Copyright ©2004 by Paul Niquette. All rights reserved. 
very pilot, it seems, will eagerly express his or her own opinion about how to deal with this puzzle. Aviation books and websites provide a wide selection of empirical solutions. A focus group with aviators might go something like this...
They cannot all be right. Let's get to work (if you don't dig trig, you can skip on down to here). Figure 1 depicts a horizontal view of an airplane in straight and level flight. It is being acted upon by four forces (shown as vectors, each having a direction as indicated by the arrows and a magnitude expressed here in pounds). By inspection of Figure 1, we see a balance of forces such that for straight and level flight, L = W, which keeps the plane aloft, and T = D, which keeps the plane moving through the sky at a steady airspeed v. That's airspeed, not groundspeed. The aerodynamic forces are proportional to the square of airspeed The most fundamental attribute of an airfoil is its "lifttodrag ratio" ('L/D', pronounced rather hurriedly 'elloverdee') and for a typical singleengine plane, 6 < L/D < 11. Figure 2 provides a horizontal view of an airplane in a left turn, still in level flight. To remain aloft, W must now be balanced by the vertical component of L, which calls for a value of L itself larger than W. The expression "Requisite Lift" gives emphasis to this reality. When b = 0^{o}, cos b = 1 and L = W, as we saw in Figure 1. At any other angle of bank, cos b < 1, which makes L > W. For example, when b = 60^{o}, cos b = 1/2, and L = 2 W. The airplane, along with the passengers inside, "feel" twice as heavy as when they were taxiing for takeoff.Vectors can be simplified by strategic selection of vantage point. Thus, in Figure 2, neither D nor T can be "seen" (T points straight out of your screen, and D points straight in). That allows us to exclude D and T in the analysis below. By inspection of Figure 2, we see that for a turn in level flight, L = W / cos b. Meanwhile, the Lift vector L is manifesting a horizontal component, L sin b, which produces a turning maneuver  a circular flight path, creating a balancing Centrifugal Force F = L sin b. Sophisticated solvers don't bother to distinguish "centrifugal force" (fugitive from the center of a circular path) and "centripetal acceleration" (that which is capable of drawing a mass toward the center), the equivalence having been settled three centuries ago by Newton's Second Law of Motion.o solve the Pilot's Nightmare puzzle, we must first figure out how long it will take to turn back to the airport at a given angle of bank b. For that we need to ascertain the RateofTurn w (a typographic approximation of omega, the traditional symbol for "angular velocity"). Elementary physics teaches variations of the expression F = W/g w v. The denominator g is the acceleration of gravity (32.2 feet per second per second), such that W/g represents the mass of the airplane flying a circular course at tangential velocity v (in feet per second) and at rateofturn w (in radians per second, where radians = 180^{o}). Since L = W / cos b and F = L sin b, then F = W sin b / cos b = W tan b. Since W/g w v = W tan b, the value of W gets canceled out. Solving for rate of turn, we have w = g tan b / v radians per second.
To turn back toward the airport and line up with the runway, the gliding flight path will require more than 180^{o} of turning. As suggested in the sketch, your strategy calls for 270^{o } in one direction (left, say) plus 90^{o} in the opposite direction (right) for a total of 360^{o} of turning, or 2 radians. That maneuver will require 2 / w seconds, and during all that time the plane will be losing altitude.n the expression w = g tan b / v, we find out merely what every pilot knows  that for a given angle of bank b, there is reciprocity between airspeed v and rate of turn w: The faster the plane flies the slower its rate of turn. "High speed maneuverablity" is an oxymoron. To make a quick turn, you must slow the airplane as well as bank steeply. There are limits, however... Fly too slow and the plane stops flying. Oh right, and as b approaches 90^{o}, cos b approaches zero. Look what that does to the requisite Lift. Since L = W / cos b, the force exerted by the wings on the airplane  and therefore the apparent weight of passengers inside  will become astronomical.The Pilot's Nightmare puzzle postulates an engine failure during flight. Without propulsion, of course, a plane will be gliding  but not necessarily diving (moviemakers take note). Figure 3 depicts a horizontal view of an unpowered airplane gliding in a straight line at airspeed v with its Direction of Flight sloping downward at Angle of Descent a. As with v in Figure 1, a is relative to the local parcel of air in which the plane is flying, not necessarily with respect to the ground (with a headwind equal to v, for example, the plane would be descending vertically). The Pilot's Nightmare puzzle, by the way, assumes landing in a tailwind, but as shown in Figure 3, the plane has Vertical Descent Rate u = v sin a (in feet per second), where u does not depend on relative wind. Whatever the windspeed, at a given airspeed, the larger the value of a the faster will be the rate of descent u and the sooner the plane will run out of altitude. Elementary trigonomery reveals that with the appropriate value of a, there can be a component of Weight aligned with the Direction of Flight, W sin a, which will exactly balance aerodynamic Drag D to keep the plane in motion. That component of weight acts as a replacement for engine Thrust T seen in Figure 1. In a gliding maneuver, there will still be a component of Weight that acts perpendicular to the wings W cos a, and that must be balanced by L. "Requisite Lift" appears again in Figure 3 to give emphasis to this reality. As you might expect, there is an optimum airspeed for gliding:
