Toll Free Order Line: 1-866-247-4568
Welcome to iPilot, please Sign In or Register

CHART SUBSCRIPTION

TOP PRODUCTS
WEATHER

 

If you're just starting the process or Learning to Fly or a veteran looking for an online resource to continue your education, you've come to the right place. Our expanded learning section has features for everyone!

Trivia Testers

The world's smallest jet engine (whether turboprop, turbofan, or turbojet) used in a "real" passenger carrying aircraft weighs in at...

Oooh, dad! Can I get one of those for my bike?
The world's smallest jet engine (whether turboprop, turbofan, or turbojet) used in a "real" passenger carrying aircraft weighs in at

  1. 19 grams
  2. five pounds
  3. 75 pounds
  4. 225 pounds

Answer: C, and if you stretch things a bit, B. Although there are very tiny engines, such as the Jet-X (choice A), which produces six ounces of thrust (and which is, strictly speaking, really a rocket motor); the JF-35 Nano-Bee, which has about 2.5 Kg of thrust; the AMT Olympus (choice B), which yields about 43 lb thrust; or the Cobra turbojet which at 6.2 pounds, produces 30 pounds of thrust, these are all almost always used by model aircraft enthususiasts. (And for those purists out there, yes, I know that pounds and kilograms are not, strictly speaking, the proper units of thrust, and it should be lbf or Newtons.) The TRS-18 used by the 12 foot long Bud Light microjet, which produces 360 lb of thrust and propels it to speeds of about 300 knots, is about a foot wide and two feet long, and weighs 75 pounds. In comparison, the next smallest jet engine that is (or would have been) used on a passenger aircraft is Williams International's EJ-22 (aka FJ-22) for the Eclipse 500 (which Eclipse recently dropped, due to its having had numerous problems). It weighs in at about 85 lb, and produces about 770 lbf of static thrust. (It came from their FJX-2 engine, which is about 100 pounds in weight.) After that, the lightest commercially produced "real" jet engines may be the F107 or F112s used in ALCM and Tomahawk missiles; they weigh about 146 pounds. If there were a smaller jet engine on say, an ultralight, it would need to produce somewhere around 50 lbf thrust. Actually, a French pilot has installed two AMT Olympus turbines in his Cri-Cri (Cricket) "ultralight". The original design has an empty weight of about 180 pounds, so by FAA CFR Title 14 Part 103 definition at least according to weight, it would be an ultralight (although it's actually not, because it exceeds the maximum 55-knot speed by quite a bit). Each engine has an installed total system weight of about seven pounds, and provides a thrust of about 43 pounds. The pilot, Nicolas Charmont, has achieved a top speed of about 150 mph (and he gets 100 mph with one engine).

Looking sharp, moving fast
Hypersonic vehicles do not generally have sharp-edged wings or noses. Why not?

  1. Shock waves above Mach 5 are parabolic, not angular, and so they don't need to be pointy. Blunter structures are stronger, not to mention cheaper to make.
  2. Sharper and "pointier" objects have proportionally greater surface area to volume ratios. Therefore, they have more area to heat up, and less volume with which to absorb that heat. Even inconel, with a 1370 deg C melting point, cannot hold up to hypersonic flight when fashioned into something sharp, skinny, and pointy. That's why the front end of the Space Shuttle looks about as sexy and suave as a sea manatee.
  3. They can be sharp and spiky, actually. That surface area to volume stuff is fuzzy thinking. The more surface area something has per unit volume, the easier it is for it to radiatively dissipate heat.
  4. Surface insulation tiles, like the curvy brick jacket of silica fiber tiles sported by the Shuttle, can't take load deformation, which is sure as heck what they'd get if they stuck out into a Mach 5 breeze.

Answer: B. There must be enough material to absorb heat. A pointy object would be much more likely to get so hot it that would just burn off.

Vector analysis isn't just tedious. Why, it can be exasperating, too!
In a constant rate climb, the lift produced by an airplane's wing is

  1. exactly equal to its weight
  2. greater than its weight
  3. less than its weight
  4. equal to its weight, but only at L/Dmax.

Answer: C. In any sustained climb, the lift produced by the wing is actually less than the weight of the airplane. How can this be so? In unaccelerated straight and level flight of course, lift equals weight. Now take an F-16, climbing at a steady rate, straight up. The wings are contributing nothing; that GE or Pratt and Whitney is doing all the work. Okay, now take your Beetle Bomb Bug Smasher, climbing like a scalded ape at a dizzying climb angle of, oh, six degrees. Does the same thing hold true? Yup. When you pull back on the controls, whatever thrust is coming from that prop, multiplied by the sine of your climb angle (which would work out here to about one tenth of it) is what's helping hoist you skywards. (So now only nine-tenths of that thrust is left to counteract drag forces, and guess what? You slow down.)

Basic Membership Required...

Please take a moment and register on iPilot. Basic Memberships are FREE and allow you to access articles, message boards, classifieds and much more! Feel free to review our Privacy Policy before registering. Already a member? Please Sign In.

Topics