GarandOwner

July 9, 2008, 01:36 AM

One great gun debate that has been brought up recently is whether bullets of a heavier weight drop faster than a lighter bullet of the same size. Many refer back to basic physics that states that in a vacuum two objects that are acted on by the same acceleration (in this case gravity) will move at the same rate regardless of mass or size. It is this concept that has led many to claim that bullets drop at the same rate regardless of their weight. However there is a key component missing in this claim. In the real world, we don't shoot in vacuums, we shoot through air. For this reason we must account for air resistance. Air resistance occurs from the flow of air around an object as it moves. In the case of bullet drop, there are two components of air resistance, a vertical resistance and a horizontal resistance. The only one that has an impact on the amount of time it takes for a bullet to hit the ground is its vertical resistance. One concept that carries over regardless of wither or not the bullet is moving through a vacuum or a fluid, is that a bullet that is shot parallel to the ground will hit the ground at the exact same moment as one that is dropped at the same time. For this reason we can examine free fall to determine if an objects weight is significant in the time it takes to hit the ground. It does use some higher level math (calculus and differential equations) but I tried to explain it simple enough that anyone can follow and understand.

First case: Neglecting air resistance

position, velocity and acceleration are all related to one another. Velocity is the derivative of position, and acceleration is the derivative of velocity, and the second derivative of position. First we will start by looking at acceleration. A bullet is generally fired no more than 6 feet above the ground, for this small variation we can neglect the changes in acceleration due to gravity since it is minimal (minimal meaning that it doesn't even change out to the one ten thousandth place holder. (.0001)) so acceleration is a constant g that is not dependent on time. As with all physics equations, we start with Newton's second law: F = ma where a force equals a mass times its acceleration. We know gravity, but we want to find it as a function of time. So we have:

ma(t) = mg

where m is mass, a(t) is the acceleration as a function of time, and g is gravity

We can see that mass cancels from both sides and we are left with:

a(t) = g

Taking the anti derivative (Integral) of acceleration we can get the velocity function. Since gravity is a constant it is a simple integral that evaluates as:

v(t) = gt + C1

Once again we can take the anti derivative to find the position as a function of time.

h(t) = 1/2gt^2 + C1t + C2

If we impose the initial conditions v(0) = Vo and h(0) = ho then this equation becomes:

h(t) = 1/2gt^2 + Vot + ho Where Vo is the initial velocity and ho is the initial position.

This function shows that an objects position is independent of its mass, so no matter the mass of the object, it will hit the ground at the exact same time. However this function has a flaw: it neglects air resistance, as we all know air resistance is significant. So let us start again including air resistance.

Second Case: bullet drop with air resistance

Air resistance is bv, where b is a constant that depends on the objects shape and the density of air, and v is the objects vertical velocity. Once again we get our basis for the equation from F = ma so this time we have:

ma(t) = mg - bv (air resistance is negative because it is an acceleration that "slows" the effect of gravity)

Once again to find velocity we must integrate with respect to time: It must be noted that this time our acceleration function has a velocity component, for this reason we can make it easier on ourseves by writing acceleration as the derivative of velocity with respect to time (dv/dt) so we have:

m (dv/dt) = mg - bv

by rearranging the equation we can make it more manageable to integrate

dv/(mg - bv) = dt/m

now we integrate each side, the left side is integrated with respect to velocity, and the right side with respect to time. So the equation becomes:

-(1/b)ln|mg - bv| = t/b +c where c is the integrating constant

we want to isolate velocity by itself so we multiply both sides by -b and take the exponential function of each side to help break down the left side this gives us:

mg - bv = e^(-bc)*e^(-bt/m)

which can be written as:

mg - bv = Ce^(-bt/m) where C = e^(-bc)

some simple algebra re-aranges the function so we have v by itself, this is:

v = (mg/b) - (C/b)e^(-bt/m)

by once again imposing the initial condition where v(0) = Vo we can solve for C. We see that when time is zero the exponetial term goes to 1 so:

Vo = (mg/b) - (C/b)*1

solving for C here we get that C = (Vo)b - mg plugging this back into our velocity equation yields:

v = (mg/b) + (Vo - (mg/b))e^(-bt/m)

It can be seen that velocity is a function that IS Dependant on mass when air resistance is included, so this means that the weight of a bullet DOES effect how fast it drops. Plugging in some simple values will show that an object of the same shape (same b value) with a larger mass WILL in fact drop FASTER than one that is the same shape but lighter.

