Possibly related to How does the load factor vary when the aircraft pitches up/down?

Could you help to settle a debate? I have recently flown a dead-stick approach with a CFI in the right seat. Turning base I was slightly high, and deliberately went past the final track, then banked to around 30 degrees and re-intercepted from the other side. During the turn, the plane was coordinated and I didn't pull up; I've merely altered back-pressure to keep the airspeed stable.

My first CFI taught me that whilst descending I can essentially steep-bank the plane provided I don't pull up to the extent that the airspeed drops (ie not a level turn). Due to the descent the load factor wouldn't increase, at least not to the extent it does in a level turn.

The new CFI was of a different opinion. Which one is true, or is it somewhere in the middle?

For the setup: In a level turn, 30 degrees would equate to 1.15G and increase the stall speed by 7%. What are these values if at the initiation of the turn, the aircraft was in a stable (call it 600fpm) descent, and the airspeed was kept constant (relaxing back pressure to not decrease or increase the speed)?

  • $\begingroup$ You might be on the wrong path in discussing load-factor. You may be more interested in stall behavior, which is indirectly related to load-factor but not synonymous. I say this because you are discussing an approach which generally means you are well below Va and thus load factor is not really relevant. What is relevant is the AOA and momentary vertical(relative to earth) acceleration. As a bit of an aside the most efficient use of availible lift for a turn to a heading is around 45deg bank and best-glide range AOA. Best bank and speed varies some based on the details of real aircraft. $\endgroup$
    – Max Power
    Commented Aug 4, 2023 at 23:24

6 Answers 6


Due to the descent the load factor wouldn't increase

I have to disagree with your first CFI, even though I suspect that he was misunderstood. Your load factor due to descent will decrease with the cosine of your flight path angle. If you were flying at 60 knots and descending at 600 fpm (which is 30.867 m/s and 3.05 m/s, respectively), your flight path angle was -5.67°, the cosine of which is 0.995. This means your load factor was reduced by 0.5%. This is so insignificant that you might as well neglect it.

Of course, in a vertical dive (cosine of 90° = 0), you can roll any way you want without having any load factor in the plane-fixed vertical direction at all.

  • $\begingroup$ I know you know this, but ... a Minor, I repeat, minor, quibble. The load factor required to maintain your aircraft pitch and bank angle is dependent on the cosine of your flight path angle (as well as your bank angle). But the actual load factor is dependent only on your AOA and Airspeed. (some inertial effects due to changes in flight path angle can also affect it.) $\endgroup$ Commented Jun 26, 2023 at 19:21
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    $\begingroup$ Thanks to you, too, I shall digest this and hopefully ask decent follow up questions. $\endgroup$ Commented Jun 26, 2023 at 20:17
  • $\begingroup$ @CharlesBretana Absolutely! But then you need to maneuver, like pitching up, to add more load factor. I tried to answer the question which is about a descending turn and the load factor change due to the descent. $\endgroup$ Commented Jun 27, 2023 at 18:52
  • $\begingroup$ @Peter, well, just changing pitch attitude, or bank angle, will not change your load factor. The aircraft trim sets the AOA, and the AOA, in combination with airspeed, determines where you are on the Vn diagram, which sets Lift & load factor. So, unless you change the trim, or move the flight control stick (yoke), the load factor will remain, (or change), to whatever aircraft AOA and velocity dictates. $\endgroup$ Commented Jun 28, 2023 at 13:39
  • Load Factor (in the sense of wing loading) is the ratio of Lift to Weight.

  • Assume for simplicity Thrust is zero. (If Thrust were not zero, we would simply have to substitute the "Excess Drag" vector, defined as (Drag - Thrust), in place of the Drag vector, in all occurrences below.)

  • Assume "coordinated" flight, meaning zero sideslip, meaning that none of the aircraft's weight is being supported by an aerodynamic sideforce generated by the air striking the side of the fuselage, the side of the vertical fin, etc.

