Sabinajiles,

You must have known that I would not be able to resist . Let's see now. Doesn't 5 bore mean that it takes 5 balls of bore size to equal a pound?

If so, then the ball weighs an incredible 1400 grains. Since the volume of a ball is proportional to the cube of the diameter of a ball and the weight of ball composed of a given fixed density (lead) is proportional to volume of the ball, we can determine the diameter of the 5-bore as follows:

The ratio of .45^3/128 grains is equal to the ratio of Dia5bore^3/1400 grains

or

dia 5 bore = 3rd root(1400*(.45^3/128)= .998 or 1 in.

Now 80 grains of 3f propels a 460 conical to a muzzle energy of approx. 1850 ft-lbs. We can say without question, that at least 1850 ft-lbs of energy resides in 80 grains of 3f. If the 5-bore can wring the same energy per unit weight at the 460 can in the .50 cal, then the energy exiting the muzzle of the 5 bore would be:

KE 5bore= (420/80)x1850 = 9712.5 ft. lbs.

Muzzle Velocity is then 1775 fps.

I would have to say, this is one mean round ball .

Regarding the BC, all roundballs made of the same material (ie lead), have the same BC. For lead, the BC is (theoretical) .075, though I have seen other estimate real world BC at .068. In any event, BC even depends on the velocity because supersonic projectiles carry a shock front which messes with the value considerably.

Here is a table for taylor, KE , and KP, out to 60 yards.(I figure they shot em close).

Range Vel. Taylor KE KP
---------------------------------------------------------
Muzzle 1775 355 9792 108,047
15 yds. 1637 327 8328 84,755
30 yds. 1509 302 7077 66,387
45 yds1391 278603952,336
60 yds1286 257 514041,090

One very mean roundball. Compare that to a 50 BMG which has muzzle values:

TaylorKO= 150
KE =13,054
KP = 121,719

Do youguys remember my last thread where I proposed minimum kp of 570 for deer? Do you reckon these two roundscould kill 200 deera piece?

Happy Hunting, Phil