Position MMC of 2 holes on round part

[Xerif.85]

Good morning,

I am confused on how to calculate the bonus tolerances of this drawing. I looked for similar examples in this forum, videos and manuals but I can find something quite similar. I am really hoping someone can help me to better understand these Position callouts.

Do you combine the A) Tolerance Zone Diameter of a Position MMC applied with no datum reference + B) the Total Location Tol Dia of a Position MMC applied to a hole? I'm referencing ASME Y14.5-2009.

For example,
Drawing - Measured value
.344 - .346
.625 - .6246
1.372 - 1.3716
.125 - .126

The interpretation for A), would be Tol = Position tolerance + (measured value - (1.372-.001) = .002 + (1.3716 - 1.371) = .0026
Position = 2 x sqrt ( (.346-.344)^2 + (.6246/2-.625/2)^2 ) = .004
Which would make the part fail, since .004 > .0026

The interpretation for B), would be Total Location Tol Dia = Tolerance + bonus = .005 + 0.001 = .006
Same position, but now the part passes since .004 < .006.

My questions are, do you use interpretation A, B or both? If both, do you sum, rest or what do I do?

Thank you so much!

 

[3DDave]

[pre]1) Do you combine the
A) Tolerance Zone Diameter of a Position MMC applied with no datum reference
+ B) the Total Location Tol Dia of a Position MMC applied to a hole?[/pre]

No. Because the only "bonus" applies to the tolerance of the feature. The other part is "datum shift." It happens to look as if they can be combined here because all the particular features are circular as is the tolerance zone. In almost no other case will this work.

[pre]Position = 2 x sqrt ( (.346-.344)^2 + (.6246/2-.625/2)^2 ) = .004[/pre] looks correct, but you didn't specify the MMC size of the hole, so no bonus can be calculated. Also, the odds the measured value for vertical spacing being exactly divided at the same time the two holes have identical horizontal offsets is minimal, so it is unlikely for this calculation to ever be useful on a real part - which is OK for artificial example purposes, but be aware not to carry it anywhere else.

[pre]Which would make the part fail, since .004 > .0026[/pre] is nonsensical as the first is just the deviation of the hole and the second part is the available datum shift. They are not related as a limit.

[pre]The interpretation for B), would be Total Location Tol Dia = Tolerance + bonus = .005 + 0.001 = .005[/pre] is unclear. The tolerance is dia 0.002; lacking an MMC size the bonus is dia 0.001 (dia 0.126 - dia 0.125) so the total is dia 0.003.

I would suggest making models with tolerances like dia 0.25 inch and overlay them on a true-position diagram with Virtual condition features. Make the features large enough so they don't disappear. See what shifting the toleranced features while not violating the datum feature virtual condition does.

 

[CheckerHater]

.125 DIA doesn't seem to have size tolerance, so the "virtual condition" is, well, too virtual?

Naturally the drawing is confusing.

"For every expert there is an equal and opposite expert"
Arthur C. Clarke Profiles of the future

 

[Xerif.85]

Sorry about the confusion, the tolerance for DIA .125 is +/.005, didn't include the drawing block because .XXX +/-.005 is very common and I included it in the interpretation for B).

Regarding "...the odds the measured value for vertical spacing being exactly divided at the same time the two holes have identical horizontal offsets is minimal...", what I did in the instrument, is to draw a center-to-center line for the two .125 holes, then create a parallel line thru the center of datum A DIA 1.372 so I am able to "clock" a Y axis, then I created a perpendicular line to Y-axis thru the center of datum A DIA 1.372 to create my X axis. Using axes, I can measure the coordinates of each hole.

The coordinates I have are (-.346,-.3123) and (-.346,+.3127), which are both deviated .0002 in Y and have the same X values.

This approach is also helpful for measuring the X and Y offsets of the center hole and correct the CNC machine.

Finally, since context is very important, I would like to add that we are reading the print from a client, we manufacture the part, and I am programing our inspection measuring system.

Thank you for your time.

 

[3DDave]

My company defaulted to other values for 3 place and rarely used +/- tolerance on holes. Usually +/- .015 .xxx title block; +.010/-/001 on holes this size. So - include named sources for all values if they aren't in the diagram.

You proved the point - the average vertical location is not evenly divided; it is shifted by 0.0002.

Your calculation was measured distance between the small holes divided by 2, not the individual measurements.

Your first measurement was .6246, now it is exactly .625.

Lucky coincidence that the CNC shifted both an amount that looks the same, but this may not be the lowest possible variation measurement in all subsequent parts.

 

[Burunduk]

The total position tolerance for each of the .125 holes is just the specified tolerance plus how much it departed from MMC size, which means 0.002+(.126-MMC).
The departure of datum feature A from its MMC is not added to the value.
The measured position value is 2×sqrt(ΔX^2+ΔY^2) where the Deltas are the deviations from the true position.
Now since datum feature A departed from its MMC (MMB), it means you can translate the measured coordinates of the actual holes as a group closer to their true positions (and reduce the Deltas accordingly) because your entire part can shift around relative to the datum axis by the amount of that departure (which is .0006 in your case, but it can be tricky to apply without an actual fixture with the MMB simulator for A).