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Assume that a 747 is flying at cruising altitude and speed. If the pilot decided (for a presumably good reason), to turn the plane 180 degrees,

  1. How quickly could the turn be completed?
  2. How wide would the turn radius be?
  3. How much altitude would be lost, if we're not assuming a level turn?

EDIT: I learned something new from both of the great answers below. I accepted the one that was easier to follow. Since both answers assumed a level flight, I have added a #3.

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  • $\begingroup$ See if that answer helps you aviation.stackexchange.com/a/1647/878 It's almost what you are looking for. $\endgroup$ – Stelios Adamantidis Aug 4 '14 at 20:16
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    $\begingroup$ I don't think he's asking about a standard turn rate, but rather how fast it could be turned around, which of course is dependent on speed, bank angle, and whether or not altitude is being maintained. This should be clarified in the question. $\endgroup$ – Bret Copeland Aug 4 '14 at 20:28
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    $\begingroup$ If I was at altitude and suddenly realized we were about to cross into airspace that might get us shot down, I would, smoothly, do 3 things simultaneously: (1) start the nose down to around 25 degrees down, (2) start a roll to a bank angle of 60-80 degrees, and (3) start the power back to idle. At the same time, I'd ask the first officer to deploy the speed brakes, which would give me great roll response. $\endgroup$ – Terry Aug 5 '14 at 0:27
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    $\begingroup$ @Terry: Great comment! This needs a few 1000 feet of altitude, but will produce the fastest heading change. The answer is different when the pilot wants to stay at the same altitude, and then I think this question is NOT a duplicate. $\endgroup$ – Peter Kämpf Aug 5 '14 at 8:13
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    $\begingroup$ @Paul - Hidebound pansies or not, I have over 3000 hours on type and can confirm that the 747 cannot take the negative G (no matter how small the %) required for the half roll to initiate a split-S turn. The Barrel roll that Tex did during the demos of the 707 is a "positive G" maneuver where the airframe remains in a state of positive G throughout. The half roll will bring you into a negative G state and the subsequent recovery to level flight will exceed the airframe limits of the 747. Its a people carrier and not an Extra 300. try it and you will die. I hope you don't work for an airline. $\endgroup$ – Mike Richardson Sep 20 '16 at 8:48
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How long, and how wide of a turn depends entirely on three things:

  1. What is the true airspeed.
  2. What is the bank angle of the turn?
  3. Is level flight maintained during the turn.

Since it's easiest to calculate, and presumably what you intended, we'll assume #3 is true (level flight). However, if you really needed to get turned around fast, turning in a descent will give you a tight turn without the extra g-forces.

Cruise Speed for a 747 depends on altitude, generation, company guidelines, etc, but we'll go with Mach 0.85 for this example, which seems to be in the ballpark. If we say it's cruising at 35,000 feet in standard atmospheric conditions, that comes out to 490 knots TAS.

Next, pick a bank angle. A reasonable bank angle for that airplane with passengers on board is 25°, and you could probably get away with 30°. Anything beyond that will cause passengers to complain. Of course, if it's an emergency you might consider something higher. Just remember, the higher the bank angle, the stronger the g-forces which will be felt by the people inside, and the aircraft itself, which has structural limitations. The G-force can be calculated using:

$$ gforce = \frac{1}{\cos(bank)} $$


For an example, let's select 25° bank since that's the most realistic.

Radius

Calculate the radius of the turn using this formula, slightly modified from Wikipedia to give nmi instead of feet:

$$ Radius\ of\ turn\ in\ nautical\ miles= \frac{velocity^2}{68579 \times \tan(bank)} $$

Which gives us:

$$ 7.51nmi = \frac{490^2}{68579 \times \tan(25°)} $$

So the turn itself would be about 15 nautical miles wide (≈ 91,000 feet) not accounting for wind.

Distance Traveled

Using basic geometry, the distance traveled for a 180° turn is $d = r\pi$ (half the circumference of a circle), so:

$$ 23.59= 7.51\pi $$

Giving us 23.59 nautical miles traveled.

