Comments on: Tuesday Trivia Challenge from Bryan Litz, Ballistics Guru

By: Thomas

Thomas — Thu, 14 Mar 2013 09:18:36 +0000

How do I use that energy to heat my house ???

By: Bryan Litz

Bryan Litz — Wed, 13 Mar 2013 14:09:43 +0000

Here’s how I figured it:

Using a standard ballistics program, we calculate the following output:

Range…. Energy……TOF
(yards)…(ft-lb)….(sec)
0……….2729….. 0.0000
500……..1324….. 0.6785
1000…… 547…… 1.6921
1500…….337…… 3.1636

Focusing on the Energy column, we can see the bullet loses 1405 ft-lb in the first 500 yards (2729 – 1324 = 1405). We’re looking for power, which is energy per time. So you divide the change in energy by the time of flight: 1405/0.6785 = 2070.74 ft-lb/second. To convert from ft-lb/second to Watts, divide by 0.738: 2070.74/.738 = 2806 Watts. So we can say an average of 2806 Watts of power is applied for the first 500 yards.

The same process is repeated for average power from 0 to 1000, and from 0 to 1500 yards. I got 1747 and 1025 respectively.

Visit the Applied Ballistics facebook page to join the discussion:
http://www.facebook.com/AppliedBallisticsLLC

-Bryan

By: mike

mike — Wed, 13 Mar 2013 02:20:55 +0000

a 1924
b 2907
c 3389
muz energy is 2729 ft pound = 3700 watt
500y energy is 1310 = 1776 watt
3700 watt – 1776 watt = 1924 watt lost in flight and so on for other range.
Just a guess! Maybe momentum and time of flight as something to do with it too?

By: Matt R.

Matt R. — Tue, 12 Mar 2013 23:53:11 +0000

Bryan,
From the book, here is what I got:
500y = 2807W
1000y = 1290W
1500y = 756W

I think that this is accurate, rounded to the nearest whole digit.

–Matt R.