Floating fastener formula for unequal distributions 3

[Tunalover]

Bruce Wilson's book DIMENSIONING AND TOLERANCING HANDBOOK, Genium Publishing, 1995 puts forward a formula to use for GD&T and the case of the floating fastener when the two joined parts have differing positional tolerances and hole sizes:
H1+H2-2F=T1+T2

I've been tearing my hair out over this formula because I don't think it always works. For example, consider if H1=.190, F=.164, and T1=.014. This would leave
H2=.152+T2 which basically says that T2 must be greater than .012 in order to make H2 bigger than F. This makes no sense! Anybody have any comments on the validity of this formula or does anybody have something to add that would make it make sense? TIA


Tunalover

 

[PeterStock]

Per ASME Y14.5-1994 Appendix B, Section B3 the floating fastener formula is

H = F + T or T = H - F

where H = the minimum hole dia
F = the maximum fastener dia
and T = the positional tolerance dia

for two parts with different hole diameters and position tolerances the same formula applies to each of the parts separately.

H1 = F + T1
H2 = F + T2

Or

T1 = H1 - F
T2 = H2 – F

You only need to “split” the tolerance between two parts when you are using fixed fasteners.


Peter Stockhausen
Senior Design Analyst (Checker)
Infotech Aerospace Services

http://www.infotechpr.net

 

[Tunalover]

PeterStock-
I'm familiar with the floating fastener formula and how it is used. I guess I'm looking for the value in Mr. Wilson's formula. Is it dangerous to use? Is it valid?

Tunalover

 

[KENAT]

Well, if you combine the 2 separate formulas Pete gives from 14.5 you get

H1+H2 = F +T1 + F +T2

Which rearanges to:

H1+H2-2F=T1+T2

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[pmarc]

In my opinion Mr. Wilson's formula for floating fastener is correct.

Tunalover, why do you think that it does not make sense that ''T2 must be greater than .012 in order to make H2 bigger than F''? You chose .014 for T1 therefore you obtained such result. You could have chosen for example T1=.025 and then T2 would have to be greater than .001 to make H2 greater than F. I do not see nothing wrong in this formula - logically and mathematically it is correct. The only limitation I can see is that positional tolerance values T1 and T2 can't be of any value - the following relationship has to be always correct: T1+T2=H1+H2-2F
But this is the logical conclusion of floating fastener case.

 

[drawoh]

tunalover,

This has come up before. thread1103-125608

That equation generates weird results if you are not careful.

JHG

 

[Tunalover]

pmarc-
Imagine if T2=.004 then H2=.152+.004=.156 which isn't even big enough to clear the fastener (F=.164). That doesn't make sense. The formula doesn't work.



Tunalover

 

[pmarc]

Tunalover,

I see what you mean and you are right.

So to be fully correct from mathematical point of view another condition has to be added to the formula H1+H2-2F=T1+T2; H1 and H2 have to be greater than F.
It is logical because otherwise it would not be possible to fit the fastener to both holes.

I think in most cases floating fastener formula is used to calculate missing positional tolerances. People automatically assume that minimum holes diameters are greater than maximum fastener diameter and they look for positional tolerance values.
You started from the other side - you assumed one of positional tolerances and were searching for the second one, additionally having a minimum size of second hole unkown.