Monday, November 26, 2018

Are jet engines high performance?

 Are jet engines high performance?  Seems a silly question.  FAR 61.31(f)(1), seems pretty straightforward: have an engine with more than 200hp?  It's high performance.  200 or less?  Then it's not.

But I've had a few jet jockeys argue that jets (and they were all talking about transport-level aircraft like 737's and larger) are not high performance because the engines are measured in pounds of thrust and not horsepower.

Meaningful distinction?  No.  Just different ways of measuring how much "oomph" an engine has, but they can be translated from one to another.

Here's the short version:

Power is Energy Delivered over time.  200 horsepower is equal to 149,200 Joules of energy / second (J/s).  If an engine is delivering 149,201 or more J/s, then it is above 200hp, and thus high performance.  The kinetic energy of the aircraft in Joules is (1/2)mv2 (m in kg and v in meters/second) so I leave it as an exercise to the reader to determine if a Boeing 737 on a takeoff role is accumulating more than 149,200 J/s.  (Hint: the answer is "yes.")

And here's the longer version:

Now bear with me here for a little math.  I'm embarrassed to admit that I actually have a degree in physics from MIT, but since I haven't used it in 30 years I need to think about this.  But with that disclaimer...

Pounds of thrust is a measure of force.  (That's actually the difference between mass and weight: weight is the force of gravity exerted on a given mass, which is why your weight is different on, say, the moon, but your mass is unchanged.)

Horsepower is a measure of power, and power in turn is the rate of energy transfer.

These are different, but related quantities.

Energy (or "work") is force multiplied by distance.   I.e.:

E = fd

Deliver that energy at a constant rate over a period of time to compute the power involved:
P = E/t = fd/t

Sanity check: 1 horsepower is 550 "foot-pounds per second" - i.e., foot (distance) x pounds (force) divided by time.  So the units here check out.

But alas, d/t is distance over time - a velocity!  So, power in relationship to speed is force times speed (v):

P = fv
So now we can relate horsepower and force: they're related by the speed.

But wait, that doesn't make sense.  A piston engine is delivering a given amount of power regardless of speed.  And a jet engine is delivering a given amount of thrust regardless of speed.  So what gives?

Let's look in the power (piston) world first.  At some given speed (v), your engine is delivering energy to the air around the aircraft (not directly to the aircraft!) at some rate of energy per unit of time, aka "power" P.  This delivery of energy, by Newton's third law, creates an equal and opposite reaction, which causes the engine (and thus the plane to which it is hopefully attached!) to move forward.

By the math above, your effective force f is P/v.  Of course, at rest (v = 0) that's an infinite instantaneous force, but that "infinite" force is for just a split second, and the aircraft starts moving.* More to the point, since we're in the "power" world rather than the "force" world for the moment, the power delivered by the piston engine is adding energy to the aircraft, which causes it to move.

As the aircraft absorbs energy, its speed increases until the engine is putting energy ("kinetic" energy - energy of motion) into your airplane at precisely the same rate that friction** is removing energy from the system; at that point, the total energy of the aircraft stops changing.  Kinetic Energy is (1/2)mv2, which is basically a reflection of the mass and the speed.  Sure, as you burn fuel, you get lighter, but over small time periods that change is insignificant.  Therefore, if your net energy is unchanging, the speed is essentially constant.

The same thinking applies in the force (turbine "pounds of thrust") world.  The power delivered to the aircraft is P = fv as described above.  So your jet engines spool up and deliver some amount of force.  While the aircraft is at rest (v = 0), the energy delivered to the aircraft is 0, so the effective instantaneous power is obviously 0***.  But alas: apply a force to a mass and v changes from zero to non-zero - the plane acquires energy (1/2 mv2) and begins to accelerate, absorbing more energy.

Delivery of energy over time, of course, is the very definition of power, so this can be measured in horsepower, if you like.  But this time we're in the "force" world: the thrust of the engine provides a force against the aircraft, causing it to start moving.  As the plane accelerates, so do the opposing forces of friction, draining away some of the energy that is being added by the engines.  (And, of course, as you climb, you are creating lift, which means creating additional force to oppose gravity).

Eventually, friction creates as strong a force opposing the engine's force (and you stop climbing as well): equilibrium of forces is achieved and the aircraft reaches a stable velocity, neither accelerating/decelerating nor climbing/descending.

The key thing to note is that the two examples I give below are PRECISELY the same, just looking at it from the point of view of power or force.  The only difference is that I labeled one "turbine" and one "piston."  I could have just as easily swapped those labels and everything would still work out just fine.

So this was a long and geeky way around to my esoteric point that "pounds of thrust" and "horsepower" are tightly coupled, related by the mass and speed of an aircraft.

More importantly, all but the smallest experimental/hobbyist jet engine is absolutely going to deliver more than 200hp of power to the aircraft to which it is attached.

And thus, any aircraft with such an engine qualifies as high-performance.

*OK, perhaps worth clarifying: imagine your aircraft is chained in place.  It won't move, period.  The engine is putting energy into the surrounding air, so this results in a force being applied to the aircraft.  The chain, however, exerts an equal and opposite force, so the aircraft doesn't move.  As a result, the air absorbs all of that energy; the chain is effectively shunting the energy that would have gone into the aircraft into the air, so the air behind the aircraft is moving a bit faster than it would have been if the plane were moving.

