Define Guage

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Robin128 · #1 Post by **Robin128** » Mon Feb 28, 2011 9:23 pm

Which is correct?

The number of equally-sized lead balls making up 1 Lb in weight which can fit in the barrel = bore/gauge.

OR

The weight of a solid sphere of lead that will fit the bore of the firearm, and is expressed as the multiplicative inverse of the sphere's weight as a fraction of a pound (e.g., a 1/12th pound ball fits a 12-gauge bore).

No Googling!!!

:roll:

#2 Post by **20series** » Mon Feb 28, 2011 9:26 pm

the second was always my understanding :shock:

Alan

Robin128 · #3 Post by **Robin128** » Mon Feb 28, 2011 9:33 pm

HMMMMM...

Any more for any more?

:evil: :evil: :lol: :lol: :lol:

John25 · #4 Post by **John25** » Mon Feb 28, 2011 9:48 pm

The latter, possibly, maybe, perhaps?

Yes - I think!

:roll:

#5 Post by **ovenpaa** » Mon Feb 28, 2011 9:57 pm

Second is my understanding as well

#6 Post by **dromia** » Mon Feb 28, 2011 9:58 pm

Whats the difference 'tween the options?

#7 Post by **Mike357** » Mon Feb 28, 2011 10:06 pm

I thought the first one was correct?

Robin128 · #8 Post by **Robin128** » Mon Feb 28, 2011 11:03 pm

Is the first one incorrect?

Have I got Dromia guessing?

:evil: :lol: :lol:

dazeroo · #9 Post by **dazeroo** » Tue Mar 01, 2011 3:31 am

Both, if my grasp of English has not left me completely.

Robin, do I have this much time to look forward to when I retire. Can't wait!!!!!!

#10 Post by **bobbob** » Tue Mar 01, 2011 6:28 am

When I got my first musket I was told it was the first one.

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