  • In a steady-state descent, we can draw a closed vector triangle between the Weight vector, the Drag vector, and the non-centripetal component of the Lift vector.

  • For any given bank angle, the ratio between the total Lift vector and the non-centripetal component of the Lift vector is fixed.

  • For brevity we'll use the term "Nclift" to mean the non-centripetal component of the Lift vector.

  • Nclift = Lift * cosine (Bank)

  • Lift = Nclift / cosine (Bank)

  • Load Factor = Nclift / (Weight * cosine Bank)

  • In a steady-state descent, (Nclift)^2 + (Drag)^2 = (Weight)^2. (This particular expression actually doesn't figure into the math below.)

  • Also, in a steady-state descent, Nclift / Weight = cosine (arctan (D / Nclift)), so Nclift = cosine (arctan (D / Nclift)) * Weight

  • The expression that we're taking the cosine of in the line above, is also the glide angle, i.e. the angle of descent.

  • Which means that in a steady-state descent, Load Factor = Lift/Weight = (Nclift / Weight)/ cosine bank = (cosine (arctan (D / Nclift))) / cosine bank = (cosine(arctan(D / (L * cosine bank))) / cosine bank

  • As we degrade the L/D ratio, and thus also the ratio of Nclift / D, we carry some of the aircraft's Weight with Drag vector rather than the Nclift vector. For any given bank angle, in a steady-state descent, lower values of L/D therefore must be associated with lower values of L, due to a decrease in airspeed, angle-of-attack, or both. So "dirtying up" the aircraft by dropping flaps, etc -- and therefore making the glide path steeper-- does force a reduction in Lift, and thus a reduction in G-loading. But how significant is this effect really, for "normal" aircraft in "normal" situations?

  • Let's assume we can vary the ratio of L/D from infinity, down to 8/1, or down to 4/1, or down to 2/1, by "dirtying up" the airplane. How do these 4 cases compare, in terms of the load factor we'd have at any given bank angle?

  • (Remember, an infinite L/D ratio in this context would simply represent a case where we did in fact have some Thrust, exactly equal to Drag, since the variable we're calling "Drag" in this answer actually is more properly described as the "Excess Drag" vector, i.e. Drag minus Thrust)

  • Using the formula Load Factor = (cosine(arctan(D / (L * cosine bank))) / cosine bank, as derived above, we find the following results:

Bank Angle L/D Angle of Descent Load Factor
0, 0, 0, 0 Inf, 8/1, 4/1, 2/1 0, 7.1, 14.0, 26.6 1, 0.99, 0.97, 0.89
30, 30, 30, 30 Inf, 8/1, 4/1, 2/1 0, 8.2, 12.2, 30.0 1.16, 1.14, 1.13, 1.00
45, 45, 45, 45 Inf, 8/1, 4/1, 2/1 0, 10.0, 19.5, 34.9 1.41, 1.39, 1.33, 1.16
60, 60, 60, 60 Inf, 8/1, 4/1, 2/1 0, 14.0, 26.6, 45.0 2.00, 1.94, 1.79, 1.41
80, 80, 80, 60 Inf, 8/1, 4/1, 2/1 0, 35.7, 55.2, 70.8 5.76, 4.67, 3.29, 1.89
85, 85, 85, 85 Inf, 8/1, 4/1, 2/1 0, 55.1, 70.8, 80.1 11.5, 6.56, 3.78, 1.97
90, 90, 90, 90 Inf, 8/1, 4/1, 2/1 Undefined, 90, 90, 90 Undefined, Inf, Inf, Inf

What can we make of these results?

For mild degradations in the L/D ratio (or more strictly speaking the L / (D -T) ratio), we don't see much decrease in the load factor associated with any given bank angle until the bank angle gets extremely steep, and we certainly don't keep the Load Factor anywhere near 1. If we degrade the L / D ratio (or more strictly speaking the L / ( D - T) ratio) from infinity all the way down to 4/1-- which is a rather extreme degradation in this ratio-- we still don't see much decrease in the Load Factor associated with any given bank angle until we look at bank angles steeper than 60 degrees. If we degrade the L / D ratio (or more strictly speaking the L / ( D - T) ratio) from infinity all the way down to 2/1, then yes, so much of the aircraft weight is supported by the Drag vector rather than the Lift vector that we can fly at an 85 degree bank angle with only a Load Factor of 1.97, which is a pretty nifty trick. But in actual practice, to achieve, such a lousy L/D ratio, we'd probably have to deploy a gigantic drogue chute behind the aircraft, so in large part we're just coming down "under canopy", conceptually speaking!

Looking over the numbers in the table, the idea that we can keep the Load Factor in a very steep bank low simply by cutting power and "dirtying up" the airplane (deploying flaps etc) to degrade the L / (D-T) ratio and increase the descent angle is clearly a fantasy in real-world situations.

And as for the idea that rather than "dirtying up" the airplane, all we need to do is "ease off" the back pressure on the stick-- that idea makes plenty of sense in a dynamic situation like a steep wingover-- nearly all of us have experienced the pleasure of seeing the nose slice down through the horizon with aircraft in a perfect 90-degee bank, while experiencing only a mild G-loading (Load Factor). But in this situation the flight path is continually curving to point more steeply downward, as the vertical speed rises-- the aircraft is in no way in any kind of steady-state condition. The dynamics of aerobatic flight are very different from the dynamics of any steady state turn, no matter how steep the bank angle or the descent path and no matter how high the sink rate, and so have little bearing on the original question that should be the focus of the discussion here-- which appears to be whether for any given bank angle, the Load Factor is significantly reduced if the aircraft is allowed to descend at some (possibly quite high) constant rate, allowing some of the weight to be supported by the Drag vector instead of the wing's Lift vector, rather than being kept at a constant altitude.

The idea of easing the G-loading by "easing the back pressure" on the stick deserves a little more attention. In the short run, a strong decrease in back pressure on the stick will always ease the G-loading, but the decrease in the wing's lift coefficient and Lift force will then make the flight path curve downward and the airspeed increase, so that the wing is once again making about as much Lift (and therefore Load Factor) as it was before the change. The only long term, sustained effect on Load Factor of a change in amount the back pressure is the pilot exerts on the control stick, lies in resulting change in the L/D ratio. The table above suggests that that will generally allow only an extremely modest long-term change in Load Factor, for flight at any given bank angle.

And finally, to try to address at least one specific question asked in the original post:

For the setup: In a level turn, 30 degrees would equate to 1.15G and increase the stall speed by 7%. What are these values if at the initiation of the turn, the aircraft was in a stable (call it 600 fpm) descent, and the airspeed was kept constant (relaxing back pressure to not decrease or increase the speed)?

You mention that you were flying a "deadstick" approach, which suggests you were flying a single-engine light airplane. Let's assume your approach was made at 70 mph. If you allowed the plane to descend at 600 fpm, that works out to a descent ratio of 10.27, which gives a descent angle of 5.56 degrees. Keeping that descent angle firmly in mind, look at the second line of the table above. Your decrease in Load Factor, compared to the constant-altitude case, is going to be on the order of 1%. You won't be able to detect any decrease in stall speed due to the fact that you are descending rather than maintaining altitude. The effect you are asking about is completely insignificant.

And please keep clear that we're not saying that if you feel the stall buffet, you should not ease off on any back pressure you are holding on the stick. Easing off the back pressure addresses the immediate problem of excessive angle-of-attack, but in the long run cannot change the L/D ratio enough to increase the steady-state descent angle enough to create any significant sustained change the steady-state Load Factor that you'll be experiencing. Keeping the stick in the new, further forward position will certainly help you avoid stalling, but that is due to the simple fact that you are flying at lower angle-of-attack (which also means a higher airspeed)-- not due to any long-term, sustained decrease in the Load Factor.

