🚀 go-pugleaf

jheath@fieldofdreams.npirs.purdue.edu (Jim Heath) writes:

## 	If we step into the Physics room for a moment perhaps we can sort this
## one out.  Remember the simple fact that energy is conserved in a system. 
## At the moment the bullet leaves the chamber, we'll call that zero elevation,
## it has a certain amount of kinetic energy from the burning of the powder.
## As it climbs, this kinetic energy is converted to potential energy at such a
## rate that, at the apex of its flight (the top), Kinetic energy =0 and potential
## energy = kinetic energy at elevation zero.  So it stops, and now it starts to
## fall.  As it falls, it accelerates and the potential energy is turned back into
## kinetic energy.  Assuming that it went straight up and fell straight down, when
## it reached the chamber of the gun on the way back down, elevation zero, it
## would have the exact same amount of kinetic energy as when it left.  
## 
## 	What does this mean?  Well, if you shot the bullet, and then stepped
## into the spot you fired the gun from, when the bullet came down, you'd be dead.
## 
## 	Of course this little exercise does ignore a few things though.  Some 
## energy will be lost due to air friction and converted to heat, though not much
## if it is a very aerodynmic bullet.  Also, the shot has to go straight up and
## straight down as well.
## 


Not so. The angle of elevation has nothing to do with it, except for the
special case where phi = 0 degrees, whereupon the whole question becomes
academic. In a system as you describe it, energy is conserved even if the
bullet is transported horizontally, i.e. the bullet hits the ground some
distance from the shooter, as long as the altitudes at the two poins are
the same.
Moreover, the air resistance is a factor very much to be reckoned with,
as any ballistics table will show.
A factory .45 handgun round with a 185grs bullet is listed in my
ballistics tables with a muzzle velocity of 1.000 fps and a corresponding
muzzle energy of 410 ft.lbs.
The same bullet, measured at 50 yds. downrange now has a velocity of 940
fps, and 890 fps @100yds. The corresponding energy levels are 360 ft.lbs
@50yds. and 325 ft.lbs @100yds. This energy is lost solely to air
resistance, and it is therefore not correct to say that energy is conserved
within the system, as a major part of it will be lost to the surroundings.
I'd guess the 'major part' to be between 40 and 70 percent, anybody care
to solve some differential equations and come up with a closer approximation?

regards,

ragnar