Here's something I've saved from here. Sylvan jump in if ya care too.
Thanks for the lesson!
ORIGINAL: Sylvan
I'd love to see how you calculated it, but it matches what I have seen elsewhere
Actually it's pretty straight forward. If the bow is a 30" draw and an 8" brace height then the power stroke is 20 1/4 inches or 1.69 ft. If you pull 70# over the entire power stroke which gives the absolute maximum stored energy possible you get 20.25 x 1.69 = 118.42 ft/lbs of stored energy. Now if the bow transfers 100% of it's energy to the forward motion of the arrow, the amount of kinetic energy on the arrow is also 118.42 ft/lbs. Now since KE = 1/2mv^2 and we know the KE and the mass of the arrow you simply solve for the unknown v = (2KE/m)^.5 So in the example I used for a 350 grain arrow it comes out to 390.3 ft/sec and a 600 grain at 298.1 ft/sec.
Even with a 0% letoff bow, which climbs to max poundage very quick and stays there, it MIGHT be doable, but would be unshootable.
Nope, not even if you could manage all that. At least not at 70# peak, 30" draw, 8" brace and 350 grain arrow.
UNLESS some new technology comes along that actually creates energy during the shot.
I think that is called a firearm. that's not true either. Turtelshell is correct, you can't create energy, the best you can do is carry a bunch of it with you. That's what we do with a firearm. We carry the stored energy in chemical form in our ammunition. In archery, we carry it around in chemical form in our muscles. Unfortunately, even the strogest of us are pretty limited.
Actually, I just noticed a dumb mistake in the explination. It's not 20.25 x 1.69 but rather it should have been 70 x 1.69 = 118.42 ft. lbs. Sorry about that!