By inspection of Figure 3, then, we see that L = W cos a, and since cos a ? 1, L is less than the aircraft weight W. The previous sentence does not have an exclamation point. It appears instead in the answer to the following question: How can the plane stay aloft if Lift is less than Weight? Aha, there is a vertical component of Drag D that actually helps to hold the airplane up! The casual solver will note in passing that when the Angle of Descent a approaches 90^{o}, cos a approaches zero making Lift totally useless for keeping the plane aloft. Meanwhile, since sin a approaches one, D approaches W, and the plane would be executing a rather steep dive at an airspeed called, somewhat ominously, "terminal velocity."It may not be obvious but if you look closely at Figure 3, you will see that in a poweroff glide at v_{0}, the lifttodrag ratio L/D = cos a / sin a or L/D = 1 / tan a. A gliding plane with an L/D = 10 will therefore have an Angle of Descent a of about 6^{o}. If the plane has a stall speed of, say, 50 nautical miles per hour (kts), minimum drag speed would be about 65 kts (v_{0} = 110 feet per second). The Vertical Descent Rate calculated as u = v_{0} sin a will be about 12 feet per second. From 1,000 feet AGL, the pilot would have not quite a minute and a half (83 seconds) before the flight is concluded. Fine, but that assumes no turning.Figure 4 depicts the situation postulated in the Pilot's Nightmare puzzle  a gliding turn. As in Figure 2, we have made a strategic choice for our vantage point. Our headon view this time, though, is looking "uphill" along the descending direction of flight. Thus, neither D nor W sin a (the deadstick replacement for T) can be "seen" in Figure 4. In place of W, which always acts downward vertically, Figure 4 shows W cos a, which acts downward perpendicular to the wings as derived in Figure 3. The Lift vector L, which acts perpendicular to the wings, is shown to have a requisite value L = W cos a / cos b. That's what it takes to keep the plane aloft. You will notice that whenever b > a, cos b ? cos a, and L > W much as we saw in Figure 2. Gliding, say, at Angle of Descent a = 6^{o} and at an Angle of Bank b = 50^{o}, the requisite Lift L = 1.5 W. The wings are carrying about half again the gross weight of the airplane, which raises the question, how does the pilot increase the value of L? Answer: increase airspeed, of course. But how much?As described above, L increases according as the square of v. To compensate for banking, we observe that the requisite Lift must be increased from its straightflight value, L = W cos a, by a ratio that looks like this: (W cos a) / (W cos a cos b) or more simply 1 / cos b. Airspeed, then must be increased from its straightflight value v_{0} by a ratio that looks like this: v^{2} / v_{0}^{2}. Thus, for a gliding turn, the requisite airspeed will be given by v = v_{0} (1 / cos b)^{1/2 } (feet per second). Hmm, this determination turns out to be independent of the Angle of Descent a as well as W (a counterintuitive reality, wouldn't you say?). ithout an engine, the only way to increase v might seem to require steepening the angle of descent a. Doing so will increase the rate of descent u all right, but that would happen anyway just as a result of increasing v. In fact, since D varies with v^{2}, the force of Drag will increase by the same ratio as Lift, 1 / cos b. Pilots are cautioned in early flying lessons that the speed at which a plane will stall increases with angleofbank by that same ratio; so too will the minimum drag speed increase. That means, the pilot can maintain the same angle of descent a in the gliding turn as for straightflight, while the rate of descent u increases with v. How's that for a counterintuitive reality? Applying simple algebra, we are now able to derive the following two expressions: w = (g / v_{0} ) (cos a sin b) / (cos b)^{1/2} (radians per second)Thus, the time to complete the turning maneuver (270^{o} + 90^{o} = 360^{o }= 2 radians) will be given by 2/ w = 2v_{0 }(cos b)^{1/2} / g cos a sin b (seconds). Multiplying that value of maneuvering time by the rate of descent u gives us the altitude lost in executing the gliding turn. Let's call that h. h = 2v_{0}^{2 }tan a / g sin b (feet of altitude)Here are specifications for my favorite airplane as exemplars for our calculations:
So, then,
using that formula for
h, what angle of bank would
you
choose? The following table gives h
for the values
of b offered in the Pilot's Nightmare
puzzle:
At b = 15^{o}, you will take over 80 seconds to complete your gliding turn and will lose over 900 feet. You can cut that altitude loss almost in half by choosing 30^{o}  and half again at 60^{o}. Obviously, the greater the value of b, the lesser the value of h. You might as well, therefore, just go ahead and roll your plane on over to a ^{o }angle of bank. You will complete your maneuver in a little over 11 seconds, with an altitude loss of only feet. Watch out, though... Any maneuver using a bank angle greater than 60^{o} is officially classified as "aerobatic" and can do structural damage to most models of light aircraft. A puzzlesolver might not be expected to know that, perhaps, but it is required knowledge for all pilots  and for good reason: At a 75^{o} angle of bank, your plane and passengers would be pulling upwards of 4 gs. Your banking stall speed will be 100 kts  twice that of level flight. The minimum drag speed you must carry will be 130 kts (215.5 feet per second). So maybe you are not about to stall; however, truth be known: The structure of the airframe would actually be less challenged if your plane did stall, thus releasing the "Grasp of Bernouli." A stalled wingfoil therefore makes sure that your airplane will arrive at the crashsite...in one piece.Best check your aircraft owner's manual before horsing around in the sky at a 75^{o} angle of bank. Besides, choosing 60^{o} only costs you 272 feet in altitude loss. Oh, but there's more to think about. Much more... ne of aviation's many counterintuitive aspects is  tahdum  "adverse yaw." For surviving the Pilot's Nightmare you must execute a gliding maneuver comprising two turns in order to line up with the runway of departure. Now, an airplane cannot change its bank angle instantaneously. The pilot applies aileron forces to institute a rolling motion, but  hey, the next sentence deserves its exclamation point: While an airplane is actually rolling, it is not turning! That's right, it is flying straight, as a consequence of adverse yaw, which, for a coordinated turn, must be balanced by applying opposite rudder. Obviously, the greater the value of b, the longer the rolling takes and the longer the turning is postponed. For our purposes here, let us assume a rather generous rollrate of 15^{o} per second (for a slowly gliding plane, remember). Rolling the plane from wings level into a bank angle b of, say, 60^{o }for the course reversal will take you about four seconds. The opposite turn requires first rolling the wings level and then rolling into the opposite bank angle to get lined up with the runway  another eight seconds. Finally, there's rolling the wings level before touchdown, which takes yet another four seconds.
y_{1 }= (b/15)(v + v_{0 }) / 2, which assumes climbing at minimum drag speed v_{0 };The total altitude lost while balancing adverse yaw can be estimated as (y_{1 }+ 2y_{2 }+ y_{3}) tan a. Figure 5 also shows cumulative horizontal heading changes, which turn out (no pun intended) to be less than the total of 360^{o} (2 radians) you must use if there were no such thing as adverse yaw (as shown in the puzzle sketch). What you want to do is take the shortest path back to the airport on a course shaped like a teardrop. By simple trigonometry, we find that there is an angle given by... = arc tan (r / y_{2 }).By inspection of Figure 5, we learn that the first turn requires a total of + radians and the second turn requires radians in the opposite direction for a total of + 2 radians to bring the plane back to the airport lined up with the runway. Finally, that distance beyond the end of runway available for climbing x is given by... x = 2 (r sin + y_{2 }cos ) + y_{3 } y_{1}Taking into account the effects of balancing adverse yaw, then, we see in this tabulation...
...that the solution to the Pilot's Nightmare puzzle is b = ^{o}  decidedly not b = 75^{o}. ut wait. Have a look at Figure 6 on the right. The elevation view depicts your loss of h feet in altitude while reversing course and lining up with the runway. Figure 6 shows another reality  that you still have to get back to the airport while gliding at Descent Angle a (the dotted red line). The longer the runway the better  net of your takeoff roll (the unused runway is, so to speak, "reaching out" to receive your crippled craft). Your altitude at any given point during the climbout will depend on your Climb Angle c along with the distance flown before enginefailure (the green line). As shown in Figure 6, c better be greater than a or you'll never make it. To determine c, you will need to know your airspeed v and rate of climb u. By the way, wind helps your cause by making your angle of climb steeper and your angle of descent shallower  and by shortening your takeoff roll. However, wind also increases your groundspeed at touchdown. Accordingly,
there are plenty of
technical factors beyond those addresed in solving the
Pilot's
Nightmare puzzle  factors that may convince
you not to execute the coursereversing
maneuver.

Epilog
The author of the Pilot's Nightmare puzzle does not claim credentials for offering advice to aviators; moreover, the solution above, while drawing upon solid principles, is necessarily incomplete. A multitude of parameters must be considered and calculated for each aircraft, for each airport, for each flight  all for an eventuality that might not ever arise in a lifetime of flying. Nevertheless, here are a few piloting policies suggested by the solution to the puzzle. While planning each flight, estimate a and c, h and x (see analysis above) for the aircraft, airport, and wind conditions.
2. Always decline a midfield departure clearance; position the plane for use of the whole runway. 3. Use the Short Field Takeoff procedure as specified in the aircraft owners manual (AOM). 4. Always set trim for Best Angle of Climb speed as given in the AOM (see 8 below). 5. Retract landing gear as soon as climb is established. 6. Retract takeoff flaps when reaching runway elevation plus h (see 1 above). 7. Scan diligently ahead during climbout for a suitable emergency landing site (ELS). 8. Upon identifying a suitable ELS (see 7 above) retrim for CruiseClimb per AOM.
...between Steps 7 and 8, execute coursereversal maneuver at a 45^{o} angleofbank b. ...after Step 8, execute emergency landing at suitable ELS (see 7 above). 