**If you're climbing, then some of that kinetic energy is also being converted into potential energy, providing another drain on the energy that the engine is putting into the aircraft.  So properly the equation here is that kinetic energy is added to the aircraft by the engine and is removed by friction and increases in potential energy.  If energy is being added faster than it is removed, the aircraft speeds up; if it is added more slowly than it is removed, the aircraft decelerates.

***That's actually power to the aircraft. Energy is conserved, so if it isn't going into the aircraft, it's all going into the surrounding air. Effectively, the power is thus conserved as well.

Saturday, November 17, 2018

Solo flight and commercial ratings progress

Solo time is pretty straightforward, right?  If you're the only one in the airplane, then it's solo.  Anyone else aboard, then it isn't (and shouldn't be logged as such).

But as you advance in your ratings from private to instrument to commercial, you start to fly more sophisticated and more expensive aircraft.  Building solo time can be difficult due to insurance requirements around minimum experience, or due to the lack of a appropriate ratings (e.g., working on a multi-engine rating after earning commercial in a single-engine airplane).  If a student simply doesn't have the requisite level of experience, then they can't rent from an FBO, placing a burden on getting solo experience.

For this reason, the FAA has created a substitute for solo in the requirements for flight experience for a commercial rating (61.129).  The wording is that you must either be solo or "performing the duties of pilot in command...with an authorized instructor on board."

To indicate this sort of a flight on MyFlightbook, you should log the flight and attach the "Duties of PIC" (DPIC) property and the "Instructor on Board" (IOB) to the flight.  (I'll refer to this below as "DPIC/IOB" time)

There is often confusion about this, and I have on more than a few occasions received an email complaining that MyFlightbook is not crediting all of the time it should be crediting, or is missing their long cross-country flight.

The idea is that this is supposed to emulate solo flight, so the instructor is there not to teach you, but rather to satisfy insurance restrictions.  You're doing all of the work as if you were in fact in solo flight.  The instructor, therefore, should essentially be sitting on their hands for the flight.  The mistake people typically make here is that they read "instructor on board" as meaning "I can log dual." 

Unfortunately, this is a misreading of the regulation.  The normative legal interpretation from the FAA can be found in the 2014 FAA "Kuhn" interpretation (or Google the phrase "FAA duties of PIC Kuhn"): you cannot simultaneously receive instruction and count flight towards 61.129(a/b)(4).

The money quote from the Kuhn letter is "Because this flight time is a substitute for solo flight time, the pilot is not receiving instruction and therefore cannot log this time as dual instruction received."

For this reason, MyFlightbook subtracts out any dual instruction that is logged from the amount of  DPIC/IOB time to determine how much time can credit towards 61.129's requirements.

Note too, that since this is a substitute for solo flight, you must not log it as solo either (after all, you're not alone in the aircraft).

MyFlightbook computes a flight's contributions towards solo requirements as follows:
  • Compute an effective DPIC/IOB time as
    MIN(DPIC/IOB, Total Time - Dual Received)
    So, for example:
    • A flight with 2 hours of DPIC/IOB, 2 hours of total time, and 2 hours of Dual yields
      MIN(2, 2 - 2) = MIN(2,0) = 0.
    • A flight with 2 hours of DPIC/IOB, 2 hours of total time, but only one hour of instruction, yields
      MIN(2, 2 - 1) = MIN(2,1) = 1 hour.
      I.e., I'm removing the hour of instruction per Kuhn, but the other hour can count as DPIC/IOB.
    • A 2 hour flight with 60 minutes of instruction and 90 minutes DPIC/IOB yields
      MIN(1.5, 2 - 1) = MIN(1.5, 1) = 1 hour.
      I.e., the flight is ambiguous, but at most one hour of this flight could possibly contribute time here, so I'll give you the hour.
  • Compute an effective solo time of
    MAX(explicitly logged solo time, effective DPIC/IOB)
    This value is what is used for total solo time requirements.  Note that this also implicitly corrects for accidental logging of both DPIC/IOB time and solo time.  I suppose it's possible that you have a flight where you fly with an IOB for a while, then the instructor gets out and you fly solo, but I also suspect that's pretty rare, since usually the whole reason you had to fly with an instructor on board is precisely because you can't fly solo.
  • Compute an effective night VFR solo time of
    MIN(effective solo time computed above, Night - IMC)
    (Note that this could be a bit conservative: e.g., if you had a 2 hour solo flight where the first hour was in day IMC and then it cleared up and the 2nd hour was at night in VMC, then the flight won't credit.  But that's a bit of a corner case and it means that the error is conservative - better to understate than overstate experience!)  Anyhow, this value is what is used for any night solo time requirements.
  • For any activities such as takeoffs/landings or a long cross country that must have been performed while solo, add them to your experience if the effective solo time is greater than 0.
  • Finally, the various 61.129(4) requirements computed using both actual solo time and the effective substituted solo time.  After all of your flights are examined and the experience is summed up, MyFlightbook will report your progress using either exclusively actual solo time, or exclusively DPIC/IOB time (whichever has more time recorded), but not both.  This is because you cannot mix-and-match: all of the solo time in 61.129 MUST be actual solo time or it must all be DPIC/IOB time (see the FAA's 2016 "Grannis" interpretation for details). 
All of the above, of course, are in the context of 61.129, and thus only apply in determining your progress towards meeting the experience requirements therein.  MyFlightbook's 8710 form, on the other hand, is a summary of experience without the context of any particular rating.  As such, the 8710 form does not include the DPIC/IOB substitution, it only looks at solo time that is recorded as such.

So by all means, go out and build your "solo" time with an instructor on board.  But don't log it as instruction received.