Shot size versus energy loss

Going to be using my old model 870 lightweight on (20-gauge) on pheasants this year. It's always done just fine, but I've always been curious about the 'real', end-result difference between, say, 4, 5 and 6 shot. All are starting between 1220 and 1300 fps with a 1-oz load (manufacturers' data). Can anybody give me any data on the energy difference downrange?

Do the bigger pellets, with their higher mass, lose velocity slower, or are they less prone to wind, etc?

If not, I will stick with 6's, which have always very well. In fact, I've always been shocked at the ranges I could reliably kill a pheasant with 2 3/4-inch #6 loads and a full or modified choke.

Do the bigger pellets, with their higher mass, lose velocity slower, or are they less prone to wind, etc? Al

slower, or are they less prone to wind, etc?

YES and for the same reason that a 16 inch naval rifle on a battleship will shoot 15 miles or so and a 30-06 rifle will only shoot 2 miles, more or less. The responsible phenomenon can be put forth with Newton's laws of motion and the equation of the Conservation of Energy. If you allow me to skip the math, the two theories simply conspire to show that in a collision with particles the velocities of small masses are affected more than those of large masses. This fact can easily be demonstrated on a pool table. Shoot the cue ball at a numbered ball with no English and on line and observe that the cue ball will stop dead. That is to say the cue ball will lose ALL of its velocity, momentum, and energy, during the collision. Now shoot a heavier ball at the same numbered ball and the heavier ball will NOT stop because of the collision. It will always keep moving after the collision but at a slower velocity, less momentum, and less energy. Now, when bullets are shot in air which is composed mostly of Nitrogen and Oxygen molecules, the collisions of the bullets with the molecules slows the smaller bullets (less mass) more so than the large bullets (more mass) in the same manner as the cue ball and the heavy ball and the number balls. It is just one of those things with which we must live. If you like this explanation ask me why momentum is the primary variable in the harvesting of game with a bullet rather than energy. If you do, you will never talk of energy again.

Larger shot will retain energy better and should drift a bit less, but if your gun is doing well with what you're using, stay with it. Could be you have a very good pattern with that load.

OK I'm asking. Post it for everyone.

Momentum (MV) versus Kinetic Energy (KE)

The momentum of a bullet is the product of its mass and its velocity (m * v). If the units of lb-ft/sec are desired, and its velocity (v) is in ft/sec and the bullet's mass (m) is in grains, then the equation for calculating momentum is M = m * v/7000.

Kinetic Energy of a bullet is the product of its mass and the square of its velocity (m * v * v). If the desired units are ft-lb, and its velocity (v) is in ft/sec and the bullet's mass (m) is in grains, then the equation for calculating kinetic energy is KE = m * v * v/450800.

A 405 grain bullet traveling at the velocity of 1292 feet per second and a 60 grain bullet traveling at 3357 f/s have the same kinetic energy (1500 f-lb); but, the 405 grain bullet has a little over two and half times (2.6:1) the momentum (74.75 lb-ft/sec versus 28.77) of the 60 grain bullet. Now we know why the 405 grain bullet or similar was used to annihilate the bison herds of North America.

In the following discussion we shall skip the math and the physics and only present homely experiments and reason to verify the truths of statements. When a bullet collides with an animal, momentum is conserved and the energy is not. Thus, the collision is inelastic.

In the following discussions assume the following: Bullets retain their shape and are of the same weight, length, and diameter. No distortion and no fragmentation of the bullets. The bullets do not exit the animal. The same animal is to be shot in the same manner and conditions.

The important phenomena in killing of a large animals are the following.

1. The pressure wave caused by the bullet. This is often called the 'shock' of the bullet or the 'shock wave.' Now theory and experimentation show that the magnitude of the pressure wave is proportional to the momentum of the bullet; i.e., doubling the momentum of the bullet doubles the peak pressure of the wave. To demonstrate this fact, drop a big rock in water and a small rock and notice which one causes the higher wave. So that, the 405 grain bullet in our example will produce a 'shock' in the animal 2.6 times greater than the 60 grain bullet. Now we know why a rabbit is blown into pieces when shot by a 30-06 Gov't.

2. We know that heavier bullets (likely with more MV) slow down less than lighter bullets during collisions (see previous posting by Mitchell or throw a rock and a ping-pong ball of the same diameter and observe which one travels farther). Therefore, we expect the 405 grain bullet to penetrate deeper into an animal than the lighter bullet even though the lighter bullet is initially faster. In our example the 405 grain bullet is expected to injure more vitals with its longer wound channel.

3. We know that heavier bullets (likely with more MV) are deflected less than lighter bullets if a bone is hit. To ascertain the truth in the statement, roll a bowling ball into bowling pins and then a soccer ball and see which ball is deflected more. Therefore, we expect our well aimed 405 grain bullet to hit more vitals than the 60 grain bullet.

There is a big difference between steel shot and lead shot. Steel needs a lot faster velocity to have the same energy / killing power. Whether you're shooting 7/8 ounce or 2 1/2 ounce of shot, the velocity is what gives you longer range.

If you have a specific shot type and size, I can look up the velocity downrange and calculate a maximun lethal range.