The HSAC had a brief discussion on our Google Groups forum over the weekend, the upshot of which is that SolidWorks can apparently be convinced to act as a wind tunnel simulator, so the aero guys are going to use it over the coming weeks to get some initial data and ideas about our design. This should save us some time and effort on refining the design, which translates into fewer models, fewer failed wind tunnel tests, and more funds to put towards more fruitful research.
In an ongoing discussion on the XKCD fora, one commenter explained that if the HSAC plane really is analogous to the F-104 I wrote about last week, our turbine isn’t going to be able to generate nearly enough thrust to overcome the drag from our wing area. The technical details are below the break for those who are interested.
gorcee of the XKCD fora wrote:
At this scale, I would worry about … the ability for the engine to actually run at supersonic speeds. […] I’m not entirely positive these 50 lb thrust engines are going to really do it for you. It’s just not a matter of thrust v. weight, even though that’s a number that gets used a lot. It’s a thrust vs. wing area issue.
Take the F-104, for instance. This had a zero lift drag coefficient of about 0.017, and an induced drag coefficient of 0.048. Now, remember, drag coefficients for aircraft are based on the wing planform area, not the frontal cross-sectional area. Why is the wing area important? Recall that the drag equation is
S is your wing planform area. Your thrust has to directly counteract this drag. Our drag coefficient is determined by design, and the density of air only really changes with altitude (and that difference is not huge in your operating envelope). So for a given airspeed, we have to consider thrust vs. wing area (wing area is also involved in lift, which is why thrust v. weight is used, but for our comparisons, let’s assume a suitable is obtained). The F-104 put out 15,600 pounds of afterburning thrust (10,000 lb dry), and had a wing area of 196 ft2. It generated 9244 lb of drag at Mach 0.7
Assume you used a perfect scale model of that aircraft, you’re looking at a wing area of about 3.7 ft2. At Mach 0.7, you’d be generating 174 lb of drag, which is more than what 3 of those engines would put out. So you’ll have to substantially reduce your drag coefficient to achieve even Mach 0.7.
The upshot of all this math is that, unless we deviate from the Starfighter’s wing plan pretty significantly, we’re going to need to have a drag coefficient that is at less four times smaller than that of the F104. I’m sure there are ways to accomplish this, but I don’t know what they are. We’ll have to wait and see what the aero guys can cook up. Many thanks for the explanation, gorcee!
In other news, I’m going to begin a weekly feature on here following my workout regimen. Being selected for astronaut candidacy is challenging – this Monster article on becoming an astronaut claims that there are 3500 candidates for 20 slots every two years – and being in top physical condition is one of the selection criteria. I’ll go into more detail when the feature begins, but suffice it to say I’ve got my work cut out for me. Tentatively, I’m going to shoot for posting “This Week in Fitness” on Sundays.