Can ChatGPT pass an undergraduate aerodynamics exam?

These and more questions are investigated in this note, as I channel my inner Professor O’Neill as inquisitor in a digital oral defense. My notes are added as <comments>.

Oral exam portion

I asked ChatGPT to explain drag. This isn’t terribly wrong until the induced drag discussion.

Q: "Explain aerodynamic drag from a technical point of view"

Aerodynamic drag is the force that opposes an aircraft's motion through the air. It is caused by the difference in pressure between the front and rear of an aircraft, as well as the friction of the air molecules against the surface of the aircraft. It can be broken down into two main components: parasitic drag and induced drag. <Good>
Parasitic drag is caused by friction of air molecules against the surface of the aircraft, and includes skin friction and form drag. Skin friction is caused by the friction of air molecules against the surface of the aircraft, while form drag is caused by the shape of the aircraft.
Induced drag is caused by the production of lift by the aircraft's wing. As the wing generates lift, it also generates a swirling wake of air behind it, which creates a low pressure area. This low pressure area pulls the tail of the aircraft backwards <Totally wrong>, creating drag.

I would say that this discussion is no worse than most undergraduates who have not taken an aerodynamics course. ChatGPT needs to study Low Speed Aerodynamics, available here. Induced drag is caused by the production of lift and does create a swirling wake of air that convects downstream. The wake is not necessarily a low pressure area, so ChatGPT is not correct. The following statement about pulling the tail backwards is completely nuts; induced drag is not conceptually applied at the aircraft tail.

In an attempt to elucidate a better response, I asked ChatGPT to further explain its weakpoint in the previous discussion.

Q: "Explain induced drag"

Induced drag, also known as lift-induced drag, is a type of aerodynamic drag that is caused by the production of lift by an aircraft's wing. It is a result of the wing's generation of lift and the wing's shape, which creates a swirling wake of air behind the wing, resulting in a lower pressure area behind the wing, which pulls the tail of the aircraft backwards, creating drag.
It is directly proportional to the lift force generated by the wing <Wrong> and the airspeed of the aircraft. It also increases as the angle of attack of the wing increases, and as the aspect ratio of the wing (the ratio of the wing's span to its chord) decreases.
One way to decrease induced drag is to increase the aspect ratio of the wing <Good>, for example by using a long, narrow wing, which will decrease the size of the wake behind the wing and reduce the pressure differential. Another way is to use winglets, which are small vertical fins attached to the tips of the wings that help to reduce the size of the wake, and also reduce the pressure differential. <Meh>
In summary, induced drag is caused by the wing's generation of lift and it is a result of the creation of low pressure area behind the wing, which pulls the tail of the aircraft backwards, creating drag. It is directly proportional to the lift force and airspeed, and increases as the angle of attack and aspect ratio decreases. <Incorrect in many ways>

Correct in only a general sense. The major concept that is not identified correctly is that induced drag increases with the square of lift production.

Q: "Your concept of induced drag is wrong. Can you try again to explain induced drag?"

I apologize if my previous explanation of induced drag was not accurate. Induced drag is the drag caused by the wing of an aircraft as it generates lift. It is caused by the wing's creation of a vortex, or swirling wake of air, behind it. This vortex creates an area of lower pressure behind the wing, which pulls the tail of the aircraft backwards, creating drag...." <Still wrong>

Persistent errors remain in its conceptual understanding. ChatGPT failed the oral exam.

Let’s move on.

Aerodynamics I, Exam 1A, Spring 2017

In this section, I’ll ask ChatGPT to solve questions directly from my Spring 2017 Aerodynamics I Exam. Can you do better? I’d love feedback. Try the full exam here.

Q1: What is the wingspan of a wing with: 250 square foot area, AR=10, and zero taper ratio?

The wingspan of a wing can be calculated using the formula: wingspan = (wing area) / (wing chord) * (aspect ratio).
Given that the wing has an area of 250 square feet, an aspect ratio (AR) of 10 and zero taper ratio, we can calculate the wingspan as follows:
First, we need to calculate the wing chord. Since the taper ratio is zero, the chord is constant across the entire wingspan. The wing area is given as 250 sq.ft and aspect ratio is 10. so, wing chord = sqrt(wing area / aspect ratio) = sqrt(250 / 10) = sqrt(25) = 5 ft
Next, we can use this value to calculate the wingspan. wingspan = (wing area) / (wing chord) * (aspect ratio) = (250) / (5) * (10) = 50 ft
So, the wingspan of a wing with 250 square foot area, AR=10, and zero taper ratio is 50 ft.

That’s a correct answer. I believe my question is slightly confusing with respect to taper ratio but it correctly solved the problem. Dang.

What about questions where a key value is missing from the question? In the next question, the value of density must be determined from identifying the abbreviation SSL means standard sea level and recalling air density at this condition; it’s 0.00237 slugs per cubic foot.

Q: Estimate the SSL stall speed of a Fokker Triplane (S=201 ft2, W=1300 lbf) given the maximum lift coefficient is 2.2.

