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. <Incorrectin 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.

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 ft^{2}, 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))

Where:

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 isapproximately -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/ft^{3}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.

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.

- ChatGPT is amazingly good at understanding questions. The responses are excellent at a glance but include significant errors in concepts and calculations.
- 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. - 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. - You will almost certainly fail if you use ChatGPT for engineering exams. The fake solutions and wrong concepts are a dead giveaway.

This video by Airframe Components provides an excellent visual understanding of the PA-28 and PA-32 spar structures. Please review this video before continuing.

The PA-28 aircraft use the NACA 65(2)-415 airfoil (c.f. Piper PA-28 Cherokee Wing Comparison: An Aerospace Engineering Perspective) with wing spar attach points as illustrated in Figure 1. With the spar at the maximum 15% thickness of a 63 inch chord, the spar height is about 9.5 inches. The moment of inertia is about 18.7 [in4] resulting in a stress of 0.253 [psi] per moment [in-lb]. However, the presence of holes significantly increases the stresses at the bolt holes by about 3x.

Aluminum has the unfortunate mechanical property that all loading -regardless of the stress magnitude- contribute to the metallic lifespan.

To be continued.

]]>To familiarize yourself with the fuel available on the aircraft, the planform view of the aircraft from the POH with fuel tank overlays is illustrated in Figure 1. The aircraft has 4 tanks: two 25 gallon main tanks and two 17 gallon tip tanks for a total of 84 gallons of fuel. The main tanks are approximately 70 inches outboard from the centerline and the tip tanks are approximately 180 inches outboard.

**Aerodynamics**: Using the process described in https://charles-oneill.com/blog/cherokee-tapered-wing-float/, the wing’s effective spanwise aerodynamic center is located at approximately 43% of the span. To remain conservative, we ignore the structural weight of the wing. The simplified configuration is given in Figure 2.

Summation of moments about the root (left side) gives the following equation. When substituting for the wing panel’s lift (L/2) and the weight of fuel, the overall moment at the wing root is

Plotting this wing root bending moment (per g) over the aircraft’s envelope provides a visual comparison of the differences in loading and operational techniques. All feasible loading and operating conditions lie within the gray shaded region. The worst case (red color) is a light 150 lbf -but perhaps not so bright- pilot with VFR minimum fuel and a maximum payload to achieve gross weight of 2900 lbs; notice that the Cherokee does NOT appear to have a zero fuel weight limitation (ZFW). On the other extreme, a light 150 lb pilot with no payload results in the conditions at lower left, where the POH loading suggestion (main 1st, then tips) is the lower black line. The green line shows the results when tips are fed first until dry and then the mains are fed.

**Conclusion 1:** Adding fuel reduces wing bending at the root. This immediately shows shows that the tip fuel is almost 5 times more effective at reducing the bending moment per gallon.

**Conclusion 2:** Each gallon of main fuel is structurally equal to a reduction in payload weight of 4 pounds.

**Conclusion 3:** Each gallon of tip fuel is structurally equal to a reduction in payload weight of 21 pounds. Adding fuel tip tanks (17 gallons) acts to reduce fatigue stresses similar to reducing the payload weight by 360 pounds. This is not a negligible amount.

**Conclusion 4**: The maximum benefit of feeding the mains before the tips is approximately a 15% reduction in bending moment.

**Statement 1:** This analysis does NOT include the effects of maneuvering speed. Please refer to the appropriate POH for guidance and remember that Va reduces at lower weights.

**Statement 2**: The 15% reduction in bending moment contributes FAR more than 15% to the aircraft lifespan. 15% is in fact a substantial number. Further analysis of this fact will be conducted later (See Cherokee 235 Tip Tanks and Fatigue).

**Conclusion:** The Piper POH’s suggestion to use the main tanks first does have a substantial structural fatigue benefit.

**Begin your turn no later than the CDI dot corresponding to one-half of the angle to intercept divided by 3 and divided by the time to pass from one CDI dot to the next.**

This solution works for turning inbound & outbound for tracking a VOR radial, a LOC/ILS, or any situation where you can count intervals leading up to a desired track.

You are on a 45 degree intercept. The needle is alive and you count 4 seconds between CDI dots.

The mental math is: *45 divided by 3 is 15….. 15 divided by 4 is about 4…. Half of 4 is 2. Turn no later than 2 dots.*

The pilot needs a quicker and more intuitive solution. We can simplify this engineering-style intercept equation with a few steps.

