There are two ways to knock down a silhouette target with a bullet strike: a hit in the leg or foot can drive it backward to a point where the foot is off the rail or the target can be tipped to a point where the center of gravity is outside the area of the foot. When either of these occurs, the target falls. To do either of these, the bullet strike has to cause the target to move. The operative quantity for imparting this required motion is momentum, NOT energy. There is widespread belief that that kinetic energy is important in terminal bullet performance, and in the case of live targets such as game animals and self defense situations, I am in total agreement. In the case of trying to knock down silhouettes, kinetic energy, except as it relates to momentum, is irrelevant. I know that this goes against the grain of widely held beliefs but I think that but I think that this can be proven by the experiment below. It can be actually performed in the field (the first part anyway) or in the imagination as a thought experiment.
Walk up to a ram standing on its rail and put your finger tips against it. Then move your arm such that your hand moves forward one foot in one second. What happens? Obviously, the ram falls. Now lets calculate the momentum and kinetic energy involved here. We have to know the effective weight of the arm and this could be difficult because there are a number of factors involved since it is connected to the body. It would be somewhere between the actual weight of the arm and the total weight of the entire body. I am going to pick 64 lbs so that the numbers work out even.
The momentum is the mass times the velocity. In the English system the unit of measure for mass is the slug which is 32 lbs.
mv = m x v = (64/32) x 1 = 2 slug feet per second
The kinetic energy is one half times the mass times the velocity squared.
ke = ½ mv**2 = ½ x (64/32) x 1**2 = 1 foot pound
The ram fell even though only 1 foot pound of energy has been put into the system.
Now, lets look at it from a bullets point of view. Suppose that you had a hiccup with your powder measure and short loaded a case and it slipped through that way. Suppose that you had a trigger hiccup with that round when firing at a ram and the shot went off a little high. Luck is with you and the 168gr bullet arcs down from the heights and with its last gasp, hits dead center on target at only 36.51 fps remaining velocity. You could throw a bullet that fast. Would the target fall? Not likely. Yet if you calculate the energy, it is 1 ft/lb, same as the arm above. The momentum here is only 0.0274 slug feet per second. Two collisions with a ram, same kinetic energy in both cases. The ram that fell involved way more momentum. This indicates that momentum, not energy is the operative entity is causing the target to move and thereby fall.
The problems with trying to define a precise value for the momentum necessary to topple a ram are that the collision between a ram and a bullet is not a pure free body collision and there are differences in the targets themselves. The ram has feet that are standing on a rail or some other kind of target support. Friction between the feet and the rail can inhibit the ability of the target to move when it is hit. The target can lean toward or away from the firing line making it harder or easier to knock down. Even though a precise number cant be derived we can try to be in the ball park. Knocking down steel rams is a balancing act between having enough power to push them over (momentum on the target) and avoiding recoil (momentum on the shooter). Silhouette shooters don't like hitting rams and not having them fall. They also don't like missing targets because of the cumulative effects of recoil. We need to use enough gun to feel reasonably sure that the rams will go without going overboard and suffering from recoil.
Based upon my own experiences in rifle silhouette, I think that an MV of s a little more than 1 slug foot per second should do the job on rams.
I have been shooting for several years and have never lost a ram using the .30 cal Sierra 168 MK at 2600 at the muzzle although I have seen it fail once while I was spoting. The hit was very low in the belly and the ram was leaning toward the firing line. It rocked but did not fall. The .30cal 168 MK starting at 2600 should have a velocity of about 1620 fps at 500M which would be a momentum of 1.2150 slug feet per second.
Since I was going to the matches anyway, I decided to try out hunter rifle i.e. two shoots for one trip. I use a 700 in .243 since it makes weight, is accurate and fun to shoot even though the caliber is light. The Sierra 100gr JSP BT at 3000 fps is the ammo that I use. At 500M the velocity is down to 1866 fps meaning that the bullet is carrying a momentum of 0.8330 slug feet per second.
In four matches I hit a total of 17 rams of which eight fell meaning that an mv of 0.8330 is about 47% effective on rams. My spotter called all of the hits and showed them to me in the scope before the target setters got to the ram line to reset and paint the targets. (I am going to rebarrel it to a .260.)
This is a small sampling of data, of course, and it does not "prove" anything. Viewed as an indicator, I believe that the data shows the 0.8330 is "less than half" and 1.2150 is "almost all."
Drue
