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GDT positioning 1
- Thread starter rwild
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1
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rwild:
Yes you can and it is probably preferable in this situation.
You drawing should have a basic dimension from B & C and also between slots but I am going to assume that the sketch was probably for our discussion. To top feature control frame also should have a positional symbol identical to the bottom.
Now that I said that, here goes on the tolerance.
To arrive at the tolerance for the vertical position of .020, one must calculate it virtual condition size which is .020 smaller than MMC size (smallest slot length - .405) which equals .385. Subtract that from the actual size of the slot. Let's say the actual slot is .412, so .412 - .385 = .027. That is the total tolerance. Divide that by 2 and we get a +/- .0135 tolerance vertically for that particular size.
To arrive at the tolerance for the horizontal position of .010, one must calculate it virtual condition size which is .010 smaller than MMC size (smallest slot width - .276) which equals .266. Subtract that from the actual width of the slot. I am going to use an actual slot of .282, so .282 - .266 = .016. That is the total tolerance. Divide that by 2 and we get a +/- .008 tolerance horizontally for that particular size.
Hope this was helpful.
Dave D.
Yes you can and it is probably preferable in this situation.
You drawing should have a basic dimension from B & C and also between slots but I am going to assume that the sketch was probably for our discussion. To top feature control frame also should have a positional symbol identical to the bottom.
Now that I said that, here goes on the tolerance.
To arrive at the tolerance for the vertical position of .020, one must calculate it virtual condition size which is .020 smaller than MMC size (smallest slot length - .405) which equals .385. Subtract that from the actual size of the slot. Let's say the actual slot is .412, so .412 - .385 = .027. That is the total tolerance. Divide that by 2 and we get a +/- .0135 tolerance vertically for that particular size.
To arrive at the tolerance for the horizontal position of .010, one must calculate it virtual condition size which is .010 smaller than MMC size (smallest slot width - .276) which equals .266. Subtract that from the actual width of the slot. I am going to use an actual slot of .282, so .282 - .266 = .016. That is the total tolerance. Divide that by 2 and we get a +/- .008 tolerance horizontally for that particular size.
Hope this was helpful.
Dave D.
Hi Dave,
I have doubt here, Do we need to have "BOUNDARY" written beneath this FCFs.?
The dimension .281+/-.005 is applied for the two flat surfaces, here I can see that the mid plane is controlled with the positional tolearnce,
where as the other positional tolerance applied to the dimesion .410+/-.005, there is no opposed flat surfaces here, it is rounded surfaces here. Does this apply to mid plane here also? Dont we need a BOUNDARY concept here. ?
I have doubt here, Do we need to have "BOUNDARY" written beneath this FCFs.?
The dimension .281+/-.005 is applied for the two flat surfaces, here I can see that the mid plane is controlled with the positional tolearnce,
where as the other positional tolerance applied to the dimesion .410+/-.005, there is no opposed flat surfaces here, it is rounded surfaces here. Does this apply to mid plane here also? Dont we need a BOUNDARY concept here. ?
- Thread starter
- #5
I think Madhu454 is right, we need to have BOUNDARY written beneath the FCF's. This specific case of round-end slot is dealt with in Y14.5 using the boundary position concept. See section 5.10.1 c) and Fig. 5-47 in '94. The two Position tolerances are used to calculate an overall virtual condition boundary, and the entire feature must conform to this boundary.
Tolerance zones for the center geometry (center plane and center point) of the width and length of the slot are not defined. I would say that the method Dave suggests is a practical work-around. It would provide useful (but not official) numbers if the form error of the slot was small compared to the position tolerances.
Evan Janeshewski
Axymetrix Quality Engineering Inc.
Tolerance zones for the center geometry (center plane and center point) of the width and length of the slot are not defined. I would say that the method Dave suggests is a practical work-around. It would provide useful (but not official) numbers if the form error of the slot was small compared to the position tolerances.
Evan Janeshewski
Axymetrix Quality Engineering Inc.
Mahhu454:
That is an interesting question so I will reflect both the 94 edition and also the 2009.
94 edition
5.10.1 (c) "To invoke this concept, the term BOUNDARY is placed below the feature control frames."
Fig. 5-47 does show the BOUNDARY below the feature control frame.
The standard does not mandate it but emphasizes it for non-round features of size.
2009 edition
7.4.5.1 (c) "The term BOUNDARY may be placed beneath the feature control frames, but is not required."
Fig. 7-34 reflects exactly the same as the sketch with the term BOUNDARY.
Without a checking fixture, we must confirm the boundary has not be breached. On the .281 width, we must check the full width of the 2 flats using our calculated tolerance. In other words, contact on various spots along its length and then use the calculated tolerance, in our case, of +/- .008.
On the radius ends, we must confirm it at its furthest length or in the center using a calculated tolerance, in our case, of +/- .0135.
Of course, a checking fixture is best here.
Dave D.
That is an interesting question so I will reflect both the 94 edition and also the 2009.
94 edition
5.10.1 (c) "To invoke this concept, the term BOUNDARY is placed below the feature control frames."
Fig. 5-47 does show the BOUNDARY below the feature control frame.
The standard does not mandate it but emphasizes it for non-round features of size.
2009 edition
7.4.5.1 (c) "The term BOUNDARY may be placed beneath the feature control frames, but is not required."
Fig. 7-34 reflects exactly the same as the sketch with the term BOUNDARY.
Without a checking fixture, we must confirm the boundary has not be breached. On the .281 width, we must check the full width of the 2 flats using our calculated tolerance. In other words, contact on various spots along its length and then use the calculated tolerance, in our case, of +/- .008.
On the radius ends, we must confirm it at its furthest length or in the center using a calculated tolerance, in our case, of +/- .0135.
Of course, a checking fixture is best here.
Dave D.
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