True Position Interpretation

[JacobCheverie]

I am trying to clarify how to evaluate this true position. I have a slot that is positionally toleranced to a planar primary datum, cylindrical secondary datum, and cylindrical tertiary datum. There is a basic angle from the tertiary datum to the center of the slot.

I assume that this position is comprised of (a) how centered the slot is on the secondary datum, and (b) how close to the basic angle the slot is from the tertiary datum.

My question is this: Assume I can measure the error from both (a) and (b). How do I combine that information to obtain my result?

Obviously I will level to A, rotate to C and origin XY to B and Z to A. My coordinate system will be on A, centered on B, and clocked to C. This Position establishes a planar tolerance zone opened up at the basic angle on the print, and I am to check whether or not the midplane of the slot falls in the zone.

This wouldn't be the standard 2D/3D Cartesian 2*sqrt(dx^2 + dy^2 + dz^2) nor is it the 2D polar 2*sqrt(r^2 + r0^2 - 2rr0cos(dA)) true position formula....

Position.jpg

[marcc]

To me basic angle is for rotating alignment after you do what you wrote above. Once alignment is rotated perfect slot centers would be at Y=0 and you evaluate them in Y direction only.

[JacobCheverie]

The position of the slot is relative to the coordinate system created by ABC. System ABC is at a basic angle relative to the center of the slot, so Y = 0 does not apply anymore. We are now trying to position a plane in a rotated coordinate system. That is like me asking you "Where in the Cartesian plane is the line y = 3x + 4?" It doesn't make much sense...

[Schlag]

Level to A. Rotate to line B-C, Rotate back to the BASIC angle, 0.00 for X and Y on DATUM B. If a Datum can constrain a degree of freedom it has to.

[NinjaBadger]

I agree with marcc and schlag.


Align ABC and rotate back basic angle - evaluate slot position in Y (as drawn). If you're worried about rotating back by the basic don't be, in that rotated co-ord system datum C doesn't have any angular deviation (i.e. it's still your tertiary datum / clocking feature)

[JacobCheverie]

My issue is with rotating back by the basic angle. I am not truly in ABC anymore, and that is where the position is dimensioned to.

But what you are saying makes sense logically and practically. Any deviation in the width from the basic angle will appear when positioning it from a rotated ABC, along with any translational deviation from the implied zero.

Thank you NinjaBadger Schlag marcc

[NinjaBadger]

Think of it this way...

If you had say 8 holes on a [100] basic bolt circle diameter called out [POS|Ø0.25|A|B|C] where A was a plane, B was diameter (similar to your part) and C was a hole at TDC (on the print) - you'd have no issue creating ABC and reporting the position of the holes.

The basic nominals for 4 holes would be 50,0 (or some combination of) but the ones at 45° would have basic nominals of 35.355, 35.355.

But there is no harm in rotating to each hole in turn (by the THEO angle AFTER establishing ABC) and reporting it that way, so any X deviation is circumferential and any Y deviation is radial. Your tolerance zone is still in the same place in relation to the datum features, it's just the reported basic values which have changed.

[Schlag]

That's really what BASIC means. When you rotate to B-C and rotate "back" to the BASIC angle that is exactly what the Feature control frame is asking.

[VinniUSMC]

If you just rotate to C, then you are not in ABC. You are in a skewed reference frame. Rotating off of C by the basic angle puts your alignment to the DRF as intended.

Datum C is not 0. Datum C is XX degrees off of 0.

[Matthew D. Hoedeman]

if they had wanted "C" to be at zero, then the entire print would have been rotated so that "C" was on the same horizontal "line" as "B" instead of at an angle on the print.

[JacobCheverie]

What geometric control is tied to the basic angle?

[VinniUSMC]

The location of C, and everything to ABC. As Matt said, it's oriented the way it's oriented for a reason. Well, if you take the assumption that the drafter has a clue what they are doing as true.

[JacobCheverie]

Thanks everybody for clearing this up. The way that I must measure this is not the way that I read the print. As I have stated, it must come with more experience and issues like this coupled with the help of others.

[VinniUSMC]

The problem is that, if you just rotate to C, then the basic dimensions on the print are no longer relevant. The basics on the print rely on the datum reference frame to be oriented as shown. Yes, PC-DMIS XactMeasure will show the DRF skewed, if you select FIT TO DATUMS "YES". But, then you should change that to "Use Current Alignment" if it skews, so that your Basic dimensions match the print.

[JacobCheverie]

VinniUSMC Thanks, the problem is that my current alignment matched a proper fit to the datums so that toggle would not have changed the result. But now, from what everybody is saying, I must set my alignment as usual together with the basic rotation. Then I would use current alignment to get a meaningful report.

[CMM_ENG21]

Just an FYI, for the feature to be "3D Cartesian 2*sqrt(dx^2 + dy^2 + dz^2)" this feature would have to be spherical or the hole would need to be on an angle.......holes that sit in one plane and BASICS come from 2 sides are 2d therefore; 2*sqrt(dx^2 + dy^2 )" or whatever 2 axis'....however, holes measured in free-space using "3D Cartesian 2*sqrt(dx^2 + dy^2 + dz^2)" isn't very accurate...but a sphere with its center being what it is 520º makes its centoid ALWAYS in center...so "3D Cartesian 2*sqrt(dx^2 + dy^2 + dz^2)" always applies..

[JacobCheverie]

CMM_ENG21 Thanks, I know what you're saying. I was just putting in the general equation instead of typing the equation twice with a simple term difference, that's why I said 2D/3D.

[louisd]

Keep in mind that true position is also assessing the perpendicularity of that slot's midplane back to A, and both sides are to be considered with simultaneity.