Profile and Datum Question

[JLang17]

Please read:

thread1103-172988

Assuming the said cutout is sized with basic dimensions:

-Would you locate it, with basic dimensions, to the center of the cutout or to the edges of the cutout? (are both ways allowed?) I ask because it seems if it were located to center, then it could spin freely and need a datum reference to lock it in place. But if it's dimensioned to the edges, then you wouldn't need more datum references other than A? I guess I'm looking for clarity as to what referencing a full DRF would accomplish.

-Now my other question (which isn't related to the referenced thread, but possibly applicable). Can you locate a datum feature (with basic dimensions) from non-datum features? I feel like it's an obvious "Yes, you have to in many cases!" but I have yet to find something that covers this. Otherwise it seems you'd have to create a new Datum for the sole purpose of locating another Datum.

 

[JLang17]

If I call out the pattern as Datum B and I want to locate another feature horizontally to Datum B, I measure it from the vertical centerplane between the two holes? Or from one of the holes itself?

 

[DesignBiz]

The dimension to the other feature can be dimensioned from one of the holes in the pattern. The centerplane of the pattern involves the inspection setup regarding the datum established by the pattern.

DesignBiz

 

[JLang17]

Ok, thanks for all the help!

 

[JLang17]

Instead of creating a new topic, I'll post this question in here. This should be simple.

Picture a cube room. Datum plane A is the floor, datum plane B is a wall and datum plane C is a wall adjacent to B - creating 3 mutually perpendicular planes. I have a basic dimension from C to the wall opposite C, and a profile of surface FCF on that wall. Ideally this wall will be squared up with the rest of the "room". I'm having a hard time picking the datums to reference.

Here are my two ideas:

1- reference A|C. Perpendicular to A and rotated to be parallel with C. Located with the basic dimension.

2- reference only C. Parallel to C would by default make it perpendicular to A. Located with the basic dimension.

Is this correct? In which case referencing A would just be redundant and I'd go with option 2.

 

[KENAT]

Assuming you selected them correctly originally I'd use ABC, maybe ACB. This gives you a complete datum structure to fully constrain your degrees of freedom and fully define your wall from.

It really needs to be decided based on function, which you don't really give details on.

Take a look at ASME Y14.5M-1994 sectopm 4.4. Or maybe someone else here can give a good explanation.

KENAT,

Have you reminded yourself of faq731-376 recently, or taken a look at posting policies:

http://eng-tips.com/market.cfm?

What is Engineering anyway: faq1088-1484

 

[JLang17]

I can't figure out how referencing A and/or B applies to this situation.

Referecing C locks the tol. zone parallel to C, which in turn locks it perpendicular to A and perpendicular to B, and referencing C also gives you the location using the basic dimension.

Now we can translate toward and away from B, and up and down from A, but how is that applicable? All it's doing is moving within itself.

 

[ewh]

You can reference only C, but that will increase the tolerances relative to A & B (tolerance relative to C + tolerances of C relative to A & B).

"Good to know you got shoes to wear when you find the floor." - [small]Robert Hunter[/small]

 

[JLang17]

Can you explain that a bit more?

What changes from only referencing C to referencing ABC?

 

[ringster]

JLand17,


I would say minimizing the setup for verification.

Reminds me of an old drawing in the 70's of a cubical part dimensions sufficient to provide oveall size and the note ALL SURFACES TO BE PERPENDICULAR AND PERPENDICULAR WITHIN .XXX. Sounded ridiculous then and still does.

 

[ringster]

Correction to last post should have read ...PERPENDICULAR AND PARALLEL......

 

[axym]

JLang17,

To understand the difference between referencing ABC and referencing only C (or any of the other permutations), you have to imagine what would happen with a real (imperfect) room. A, B and C are all nominally perpendicular to each other, but in reality they will not be perfectly perpendicular - there will be some variation. You haven't mentioned what the perpendicularity tolerances between A, B, and C are, but there would have to be some tolerance.

When the profile tolerance is applied to the fourth wall, its the orientation of the tolerance zone in the real (imperfect) room depends on the datum referencing. If ABC is referenced, the profile zone will be oriented exactly perpendicular to the floor (which means that the zone is not exactly parallel to wall C). If only C is referenced, the profile zone is oriented exactly parallel to C, which means that the zone is not exactly perpendicular to the floor. There are similar effects with wall B as well.

If the perpendicularity tolerances between the datum features A, B, and C were extremely small compared with the profile tolerance, then the difference between the different datum references would also be very small. But if the perpendicularity of the datum features was comparable to the profile tolerance, the effects of different datum references would be very significant.

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[ringster]

JLang17,

Per your post of 4-7=09, the only relationship you have rewquired is a parallel relationship to C. If that is the case B and A would have no bearing, influence or necessity in the FCF.

 

[DesignBiz]

JLang17,

Maybe I am over simplifying the control but I do see you are talking about a cube.