For those over zealous mathematicians/physicists/engineers out there, you can integrate again and get the position function.

First case: Neglecting air resistance

position, velocity and acceleration are all related to one another. Velocity is the derivative of position, and acceleration is the derivative of velocity, and the second derivative of position. First we will start by looking at acceleration. A bullet is generally fired no more than 6 feet above the ground, for this small variation we can neglect the changes in acceleration due to gravity since it is minimal (minimal meaning that it doesn't even change out to the one ten thousandth place holder. (.0001)) so acceleration is a constant g that is not dependent on time. As with all physics equations, we start with Newton's second law: F = ma where a force equals a mass times its acceleration. We know gravity, but we want to find it as a function of time. So we have:

ma(t) = mg

where m is mass, a(t) is the acceleration as a function of time, and g is gravity

We can see that mass cancels from both sides and we are left with:

a(t) = g

Taking the anti derivative (Integral) of acceleration we can get the velocity function. Since gravity is a constant it is a simple integral that evaluates as:

v(t) = gt + C1

Once again we can take the anti derivative to find the position as a function of time.

h(t) = 1/2gt^2 + C1t + C2

If we impose the initial conditions v(0) = Vo and h(0) = ho then this equation becomes:

h(t) = 1/2gt^2 + Vot + ho Where Vo is the initial velocity and ho is the initial position.

This function shows that an objects position is independent of its mass, so no matter the mass of the object, it will hit the ground at the exact same time. However this function has a flaw: it neglects air resistance, as we all know air resistance is significant. So let us start again including air resistance.

Second Case: bullet drop with air resistance

Air resistance is bv, where b is a constant that depends on the objects shape and the density of air, and v is the objects vertical velocity. Once again we get our basis for the equation from F = ma so this time we have:

ma(t) = mg - bv (air resistance is negative because it is an acceleration that "slows" the effect of gravity)

Once again to find velocity we must integrate with respect to time: It must be noted that this time our acceleration function has a velocity component, for this reason we can make it easier on ourseves by writing acceleration as the derivative of velocity with respect to time (dv/dt) so we have:

m (dv/dt) = mg - bv

by rearranging the equation we can make it more manageable to integrate

dv/(mg - bv) = dt/m

now we integrate each side, the left side is integrated with respect to velocity, and the right side with respect to time. So the equation becomes:

-(1/b)ln|mg - bv| = t/b +c where c is the integrating constant

we want to isolate velocity by itself so we multiply both sides by -b and take the exponential function of each side to help break down the left side this gives us:

mg - bv = e^(-bc)*e^(-bt/m)

which can be written as:

mg - bv = Ce^(-bt/m) where C = e^(-bc)

some simple algebra re-aranges the function so we have v by itself, this is:

v = (mg/b) - (C/b)e^(-bt/m)

by once again imposing the initial condition where v(0) = Vo we can solve for C. We see that when time is zero the exponetial term goes to 1 so:

Vo = (mg/b) - (C/b)*1

solving for C here we get that C = (Vo)b - mg plugging this back into our velocity equation yields:

v = (mg/b) + (Vo - (mg/b))e^(-bt/m)

It can be seen that velocity is a function that IS Dependant on mass when air resistance is included, so this means that the weight of a bullet DOES effect how fast it drops. Plugging in some simple values will show that an object of the same shape (same b value) with a larger mass WILL in fact drop FASTER than one that is the same shape but lighter.

For those over zealous mathematicians/physicists/engineers out there, you can integrate again and get the position function.