  • $\begingroup$ I think it finally clicked, thank you! My presumption thus far was that most of the other replies were technically correct but merely keeping to look at graphs didn't change my understanding of them. This does, and is much appreciated! I'll leave it for a short while but am very much inclined to accept it as the answer to my question. Thank you! $\endgroup$ Commented Jun 30, 2023 at 9:59
  • $\begingroup$ Re "But in actual practice, to achieve, such a lousy L/D ratio, we'd probably have to deploy a gigantic drogue chute behind the aircraft, so in large part we're just coming down "under canopy", conceptually speaking!" -- I guess for some reason I was imagining holding the a-o-a and lift coefficient constant at this particular point in the thought experiment, and degrading the L/D ratio only by increasing the drag coefficient. Obviously an aerobatic plane flying straight down at term. vel. (with wings at 0-lift a-o-a) is accomplishing an even bigger decrease in L/D, but in a different way! $\endgroup$ Commented Jun 30, 2023 at 14:17
  • $\begingroup$ The point being that some parts of this answer may not be phrased quite right to be fully accurate in the context of extremely low angles-of-attack and lift coefficients, and near-vertical flight paths, when the wing truly is greatly unloaded by virtue of the forward position of the control stick or yoke. Will give it some time for ideas to "settle" a bit more before diving in with any changes-- $\endgroup$ Commented Jun 30, 2023 at 14:20
  • $\begingroup$ Another thought on the idea of decreasing the G-load (Load Factor) by relaxing the aft pressure on the stick, and my point that (for non-extreme descent angles) most of this decrease can only be temporary and goes away when the airspeed rises-- the reverse thought experiment is to think of starting in a constant -altitude turn and then increasing the G-loading by increasing the back pressure on the stick, while holding the bank angle constant. (ctd) $\endgroup$ Commented Jun 30, 2023 at 21:54
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    $\begingroup$ If you look at the 4/1 values for "Angle of Descent" in the first three rows of the table, it's clear that there's an error in the first or second row. Must have been a typo; an actual math error would show up in the "Load Factor" column as well. Planning to run through the math again and edit to correct in the next few days. $\endgroup$ Commented Jul 4, 2023 at 14:24

The load factor is not really the issue here. It is strictly determined by Angle of Attack (AOA) and Airspeed, (i.e., where you are on the Vn diagram. What is important is AOA. If you lower the nose, (pitch attitude) your vertical velocity will increase, and the loss of potential energy will cause your kinetic energy (airspeed) to increase unless you reduce power significantly. Increasing airspeed allows you to pull more Gs (load factor) at the same AOA, but it does not require that you do so.

If you don't increase your airspeed but want to maintain pitch attitude at a higher bank angle, to generate the higher load factor required for the bank angle, you will need to pull back on the stick or yoke (more back pressure) to increase the AOA to do so.

It is this increased AOA (putting you closer to the stall AOA) that is the problem. The closer to stall (max) AOA you are, the higher of risk of inadvertent stall, and spin (Loss of control) you are.

  • $\begingroup$ I'll need a moment to digest the answers and will report back once I've hopefully understood enough to ask decent enough follow-up questions if there are any. $\endgroup$ Commented Jun 26, 2023 at 20:17
  • $\begingroup$ @quiet flyer, I don't understand your confusion. Increasing airspeed gives you more airflow over the wing. Load factor is Lift / Weight (mass). Lift equation has Velocity squared in it, no? What am I missing in your comment? Are you confused by the word "require"? $\endgroup$ Commented Jun 29, 2023 at 12:12
  • $\begingroup$ If that is your issue, my use of the word "require" is just to make the point that the pilot is controlling the plane's AOA - with the control stick (yoke). Just because the nose goes up or down does not change AOA, (or load factor) all by itself. Indeed, left to itself, as airspeed increases, the AOA will remain more or less constant as determined by aircraft trim, and Load Factor will increase proportionally (as velocity squared). $\endgroup$ Commented Jun 29, 2023 at 12:17
  • $\begingroup$ Maybe the op meant to ask "How does the Load Factor required to maintain aircraft attitude, (pitch & bank), change for an aircraft in a descending turn?" $\endgroup$ Commented Jun 29, 2023 at 12:29
  • $\begingroup$ @quietFlyer... Yes, my Bad, you're correct, I misread your comment. As you stated, "for a given AoA, .... load factor is entirely determined by airspeed.". I deleted that comment... $\endgroup$ Commented Jun 30, 2023 at 21:59

