Parker Trojan Choke question.
I have a 1916 Trojan that someone cut 2" off the muzzle end.
According the barrel measurements, it seems to come out as IC anc C for the chokes. [I know- only patterning them will really tell! roger that.]
I am on a trip and 2 weeks away from the shotgun at this time, but my records make me curious/confused- and I thought I'd ask the collective for some insight.
When shouldering a Trojan [or any other Parker], is the right barrel always more 'open' than the left barrel?
On my JP Sauer and LC Smith [both uncut] the right is modified and the left is full.
However, in my records I have the Trojan down as Improved Cylinder for right and Cylinder for left [going tight to open right to left].
Now, I know the barrels are cut, and that throws things off. I don't know if the choke was adjusted after the barrels were cut either. And they haven't been patterned.
I had also read that the Parker Trojan's choke started about 4" back from muzzle, so a barrel that was chopped to be 28" would have lost 2", and have some choke.
What I am trying to figure out is what the likelihood is of me inverting the choke pattern r/l when I recorded this, vs the likelihood that the Parker has a right that is tighter than left- unlike my other SxS shotguns.
Anyone able to say R = more open than L in normal Parkers? or the reverse?
Thanks
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