Download flightlab for macOS, Windows, or Linux. Unzip it and read the Start Here file. We’ll be using this for many of the homework problems and for our project. I’ll be updating the code periodically, but every time you launch flightlab it will automatically check for and download any updates.
Open the workbench and choose New starter design. Rename the project RC-1 canonical.
RC-1 is the name we’ll give to a (deliberately medicore)
RC plane design that we’ll reuse throughout most of the homeworks.
Use the main wing as the coefficient reference.
All coordinates use x aft, y right, and z up. Horizontal surfaces are mirrored.
| Surface and role | Station 1: x, y, z, chord, twist | Station 2: x, y, z, chord, twist |
|---|---|---|
| Main wing; wing; fixed | 0, 0, 0, 0.160, 0° | 0, 0.600, 0, 0.160, 0° |
| Horizontal tail; tail; elevator | 0.565, 0, 0.010, 0.100, 0° | 0.565, 0.170, 0.010, 0.100, 0° |
| Vertical tail; fin; fixed, not mirrored | 0.560, 0, 0.010, 0.120, 0° | 0.560, 0, 0.160, 0.120, 0° |
Assign NACA 2412 to the wing and NACA 0009 to both tails. For the horizontal tail, use an elevator hinge at 75% chord and limits of ±20°. Check that the workbench reports wing area 0.192 m², span 1.200 m, mean chord 0.160 m, and aspect ratio 7.50.
| Body | Model | Length | Width | Height/diameter | x nose |
|---|---|---|---|---|---|
| Fuselage pod | streamlined_body |
0.400 m | 0.110 m | height 0.100 m | −0.130 m |
| Tail boom | streamlined_body |
0.295 m | — | diameter 0.010 m | 0.270 m |
Streamlined body uses the simple empirical drag estimation methods discussed in class. For now, in the mass menu use one placeholder point mass for the entire aircraft of 0.750 kg at x=0.048 m. In Propulsion, turn off Include battery and propulsor hardware in mass properties so the starter catalog components are not double counted. Create these 1-g cases at 1,400 m under flight cases:
| Case | Speed | Interference | Protuberance | Transition |
|---|---|---|---|---|
| Launch | 9.1 m/s | 0.05 | 0.10 | natural |
| Cruise | 12.0 m/s | 0.05 | 0.10 | natural |
| Fast | 16.0 m/s | 0.05 | 0.10 | natural |
Choose natural in the transition column. The workbench will show upper and lower transition
x/c as 1.0, meaning no artificial boundary layer trip is imposed.
You can leave everything else at their defaults.
Run Analysis for Launch, Cruise, and Fast. Trim means static equilibrium: lift equals the required weight and the pitching moment is zero. The solver is changing the angle of the elevator so that the pitching moment of the aircraft is zero (notice how the Cm curve crossse zero at the trim point). For all three cases report trimmed angle of attack, lift coefficient, pitch-control deflection, and L/D.
Take a look at the relative contributions to parasitic drag.
Below are some specs for two very different aircraft and flight conditions. You won’t be using flightlab for this problem, just hand calcs or a short script using basic relationships from the textbook.
| Quantity | RC-1 | 787-like transport |
|---|---|---|
| Mass | 0.750 kg | 230,000 kg at beginning of cruise |
| Altitude | 1,400 m | 10,000 m |
| Speed | 12.0 m/s TAS | Mach 0.85 |
| Wing area | 0.192 m² | 377.0 m² |
| Mean chord | 0.160 m | 6.27 m |
Compare Mach number, chord Reynolds number, and required level-flight lift coefficient for both aircraft at the conditions above. Briefly note which parameters are relatively similary and which are massively different between the two operating points.
An electric RC aircraft mass stays effectively constant as a battery discharges. A long-range transport, on the other hand, may burn a substantial fraction of its mass during cruise. In this problem we will investigate how that changes the desired cruise altitude.
Suppose the transport burns fuel until its mass is 170,000 kg while its geometry and configuration remain unchanged. If it stays at 10,000 m and Mach 0.85, calculate its new required lift coefficient.
A simple idealized cruise schedule instead holds Mach number and lift coefficient at their beginning-of-cruise values. Using the ideal-gas and speed-of-sound relationships. you can show that dynaic pressure is
\[q=\frac{1}{2}\rho(Ma)^2=\frac{\gamma}{2}pM^2,\]where $\gamma$ is the ratio of specific heats and has a value of 1.4 for air. Thus, at fixed Mach number, dynamic pressure is proportional to atmospheric pressure. Calculate the altitude at the end of cruise when the mass has decreased to 170,000 kg (in order to maintain the initial lift coefficient).
Consider why a real transport follows discrete step climbs rather than a continuous ideal schedule.