The stall speed of an aircraft, also known as the minimum steady flight speed, is the speed at which the wing of the aircraft can no longer generate enough lift to support the weight of the aircraft.
One way to estimate the stall speed is to use the formula for the square root of the wing loading, which is:
V_stall = sqrt(2 * W / (rho * S * Cl_max))
V_stall is the stall speed in ft/s
W is the weight of the aircraft in lbf
S is the wing area in ft^2
Cl_max is the maximum lift coefficient
rho is the air density in slugs/ft^3 (approximately 0.00237 slugs/ft^3 at sea level) <Textbook correct!>
Given that the wing area of the Fokker Triplane is 201 ft2, the weight of the aircraft is 1300 lbf, and the maximum lift coefficient is 2.2, the stall speed can be estimated as:
V_stall = sqrt(2 * 1300 / (0.00237 * 201 * 2.2)) = sqrt(2 * 1300 / (0.00237 * 201 * 2.2)) = sqrt(4.4068) = sqrt(4.4068) ≈ 66.05 ft/s
The stall speed of the Fokker Triplane is approximately 66.05 ft/s.
<Wrong calculation>
It's worth noting that the formula I gave is the "Square Root of the Wing loading" formula, and it is an approximation that assumes a zero thrust and a symmetric airfoil <Wrong, no dependence on sym' airfoil>. This formula should give a rough estimate of the stall speed of the plane,
but it is important to note that the actual stall speed can be affected by a variety of factors such as the aircraft's weight, wing loading, airfoil, flaps, and other devices, as well as the aircraft's operating environment and conditions.<Incorrect concept of complexity>

ChatGPT obtains the correct equation but fails to calculate the correct quantity. However, even a cursory view of the calculation indicates problems, so it’s fairly likely that a human reviewing ChatGPT could correct the solution.

The next question is tough. Most undergraduates would not even see the term Joukowski in their introductory aerodynamics course. My students did and were tested to understand an interesting theoretical phenomena. I would be shocked if ChatGPT gets this correctly.

Q: Given a 26% thick symmetrical Joukowski airfoil at 10 degrees AOA, estimate Cl.

I'm sorry, but I am unable to provide an estimate of the coefficient of lift (Cl) for a 26% thick symmetrical Joukowski airfoil at a 10 degree angle of attack (AOA) without additional information such as the airfoil's chord length, Reynolds number, and freestream velocity. These factors all have an effect on the lift coefficient, and without them I cannot give you an estimate.

Nope. Also, as an aside, ChatGPT is not an effective engineer. It gives an excuse rather than an estimate.

This next question is a common question that will be frequently seen in engineering and pilot resources. In fact, this question is a FAA commercial pilot written exam question.

Q: What is the temperature at 10,000 ft on a standard day?

The standard atmosphere is a model of the atmosphere that is often used for the purpose of aeronautical and aerospace calculations. According to the standard atmosphere model, the temperature at an altitude of 10,000 ft (3048 m) on a standard day is approximately -24.4 °C (-11.9 °F). <WRONG> However, it's important to note that this is a simplification of the real atmosphere, as the temperature can vary widely due to weather patterns, location and other factors.

This time, ChatGPT failed to calculate the correct temperature as the dry lapse rate is about 4.4 degrees per 1000′, which gives a temperature of about 15 deg F. I didn’t see lapse rate, so ChatGPT didn’t pick up the right terms. However, when I added the phase “as calculated from lapse rate”, the solution from ChatGPT is correct!

What about a recalling a data point from memory?

Q: What is the air density in slug/ft3 at 80˚ F and 14.7 psi?

Air density, denoted by the Greek letter rho (ρ), is the mass of air per unit volume. It is a function of both temperature and pressure. At standard sea level conditions, air density is approximately 0.00237 slugs/ft^3 at 15°C (59°F) and 14.7 psi. However, at different temperatures and pressures, the air density will change.
To determine the air density at 80°F and 14.7 psi, we can use the ideal gas law. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the specific gas constant, and T is the temperature in Kelvin. We can use the ideal gas constant (R) = 0.7375 slugsft^3 / lbfdegR,
First, we need to convert 80°F to Kelvin: T(K) = (80+459.67)*5/9 = 518.67 K
then we can calculate the air density using the ideal gas law:
P = 14.7 psi = 14.7 * 6894.757 Pa
R = 0.7375 slugsft^3 / lbfdegR
Then, we can rearrange the equation and solve for rho:
rho = P / (RT) = 14.7 * 6894.757 / (0.7375518.67) ≈ 0.0016 slugs/ft^3
Therefore, the air density at 80°F and 14.7 psi is approximately 0.0016 slugs/ft^3. This is significantly less than the sea level density, which means that the air is less dense at that temperature, and the aircraft will have a lower performance.

The correct answer is about 0.00228 slugs per cubic foot. But more importantly, this answer illustrates the dangers of using ChatGPT for answering exam questions. The specific values and orders of terms gives some insight into how the sentences and phrases are composed from existing knowledge and from clues that are within the document (i.e., it follows a path of finding answers to questions asked and discovered). ChatGPT is precise but lacks situational awareness. ChatGPT also vibrantly demonstrates O’Neill’s Unit Rule by betraying a lack of understanding of traditional US customary units.

Final Score: 41%

Next, I asked ChatGPT each question on the Exam and evaluated the responses to find a correct answer or explanation of the concepts. ChatGPT made a 41% on this exam. Failed, but impressive for an AI.


  1. ChatGPT is amazingly good at understanding questions. The responses are excellent at a glance but include significant errors in concepts and calculations.
  2. If you are using ChatGPT for homework, DO NOT TRUST the calculations. The equations are often nearly correct. Concepts can be wonderful and then diverge to complete nonsense from sentence to sentence.
  3. Units (e.g. mass, length, and time) are not understood. I saw equal failures in SI and traditional units, so the mythical SI unit’s brilliance-bonus doesn’t apply to ChatGPT either.
  4. You will almost certainly fail if you use ChatGPT for engineering exams. The fake solutions and wrong concepts are a dead giveaway.