For just what you are doing it doesnt sound like you need a full DRF to accomplish it. You could pick the floor to be Datum A. Use a flat control on A. Use perpendicularity to A for "squaring" it up as you say. This would control orientation to A for all 3 walls. You could use combinations of parallesim and perpendicularity referenced to a single Datum. You could use another datum like C to use a basic dim to the opposite wall with a profile callout, accompanied with parallelism and/or perpendicularity for the other walls. Your post seems to be concerned with orientation.

As mentioned earilier it would depend more on functionality to make a call for a DRF.

Would any of the "walls" fit inside of or outside of another part? which if any of the walls intereface with other parts? questions (and others)you could ask yourself to understand functionality and/or interface with other parts. Then you have a better understanding of what DRF(s) are required.

DesignBiz

 

[JLang17]

Let me be sure I'm understanding something correctly... When we reference datums in an FCF we are referencing the datum simulators, which in this case are planes derived from the highest points on the datum features (walls) - thus creating the most nominal "wall" possible. Apply this to 3 walls and we have our DRF.

Earlier I was assuming the DRF was perfectly squared up, but I guess it's not huh? It falls somewhere between the datum features and their theoretically perfect planes.

If this is all correct, I think I understand the answers to my question.

 

[axym]

JLang17,

I think you're getting closer.

Datum feature simulators touch the highest ponts on the datum features, but that's not the whole story. Another key requirement is that the simulators are perfectly oriented relative to one another. So the simulator for the primary datum feature makes full contact, but the simulators for the secondary and tertiary do not.

If we were to make a simulator set for the ABC datum referencing sequence, it would be three perfectly flat plates that are all exactly perpendicular to each other. One plate would make full (3-point) contact with the floor, 2-point contact with wall B and one-point contact with wall C. The DRF is a coordinate system defined in the 3 plates.

If the datum referencing sequence was different (say CAB), the simulator set would be the same but it would contact the floor and walls differently. There would be 3-point contact with wall C, 2-point contact with the floor and 1-point contact with wall B.

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[DesignBiz]

The question posed assumed that the floor and 2 walls had controls to “square them up”.

A DRF is determined by identified theoretical datums on a drawing (or CAD model). In this case “walls” are being used as datums. Datum simulators establish those datums in an inspection setup.

Lets say an inspection setup is made relating 3 datums (gage plates) oriented to each other. They are imperfect yet are made to more accurate gage tolerances (they establish the "theoretical datums"). The part features are manufactured at “production tolerances”.

The inspection setup for a particular DRF would orient and immobilize the part by contacting the resultant manufactured features to datum feature simulators (surfaces on the manufactured part in contact with the datum simulators of the setup). (E.g. High points on manufactured floor to surface plate for Datum A and so forth).

If the manufactured part is oriented with “surface A” contacting the ”surface plate simulating Datum A” in the inspection setup; then perpendicularity of the walls to the floor could be inspected to a single dautm. A DRF (3 mutually perpendicular planes)wouldnt be needed.

You can turn the part over in order to contact “wall B” to the “surface plate which establishes Datum B”.
At this point you could verify perpendicularity of the walls. (wall B, wall C and the wall opposite wall C to each other).

When you consider a “basic dimension” from wall C to the opposite wall, you will need to check for location. A profile FCF could be used to control location and orientation.

In this case I would probably use datum C as primary (the dimension being verified is from C to the opposite wall); datum B as secondary (a perpendicular wall to B); and A as tertiary. (To me changing B and A around wouldnt have much impact considering the dimension being verified).

When the word “nominal” is used, it makes me think of size. In respect to a DRF you are establishing 3 mutually perpendicular planes with an origin. No size is involved for this.

It would be true that the simulated DRF (inspection setup) is not perfect. However, the drawing DRF does consist of "true geometric counter parts" of features, which would be considered "theorectically exact".




DesignBiz

 

[JLang17]

Ok, I think I understand this better, but I'm still confused as to what referencing A or B does in this situation. If C is primary, then the tol. zone is located from and locked parallel to datum simulator C. As a result it is perpendicular to simulators B and A. There's no need to locate from B and A since it would just be locating within itself so to speak. Same for rotation, it would just be rotating within itself.

If this were not a wall, but say a cutout on the wall, I can understand referencing the whole DRF - it would locate and constrain the cutout on the wall, but for the wall itself I don't see the need for referencing A or B.

 

[ewh]

Say datum A is flat within .010 and datum B is perpendicular to datum A within .015, while datum C is perpendicular to datums A and B within .020.
You lock the additional surface to Datum C. If the location relative to A and B are of no concern, this is fine and you can control it fairly tightly (though not tighter than the flatness of datum C). If it is a concern, then you have to allow for those accumulated tolerances controlling A and B.

"Good to know you got shoes to wear when you find the floor." - [small]Robert Hunter[/small]

 

[JLang17]

Alright, let's say I was also concerned about perpendicularity to B. I reference C|B in the FCF with a profile tolerance of .010. Does this mean that the tolerance zone of .010 is parallel to C, but within the zone, the wall is rotated as perpendicular as possible to datum feature B?

 

[ringster]

JLang17,

Once the functional requirements have been determined, the proper datum features and relationships can be specified. Otherwise, you can spend a lot of time speculating on what might be an infinite number variations.