B.6 Coaxial Feature 4

[aniiben]

Does anyone have any idea on how to understand this formula:

H1 +H2 =F1 +F2 +T1 +T2

Trying to get a full picture "where din formula is coming from" and how it is driven from the applicable virtual conditions boundaries.

 

[drawoh]

aniiben,

Here is how I do it. Click on the section on GD&T Positional Tolerances.

You are sticking round objects through round holes. The holes are located by postional tolerances. If the round objects are classified as screws (they could be press-fit pins or fabricated features), they are located by positional tolerances too.



--
JHG

 

[pmarc]

aniiben,

Here is my take on how the B.6 formula could be derived.

https://files.engineering.com/getfi...c-81e2-6deb5f0752be&file=coaxial_features.JPG

Notice that this is nice example of a case where simple comparison of extreme boundaries/virtual conditions (for datum features and toleranced features) doesn't work. It is all because the MMB size of datum feature A of the pocket (20.0) doesn't equal the MMB size of datum feature A of the pin (19.89).

 

[3DDave]

Consider if H1 and F1 are exactly aligned/coaxial.

Then H2 - F2 = T1, which is the maximum diametral clearance between the features under those conditions.

Now consider if H2 and F2 are exactly aligned/coaxial.

Then H1 - F1 = T2, which is the maximum diametral datum shift under those conditions.

Add the equations to account for both conditions:

H1 + H2 - F1 - F2 = T1 + T2 and then rearrange a bit

H1 + H2 = F1 + F2 + T1 + T2.

The following restrictions apply:
H1, H2, F1, and F2 are the MMC diameters (in the Appendix)
H1 >= F1 (not explicit in the Appendix)
H2 >= F2 (not explicit in the Appendix)

 

[aniiben]

Thank you pmarc. I appreciate your help

Thank you 3DDave. Thank you for your support.

 

[aniiben]

Also, last but not least
Thank you drawoh for the additional material for GD&T training materials "Calculating Locational Tolerances"