The first CFI is technically correct; you can safely use steep bank angles to descend quickly, taking care not to load the wings any more than required to sustain the turn, but using steep banks to recover a bad base leg setup is not a good idea for a new pilot.

Better to use a slipping turn, or just fly the question mark pattern you used, being careful to keep the turns coordinated and not succumbing to the urge to point the nose at the runway with rudder (a practice that has killed many pilots). If you have passengers on board, you aren't going to be using steep turns or slipping turns, unless you know your passenger is ok with them (they can terrify passengers).

When I tow gliders (Piper Pawnee), I use extremely steep bank angles for the turns to base and final, mostly to maximize my visibility to gliders in the circuit at or near my level, by presenting the wings in planform to them, and I'll roll to near 90 deg and just let the thing fall, pulling just enough pitch to sustain the turn rate. I won't even be at 1G in that case, and with the wings largely unloaded there's little risk of stalling.

But think of it as an advanced maneuver to do when you have experience, and not something you do with passengers.

  • $\begingroup$ Not sure if 300 hrs counts as new, but ok. I didn't necessarily "recover a bad base leg" but used geometry before employing additional tools such as flaps or slips. I don't understand your second para, in how you would propose to fly a slipping turn whilst carefully keeping the turns coordinated. Could you elaborate? I'd still be interested in the technical background of the manoeuvres that I and you in a way described, and how that lines up with the previous two responses. $\endgroup$ Commented Jun 27, 2023 at 14:09
  • $\begingroup$ Most ppl do question mark turns to final because the were late with the turn initiation. That's what I was referring to. What you were doing was S turning, only once, which is really a normal maneuver on a forced approach. A slipping turn is definitely uncoordinated, just another version of side slip, but you are keeping AOA low while doing it to reduce stall risk. If you fly an Ercoupe without rudder pedals, S turning to lose alt is the only option on any approach. $\endgroup$
    – John K
    Commented Jun 27, 2023 at 15:22
  • $\begingroup$ Thanks for the clarification; slipping and coordinated didn't quite make sense in your post. $\endgroup$ Commented Jun 27, 2023 at 16:05
  • $\begingroup$ I'm having trouble with this answer. You state "by presenting the wings in planform to them, and I'll roll to near 90 deg and just let the thing fall, pulling just enough pitch to sustain the turn rate." -- could it be the case that you are not pulling back on the stick enough to hold the glide path angle constant? Exactly as is arguably the case as you reach a 90-degree bank angle at the apex of a wingover, as is the nose is dropping through the horizon? That's a totally different situation than a steady-state maneuver. The G-loading can be (temporarily) much lower in the non-steady- $\endgroup$ Commented Jun 29, 2023 at 7:54
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    $\begingroup$ As you go steeper and steeper, more of the lift vector is oriented laterally. At 90 deg, there is no vertical component at all, and the airplane becomes a ballistic object, unless you force the fuselage to operate at an angle of attack and make lift, using top rudder. $\endgroup$
    – John K
    Commented Jun 30, 2023 at 17:08

I've merely altered back pressure to keep airspeed stable

This question is interesting as at also touches on steep turn and emergency descent turn techniques.

When one banks 30 degrees, vertical lift is reduced by cosine 30 × G = .866 If airspeed remains the same, angle of attack must be increased by around 15 percent, resulting in a load factor or 1.15 G.

