Basic Tolerance Question.

[ronancorcoran10]

Sorry in advance for this basic question but I'm confused.

So for a Class IV transitional fit the tolerance on a 1'' hole and shaft is as follows according to this link(http://www.engineersedge.com/class_iv.htm)
Basic diameter = 1'' Clearance 0.0002''-0.0018''
Hole Dia = 1'' to 1.0009'' difference =0.0009''
Hole Dia =1''+.0009''/-0.0000''
Shaft Dia= 0.9991''-0.9998'' difference =0.0007''
Shaft Dia=0.9991''+0.0007''/-0.0000''

So a machinist has to be more precise on the shaft as there is only .0007'' to play with while the hole has 0.0009'' to play with.

why are the tolerances not done like this below.
Basic diameter = 1'' Clearance 0.0002''-0.0018''
Hole Dia = 1'' to 1.0008'' difference =0.0008''
Hole Dia =1''+.0008''/-0.0000''
Shaft Dia= 0.9990''-0.9998'' difference =0.0008''
Shaft Dia=0.9990''+0.0008''/0.0000''

So a machinist has .0008'' to play with on both shaft and hole.

I don't understand why they are allowing greater range of error on the Hole diameter, why isn't the allowable error the same for shaft and hole.

Thanks

Ronan

[Kelly_Bramble]

It is easier to manufacture an external feature of size than an internal feature of size. One can use short rigid tooling vs longer cantilevered tooling - less deflection, more access and more predictability.

The ANSI tolerance tables you're referring to reflect functionality as well as manufacturability..

Ultimately end item products and components need to function as intended and if the design can be facilitated to embrace ease of manufacture - everybody wins.

More here: Design for Manufacturability and Assembly (DFM / DFMA)

Sorry in advance for this basic question but I'm confused.

So for a Class IV transitional fit the tolerance on a 1'' hole and shaft is as follows according to this link(http://www.engineersedge.com/class_iv.htm)
Basic diameter = 1'' Clearance 0.0002''-0.0018''
Hole Dia = 1'' to 1.0009'' difference =0.0009''
Hole Dia =1''+.0009''/-0.0000''
Shaft Dia= 0.9991''-0.9998'' difference =0.0007''
Shaft Dia=0.9991''+0.0007''/-0.0000''

So a machinist has to be more precise on the shaft as there is only .0007'' to play with while the hole has 0.0009'' to play with.

why are the tolerances not done like this below.
Basic diameter = 1'' Clearance 0.0002''-0.0018''
Hole Dia = 1'' to 1.0008'' difference =0.0008''
Hole Dia =1''+.0008''/-0.0000''
Shaft Dia= 0.9990''-0.9998'' difference =0.0008''
Shaft Dia=0.9990''+0.0008''/0.0000''

So a machinist has .0008'' to play with on both shaft and hole.

I don't understand why they are allowing greater range of error on the Hole diameter, why isn't the allowable error the same for shaft and hole.

Thanks

Ronan

[ronancorcoran10]

Thanks for the information and the quick reply!

It is easier to manufacture an external feature of size than an internal feature of size. One can use short rigid tooling vs longer cantilevered tooling - less deflection, more access and more predictability.

The ANSI tolerance tables you're referring to reflect functionality as well as manufacturability..

Ultimately end item products and components need to function as intended and if the design can be facilitated to embrace ease of manufacture - everybody wins.

More here: Design for Manufacturability and Assembly (DFM / DFMA)