Duration

$$ time\ in\ minutes = \frac{distance \times 60}{TAS} $$

Therefore, at 490 knots (nautical miles per hour) and 25° bank, it would take about 2 minutes, 53 seconds to complete the turn.

$$ 2.89 = \frac{23.59 \times 60}{490} $$

You could go through and calculate the results for any speed and bank angle you want. This is true regardless of the type of airplane in question.


Turning in a Descent

Your edited question asks how much altitude would be lost if it were not a level turn. There is also no single answer to this question because it depends entirely on how you perform the maneuver.

One reason you may have to descend in a turn is because the increased lift results un increased induced drag, and the aircraft engines may not have enough power to compensate, resulting in the decreased airspeed, and possibly a stall. In this case, the equations for calculating the turn radius are exactly the same as in level flight. The amount of altitude lost will depend on the descent rate required in order maintain airspeed, which will vary depending on the engine power available, the weight of the aircraft, and the drag curve at the given speed and AoA. Peter Kämpf's answer gives an example of what this might look like for a 747 in a 1.5g turn.

Another version of a turning descent would be an accelerated descent. The advantage in this case it that it would allow turns to be made without additional g-force. The disadvantage is that you'll be accelerating downwards instead of descending at a constant rate. This can get out of hand very quickly and would only be good for very short duration turns.

To illustrate this disastrous option, let's look at what would happen in a 1g turn at the same 25° and 490 knots TAS. Since the math is more complicated, it'll be easier to talk about this portion in metric units. Here's a conversion table:

1 nautical mile = 1852 meters
1 knot = 0.514444 meters per second
1 foot = 0.3048 meters

First, in order to maintain a constant 1g, the lift vector is simply rotated in the turn (instead of rotated and increased in order to maintain altitude). Therefore, the magnitude of our lift vector will be equal to the acceleration due to gravity near the Earth's surface. We'll round that to 9.8 m/s/s.

Inward Acceleration

Our acceleration towards the center of the turn $a_c$ (the portion of our lift vector pointed inwards instead of up) can be determined using the following equation where $\theta$ is the bank angle:

$$ \sin(\theta) = \frac{a_c}{g} $$

Solving for $a_c$:

$$ a_c = g\sin(\theta) $$

Therefore

$$ 4.142 = 9.8\sin(25°) $$

Radius

So 4.142 meters per second per second is how fast, at 25° of bank, and 1g, we'll be accelerating towards the center of the turn. With that, information, along with our known velocity, we can calculate the radius of the turn using this equation where $v$ is our velocity in meters per second, and $R$ is the radius in meters:

$$ R = \frac{v^2}{a_c} $$

Plug in our numbers:

$$ 15341 = \frac{(490 \times 0.514444)^2}{4.142} $$

This comes out to a radius of 15341 meters (8.28 nautical miles).

Distance Traveled

Now that we have our radius, we can calculate how long it will take us to turn 180°. This part is the same equation as before.

$$ 48195 = 15341\pi $$

Giving us 48195 meters (26 nautical miles).

Duration

Distance divided by velocity gives us duration.

$$ t = \frac{d}{v} $$

In our case:

$$ 191.2\ seconds = \frac{48195}{490 \times 0.514444} $$

So, it'll take 3 minutes 11 seconds to complete the turn.

Downward Acceleration

The last value we need before we can calculate the altitude lost in the turn is to calculate how fast we're going to be accelerating towards the ground in this turn. Lets first calculate the upward portion of our lift vector:

$$ a_u = g\cos(\theta) $$

Therefore

$$ 8.882 = 9.8\cos(25°) $$

Our lift vector is accelerating us upwards at 8.882 m/s/s, while gravity is trying to pull us down at -9.8 m/s/s. That comes out to a net vector of -0.918 m/s/s.

