7
$\begingroup$

Reading the manual for the T6b (turboprop aircraft single engine), it says different altitude losses during a 360 degree turn, holding 125 kts (the best glide for range speed), power is either dead, so propellers are feathered or is simulated for exercise purposes at 4-6%, which simulates the propellers being feathered.

30° AOB 2000 ft,
45° AOB 1500 ft,
60° AOB 1000 ft

Why does it seem to have less effect on altitude loss at higher bank angles? Isn’t there less effective lift to compensate for weight during a turn, forcing the aircraft to lose more altitude? I think it’s something to do with the rate of turn. The faster you are turning, the sooner the loss of altitude gets to stop. I’m not sure. How is it possible to lose less altitude with less lift available?

$\endgroup$
9
  • 1
    $\begingroup$ YamchaAviator, would you happen to have an on-line source for that manual? $\endgroup$ Commented Oct 7, 2022 at 2:16
  • 2
    $\begingroup$ It probably means it completes the 360 turn faster. So less bank you lose altitude less. However it takes longer to complete a 360 deg turn. So presumably based on this information your better off with a 60 deg quick turn to save energy then babying it. It’s two competing factors. Increased rate of decent due to high bank vs taking longer to compete the turn. With this aircraft and configuration it says the later is more optimal. $\endgroup$ Commented Oct 7, 2022 at 3:16
  • 2
    $\begingroup$ It’s NOT saying the aircraft glides better banked. It’s saying less altitude is lost with out power, by completing the turn as quickly as possible roughly. $\endgroup$ Commented Oct 7, 2022 at 3:20
  • $\begingroup$ Also too, if you needed to lose altitude sometimes I beleive a 360 degree turn is used before turning on to final for an un powered landing, similar to the space shuttle. So you can select an bank angle to better precisely achieve the desired altitude. $\endgroup$ Commented Oct 7, 2022 at 3:25
  • $\begingroup$ You have a random altitude. You have excess altitude. You are with in range or even at the start of the unpowered final approach. This would be a case you could do a 360 turn to arrive back at final approach at the lower altitude. By using either a 30, 40, or 60 deg bank angle you can estimate the altitude you will lose. And arrive at the right height. Basically glided landings always rely on excess energy and altitude. But can never have under which is the case you can not make the runway. $\endgroup$ Commented Oct 7, 2022 at 3:27

4 Answers 4

12
$\begingroup$

If you have a shallow bank angle, your load factor is low, and thus the aircraft needs to generate less lift. Less lift means less drag, and so a shallow bank angle seems better than a steep bank angle.

However, a shallow bank angle makes for a wide turn. The distance travelled to complete a 360 degrees turn is much longer at shallow bank angles than at steep bank angles. And when you have to travel further, you lose more altitude.

It is more favourable to accept the higher drag penalty of a steep bank angle over a short turn circle circumference, than to coast over a long turn circle with a low drag penalty.

The distance travelled in a 360 degree turn can be found with:

$d = 2\pi \frac{v^2}{g \tan(\phi)}$

with:

  • $d$ distance travelled
  • $v$ true airspeed
  • $g$ gravitational acceleration (~9.81 m/s2
  • $\phi$ bank angle

( you have to use S.I. units)

This shows that at 30 degrees bank angle, you travel 3 times the distance to complete a full circle compared to a full circle turn at 60 degrees bank angle.

$\endgroup$
1
  • $\begingroup$ Thats it right there. Awesome. Thank you! $\endgroup$ Commented Oct 11, 2022 at 2:06
5
$\begingroup$

The rate of turn is higher with increased bank angle at a constant speed. At 45deg the rate of turn is 60% higher than at 30deg bank, so there is less time descending in a 360deg turn so less altitude is lost.

enter image description here

$\endgroup$
3
$\begingroup$

There is something inherently dangerous about trying to hold a given airspeed and a steeper and steeper bank.

This a a case where one should consult with a qualified expert on that type of aircraft and, better yet, test these types of turns at altitude with one.

Vbg is typically around 1.3 stall. In order to maintain adequate vertical lift with a 60 degree bank (2G), airspeed must be at least 1.4 stall, 45 degrees is 1.18, 30 degrees is 1.07.

With less vertical lift (inadequate airspeed), the aircraft will sink and pitch down, increasing airspeed. Trying to maintain adequate vertical lift in steeper banks with higher and higher AoA (and constant airspeed) not only decreases the lift to drag ratio but (obviously) increases the chance of stalling. The safest way to turn at a higher angle is to increase airspeed.

Again, expert advice is recommended here, actual flight data would be preferred.

$\endgroup$
14
  • $\begingroup$ I would say that at 60 degrees, stall speed increases by 1.4 (as does the airspeed for any other given AoA). In the T6b specifically, would this raise the stall speed dangerously close to the wings-level best glide speed? That seems to be the crucial question of interest. Also, whether the wings-level best glide speed is way below 1.4 x the wings-level min. sink rate speed-- that also would seem unlikely to end well. All of which I think is basically your point. So does this airplane have an unusually high V (best glide), orvwhat $\endgroup$ Commented Oct 6, 2022 at 22:38
  • $\begingroup$ Steep turns and emergency descents are done at much higher airspeeds than Vbg. $\endgroup$ Commented Oct 6, 2022 at 22:44
  • $\begingroup$ Do you think questioner (or actual manual) erred in stating 125 knots is Vbg? Hmmm.. $\endgroup$ Commented Oct 6, 2022 at 22:46
  • 1
    $\begingroup$ @Max -- You may well be on to something important to solving our little puzzle here, but I suspect you meant to say KCAS, rather than KTAS? Surely KTAS is not a consistent guide to stall speed, across a range of various density altitudes, etc. $\endgroup$ Commented Oct 7, 2022 at 10:55
  • 1
    $\begingroup$ @Max -- only if you are dealing with one particular place and time in the atmosphere (fixed density altitude). KTAS at stall is very different at sea level then at 20,000'! $\endgroup$ Commented Oct 7, 2022 at 22:20
3
$\begingroup$

In the long run, sink rate is not due to a "shortage of lift" but rather to a shortage of thrust.1 If the angle-of-attack were held constant, then in the very short run, increasing the bank drives a brief shortage of (the vertical component of) lift-- this is involved with the dynamics of causing the flight path to curve down as the airplane accelerates (gains airspeed)-- but that's not what you need to be focusing on. After a very few seconds, the aircraft will have gained enough airspeed to restore the vertical component of lift to essentially its "normal" value. So... it's complicated.

Furthermore, consider this-- in your problem, the specified airspeed is assumed to be fixed. This means that the angle-of-attack is certainly not fixed. As the pilot enters the turn, to hold the airspeed constant, he must be increasing the angle-of-attack sufficiently that the vertical component of lift stays (very nearly) constant throughout the turn entry. So there's no shortage of the (vertical component of) lift to speak of-- not even temporarily, as the aircraft is entering the turn.

So the steeper bank turns are being flown at a higher angle-of-attack than the shallower turns. But that's never going to offset the effects of the steeper bank angle enough to actually cause a decrease in sink rate. The key factor is rather that you complete the turn faster when flying at a steeper bank angle, so the altitude loss per degree of turn is decreased.

Nonetheless, the fact that the steeper bank turns are being flown at a higher angle-of-attack does bias the results for minimum altitude loss per degree of turn toward a somewhat higher bank angle than we'd obtain if we specified that the angle-of-attack were held constant (and the steeper bank turns were therefore being flown at a higher airspeed than the shallower turns)--at least so long as we remain on the "front side" of the drag curve.2

Edit:

Actually I initially missed the part about the specified airspeed being the exactly the same as the wings-level best glide speed. So we're not deep on the "back side" of the drag curve, but we're not way on the "front side" either. Rather, we are (in the wings-level case) right at the very bottom of the drag curve (i.e. drag is at a minimum).

A key point is that when you are turning, you don't care how far forward you go, you just want to minimize sink rate and maximize turn rate, so there's no reason to fly at best glide speed, you instead want to fly at the min sink rate speed for that bank angle, or arguably even a bit slower, to maximize the turn rate, and therefore minimize the altitude lost per degree of turn.

So starting from the wings-level best L/D (max glide) speed and angle-of-attack, increasing the angle-of-attack as we enter a turn is initially beneficial, in terms of minimizing the altitude lost per degree of turn.

At some point, however, if we continue to increase the bank angle while holding the airspeed constant, we will end up increasing the angle-of-attack so much that we are mushing along near the stall at such a high angle-of-attack that the sink rate, and therefore the altitude lost per circle, will be severely increased. (In fact, if the bank angle becomes steep enough in relation to the airspeed, we actually reach the stall angle-of-attack if we attempt to hold the airspeed constant by continuing to move the stick or yoke aft-- see this related answer to the present question.)

