. My attempts to argue against this with...
• "How can the best value of, say, 0.200 + 0.004 be 0.200 when that value is on the verge of being out of spec - and therefore difficult and costly to make?" or
• "If the in-tolerance range of sizes is 0.200 to 0.204, don't you think most of them would follow normal Gaussian distribution and be 0.202?"
...usually fall on deaf ears.
Q: Is my understanding of this essentially correct or is it flawed?
Best Regards
• Nominal value was of the largest pin with a negative tolerance,
• Nominal value was of the smallest hole with a positive tolerance and
• Nominal value for the location of a hole was the average value with a ± tolerance.
...as seen for most dimensions on this section of an Army Ordnance blueprint:
But many I come into contact with feel the nominal values on this and other Ordnance blueprints are something special and represent the optimum/best/ideal/preferred/optimal/most desired value - and not merely the MMC. My attempts to argue against this with...
• "How can the best value of, say, 0.200 + 0.004 be 0.200 when that value is on the verge of being out of spec - and therefore difficult and costly to make?" or
• "If the in-tolerance range of sizes is 0.200 to 0.204, don't you think most of them would follow normal Gaussian distribution and be 0.202?"
...usually fall on deaf ears.
Q: Is my understanding of this essentially correct or is it flawed?
Best Regards
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