Min Distance Calculations 2

[pmarc]

If you don't mind, I would like to hear some answers to the question from the attachment.
Please try to imagine really the worst-case than can ever happen for this type of part.
Drawing is in accordance with ASME Y14.5-2009.

http://files.engineering.com/getfil...-487f-92f3-159e42fe7896&file=Min_distance.png

 

[powerhound]

Looks like mathematically both should be 4.85 but I think I know where you're going with this.

John Acosta, GDTP S-0731
Engineering Technician
Inventor 2013
Mastercam X6
Smartcam 11.1
SSG, U.S. Army
Taji, Iraq OIF II

 

[ewh]

I get 4.8

“Know the rules well, so you can break them effectively.”
-Dalai Lama XIV

 

[CheckerHater]

4.8 in both cases, but I was in the hurry to get home from work.
Back to the old drawing board!

 

[Keith1029]

I got 4.8. Then I did it *properly* and got 4.7.

I imagined the OD at LMC but with a bad bow in it so that the envelope was at MMC. The hole is then made at LMC and offset at the .1 limit from coaxial with derived A. Then I did this math:

1. Distance from Midline (A) to MMC boundary = 20.1/2 = 10.05
2. Shortest distance from midline (A) to OD edge = 19.9 - 10.05 = 9.85
3. Maximum distance from midline (A) to edge of ID hole = (10.1/2)+.1 = 5.15
4. Thinnest wall section therefore is (2)-(3) = 9.85-5.15 = 4.7

Correct?

 

[Keith1029]

The above was just for case one; haven't taken a good look at case 2 yet to see if it could differ.

 

[greenimi]

4.8 on both cases for me (but, I am at the beginning of learning this stuff namely GD and T)
What is the right answer pmarc?

 

[jimbod20]

Case #1:
Outer boundary of inner hole = (size at LMC + position tolerance at LMC) = 10.1+0.2 = 10.3. This is the virtual condition boundary. Min wall = 4.8. (19.9-10.3)/2.

Case #2:
Outer boundary of inner hole = (size at LMC + position tolerance at MMC + bonus) = (10.1+0.0+0.2) = 10.3. This is a resultant condition. Min wall = 4.8 (19.9-10.3)/2.


jim

 

[Belanger]

I also get 4.8 for both, but I wonder if pmarc (the OP) might be thinking about an angular issue.

If the OD bends one way and the ID bends the other way, the actual local size of each will sit at a different angle, which gets into trig. That might create a singularity point where the distance X is different?

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

 

[Keith1029]

Could someone point out the error in the methodology of my second calculation (resulting in the 4.7)? I initially calculated based on the same formula as Jim, but I don't seem to see the error in my secondary method of calculation. I believe the difference comes from the fact that the actual local axis of the OD becomes offset from the Datum axis (A) (taken from the datum simulator of a cylinder enclosing the maximum envelope). Is this a case of me applying the datum rules incorrectly?

 

[jimbod20]

Keith,

That assumes my answer is correct. I asked a 'similar question' the other day and spent time to study in detail, producing the answer I provided.

jim

 

[pmarc]

Keith,
Your methodology is in line with mine.
4.7 is the result I get for both cases.

I had a plan to post a sketch explaining the calculations, but I think your description does the job perfectly. Thanks.

 

[jimbod20]

Have a look at figures 2-12, 2-13 and 2-14 in ASME Y14.5-2009.

jim

 

[ewh]

Good lesson!

“Know the rules well, so you can break them effectively.”
-Dalai Lama XIV

 

[greenimi]

[I had a plan to post a sketch explaining the calculations, but I think your description does the job perfectly. Thanks.]

pmarc,
Can you post that sketch?-- my brain does not work today---

 

[pmarc]

Here is the sketch. The bottom picture matches to both cases:

http://files.engineering.com/getfil...fb2-b231-223c713e548f&file=Min_distance_2.PNG

Does it make sense?

 

[SDETERS]

I am sorry I must have missed it how did you get "A" .2

 

[pmarc]

0.2 simply comes from substracting 19.9 (LMC size of OD) from 20.1 (MMC size of OD). This 0.2 is the maximum amount of straightness error allowable for OD.

 

[SDETERS]

Ok I found out where I was going wrong I think. It took me a while to view this. With the local part feature of size is at 19.9 this does not mean that the 19.9 diameter is centered on the "A" Datum axis. "A" Datum Axis is established by 20.1MMC OD. Per the sketch above.

 

[pmarc]

Bingo, SDETERS.
Due to the possible bow of the cylinder = 0.2, the 19.9 diameter is not centered at datum axis A. The amount of shift = half of 0.2 = 0.1. And this 0.1 is exactly the difference between 4.8 and 4.7.