But wait, lift is less than weight in a climb (or descent), as Peter Kampf pointed out. In a shallow descent lift requirement is less than 1 percent less.

So, if one is approaching at 65 knots and raises their stall speed from 50 to 53 knots, there is still a 12 knot margin.

OK, unless a 12 knot tailwind gust comes along. But you're still better off than the one trying a short field approach at 60 knots.

What is being done here is called an "S" turn, and can indeed be used if the approach is too high. Excellent that you are coordinated!

But know please in banked turns one can also use increasd speed, rather than increasing AoA, to satisfy the lift requirement. So, especially low and slow, dip the nose a bit if making a 30 degree turn. If the plane is staticly stable, it should automatically speed up a little even if you do not re-trim.

So, try it with your CFI. But you may find the instructor happier if you set up you landing perfectly starting with the downwind, right into a stable, straight final approach.

  • $\begingroup$ The 172 answer may be of interest. Also, I have some ideas on resurrecting a closed question, is that ok? $\endgroup$ Commented Aug 23, 2023 at 4:56
  • $\begingroup$ @ThomasPerry yes, that is fine. $\endgroup$ Commented Aug 23, 2023 at 10:39

This answer considers banking during gliding flight while maintaining a specific rate of descent. In addition, power needed to maintain the rate of descent is examined. The airplane evaluated was similar to a 172 having a flight weight of 2550 lb and wing loading of 14.7 lb/sq-ft. The plane was flown on paper assuming a near-idle, initially neutral, wind-milling propeller at zero thrust. The on-paper conditions easily accommodate powered descent within the constraints imposed. The assisting use of flaps is demonstrated by a low-wing concept aircraft.

The question specifically asks –

What is the load factor in a descending turn?

As posted, the question states two scenarios regarding a descending turn. The aircraft is described in a power-off, coordinated, descending turn, angle of bank is 30 degrees and gliding at a 600 fpm rate of descent. Airspeed is kept constant. The question as qualified is very interesting given the constraints regarding angle of bank and rate of descent. This problem is most easily evaluated by examination of the consequence of forces on the airplane in a banking turn within the gliding-flight definition posed in the question. In short, what is the aircraft loading factor, n, in a descending, gliding turn, and is that number easily calculated? - forces in gliding flight

The basic considerations involve forces on the airplane and attitude of the wing. A slightly adjusted sketch by Rina (above) is used to illustrate these forces. Here, the plane is gliding at lower velocity with approximate 15-degree left bank. Please note elements depicted in the sketch – the aircraft is not in, but is close to, level-wings gliding flight. The banked-aircraft flight path is gently curved, but instantaneously, flight is directed along the tangent to the flight-path. Nevertheless, two aspects of flight happen together for the gliding turn. First are forces on the aircraft as though incurred in a simple, wings-level glide. These forces all act through the aircraft center of gravity. They are depicted as $L_g$ and $D_g$ for gliding lift and drag, respectively. Together, these directed forces are in a vertical plane including the plane’s center of gravity and flight-path tangent. In sum, these forces, $L_g$ and $D_g$, oppose weight of the aircraft, $W$. Note that at the center of gravity located within the aircraft, co-ordinate axes x, y, and z intersect. Stating again, the flight path is directed through the center of gravity, as is spatial orientation of the aircraft. Note that in simple, wings-level, gliding flight, lift in the vertical plane, $L_g$, is directed forward of vertical by the gliding angle, $\gamma$, of the flight path.

