This page exists to log my visits to the surviving Lockheed A-12 aircraft. Photos are in serial number order.
60-6925 in NYC: no photo
A12 60-6930 in Huntsville, AL
A12 60-6933 at the San Diego Air & Space Museum
A12 60-6937 in Birmingham, AL
A12 60-6938 in Mobile, AL
Today, we manipulate bits to resurrect a hobbled 15 year old GPS receiver. In the process, I learned about WAAS and s-records to successfully update a client’s Magellan Sportrak GPS unit.
Today, my MS student Mr. Christopher Simpson successfully defended his thesis:
CONTROL SURFACE HINGE MOMENT PREDICTION
USING COMPUTATIONAL FLUID
DYNAMICS
The work demonstrated several key concepts necessary for the use of CFD in rapid aircraft prototyping of aircraft control surfaces. The thesis evaluated both 2D and 3D geometries using NASA LaRC’s FUN3D computational fluid dynamics software.
Christopher also conducted unsteady and adjoint refined solutions.
The final version is available here.
University of Alabama AEM 617 students visit the Southern Museum of Flight’s A-12
Today, we get a behind the scenes visit of the Southern Museum of Flight in Birmingham, AL to discover the complexity of aircraft systems. In the process, my class of graduate students learned to see aircraft in a different light.
Special thanks to the museum director Dr. Brian Barsanti. My students received a rare treat, a guided tour with both a professor of history and a professor of aerospace engineering.
Landing on Mars is a difficult engineering problem. A recent conceptual project demonstrated this reality through the atmospheric reentry profile. In other words, assuming that your spacecraft approaches Mars at 200 kilometers and 6.5 kilometers per seconds, what is the altitude-velocity profile during the reentry trip?
Complicating the situation is an observation that your vehicle’s aerodynamics strongly affect the profile. In particular, increasing the life decreases the observed g-load, but leads to atmospheric skip when the lift is not controlled.
A ballistic reentry vehicle should expect a maximum g-load above 10 in the 50 km altitude region. This conceptual analysis strongly suggests that manned reentry to Mars will require hypersonic aerodynamics generation of a lift over drag (L/D) ratio in the 0.5 range.
The graduate level AEM 614 Wing and Airfoil Theory class at the University of Alabama was resurrected, developed, and taught in the fall of 2015.
The course used two textbooks:
Flight Vehicle Aerodynamics, Mark Drela, MIT Press, 2014
Aerodynamic Design of Transport Aircraft, Ed Obert, IOS Press, 2009
Student Evaluations: 15Charles O’Neill (AEM 614-001 Airfoil And Wing Theory)
Aerospace engineers encounter vortex methods as undergraduates. The famous thin airfoil theory is a continuous vortex sheet. The famous XFOIL is a panel method with an IBL solver for viscous effects.
Q: Can a simple discrete vortex method capture thickness?
A: Yes, but I would recommend a linear vortex method instead.
NACA 4412 at 10 degrees: 200 Elements
NACA 4412 at 10 degrees: 16 Elements
The source code and executable are available here.
The visualization is a pure Fortran library exporting to scalable vector graphics (.svg).
\(\)In the spring, a colleague asked a question regarding the computation of circulation about an airfoil. In response, I created a comparison of methods to calculate the circulation about an NACA 0012 airfoil at 10 degrees angle of attack, Mach 0.16 and a Reynolds number of 6 million. Data was generated with the FUN3D CFD solver with an SA turbulence model. The visualization is through TecPlot360. The pressure coefficient, \(C_p \), field is given below:
The definition of vorticity is: $$ \omega = \nabla\times V$$ In 2D, expanding gives $$\omega = \frac{dV_y}{dx} – \frac{dV_x}{dy} $$Computing the vorticity in the domain gives the following field.
Three methods will be used to determine the circulation \(\Gamma \) about the airfoil.
This computation gives a circulation of \(\Gamma = 1.01\).
Warning: Don’t make this fundamental error in determining the density of moist air.
Recently, I’ve been preparing a graduate level course on Wing and Airfoil Theory. One lecture delves into modeling the real atmosphere including humid, wet, moist air. Specifically, I wanted to calculate the density of humid air. Visit my Atmosphere page for more information and tools.
[Update 1st March 2019] A moist air density calculator is at the top of this page.
My first thought was to grab a psychometric chart from my wife’s ASHRAE handbook which is available in pdf format at: ASHRAE Psychometric Chart No. 1
The chart plots lines of density, dry-bulb temperature, wet-bulb temperature, enthalpy, and relative humidity. As an example, at 90 F and 90% relative humidity, the chart indicates a volume per cubic foot of dry air of approximately 14.48. This gives a density of 0.00214 slugs per cubic foot.
However, this density is not the density of humid air. I didn’t catch the mistake until verifying my own independent calculating routine.
\(\) My derivation started with an ideal gas approximation of dry air and water vapor. The total pressure \(p\) is the summation of partial pressures for dry air \(p_d\) and water vapor \(p_w\).
$$\rho = \frac{p_d}{R_d T_d}+\frac{p_v}{R_v T_v}$$
Since the temperatures are identical, the density is:
$$\rho = \frac{p_d M_d + p_v M_v}{\bar R T}$$
Given that the relative humidity relates the partial pressure of water vapor to the saturation pressure of water pressure \(p_v = \phi p_s \), the density formula is rearranged to
$$\rho = \frac{p M_d + \phi p_s \left( M_v – M_d \right)}{\bar R T}$$
Since the molecular mass weight of dry air is \( 28.97 \frac{lbm}{lbmol} \) while water vapor is lower at \( 18.00 \frac{lbm}{lbmol} \), the density of wet air is always lower than dry air.
The reduction in magnitude depends on the saturation pressure of water vapor. For the region of interest to aerodynamics, the Arden-Buck equation provides a curve fit approximation to the partial pressure of water vapor. Converted to standard units, the partial pressure in psi given a temperature in R is:
$$p_s(T) = 0.08865 \exp\left(\frac{-0.002369 (T-8375.65)(T-491.67)}{T-28.818}\right)$$
Combining the density and partial pressure formulas gives an alternative estimate of the density of air at 90 F and 90% humidity of 0.002203 slugs per cubic foot.
So what gives? Who is correct? Why is the gold-standard ASHRAE chart suggesting a lower density? The key words are: volume per cubic foot of dry air. 
The ASHRAE chart gives the density of only the air and NOT the water vapor. The ASHRAE chart clearly states that only the dry air component is considered.
Adding in the water vapor requires finding the humidity ratio.
For the 90 F and 90% case, the humidity ratio is 0.028. Each 100 pounds of dry air has 2.8 pounds of water. Correcting the previous ASHRAE density gives:
$$ \rho = (1+0.028) 0.00214 \frac{slugs}{ft^3} $$ which is 0.002207 slugs per cubic foot.
Mystery solved.
Warning: There are websites out there inadvertently giving the wrong density of wet air. One example is denysschen.com. A single sentence would correct the confusion. On the other hand, the site at hvac-calculator.com appears to give the correct value.
I successfully retrieved my Liebelle glider from the top of a 30 foot tall tree. I recommend this method over pulling the plane down through the tree with string.
Here are the supplies:
The requirements were specific. I didn’t want to destroy the plane. The tree was too small to climb; the plane was in the top branch. The solution was to lift the plane out vertically with the pole.
Maneuvering the pole to connect the fuselage into the cup
Lifting the plane
Success!
Lessons learned: