I wrote “Flying around
We’ll do everything – in very unAmerican fashion – in metric units (the SI system). To convert knots to meters/second, we have m/s = knots × 0.5144444.
We’re interested in level turns which means that, whatever the bank angle, the vertical component of lift must be 1 G, or an acceleration of g = 9.80665 m/s. A bank angle of α° gives a horizontal component of lift equal to g × tan(α) – at 0° bank (straight and level), the horizontal component of lift is g × tan(0) = 0, a bank angle of 45° gives a horizontal component of lift equal to g × tan(45°) = 1 g, and at 60°, the horizontal component of lift is g × tan(60°) = 1.7 g.
The horizontal component of lift will pull us through the turn. In a steady, level turn, the horizontal component of lift equals the centripetal acceleration, a, which is given by a = ω²r = v²/r, where ω (Greek omega) is the angular velocity and r is the radius of the turn; v is the linear speed.
So, if we’re flying at a speed of v m/s and we make
a level banked turn of α°,
we have v²/r = g
× tan(α). Solving for r,
the radius of the turn, we get
r = v²/(g × tan(α))
And, if we fill in all the constants and do the conversion from knots into m/s (0.51) and meters into feet (3.28), we get:
f =
3.28 × (0.51 v)²/(9.81 × tan(α)) = 0.0885 k² / tan(α)
where f is the turn radius in feet, k is the speed in knots and α, the bank angle in degrees.
Let’s give that some meaning. Here’s a table with various speeds (in knots) and bank angles (in degrees), giving the turn radius (in feet):
Turn radius in feet 
Bank angle in degrees 

20 
30 
40 
50 
60 
70 

Speed in knots 
60 
876 
552 
380 
267 
184 
116 
70 
1192 
751 
517 
364 
250 
158 

80 
1557 
981 
675 
475 
327 
206 

90 
1970 
1242 
855 
602 
414 
261 

100 
2433 
1534 
1055 
743 
511 
322 

110 
2943 
1856 
1277 
899 
619 
390 

120 
3503 
2208 
1519 
1070 
736 
464 

130 
4111 
2592 
1783 
1256 
864 
545 

140 
4768 
3006 
2068 
1456 
1002 
632 
So, if you can reliably fly a 45°degree banked turn, at 70 knots you can do a 180 in about 600 feet (remember that the radius of the turn is half the diameter), while at 110 knots, you’re going to need more than 2000 feet. That’s a dramatic difference!
At a bank angle of 60°, you’ll pull 2 Gs, and you need 500 feet at 70 knots and 1020 at 100 knots. Note that speed affects the turn radius much more than bank angle.
Now we can ask ourselves, what is the optimum combination of speed and bank angle for a minimumradius turn? We’d have to slow down as much as possible and bank as much as possible. But, if you go slower, the maximum bank angle you can use goes down because you’ll stall if you pull too many Gs. Stall speed is given by the equation
_{}
(adapted from Kershner’s Advanced Pilot Manual). Since _{}, S and ρ are constants, we’ll just replace them by C. For a given aircraft, the denominator in the equation is a constant, so stall speed is proportional to √W, the weight of the aircraft and the number of Gs you pull in the turn (which, effectively, is a multiplicator for the weight of the aircraft). In horizontal flight, G = 1, so the formula reduces to the one given by Kershner.
If the stall speed of your aircraft at some given weight is 50 knots, for example, and you pull a 60°, 2G turn, stall speed becomes_{}_{}, or 1.41 × 50 = 70 knots. Now, assume you choose a safe minimum flying speed that is, say, 1.3 times the stall speed. This minimum speed will grow as the bank angle increase – because you want to continue to fly at 1.3 times the stall speed.
G relates to
bank angle α
(in a level, constantspeed turn) like this: G = 1/cos(α).
We’ll substitute this in the stall speed equation and then substitute
that result in the turn radius equation, giving:
_{}
Just two variables remain, r, the turn radius, and α, the bank angle; sin(α) increases steadily with increasing bank angle, albeit with diminishing speed, to reach its maximum at a 90° angle. The minimum radius turn is thus achieved at a 90° bank. But you’ll be pulling infinitely many Gs and your airspeed will have to be astronomical to prevent the stall. No aircraft is built to do that. But the fact remains that the steepest possible turn gives the best turn radius.
The maximum steepness of the turn depends on the number of Gs it causes and thus on the structural forces the aircraft can withstand. The Cessna R182 I fly, for example, is certified for a maximum of 3.8 Gs, which corresponds to a bank angle of 75°. The stall speed in level flight is 55 knots (forward CG). At 3.8 Gs, that becomes 55· √3.8 = 107 knots. The radius of the resulting turn has a radius of just 273 feet (a diameter of 546 feet).
Should we try that? Better not!
Banks of more than
60 degrees count as aerobatic maneuvers – you can’t do them without a parachute
and you certainly can’t (and shouldn’t) do them over the
So here’s what I
do:
At a 60degree bank, our calculations give: G force is 2, minimum safe airspeed becomes 55·1.3·√2 = 101 knots; turn
radius is 0.0885·101²
/ tan(60°) = 522 feet: diameter 1044 feet. I feel perfectly comfortable with this. In fact, I usually fly these 60degree banked
turns between 100 and 120 knots and that gives me plenty of room to turn just
south of (i.e., before reaching) Roosevelt Island.
The POH for the
R182 gives a significant difference between indicated and calibrated airspeeds for the
stall: with forward CG, indicated is 39
KIAS, while calibrated is 54 KCAS; with rearward CG it’s 41 KIAS and 55
KCAS. For the calculations above we obviously
need true airspeed, and calibrated is closer to that than indicated. But when you fly, you have to rely on
indicated airspeed. You and I are likely
to monitor the airspeed before the
turn, while still straight and level at or above 100 KIAS. At these speeds, and in level flight, the
difference between indicated and calibrated airspeed is small.
The POH for my
R182 (1981) gives a stallspeed table for bank angles of 0°, 30°, 45°and 60°
and both indicated and calibrated airspeeds.
For flapsup, forward CG, 3100 lbs, Figure 53 gives:
Bank 
KIAS 
KCAS 
KIAS/KCAS 
0° 
41 
55 
.75 
30° 
44 
59 
.75 
45° 
49 
65 
.75 
60° 
58 
78 
.74 
Note that the
error in indicated airspeed is the same, for every entry in the table. My hypothesis is that the error is almost
entirely a function of angle of attack
and, at the stall, this, of course, is
precisely the critical angle of attack
and, therefore, the same for each table entry.
Using the airspeed
calibration chart, which is for level
flight, 58 knots indicated
corresponds to 64 knots calibrated,
which isn’t quite enough to prevent a stall even at 45 degrees of bank. Again, this emphasizes the need for a healthy
safety margin and for knowing what you’re doing before attempting any kind of
flying anywhere near the performance limits of your airplane.
While you turn, a wind from the side is going to widen or narrow the turn, depending on whether you turn with the wind (bad idea in the East River) or against it (good idea). How much you’re going to be displaced by it depends on how long you spend in the turn. Time is distance divided by speed, so the time in a 180° turn is πr/v. If we use meters, and meters/second, we get time in seconds. Here’s the table again, with the number of seconds to complete a 180 (please note that some combinations of bank angle and speed don’t make sense – there are probably few aircraft that can do a level, 70° banked turn at 60 knots without falling out of the sky):
Seconds to complete a 180 
Bank angle in degrees 