For a simple gliding bank of 15 degrees, little else is changed from a wings-level glide except the orientation of lift, which is now directed off-vertical 15 degrees to the left. The illustration provides an estimated view of flight for our simulated 172 with these conditions. Note the angle of bank, $\phi$, and the lift-oriented vertical plane for a wings-level glide, mentioned above, is also tilted by angle $\phi$, 15 degrees leftward. This tilted plane also includes the angle of attack, $\alpha$, for the wing, and flight-path tangent. The flight-path angle, $\gamma$, is still in the real vertical plane, however. The weight of the aircraft, $W$, must now be borne by increased lift $L$ (and drag) incurred in the banked turn. This increase in lift is directly related to the bank angle and is determined in a manner similar to that in level flight. Also note force $F_c$, the centripetal force incurred by the horizontal component of lift, is directing the airplane in a leftward curved path. Lift in the glide, plus centrifugal force (the exact counterpoise of centripetal force in the turn), increase the force of lift incurred in the turn, and consequently increase lift-force loading on the wing. For conditions shown in the illustration, the proportion by which wing loading will increase, is easily calculated. This value is given simply as,

$$n = cos\gamma/cos\phi .$$

However, to calculate this number, given our angle of bank, $\phi$, in descending flight, we must also know our flight-path angle of descent, $\gamma$. We can estimate this angle simply as the arcsin($v_d$/$V$) were $v_d$ is the vertical velocity on descent, and $V$ is our flight velocity. For example, descent at 600 fpm is the same as descent at 10 fps, and a flight velocity (air speed) at 66 kts is about 112 fps. Consequently, the glide slope to be maintained is about 11.2 which is slightly better than the lift-drag ratio of about 10.2 for our airplane in these flight conditions. This means that power must be added, even though slight, for the aircraft to maintain 600 fpm descent at 66 kts, especially while gliding in a 30 degree bank. In other words, our airplane has too much drag at 66 kts to match a glide slope of 11.2. So a bit of power has to be added to flatten our glide and lessen our descent to 600 fpm. The instrument indicated pitch, noted in the analysis, is about +2.3 degrees.

Free-air Flight Performance of the Airplane

Free-air flight considers the airplane flying in a still-air environment. Most of the aerodynamic calculations assessing flight performance are for the plane in its parcel of air. In the current case, we will not consider wind. For the airplane under these conditions, the rate of descent in feet-per-minute is easily determined as the flight velocity in feet per second (fps) divided by the lift-drag ratio, $C_L/C_D$, of the airplane, the result of which in fps is multiplied by 60 to obtain rate of descent in fpm. Consequently, 60sec x 112 fps/10.2, gives the resulting rate of descent, 658 fpm. In a 30 degree bank at the same flight velocity, if our lift-drag ratio is about 9.3, our resulting rate of descent would be increased to about 720 fpm. The estimated performance of the simulated 172 shows it will not achieve an on-glide rate of descent of 600 fpm across its active range of flight velocities without the addition of power. Consequently, an attempt to pull up and lessen the rate of descent without the addition of power will exhaust airspeed and result in a stall. This statement is made simply on the basis of the estimated performance of the airplane, and in the interest of safety. In the abstract, let’s see how this was determined.

First, we should look at the 172 glide performance and the flight-required glide performance across the range of active flight velocities needed to maintain a rate of descent of 600 fpm. In each of the graphs being shown, the blue line marked by open circles will represent the flight requirement at a rate of descent of 600 fpm, darker blue line with solid circles is the simulated 172 performance in a wings-level descending glide, and red line with open circles is the simulated 172 performance in a descending glide at a 30 degree bank. An estimated polar for the simulated 172 was used in all of these determinations. The estimated drag was slightly increased in the assessment in order to conservatively portray the required application of additional power (or thrust). -

corrected image for flight conditions

As shown in the first graph (left, upper), the simulated 172 is able to come close to, but unable to match, the glide ratio needed to maintain a rate of descent at 600 fpm. The shown minimum rate of descent in a wings-level glide is about 650 fpm at a flight velocity of 66 to 71 kts. The second graph (left, lower) shows the sustained maximum drag that cannot be exceeded to maintain a rate of descent of 600 fpm. In a wings-level glide, the simulated 172 clearly exceeds this maximum across its range of performance flight velocities. The third graph (right) shows the difference between these two measures, namely, the required thrust necessary to flatten the glide slope of the 172 to match a 600 fpm rate of descent. In other words, in a wings-level glide at its relative minimum gliding velocity of 66 to about 71 kts, the estimated thrust necessary to flatten the glide slope to achieve a 600 fpm rate of descent, is about 16.5 lbs. This is just a very slight advance on the throttle, just enough (with a little trim) to see the rate of descent decline to 600 fpm.

Parallel conditions for a gliding, 30 degree bank, show the simulated 172 estimated rate of descent is relatively constant at 700 fpm or so in the range of 66 to 89 kts. Consequently, the necessary thrust to achieve 600 fpm in a 30 degree bank is a modest 68 to 80 lbs. This simulated 172, by nature of these gentle flying requirements, seems an easy and relatively forgiving airplane to fly.

Do all airplanes fly that way?

We might wonder, are other planes as easy to fly? Obviously, some are more challenging. Let’s take a look at a low-wing conceptual example somewhat similar to the 172, but being slightly heavier with a wing loading or 16.5 lb/sq-ft. We are, again, considering in parallel, the same aspects for descent at 600 fpm, exactly as done for the 172, but we also consider the use of flaps in a 30 degree gliding turn. These aspects are indicated just as in the graphic, above, but use of flaps is shown with lighter red features. -

corrected image for flight conditions

When a similar comparison is made, we can see the concept plane has much lower drag at lower flight velocities and, consequently, a better lift-drag ratio. These lower-drag factors can present difficulties if trying to bring the aircraft to a 600 fpm rate of descent at flight velocities below about 105 kts. The plane is simply too aerodynamically clean for the 600 fpm descent glide slope. Whether in a wings-level glide, or gliding 30 degree bank, the plane can be managed nicely for a 600 fpm descent at airspeeds near 120 kts, or somewhat above. But at velocities below 105 kts, more drag is needed to manage the glide. Flap deployment will increase drag and should be helpful; let’s see what happens in a 30 degree gliding bank.

For safe deployment in a gliding bank below 90 kts, flaps provide favorable on-glide flight conditions. The added drag provides needed assistance in managing the glide and rate of descent. Note that at 66 kts, on-glide conditions at 600 fpm descent are little changed, yet more favorable as power is slightly needed, or not at all, to maintain the rate of descent. Consequently, at 66 kts the application of slight power is indicated, but below 90 kts, little or no power is indicated. With an adjustable pitch prop, managing drag and maintaining a descent at 600 fpm across this range of flight speeds is not a problem, particularly with the use of flaps below 90 kts. Above 110 kts flaps are not required, but the gentle application of power is necessary to flatten the in-flight glide slope to maintain a descent of 600 fpm. Flying this plane without flaps, the pilot must be well ahead of their game. The use of flaps, however, makes flying this plane much more manageable.

  • $\begingroup$ Thanks for the amount of detail. I may be too stupid to understand on how it answers the original question though - quite a few points appear to not have anything to do with the question. I may need somebody to explain the relevance of the information, however well presented it appears at first glance. $\endgroup$ Commented Aug 30, 2023 at 8:52
  • $\begingroup$ @ExternalUse You're a smart fellow! Specifically the load factor in a descending turn is easily calculated. You provided a setup & shown are flight performance characteristics of a Cessna 172 indicating you could not maintain a 600 fpm descent without slight power. This was simply a general instance along w/comparison to another aircraft. Aspects shown are for various constant flight speeds. I don’t know what your plane was; this was just an example. The load-factor derivation & glide-slope relationship came from aspects given in the illustration, in case you wanted to investigate further… $\endgroup$ Commented Aug 31, 2023 at 18:38
  • $\begingroup$ @ExternalUse I definitely appreciate your comments, though. They are very helpful! $\endgroup$ Commented Sep 1, 2023 at 18:12

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