**www.capebuffalo.co.za**

**Ballistics**

**Energy Calculations**

**Foot pounds**

It is amazing how many authors write about the subject and have it all wrong. I will give a brief outline on the energy foot pounds calculation theory and calculations.

Any questions can be emailed to me and I will explain further : info@capebuffalo.co.za

**Energy**

**(of motion)**
- Kinetic energy or force carried by a bullet at that point in its trajectory.
In common use and popular shooting literature it is expressed in foot-pounds,
one ft/lbs being the amount of force required to lift a one-pound weight one
foot above the ground. Formula: Energy ( in ft/lbs ) equals bullet weight
( in grains ) multiplied by the velocity ( in feet per second ) squared, divided
by 450240. Often wrongly equated with killing power, energy is not a
reliable gauge of this, as it does not take into account penetration or bullet
performance.

__ENERGY UNITS__

One foot-pound is the amount of energy expended when a force of one pound acts through a distance of one foot along the direction of the force.

1 foot-pound is equivalent to:

- 1.3558179483314004 joules (exactly)
- 13558179.483314004 ergs (exactly)
- 12 inch-pound force (exactly)
- 192 inch-ounce force (exactly)
- 0.001285067 British Thermal Unit
- 0.323832 calorie
- 0.000323832 "food calorie"
- 32.174049 foot-poundals

For example - one would calculate the kinetic energy of an 80 kg mass traveling at 18 meters per second (40 mph) as:

Kinetic Energy = .5*80*18*18 = 12,960 joules

Note that the kinetic energy increases with the square of the speed. This means, for example, that an object traveling twice as fast will have four times as much kinetic energy.Thus, the kinetic energy can be calculated using the formula:

*
E*_{k}__
is the kinetic energy in __
joules__ , __**
m is the
mass in kilograms, and , v**

To calculate the foot-pound force (ft·lb_{f}) of energy for a bullet
in English units of measure; given *bullet weight in grains*
and

W = ½ m·v² = energy[ft·lb_{f}]
= weight[gr_{f}] × (velocity[ft/s])² / (2 × 32.1739 ft/s² × 7000 gr_{f}/lb_{f}).

**Playing with
numbers- Energy, Velocity, Weight and Momentum.**

The lighter something is, the easier it is to accelerate it to a high speed. By shaving a few grains off a bullet's weight, muzzle velocity can be increased, and this gives a big increase in muzzle energy, since

**E**nergy
= **½** x **M**ass x **V**elocity^{2}

or

**E****nergy
(ftlbs) = [(Velocity (fps))**^{2
}**x Weight (grains)] ÷ 450,240**

__Please refer
to reloading manuals at all times, Nosler provides for excellent reading
material.__

email us: info@capebuffalo.co.za