It seems odd that the aircraft has apparently not yet reached that unfavorable part of the flight envelope when the bank angle has been increased all the way to 60 degrees, with no increase in airspeed beyond the wings-level best glide angle (best L/D) airspeed. For sailplanes and light airplanes, I've typically seen bank angles around 45 degrees recommended to complete a turn with minimum altitude loss, while flying (as best as I recall) at the wings-level best-glide airspeed.

Perhaps there is an unusually wide spread between the wings-level airspeeds (and angles-of-attack) for max glide (best L/D), and for min sink rate, in this aircraft. Does the manual publish the airspeed for minimum sink rate in wings-level flight, for the same aircraft weight at which the wings-level best glide speed is 125 knots?

Also, consider this-- if we are free to choose any bank angle, is keeping the airspeed right at the wings-level best glide speed really always the very best strategy for minimizing the altitude lost per degree of turn, or at least a close approximation of such? Enquiring minds want to know...

Footnotes:

  1. Yes, as has been covered elsewhere on this site, the geometry of the force vector triangle is such that (the vertical component of) lift must be slightly less than weight whenever the flight path is descending (or climbing!) in a steady-state (constant-airspeed) condition, but it's a misconception to think that this discrepancy is what is causing the flight path to descend (or to climb!). Rather, this discrepancy simply very slightly decreases the airspeed at which the aircraft will be in equilibrium (i.e. the airspeed will be constant), for any given angle-of-attack and bank angle and glide angle (or climb angle!). In other words this dynamic is very much a sideshow, not the main event.

  2. If the aircraft were initially being flown in slow, "mushing" flight, on the "back side" of the drag curve, then any further increase in angle-of-attack would clearly be detrimental, and if we specified that the airspeed were held constant, we would bias the results toward a lower optimum bank angle for minimum altitude loss per degree of turn, than if we we specified that the angle-of-attack were held constant. Exactly where on the drag curve we'd need to (initially) be flying so that we'd end up computing the same optimal bank angle to minimize altitude loss per degree of turn regardless of whether we held airspeed constant or held angle-of-attack constant is not completely obvious (and perhaps grounds for an interesting exercise or new ASE question.). It appears that this "break-even" point would correspond to a starting airspeed somewhat slower than the wings-level max-glide (best L/D) airspeed, but well above the wings-level min sink rate airspeed.

$\endgroup$
9
  • 1
    $\begingroup$ I was testing my Pazmany PL-2 homebuilt to see the minimum altitude loss I could achieve following engine failure. I'd dive with engine at idle, pull up into the climb attitude, and as speed decayed past normal climb speed, I'd crank and bank. 45 deg bank got me around 180 deg in something like 250 ft IIRC. But on one try I pulled with a bit too much gusto and zowie it snapped over the top and I found myself pointed at the ground in the space of about a second. That was sobering. $\endgroup$
    – John K
    Commented Oct 7, 2022 at 1:56
  • 1
    $\begingroup$ The notion of your first paragraph should be required reading for all pilots, even the most experienced. The belief that a climb is a result of surplus lift is absolutely pervasive (easily 95+%) among pilots at all levels. Established steady climbs and steady descents are 1.0G maneuvers that do not change the lift produced or the wing loading once established. Nearly every lift/weight/thrust/drag drawing ever used in flight instruction are not just incorrect but misleading. And yet as a community we act genuinely puzzled about the frequency of stall/spin fatalities in GA. It is a travesty. $\endgroup$
    – Max R
    Commented Oct 7, 2022 at 3:00
  • 2
    $\begingroup$ @JohnK I agree that 45° bank is the magic number for “the impossible turn.” It’s been a fascination of mine for years and I practice/experiment a lot… just this year I have easily over an hour of gliding in 45° bank with the stall horn blaring. In addition to very deliberate use of the rudder, I’ve found another trick--I nose over abruptly so that my transition from climb speed to stall speed is done at about .5G. My goal is to perform as many degrees of turn as possible with less lift demand and thus lower stall speed. I now can teardrop my 3100 lb plane in less than 250 feet. $\endgroup$
    – Max R
    Commented Oct 7, 2022 at 3:19
  • 1
    $\begingroup$ I wonder if one or more of the altitude-loss-per-turn figures in the manual might simply be wrong, at least given the apparent specified constraints (125, clean.) $\endgroup$ Commented Oct 7, 2022 at 12:01
  • 1
    $\begingroup$ @quietflyer I would not rule this out at all. Cessna 182 RGs have a significant (10 knot) typo error in several entries in the POH stall table, and surprisingly has never revised the POH. $\endgroup$
    – Max R
    Commented Oct 7, 2022 at 14:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .