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Limits and Fits

5 REPLIES 5
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Message 1 of 6
SKinzel
1819 Views, 5 Replies

Limits and Fits

So I was double checking some fits tolerances in Inventor 2012 64 bit  (an idw file) and seem to have found an incorrect tolerance.  s6 comes up with +.0031/+.0023 for a shaft that is 3.125 diameter.  According to the standard I have it should be +.0029/+.0022.  When I check for an FN2 fit which calls out for the shaft tolerance of s6 using the Limits and Fits Mechanical Calculator in the assembly file the correct size of +.0029/+.0022 is reported.

 

Does anybody know of a quick way to double check the actual tolerance values used for different fit specifications?  I'd like to be able to trust my drawing file to use the correct tolerance but it looks like I'll need to double check it if I use the "Limits/Fits" tolerance method in the dimensions.

 

 

Stuart Kinzel
Inventor 2013-64bit, HP EliteBook8740w Intel Core i5CPU 2.67 GHz
8GB memory
Windows 7 64bit
5 REPLIES 5
Message 2 of 6
Doug_DuPont
in reply to: SKinzel

Acording to my info the +.0031/+.0023 is correct for s6 on a 3.125" shaft

Douglas DuPont
Inventor 2016 Pro, Vault 2016 Pro
Quadro M4000
Windows 10 64 Bit
Message 3 of 6
SKinzel
in reply to: Doug_DuPont

What is the origin of your info?  I've checked a couple of different sources and they have all stated that s6 for 3.125 should be +.0029/+.0022.  The sources I have checked include Machinery Handbook 25th Ed., my copy of the Cylindrical Fits standard (which is quite old), the fit calculator within Inventor itself and a calculation spreadsheet MITCalc that has many useful calculations.  I checked the other fits such as t6 and u6 and the Inventor dimension values agree with my sources they just don't agree on the s6.

 

I should mention that what prompted this is a customer of ours questioning the numbers I used.  I haven't heard back what they say the tolerance should be.  I wouldn't be suprised if my copy of the standard is out of date but I would be suprised if teh other sources were out of date.

 

I'm attaching some images.  The first is the window for teh fit calculator in the Design Tab of an assembly and the second is the dimension dialog box from editing a dimension in a .idw.

Stuart Kinzel
Inventor 2013-64bit, HP EliteBook8740w Intel Core i5CPU 2.67 GHz
8GB memory
Windows 7 64bit
Message 4 of 6
hosford
in reply to: SKinzel

Table 11. ANSI Standard Force and Shrink Fits

The inch specification in the machinerys handbook (28th edition) is ANSI B4.1 1967 (R2004)

1.97" - 3.15"+2.9 / +2.2 (thousandths of an inch)

 

Table 3. American National Standard Preferred Hole Basis Metric Transition and Interference Fits

The metric specification in the machinerys handbook (28th edition) is ANSI  B4.2 1978 (2004)

80mm basic +.078mm / +.059mm (+3.1 / +2.3 (thousandths of an inch))

 

 

To me these are two different fits, its interesting. wonder who is right/wrong.

I can tell you I always specify the fit I want manually, but then again I dont do that much with standard fits.

Thaddeus Hosford
NUC9i9QNX i9-9980HK, Win 10 Pro 64
Nvidia GTX 1650
Inventor 2021
Message 5 of 6
Doug_DuPont
in reply to: SKinzel

 

We use

The metric specification in the machinerys handbook (28th edition) is ANSI  B4.2 1978 (2004)

80mm basic +.078mm / +.059mm (+3.1 / +2.3 (thousandths of an inch))

Douglas DuPont
Inventor 2016 Pro, Vault 2016 Pro
Quadro M4000
Windows 10 64 Bit
Message 6 of 6
SKinzel
in reply to: Doug_DuPont

Thanks.  At least I'm not crazy.  Well, at least not in this instance Smiley Happy

 

Looks like there is some inconsistency between the "hard metric" standard and the "hard inch" standard specification.  Thanks everybody.

Stuart Kinzel
Inventor 2013-64bit, HP EliteBook8740w Intel Core i5CPU 2.67 GHz
8GB memory
Windows 7 64bit

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