Altitude Lost

This part requires a little bit of calculus because our descent rate is accelerating. The integral of acceleration is velocity ($v = at$), and the integral of velocity is distance:

$$ d = \frac{at^2}{2} $$

So, lets calculate the change in vertical distance (altitude):

$$ -16780 = \frac{-0.918 \times 191.2^2}{2} $$

So, theoretically, our airplane lost 16,780 meters of altitude (55,052 feet). Of course, since we were only 35,000 feet in the air to begin with, this is rather bad news for us. Besides hitting the ground, you would also come close to risking structural damage to the airframe because in this example, 490 knots is considered the speed over the ground, but the total velocity would be higher (about 596 knots) by the end of the turn due to the sink rate.

You'll also notice that, assuming we had the altitude to lose in the first place, the turn took longer to complete than in level flight. This is because the magnitude of the lift vector was smaller.

You can feel free to experiment with other speeds and bank angles, and you could also experiment with higher g turns (simply replace $g$ in the equations with $2g$ or similar). In some cases higher than 1g, you may actually gain altitude, though it's unlikely a 747 would be able to sustain a high-g maneuver which gains altitude.

As a quick second example, consider a 2g turn with 80° of bank at 400 knots:

$$ a_c = 19.302 = 2g\sin(80°) \\ R = 2193.8 = \frac{400 \times 0.514444}{a_c} \\ d = 6892 = R\pi \\ t = 33.5 seconds \\ a_u = 3.404 = 2g\cos(80°) $$

Giving a total altitude lost of 3589 meters (11,775 feet) over 33.5 seconds. However, at the end of the turn, you would have a sink rate of 214 meters (703 feet) per second. That's over 42,000 feet per minute. It would likely be possible to recover from that in the altitude remaining, but it would not be pleasant.

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    $\begingroup$ Try to calculate the instationary turn with maximum lift and use the missing thrust to calculate the sink rate. This will give more realistic results. $\endgroup$ – Peter Kämpf Aug 9 '14 at 1:35
  • $\begingroup$ @PeterKämpf I agree that the contrived example is unrealistic in the sense that the maneuver would never be performed like that, but the results I gave seem theoretically accurate. If you never increase lift to match gravity, you'll continue in an accelerating descent. If you assume a constant rate descent, then the equations are exactly the same as in level flight... I should probably point that out in the answer. If that doesn't seem right to you, feel free to correct or give insight. I'm not an aerospace engineer or physicist. $\endgroup$ – Bret Copeland Aug 9 '14 at 21:01
  • $\begingroup$ In a descending turn you can pull more gs than in a steady turn, because the energy gained by descending compensates the drag increase over the drag of the steady case. Here we run probably into buffeting limits already in the static case, so there will be a massive drag increase for little gain in lift. But the turn will be tighter if flown just like Terry described it. $\endgroup$ – Peter Kämpf Aug 9 '14 at 22:20
  • $\begingroup$ @PeterKämpf just so I understand the point you're making, essentially what you're saying is: you could perform a steep turn where upward lift equals gravity, but due to the extra induced drag, the engines wouldn't be able to maintain airspeed; however, you could trade potential energy for airspeed via a constant rate descent in order to perform the maneuver. Correct? Yes, that does seem like a more realistic way to do it. I'll update my answer to bring that up and point people to the example in your answer. $\endgroup$ – Bret Copeland Aug 10 '14 at 4:07
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To calculate turn rates, best start with a load factor $n_z$ or roll angle $\Phi$ and calculate all other parameters using these formulas: $$n_z = \frac{1}{cos \,\Phi}$$ Radius: $$R = \frac{v^2}{g\cdot tan\,\Phi}$$ Angular velocity (rad/sec): $$\Omega = \frac{v}{R} = \frac{g\cdot tan\,\Phi}{v}$$

An airliner in cruise will fly close to the local maximum lift, so it will not be able to sustain a steep turn. Pulling more than a few percent of 1 g will cause stronger shocks on the upper wing which will cause a steep drag increase and can even stall the airplane. This is called a high-speed stall. Fortunately, things get better extremely quickly when the airplane slows down a little. If it slows down too much, it will, however, enter into a low-speed stall, because both are separated by a small speed range when flying in low density air at high Mach number.

