# What were the forces imparted by the chicken gun?

How hard does the chicken gun hit? Putting aside the apocryphal story of the British engineers who forget to defrost the chicken before firing it, what is the peak force the average chicken imparts?

(Of course this is a function of speed and impact site, I'm just looking for a data point or two.)

• Wikipedia says 270kts typically. Jan 30, 2023 at 3:08
• @JohnK, I saw that, but the peak forces will be a function of how compliant the chicken is and the wikipedia article doesn't cover that. Jan 30, 2023 at 11:59

According to the relevant rules (for example 14 CFR 25.571):

The airplane must be capable of successfully completing a flight during which likely structural damage occurs as a result of -

1. Impact with a 4-pound bird when the velocity of the airplane relative to the bird along the airplane's flight path is equal to Vc at sea level or 0.85Vc at 8,000 feet, whichever is more critical;

Note that in the calculation of the kinetic energy, the mass of the bird (4lb, 1.8kg) enters in the equation together with the designed cruising speed of the airplane Vc (and obviously not the bird's flying speed).

That value of kinetic energy should give you an idea of how piercing the impact can be. Translating the value of kinetic energy into a force is quite complicated since the force depends on the structural properties both of the bird and of the hit structure. That's why chicken guns are still in use despite the fancy simulation tools run during the design phase.

Just as example, this is how it looks like reality vs. simulation (all pictures from this paper):

Getting a ballpark estimate of the impact forces might be done using an energy approach with some simplification; we suppose that:

• the airplane's structure deforms in a spring-like way; actually after the impact the structure remains deformed and it might even be broken, so it doesn't really behave like a spring;
• the impacting body does not disintegrate i.e. its kinetic energy is all transferred in deformation of the spring; this is also not perfectly true.

We have:

$$E_\mathrm{k}=\frac12 mV^2=E_\mathrm{e}=\frac12 kx^2$$

$$F=kx$$

Where $$k$$ is the rigidity of the spring, $$x$$ its deformation and $$F$$ the force needed to deform it.

Equating and extracting $$F$$ we get:

$$F=\frac{mV^2}{x}$$

Let's substitute some numbers:

• $$m=1.8~\mathrm{kg}$$
• $$V=100~\mathrm{m/s}$$
• $$x=10~\mathrm{cm}=0.1~\mathrm{m}$$

$$F=180~\mathrm{kN}$$

• That's really interesting, but I need the force, not the pressure. (We're trying to determine the reaction forces on the wing/fuselage attachment anchors if a birdstrike on the wingtip occurs.) Through your answer, I found DTIC_ADA021142, which includes pressures vs distance, and therefore can integrate to get force. Any chance you could update your answer with data from that, so I can mark the answer as correct? Jan 31, 2023 at 11:14
• @KennSebesta: done. Feb 1, 2023 at 16:33
• Thinking of it as a spring is insightful. This works out to a 50000G deceleration, which leaves me awestruck, but that's the math. This makes me think that the chicken-gun test is inadequate to answer my question for the airplane design, since considering these astronomical forces it's much more likely that the LE of the wing will cleave the bird in two. Feb 1, 2023 at 23:20
• My next SE.Aviation question: "how much force does it take to chop a bird in half with a NACA 4-digit airfoil?" Feb 1, 2023 at 23:20

Rewatched the MythBusters episode to get their speeds and data. Now, it's time to do some math.

$$m = \frac{4~\mathrm{lb}}{32.174~\mathrm{ft/s}^2} = 0.124~\mathrm{slug}$$

$$V = 140~\frac{\mathrm{mi}}{\mathrm{h}} = 205.3~\frac{\mathrm{ft}}{\mathrm{s}}$$

$$t = 0.007~\mathrm{s}$$

In both cases, $$a = V1/t = 29333~\mathrm{ft/s}^2$$

Since $$F_\mathrm{avg} = ma$$, $$F_\mathrm{avg} = 0.124~\mathrm{slug} \cdot 29333~\mathrm{ft/s}^2 = 3647~\mathrm{lb}$$

Separately, if we consider the numbers in the wikipedia article you posted, $$V = 350~\mathrm{mi/h} = 513~\mathrm{ft/s}$$. If we assume the same rate of energy dissipation as the MythBusters test and a 4lb chicken, then:

$$k_\mathrm{MB} = \frac{1}{2}mv^2 = \frac{1}{2}(0.124~\mathrm{slug}) \cdot (205.3~\mathrm{ft/s})^2 = 2613~\mathrm{ft}\cdot\mathrm{lb}$$

$$\frac{k_\mathrm{MB}}{t_\mathrm{MB}} = \frac{2613~\mathrm{ft}\cdot\mathrm{lb}}{0.007~\mathrm{s}} = 373,311~\mathrm{ft}\cdot\mathrm{lb/s}$$

$$k_\mathrm{wiki} = \frac{1}{2}(0.124~\mathrm{slug})(513~\mathrm{ft/s})^2 = 16316~\mathrm{ft}\cdot\mathrm{lb}$$

$$t_\mathrm{wiki} = \frac{k_\mathrm{wiki}}{\frac{k_\mathrm{MB}}{t_\mathrm{MB}}} = \frac{16316~\mathrm{ft}\cdot\mathrm{lb}}{373311~\mathrm{ft}\cdot\mathrm{lb/s}} = 0.043~\mathrm{s}$$

Then, $$a = V/t = \frac{513~\mathrm{ft/s}}{0.043~\mathrm{s}} = 11737~\mathrm{ft/s}^2$$

and $$F = ma = 0.124~\mathrm{slug} \cdot 11737~\mathrm{ft/s}^2 = 1455~\mathrm{lb}$$

If instead we assume the deceleration time is constant:

$$a_\mathrm{wiki} = \frac{513~\mathrm{ft/s}}{0.007~\mathrm{s}} = 73285~\mathrm{ft/s}^2$$

$$F_\mathrm{wiki} = 0.124~\mathrm{slug} \cdot 73285~\mathrm{ft/s}^2 = 9087~\mathrm{lb}$$

And interestingly enough, based on the MythBusters' testing, a frozen and thawed chicken would impart the same force.

• Re last sentence-- how could that be true? Any way to explain a bit to give some insight on how that could be true? Is the idea that with the frozen bird, the windshield shatters before the before the full force can be applied, somehow somewhat akin to the advantage of a bullet that "mushrooms" quickly rather than maintains original pointed shape for longer? Just grasping at straws here, and curious, feel free to ignore, or respond/edit, as you see fit-- Jan 30, 2023 at 20:41
• Force is a tricky metric when evaluating collisions. Assuming the entire chicken is eventually "absorbed" by the aircraft (all pieces of the chicken eventually come to rest in the aircraft's inertial reference frame), the total energy (mv^2/2) and momentum (mv) transferred are equal, since m and v are equal. But force is not. The thawed chicken deforms on impact, dramatically increasing the collision times and correspondingly decreasing the, let's say very casually, "average" instantaneous forces. Some energy is also dissipated internally via deformation heating, unlike the frozen chicken. Jan 30, 2023 at 20:57
• I haven't seen the MythBusters episode, but in general MythBusters did incredibly poor science. It was entertainment, not science (and I quite liked the show). Another factor to consider is the area over which force is applied: the thawed and deforming chicken will spread out the applied force over a larger area than the frozen one. That factor may be more important than the duration of impact. Jan 30, 2023 at 21:08
• In the first 2 "Chicken Cannon" episodes, the Myth Busters massively missed the mark. The third episode with the layers of glass panes, they got it right. Shooting into an iron plate doesn't model a bird hitting a windshield well at all... a 4-pound water balloon dropped on you from 6 feet would hurt a little (and soak your clothes), while a 4-pound block of ice dropped from that height would cause real damage. Same deal with thawed vs frozen chickens.
– Ralph J
Jan 30, 2023 at 22:02
• @FlyJunior172: thanks for doing this. Could we ask for it to be done in metric, so a wider audience can review the math, and not trip up on lbf vs lbm problems? Jan 30, 2023 at 22:09