Thank you gentlemen, your comments are helpful, but still leave me in need of more.
Regarding copying a shaft that shoots well, I will do that, but after I have a working spine to begin with. The initial problem with this method is that I am almost a completely incompetent archer. Right now I am all over the place, so I cannot find a shaft that shoots well. Once I have some that I know are the same in the ball park, I can work on my from, develop some consistency, fine tune that by judging spine up or down based on missing left v. right, then coping as you suggest the most appropriate spine.
Regarding, 26" with 2lbs, I’m happy to work with that if I can. But I do not quite now how. Jim Hills' table for that says to get a 26" arrow to spine at 35lbs with a 2lb weight, you want a deflection of .743. If that number is not a distance in inches, then I do not know what it is. I could make an arrow 26" and put 2 lbs on it. That will give me both a distance and an arc. I might be able to do something with that, but I am not sure what.
If we could add 5 inches to a 21” arrow spined at 35, with 5lbs per inch, then the longer arrow should then spine at 10lbs. I do not know what to do with that. Hill’s table does not go that low. I could not spine at 26” arrow at 10lbs then cut it down to 21”.
If length/deflection=spine then 20”/deflection=35lbs gives me deflection= .58 or 37/64. I can measure that if it is in inches. But I do not know if it is. Is it only good for 2 lbs? If I use 1lbs would that be 18.5/64"?
Or is this formula only good for 26” and 2lbs and 5lbs/1” only as estimate that works on the margins?
Thanks again.