## Tuesday Trivia Challenge from Bryan Litz, Ballistics Guru

Here’s a Ballistics Trivia challenge, put together by Bryan Litz of Applied Ballistics LLC. Bryan is Berger Bullets’ Ballistician and the author of Applied Ballistics for Long Range Shooting. Bryan posed the following Ballistics Question about Kinetic Energy and Aerodynamic Drag:

Consider a .30 caliber 175 grain bullet with a G7 BC of .259 (Berger 175 OTM) fired level at a muzzle velocity of 2650 fps in standard (ICAO) sea level conditions.

As this bullet flies downrange, it loses velocity due to aerodynamic drag. As the velocity of the bullet decays, so does its

Kinetic Energy(in ft-lbs). The Kinetic Energy lost by the bullet in a given amount of time can be defined in terms of power.Another way to think about this is that the aerodynamic drag on the bullet can be expressed in terms of power, calculated from the projectile’s change in Kinetic Energy over flight time.

Question: How much power (expressed in Watts) is applied to the bullet by aerodynamic drag on average over:

A) 500 yards?

B) 1000 yards?

C) 1500 yards?Guesses are welcome, but this one can be calculated exactly.

### Similar Posts:

- G1 vs. G7 Ballistic Coefficients — What You Need to Know
- Coefficient Conundrum: G1 vs. G7, Which BC Should You Use
- Bullet Flight Video Reveals Shock Wave and Bullet Base Drag
- Great Video Series with Bryan Litz Explains Long Range Shooting
- Applied Ballistics Expands Bullet Library to 533 Bullets

Share the post "Tuesday Trivia Challenge from Bryan Litz, Ballistics Guru"

Tags: ballistics, Bryan Litz, Drag, G7 BC, Kinetic Energy Bullet

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.

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?

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

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