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Rotationally Symmetrical Tank Volume

This note demonstrates how to quickly find the volume of a rotationally symmetrical enclosure. I find this approach useful for estimating and designing tank volumes. As an example, you can quickly find the volume of a toroidal tank with a 1 inch circle revolved about a line 2 inch from the circle’s center is slightly under 10 cubic inches (precisely π squared).

The equation is simple: Volume = 2π·A·L, where A is the area of the revolved cross section and L is the distance from the rotational axis to the center-of-gravity of the cross sectional area. For the torus, the area is 2π·πD2/4·L which simplifies to π2D2L/2. Neat and fast.

Here’s a design application: develop an equation for elliptical ended cylindrical tanks to hold precisely G gallons. Using the revolution equation in the figure below gives a total volume of V=πR2(h+4/3·b).

Now conversion to gallons is a linear transformation with 231 in3 per gallon.

This approach is also useful for crazy, bizarre, strange, and otherwise normal volumes of revolution. For instance, you can now find the volume of a specialty triangular o-ring seal.

The triangular cross sectional area is half the base times height, so 50 mm2. The center of gravity of a triangle is 2/3 the way from the tip to the base, so that’s 6.6 mm. The total distance from the axis to the CG is 20+6.6 = 26.6 mm. Multiply and you obtain: Volume = 2π 50 * 26.6 = 8377 mm3

Why does this work?

The key trick here is recognizing that in cylindrical coordinates, the differential volume is dV=r·dr·dθ·dz, which appears to contain the moment equation dM=x·dx in a 2D Cartesian frame. A derivation is presented below.

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.

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Cherokee 235 Tip Tanks and Fatigue

A previous analysis indicated that the POH fuel-tank recommendations reduced the wing root bending moment. This note attempts to quantify the lifespan impact of the estimated maximum 15% reduction in bending moment.

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.

Figure 1: PA-28 Wing Spar at the Root

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

To be continued.

Cherokee 235 Tip Tanks & Wing Bending Moments

The Piper Cherokee PA-28-235 recommends the use of the tip tanks last. The purpose of this note is to estimate the reduction in wing root bending moments associated with this strategy. The PA28 model has a known AD history with wing root cracks; reducing the bending reduces the stress and increases the structural life. The POH for a C model says:

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.

Figure 1: Planform View

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.

Figure 2: Simplified Summation of Moments

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.

Figure 3: Bending Moments and Loading Cases

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.

Wind Clover Airspeed Calibration

The purpose of this page is to develop and distribute a simplified airspeed indicator calibration technique and computer program tool. Calibration from indicated (KIAS) to calibrated (KCAS) is required for certified and experimental aircraft (c.f FAR 23.1323 and FAR 25.1323). There are many techniques and flight test approaches available; however, the mathematics of generating a calibration chart or card can be daunting. This page provides a FREE self-contained airspeed calibration tool for Windows computers useful for subsonic aircraft with minimal calculation and with minimal equipment.

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.

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Singularity Films Presents…

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.

Classical Partial Differential Equations (PDE) Cheatsheet

The study of classical PDEs is a useful and typical course for engineers and scientists to both appreciate and understand the behavior of physical systems.

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

Decibel (dB) Simplification.

Or… How I learned to stop worrying and count the zeros

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.

  1. Divide by 10. “4”
  2. 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.

Robust Communications:

Q: How can I improve my communications?

A: This is an excellent question. In this note, you will learn some fundamentals and tools of effective and robust communications.

Part 1: Speaking

First, listen to this lecture.

A brief aside: My experience as a professor taught me that its easy to talk about what I just spent hours preparing. In fact, 4 years of teaching made me FAR better at engineering than 4 years of engineering school. Why is that? I really believe the difference is engagement. I had to be ready to engage a conversation spontaneously. This meant that I had to know and understand.

If in the future I teach another class, I would have students make their own set of formal notes and example problems with solutions. No traditional homework.

Part II: Writing

There is a spectrum of communication effectiveness. Listen to this lecture.

Send me your comments. I look forward to hearing from you at oneill@aerofluids.com

Mystery Hs-126 Photo

The following Henschel Hs-126 is from our family’s photo archive. This photo was attributed to my grandmother’s brother W.E. “Bud” Hills of the 101st Airborne in WW2. We suspected the photo was taken in France or Germany. There are no annotations.

Henschel 126; Hs 126; WW2

The aircraft is a Henschel Hs 126, an observation aircraft developed in the late 1930s and essentially obsolete and out of production by the early 1940s. The Hs 126 is a surprisingly large platform given the mission. One successor, the amazing Fiesler 156 Storch outperforms the Hs 126 for short, rough, and unprepared flight operations. The other successor, the Fw 189 “Flying Eye” was superior for observation. The aircraft type was rarely used on the Western front after 1940, most were sent to the Eastern Front.

The markings are 5F+GH. Using public sources of German squadrons and locations, we discovered that this aircraft (“G”) was assigned to the 14th Reconnaissance wing (Aufkl. Gr. 14) 1st Staffel (“H”). This would be 1.(H)/14 with known locations in France from 1940 to 1941. The unit was sent to the Eastern front in Feb 1941 and disbanded while back in France in 1942. The Short Range Reconnaissance wing (Naraufkl. Gr. 14) was created in 1943 but was stationed outside of France. Any use of Hs 126 aircraft in 1944 would be unlikely as Me-109Gs were assigned to the unit. Operationally, the Hs 126 was not feasible in 1944.

This leads to the strong possibility that the photo was taken in 1940 during or after the Battle of France. Thus, the photo was brought back by my relative and not taken by my relative. There are other possibilities, but this is the most likely.

Question: What is the make, model and year of the car in the background? Where is the house style/architecture usually seen? Send comments to oneill@aerofluids.com