How to Convert Whitworth Wrench Sizes to AF

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SUMMARY

This discussion focuses on converting Whitworth wrench sizes to across the flats (AF) sizes, specifically addressing the conversion of 5/32 W. While no direct mathematical formula exists for this conversion, a general approximation is provided: AF ≈ Diameter + (5.5 * Pitch). The conversation also touches on the historical context of Whitworth standards, noting that these wrenches were originally designed for square nuts, which differ from modern hexagonal nuts. Additionally, the discussion highlights the existence of other measurement systems such as BA and the evolution of wrench design.

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  • Understanding of Whitworth and AF wrench standards
  • Basic knowledge of thread pitch and diameter
  • Familiarity with historical context of wrench design
  • Awareness of other measurement systems like BA
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  • Research the historical development of Whitworth standards and their applications
  • Learn about the conversion methods between different wrench measurement systems
  • Explore the significance of thread pitch in wrench sizing
  • Investigate the evolution of wrench designs from square to hexagonal shapes
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Mechanical engineers, vintage tool collectors, and anyone interested in the historical and technical aspects of wrench sizing and conversion.

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TL;DR
Help with Whitworth and other systems of measurement for wrench sizing?
Hi I'm trying to understand how to convert Whitworth Wrench sizes to across the flats (AF) sizes.

I understand what Whitworth is, but I can't find any mathematical formula how to convert and perform the calculation yourself. When I look online, all I see are conversion tables, no formulas.

I mainly ask because I'm trying to figure out what 5/32 W is in AF size. None of the tables I find online show a conversion for 5/32 W. This led me to believe it was a size not allowed or specified, but when I search online I can find some old 5/32 W wrenches from like the 1930s and 1940s.

I was also wondering if there are other systems of measurement for wrenches? I know if the BA sizes like BA0, BA1 etc.

Whitworth, BA, Inch, mm, any other systems or standards of measurement?

I know there were other systems for pitch angle, but I'm mostly interested in wrench or spanner sizes.
 
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What makes you think there is a formula of conversion.
AF wrenches are marked to fit the nut or bolt size.
Whitworth wrenches are marked to fit the thread bar stock size.
 
There is no direct exact mathematical relationship.
Whitworth simply listed the standard sizes to be used.
The nuts used then were square, not hexagonal.
The nuts were cut from square or rectangular bar stock, then drilled and tapped.
 
Whitworth was clearly rational, so there must be some rhyme and reason in the Whitworth standard, or it would not have survived.

Given; the Diameter of the bolt; and the Pitch of the thread = 1 / tpi ;
The nut thickness will be a few thou less than the bolt Diameter.
The spanner, AF ≈ Diameter + ( 5.5 * Pitch ).
That seems to hold within a few percent for 1/4" BSW and above.

The 1/8" BSW and the horrible 3/16" BSW appear to be later additions, so Whitworth cannot himself be held responsible for those deviants.

Most of the square BSW nuts I come across on old machinery were sheared from a rectangular bar, then punched before being threaded. If a nut was oversized, it was hit with a hammer to bring it down to size before being threaded. The processes used to manufacture the nuts can be seen by the marks on their surface. Most early hand-made nuts were not square, they ended up being slightly rhombic, each being an individual record of the man who made it.

Spanner tolerance is not as critical with square nuts as it is with hexagonal. Most ancient spanners have a 45° offset for square nuts, rather than the 30° needed for hexagonal nuts. I believe the change from square nuts to hexagonal nuts was primarily to strengthen and reduce the weight of the spanner, which led to more compact fasteners and spanners.
 
Baluncore said:
Whitworth was clearly rational, so there must be some rhyme and reason in the Whitworth standard, or it would not have survived.
Possibly because there was not a standard before.
 

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