20 
30 
40 
50 
60 
70 

Speed in knots 
60 
27 
17 
12 
8 
6 
4 
70 
32 
20 
14 
10 
7 
4 

80 
36 
23 
16 
11 
8 
5 

90 
41 
26 
18 
12 
9 
5 

100 
45 
29 
20 
14 
10 
6 

110 
50 
31 
22 
15 
10 
7 

120 
54 
34 
24 
17 
11 
7 

130 
59 
37 
26 
18 
12 
8 

140 
63 
40 
27 
19 
13 
8 
Now, isn’t it interesting that, the faster you go, the longer the turn takes?
Using the conversion above from knots to m/s, we see that a 10knot wind is 10 × 0.51 or 5 m/s, so, if you’re exposed to it for 10 seconds, you’ll be pushed some 50 meters or 150 feet by the wind. A kknot wind will increase or decrease the diameter of an ssecond 180 by k × s × 1.688 feet.
If you make a 45degree banked turn at 110 knots with a wind of 14 knots in your back, the diameter of the turn becomes 2000 + 14 × 18 × 1.688 feet = 2425 feet. If you turn against the wind, it’d be 2000 – 14 × 18 × 1.688 feet = 1575 feet; that’s just two thirds, not bad.
Another thing to notice from the table is that, the slower you go, the shorter the turn takes, so the shorter you’re exposed to wind pushing you off course.
So imagine doing a slightly shallower 40degree banked turn, but now at 80 knots with a wind of 14 knots in your back, the diameter of the turn becomes 1350 + 14 × 16 × 1.688 feet = 1728 feet. If you turn against the wind, it’d be 1350 – 14 × 16 × 1.688 feet = 972 feet; that’s about half and it’s quite a bit better than the steeper turn at 110 knots.
So bear this in mind: Speed matters much more than bank angle if you need to make a tight turn, so slow down and don’t overdo the bank angle to keep you from stalling, drop some flaps if necessary. If there is more than a few knots wind, think about which way you should turn to take advantage of it. Wind can make a significant difference.
And, again, don’t practise those steep turns anywhere near something hard and unyielding.
December 2006, Sape Mullender