Thus the pilot solution in words is: **Begin your turn no later than the CDI dot corresponding to one-half of the angle to intercept divided by 3 and divided by the time to pass from one CDI dot to the next**.

Full disclosure: I did not initially develop this intercept solution based on the mathematics. Rather, I experimentally flew many intercepts and developed the following rule of thumb. By a complete stroke of luck, the experimental solution happens to be exactly identical to the mathematical solution above.

This solution is only meant for intuition and the mathematical understanding of intercepts. You **must **follow ATC, the appropriate regulations and standard operating procedures when flying. Find and train with an instructor, preferably with a CFII. No exceptions!

**Requirements**: You need an indicated airspeed, a GPS with track and groundspeed readouts, a thermometer, and your altitude. You will need to fly three independent headings (approximately 120 degrees apart) for each data point. You will need to download and enter your data into the windclover program.

**Non-requirements**: You do NOT need any ground references. You do NOT need to know your precise heading or magnetic variation. You do NOT need accurate timing or any clock. You do NOT need to calculate your true airspeed. You do NOT need an aerospace engineering background or on-board flight test engineers or hardware.

**Cloverleaf Flight Profile:** You will need to fly three lines that are approximately 120 degrees apart (e.g. 100, 220, 320). Maintain a constant heading, altitude, and indicated airspeed. Using your GPS, record your ground speed and track. Enter these data values into heading columns #1, #2, and #3. The program determines the wind direction/speed and the calibration from KIAS to KCAS.

**Theory**: The theory and algorithm are summarized below as developed in a flight test engineering course’s class notes. It is important to note that there is a background routine for calculating true airspeed from calibrated airspeed and the atmospheric conditions. Also note that this theory is not necessary to use the above tool. For more information, contact Dr. O’Neill.

This is an engineering structures parody of the X Files from 1999 found in my class notes. April Fools… once every 20 years is about right for this engineering joke.

So why is this related to structural engineering? Well, there is a shortcut method using so-called singularity functions to calculate the moment and shear in beams. Refer to any classical engineering textbook. As my initial undergraduate instructor in a structural analysis course, Dr. Wolf Yeigh, would say, *it’s a good tool to have in your pocket*. FYI, that was an intense course and professor, but one that I’m really lucky and glad to have taken. And yes, I made an A.

Using a Turn-Twist strategy, once on the arc, the heading to fly is tangent to the arc. This makes the no-wind control law: Turn to heading = Radial plus 90 deg when CW or Radial minus 90 deg when CCW. Unfortunately, this control strategy contains inherent divergence; in other words, the aircraft always tracks outside the desired arc (Figure 2, left). With a Turn 10/Twist 10 step, the cross track error is 1.5%. For example, a 20 nm arc with 10 degree radial steps, would give 0.3 nm error every step.

*Is there a correction to exactly remain on the arc given a Turn-Twist step?* Yes, and amazingly enough, the result is exact and a trigonometric identity. The right portion of Figure 2 derives a correction angle (gamma) such that the exact track is from point A to point B, both on the same arc. The result is that exactly half the Turn/Twist angle is applied inside the normal +-90 heading.

For example, using a Turn 10/Twist 10 in a counter clockwise direction at the R-040 would require the heading be 305 degrees; this heading will precisely keep you on the exact DME arc at the R-030 radial.

Warning: The normal flying caveats apply: 1) This is only meant for insight and is not meant as instruction or as a change to your specific flight operations manual, 2) wind will require varying correction angles, and 3) aviate, navigate, communicate.

]]>As the (former) instructor of a course in PDEs, I reviewed classical solution techniques in a lecture titled A brief history of GES 554 PDE to prepare students for their final exam. This lecture makes an excellent refresher or rapid introduction.

If you want to review the entire 50 lecture course, visit here. Feel free to call it *The Brief History of the World of PDEs in 50 Parts*.

