Enfield barrel question

B

blu97

New member
I have a very nice enfield no 4 barrel i would like to know the twist rate of.

The barrel is 25.5 inches long and with a cleaning rod I got 2.25 turns.

I'm not sure how to figure the rate.
 
25.5 divided by 2.25 = 1 turn every 11.355 inches.
I found this info online for the No.1 Mk III, I shouldn't imagine they changed the rifling just to go to the No.4


S.M.L.E BRITISH
NO.1 Mark III Short Lee Enfield.
Adopted: 1907
Cal: .303
Length: 44.5"
Wt: 8.6lb
Action: Turnbolt
Bolt: 2 Piece non rotating head
Mag: Detach Bx. Stagg. Column
Capacity: 10 rnds
Bbl length: 25.2"
Bore Dia: 303"
Twist: Left
Rate: 1 in 10 inches


EDIT: Link to Lee Enfield No.4
 
Last edited:
I'm assuming you are talking the Model 1917 Enfield in Cal 30 (30-06)

The normal barrel was 26 inches, five grove, left hand 1 in 10.

Some of the 1917s were rebarreled in WWII with two grove right hand twist the same as the M1903 A3.

Hope this helps,
 
Blu97 said it was a No. 4 barrel, not a M1917 barrel. Anyway the No. 4 twist rate is normally 1 turn in 10 inches (1:10), left hand. Some Savage barrels are 1:10.5 and some Long Branch barrels are right hand twist, but neither matters for practical purposes.

The usual way of measuring twist is to take a cleaning rod and use a patch tight enough to be sure the rod turns with the rifling. Then insert the rod into the barrel, and mark one side of the rod at the muzzle. Pull the rod out, letting it turn with the rifling. When the mark is once again on the same side, measure the difference between it and the muzzle. That is the rifling twist. For rifles of .30 or .303, it is usually 1 turn in 10 inches, so the distance measured should be 10".

Jim
 
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