Well I have no free spinning rod sooooo....

The .54 Cal rod length from muzzle bore to end of rod inserted is 33.25 inches- for 180 degrees using the front sight as 0 degrees - I got measurements of 26.75, 27.25 & 26.825 - Avg of 26.94167 and times 2 = 53.8833 ~ that gives me a 1:54 twist?

Hmmm a .54 with a 1:54 twist rate?

The .45 cal - rod length from muzzle bore to end of rod inserted is 34.5 inches and for 180 degree using the front sight as 0 degrees - I got measurements of 25.75, 25, & 26 - Avg 25.58833 and times 2 = 51.16667 that gives me a 1:51?

Both are very similar it appears. Unfortunately the person who made both these rifles died quite some time ago. He was on consignment to build my father a pair of 50 cal pistols to which he wrote to dad where complete , jsut need to be test fired, but dad never got them. His wife never responed to the phone calls or letters. He lived in California I do beleive. Too bad.

Now that I think about it - I think I do have a free spinning rod when I brought all this home from my dad's house but I need to find it.
I used the range ram rod with a brush and a patch and made it as tight as possible. I then inserted the rod all the way down as far as it would go. I then taped the starting point at the muzzle end creating a 'flag' like mountain magic did. I marked 180 with marker on the bottom of the barrel so I could take out guessing. I placed this on a gun rest to which I could draw out the ram rod trying my best to let it spin naturally. It took a bit of practice before I took measurements as from the beginning I was all over the place.....dang rookie!
Think I will re-do this once I find or buy a new spinning ram rod and use my lead sled which is up north.

So that is what I came up with......comments?

DAve...JW

UPDATE:
Ps after using the correct rod both 54 and 45 caliber bores are 1:60 twist. Use a spinning handle rod -- only way to go....