According to Lyman's 47th reloading handbook, they tested their #515142 bullet in a Shiloh Sharp's 50-70 rifle with a 1 in 48 inch twist. 

I have one of these moulds. As cast bullet diameter is 0.514. The bullet length is 1.051. So wouldn't the Greenhill value be: 

150*0.514*0.514/1.051 or 37.7? 

I'm aware that some would prefer a constant of 125 but for the moment let's stick to 150. 

So if the gun has a twist of 1 in 48 but the bullet needs a twist of at least 1 in 38, doesn't this mean that this bullet should, theoretically, be inaccurate in the rifle? 

Said the other way around, couldn't you compute the necessary bullet length for proper stabilization in a 1 in 48 twist as: 

bullet length = 150*diameter * diameter / 48 
= 150 * 0.514 * 0.514 /48 = 0.826 inches? 

Humor me. Maybe you don't like or believe in these calculations but I'm trying to figure out if I've got this idea straight. Maybe it is a nutty idea. Maybe it is genious. BUT I'm not trying to debate the merits of that I'm just trying to figure out if I understand this properly. 

One of many points that is confusing me is if I understand this idea correctly why would you want a 1 in 48 twist? Because doesn't that rate of twist require a short, comparatively light weight bullet for the 50-70?

I just got done trying new smokeless loads in my 1874's from 1872 and 1873 and one keyholed at 100 yds.  After getting home I measured the bore and in was .50x land but .535 groove.  How do you match a bullet up with that?

Slightly off topic but......

Charlie