[Phys-l] Kepler problem in disguise

[John Denker <jsd@av8n.com>]

Hi Folks --

Here is a little bit of physics that you may find
amusing.

I could tell the story backwards, starting from
the theory ... but that would make me sound much
smarter than I really am, so let me tell it the
way it really happened.

This is a genuine real-world problem that came to
me in my capacity as a flight instructor.  In
particular, it came from a guy who is preparing
to get his own certified flight instructor (CFI)
certificate.  The CFI checkride covers the same
maneuvers as the commercial pilot checkride,
except that the candidate flies from the right
seat and must _explain_ each maneuver as he
goes along.  There is also a long, detailed
preflight oral exam.  The FAA takes issuing CFI 
certificates very seriously, as well they should.

The topic for today is a maneuver called "turns
on a pylon" which is roughly half of the maneuver
"eights on pylons".  You start by choosing a
reference point on the ground (the pylon).  Then
draw an imaginary line from one wingtip to the 
other, that is, a pointer parallel to the 
conventional "Y" axis of the airframe.  The 
objective of the maneuver is fly around the pylon 
while keeping the pointer pointed at the base 
of the pylon.  Another requirement is to maintain 
coordinated flight, i.e. zero slip angle, i.e. 
airflow aligned with the X-axis of the airframe.

For simplicity, we assume the aircraft moves 
with constant airspeed.  This is usually a good 
approximation.

The requirements heavily constrain the problem.
At any given point in space, the heading and the
bank angle are constrained by the requirement to
keep the wing pointed at the pylon.  In the case
where there is no wind to worry about, it is 
easy to show that there is only one altitude at
which the maneuver can be flown.  This is called
the _pivotal altitude_.

For homework, show that the pivotal altitude is
given by V^2/g, where V is the true airspeed.

In the no-wind case, the ground track is a circle,
with the pylon in the middle.

====

The maneuver gets much more interesting in the
presence of wind.

Explaining the maneuver is complicated by the
fact that the explanations you find in books
(excluding books written by me) are very, very 
incomplete and/or outright wrong.

To figure out what was going on, I did some
experiments using a real airplane, then went
home and wrote a computer program to simulate
the maneuver.  For instructional purposes I
replaced the program with a spreadsheet, 
since students seem to be less intimidated
by spreadsheets than by c++ programs.  Also,
even though doing sophisticated graphics in 
a spreadsheet is hard, doing simple graphics
is easy.  See
  http://www.av8n.com/fly/img48/turn-on-pylon.png

One thing that leaped out at me is that in
the presence of wind, the ground track is an 
ellipse with the pylon at one focus.  The long 
axis of the ellipse extends in the _crosswind_ 
direction. 

   This often comes as a surprise to the 
   student -- and to the FAA inspector who 
   is examining my CFI candidate -- because 
   most books show the pattern as an oval
   extending in the upwind/downwind direction, 
   with the pylon in the middle or perhaps
   near the upwind end of the oval.

According to my simulation, the ground track
really is an ellipse.  The simulation is
accurate to one part per million.  This 
motivated me to look more closely at the
theory;  see below.

First, though, let us look more closely at the 
ground track.  In particular, let us consider 
the radius vector (from pylon to airplane) and 
project that vector onto the ground.  The
projection sweeps out equal areas in equal
times.

This should come as no surprise, because we
have a _central force_.  Any central force
conserves angular momentum, and conservation
of angular momentum implies equal times (and
vice versa).

This means that the groundspeed is higher when 
the airplane is near the pylon, and lower when
it is farther away.  If you look at the figure,
this makes sense.  It is also true, by geometry,
that when the airplane is near the pylon it is
more steeply banked.  These two facts qualitatively
agree with each other, since it is well known that
when your groundspeed is high you need a steeper 
bank in order to make a turn of any given radius.

However, the geometric effect mentioned in the
previous paragraph is not sufficient to make
the maneuver work out right.  In fact, when
the groundspeed is high, you need even more
bank than the geometry of the ground track can
supply.  To make things work out right, you
need to fly at a higher altitude, higher than
the no-wind pivotal altitude, so that the wing
(still pointing at the base of the pylon) will
be more steeply banked.  On the other side of 
the same coin, at the upwind end of the ellipse, 
where the groundspeed is lowest, you need to 
fly lower than the no-wind pivotal altitude.  
All in all, it's a complicated heavily constrained 
three dimensional problem.

Note that this analysis uses two different
reference frames.  Some velocities are measured
relative to the ground (groundspeeds) while other
velocities are measured relative to the airmass
(airspeeds).  This is a trap for the unwary.  It
is however natural to the problem, since the zero
slip condition is naturally expressed relative to
the airmass, as is the assumption that the airspeed
does not change;  meanwhile the requirement that
the Y-axis points at the pylon naturally applies
in the ground-based reference system, as does the
observation that we have a central force.

It is not wrong, just tricky, to use two different
reference frames.  You just need to be careful
when switching from one frame to the other.

==

I was surprised to see that this turned out to be
exactly equivalent to the Kepler problem.  I didn't
see that coming.  It would be tempting to pretend
I was smart enough to see it, but I wasn't.

I could see that we had a central force, so we would
have conservation of angular momentum and the equal
areas law.

I did not see the perfect ellipse coming.  For one
thing, we don't have an obvious 1/r^2 law.  Only 
after solving the problem was it apparent that we
got one factor of 1/r from the variation of distance
in the ground track, and another factor of 1/r from
the variation in altitude.  I had to to solve the
problem, including finding the ellipse, before I
knew how the altitude varied.

In particular, it was not obvious how the constant 
airspeed could be reconciled with the changing
groundspeed necessary to uphold the equal-area law.

Remark:  If it's true in in this one case, it must
be true in general that a Keplerian groundspeed
in one frame corresponds to a constant airspeed
in some other frame.  This must apply, somehow,
even to cosmology problems, even to the highly 
elliptical orbit of Halley's comet.

At this point I flipped open my copy of
_The Mechanical Universe_ volume I (Frautschi,
Oleneck, Apostol, and Goodstein) which presents
a somewhat unusual way of solving the Kepler
problem.  There's a fine line separating weird 
and tricky from elegant, but I consider this one 
to be on the elegant side of the line.

They start by showing that

  r^2 (d/dt) theta = const = L/m       [1]

where L is the angular momentum and m is the mass
of the orbiting particle.  Later they use this to
eliminate r^2 in favor of (d/dt) theta in the
equation of motion.  After a few steps this gives
us 

  v/u = theta_hat + e j_hat           [2]

where theta_hat is a unit vector in the tangential
direction;  j_hat is a unit vector in the Cartesian
"j" direction;  and u is a characteristic speed in 
the problem, namely GMm/L, which basically tells 
us how much energy per unit momentum we started 
with.  For now e is little more than an arbitrary
constant of integration, but its significance will 
be apparent shortly.

I've never seen equation [2] elsewhere.  I tried
without success to find it online just now.

After a few more steps, equation [2] yields the
equation, in polar coordinates, for a Keplerian
ellipse of eccentricity e.

> From equation [2] we can see that it really is
true in general that if an observer Moe flies
past an orbit at speed e u, in the appropriate
direction, then in Moe's frame the orbiting 
particle has a constant speed |v| = u.

There some folks on this list who are experts
in mathematical physics in general and conic
sections in particular who may recognize this
as a "well known" property of Keplerian orbits,
but it was a new one on me.  In any case, it
seems worth knowing.  It might come in handy
some day.

======================

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