To give you an idea what load factor is caused by which roll angle, here is a small list:

    Ω       load factor [g]
    0°        1.0
   10°        1.0154
   20°        1.0642
   30°        1.1547
   40°        1.3054
   50°        1.5557
   60°        2.0
   70°        2.9238
   80°        5.7588

I would assume that more than 20° bank will not be possible at cruise. Given that the 747 can fly at Mach 0.85, this translates into 500 kt or 258 m/s in 30.000 ft. The radius for this turn will be 18.65 km or 10 miles. Use the inverse of the angular velocity to calculate the seconds per radian: Flying a 180° turn ($\Omega = \pi$) will take 227 s or three minutes and 47 seconds.


EDIT: Bret Copeland inspired me to add one more case. Not as extreme as Terry would fly it, but it gives you a tighter turn.

Above the turn was flown without sink. I assume that the 747 can not produce much more lift without running into compressibility problems, which would increase drag massively. If I am not concerned about losing altitude, I can use the energy gain from sinking for compensating that extra drag, and then the calculation would run like this:

Starting from a 1.5g turn with 48° bank, the plane needs to produce 50% more lift, which will only be possible when the pilot accepts heavy buffeting and pitch-down (Mach Tuck). But let's just assume that is possible. I expect that the lift curve will be well past the Mach break, and drag will at least double. Using data from this source, and assuming an aircraft mass of m = 340 t, another P = 9 MW of power are needed per second on top of what the engines supply. This is possible by sinking with $w = \frac{P}{m}$ = 26.6 m/s. This is a flight path angle of $atan(\frac{26.6}{258})$ = 6°. (I would not be surprised if the real number is closer to 10°, but I have no good aerodynamic data available right now).

The turn radius is now only 6.11 km and the 180° turn is completed after 74.4 s = 1 minute and 14 s. The altitude loss with the 6° flight path is 1980 m or 6,500 ft. A more precise solution would include the fact that the lift requirement goes down with the cosine of the flight path angle, but for a first-order estimation the numbers here are good enough.

This is not exactly treating the airplane as the handbook would suggest it, but the type has survived worse treatment before.

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if you are unconcerned with spilling the drinks in Business class you could do a hammerhead or a split-s, although the hammerhead might be a bit rough on the cabin baggage during the rollover (the FA should NOT occupy the rear bulkhead seat directly off the aisle). The radius of either turn would be zero, time would probably be about half a minute with a drop of 10-20,000 feet.

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    $\begingroup$ At altitude you will not be able to produce the lift for a hammerhead. See Terry's comment above on how to change direction fast. What you propose would result in a stall. $\endgroup$ – Peter Kämpf Aug 5 '14 at 8:46
  • $\begingroup$ A split-S shouldn't need the lift? A cruising 747 should have enough altitude. Terry's comment is quite similar; the key to a quick turn is sacrificing altitude. $\endgroup$ – MSalters Aug 5 '14 at 17:32
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    $\begingroup$ In a split S you would fly inverted. A 747 is not rated for this. Turning by exceeding aircraft limits gives not a valid answer. For an aircraft which can sustain negative gs, your answer is unconventional, but very good. $\endgroup$ – Peter Kämpf Aug 5 '14 at 17:47
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    $\begingroup$ @paul: I hope you will never be a pilot with that attitude. What Tex Johnston did was flying a barrel roll. This is a completely different maneuver from the half roll you need to fly for a split-S. Unfortunately, a Lufthansa pilot heard something that the 707 can fly rolls (just like you), and tried it during an acceptance flight. After all, he had rolled jets 10 years before Tex did, but that were Messerschmitt jets. The result: The 707 lost all engines (yes, they simply fell off!) and the pilot his life. $\endgroup$ – Peter Kämpf Aug 9 '14 at 1:19
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    $\begingroup$ @Sean aviation-safety.net/database/record.php?id=19640715-0 $\endgroup$ – Cpt Reynolds Feb 25 at 20:59

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