"Nick Maclaren" wrote in message
...

In article ,
"BAC" writes:
|
| Don't want to appear pedantic, but isn't the formula for kinetic energy
1/2
| mv2? So if v is the same for two projectiles, the energy will vary by
half
| the mass ratio, hence in your example it would be 3.5 times not 14
times?
| Apologies if incorrect :-)

Whereas I have no objection to appearing pedantic - hell, I am
a professional pedant :-)

It is m(v^2)/2. So, the ratio is (m1(v^2)/2)/(m2(v^2)/2). Cancelling
common factors, one gets m1/m2. You also may have missed the fact
that a .177 pellet has 1/14th the mass of a .22 bullet - both the
type and calibre are different.


Yes, it appears I was doubly mistaken - firstly in misapplying the formula,
and secondly by interpreting your statement "The difference other than speed
between .22 firearms and .22 air
rifles are that the bullet is c. 7 times heavier than the pellet,
so a .22 bullet carries c. 14 times the energy of a .177 pellet at
the same speed." as deducing the comparative energy as a function of the
stated difference in mass between the .22 bullet and pellet.