# How to calculate drag forces on an object

Previously, I theoretically calculated the velocity of a bb, accelerated by air pressure, when it exits a barrel. In short, I calculated my velocity to be about 150m/s. However, I wanted a more realistic speed. I looked up the drag equation and tried to apply it to get a more realistic speed, but I don't think my answer is right. This is what I used:

$F_d = \frac{1}{2} pv^2C_DA$

$p$ = mass density of fluid (air) = 1.23Kg/$m^3$

$v$ = flow velocity relative to the fluid = 150m/s

$C_D$ = drag coefficient = .47 (for a sphere)

$A$ = reference area = $\pi*(0.003m)^2$ = 2.827 * 10$^{-5}m^2$ (cross section of a 6mm bb)

$F_d$ = $\frac{1}{2} * \frac{1.23Kg}{m^3} * (\frac{150m}{s})^2 * 2.87 * 10^{-5} m^2$

$F_d$ = $\frac {.184 Kg*m}{s^2}$ = $.184N$

my answer turned out to be .18N of force. Considering that the force on the bb from the air pressure is 14N, the air friction would only slow down the bb less than 1%. Is there something I'm doing wrong because it seems that a bb slows down significantly with the distance it travels? Also, is there any way to account for the increasing external air pressure pushing back on the bb as it compresses the air while it accelerates through the barrel?

• Remember that the 14 N of force from the gun on the bullet (what is a bb anyways?) only works at the barrel exit (which I expect is your starting point in your thinking here). So here the air drag is insignificant. But from here on, there is no push to keep it up. Only air drag works for the rest of the flight, which then slows it down. it seems that a bb slows down significantly with the distance it travels I assume you have some data to be able to say this - Find out from this data what the deceleration actually is and compare with the force you found. Maybe it matches – Steeven Jan 12 '16 at 22:36

If we idealize the scenario enough, this is a simple exercise in differential equations, so let's get to work. First, we know that it's initial speed is $150 \text{ m/s}$, but that is by no means its final speed - obviously, the bb slows down as it travels through air! Let's suppose that the moment the bb exits the barrel, it is no longer being pushed (as Steevan pointed out). So, the only force acting on it is air resistance. So the question is, why does the bb slow down significantly with distance traveled - we can determine this exactly, assuming the model is correct.

Now, the model you are using (apparently) for air resistance is given as

$$F_d = \frac{1}{2} pv^2C_DA.$$

We want to see how the velocity changes as a function of distance! But we know Newton's second law, so we can write that

$$F = m \frac{dv}{dt} = m \frac{dv}{dx} \frac{dx}{dt} = m v' v$$

where $v$ is now a function of distance (this uses the chain rule - hope you're comfortable with that!).

Now, we can write our differential equation:

$$mv'v = -\frac{1}{2} pv^2 C_DA.$$

Note - there is a negative sign there because the force opposes the direction of motion. That is, the force points backwards, and the particle has a positive (forward) velocity. Simplifying, we get

$$v' = -\frac{1}{2m} pC_DAv.$$

Now this is a simple differential equation to solve: we separate variables, i.e. $\frac{v'}{v} = -\frac{1}{2m}pC_DA,$ and then doing some more chain rule magic, we end up with

$$\frac{dv}{v} = -\frac{1}{2m}pC_DA \, dx.$$

Now we can integrate both sides and find our solution:

$$\int_{v(0)}^{v(x)} \frac{dv}{v} = -\frac{1}{2m} pC_DA \int_0^x dx,$$ or $$v(x) = v(0)\exp{\left(-\frac{1}{2m} pC_DA x\right)}.$$ Finally, we can plug in the initial condition, that at $x=0$, the speed is $150 \text{ m/s}$:

$$v(x) = (150 \text{ m/s}) \exp{-\left(\frac{1}{2m} pC_DA x\right)}.$$

Finally, for a numerical answer, you may want to plug in your known constants. Unfortunately, for this you need to know the mass of the bb! For the sake of argument, let's assume a mass of $0.12 \text{ g}$, the most common mass for airsoft bbs, according to Wiki - Airsoft Pellets. So, we can now calculate the speed of the bb as it travels, knowing that $\frac{1}{2} pC_D A = 0.00817 \text{ g/m}$!

So now we have a function for velocity:

$$v(x) = (150 \text{ m/s}) \exp{(-0.0681x)}.$$

For example, to find the distance at which the speed drops by half, we would solve

$$75 \text{ m/s} = (150 \text{ m/s}) \exp{(-0.0681x)},$$

which yields a distance of approximately 10 meters.

Now you see why the bb slows down significantly with distance - it's exponential decay, which tends to decrease the quantity a large amount at first, with the amount of decrease decreasing over time (or in this case, distance).

You have a different situation when the bb is inside the barrel of the bb gun. Assuming that the bb is a tight fit in the barrel (and it should be), there is pressurized air pushing it. The air is doing expansion work on the bb as it does so. Due to this, you need to use the thermodynamic relationship for the process that is involved. If you are using a constant volume of high pressure gas to push the bb out of the barrel, the process will very likely be adiabatic (no heat transfer) because it happens so fast. If that is the case, see the following link: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html