Topics covered are:

- Motivation, classification & canonical forms
- Diffusion, Elliptic, Hyperbolic, and Transport PDEs
- Solution methods: Series, Separation of variables, Monte Carlo, finite difference, Ritz / Galerkin and Transforms
- 1 page PDE toolbox
- Laplace vs Fourier transforms for PDEs
- Sturm Liouville Theory
- Wave Equations
- Strings, Beams, and Drums
- Characteristics in transport equations
- Systems of PDEs: eigenvalues & eigenvectors
- Green’s Functions
- Calculus of Variations for PDEs

- Nearly empty light-aircraft tanks in extreme hot and humid environments with extreme temperature swings theoretically could condense approximately a couple of fluid ounces a week.
- The generation rate linearly scales with empty tank volume and humidity, but exponentially with temperature.
- Normally vented tanks substantially reduce the water influx rate, but do have a breathing mode that can pump moist air during temperature and pressure swings.
- Condensation is more likely to be a long term storage threat; Large volumes of water are more likely to be ingress of liquid water.

Water is a key enabler of life and dramatically affects the behavior of air. We call “dry air” the mixture of mostly nitrogen (80%), oxygen (20%), and trace other constituents (Ar, carbon dioxide, etc). “Wet air” is what we normally encounter and is dry air + water. “Air” could also include particles + bugs + dirt. You can learn more at my course notes here for an introduction and here for non-standard atmospheres.

The important takeaways are:

- Adding water
**decreases**air density since water has a lower molecular mass than air. (Technical note: Water has 2 Hydrogen of mass 1 plus 1 Oxygen of mass 16 for a total of 18. Air on the other hand has 80% diatomic Nitrogen of mass 28 and 20% diatomic Oxygen at mass 32 for a total of 28.97.).

- Increasing temperature substantially
**increases**the absolute water carrying capacity of air. Water vapor at 100% relative humidity consists of 0.7% of the wet air mass at 50 F and 6% at 110 F, but a whopping 15% at 140 F. This is why pilots need to be much more concerned with high humidity at high temperatures and not so much at lower temperatures (This is in addition to the temperature effects on density altitude).

The saturation pressure (Ps) in Figure 2 is generated from the Arden-Buck approximation as an exponential function of temperature. This also indicates the pressure at which boiling occurs (e.g. 212 F at sea level pressures of 14.69 psi; and 200 F at 10000 ft with a pressure of approximately 10 psi).

Let’s pick a scenario where a nearly empty 25 gallon tank completely condenses the water vapor. Plus there is a complete air exchange/recharge of hot and humid (120 F and 100% relative humidity) air once per day. How much water is generated?

The answer is about 0.25 oz (7.5 ml) per day. Consider it one cupped-hand of water, or about a 2.5″ spot of water, or 0.75 seconds of fuel at 10 gal/hr. That’s enough to fully grab your attention.

The rate scales linearly with the tank’s air volume so keeping the tank 90% full reduces the generation rate by 90%. Being in 30% humidity air reduces the rate by 70%. Doubling the elapsed days doubles the generated volume. Notice that the rate exponentially scales with temperature (ps/T is exponential).

Critically, we can bound the generation rate. For example, at 120F and 100% humidity with a 20 gallon tank, the rate should be less than 0.20 oz per day.

If you are generating more than this, chances are extremely likely that there is another mechanism responsible. Go find it.

To be continued

To be continued

]]>The dB decibel scale can often be very intimidating to others, so here’s a quick way to simplify (i.e. no logs or powers) your explanation to two steps. The fundamental point to make is that a Bell is how many zeros. **A decibel is the number of zeros multiplied by 10.**

Let’s convert a ratio to dB. Pick 100. This number has a number 1 followed by two zeros before the decimal point.

**How many zeros?**“2”.**Multiply by 10.**“20”**Say that number**. “20 dB”

Let’s reverse the process and convert dB to a ratio. Pick 40 dB.

**Divide by 10.**“4”**Four zeros before the decimal place is:**“10000… ten thousand”

How about a more complicated case. Convert 25 dB to a ratio.

**Divide by 10.**“2.5”**Two and a half zeros before the decimal place is?**“more than 100 and less than 1000”**Yes, and half a decimal place is about 3.**“So 300?”**You got it.**“25 dB is about 300”

Now, convert 564 to dB.

**How many zeros?**“Almost the number 6 followed by two zeros. So 2”**Yes, but we had a 6 in front of the zeros. 6 is worth about 75% of a decimal place.**“So 2.75?”**Exactly, now multiply by 10.**“27.5”**Say that number.**“27.5 dB”

This approach is much easier to explain than defining dB = 10 log(R) and the inverse operation using pow(10) and gives much better intuition. So, in field work, I tend to just use this approach. This may seem trivial to experts, but any trick to increasing understanding and explain-ability is worth your consideration.

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