I was just brushing up on basic aerodynamics and I have something I want to clarify about flying in the region of reverse command. I went up in my 172 the other day and noticed the following while practicing slow flight:

  • I'm on straight and level flight, and I enter slow flight (dirty) the way I usually do. I pull the power back and then I raise the nose to a level attitude slightly to bleed off some airspeed while maintaining the same altitude. When I entered the back side of the power curve, I pitched for an airspeed that was significantly above Vso, but below L/D max, which in this case happened to be 55-56 Knots. However, this time I didn't add any power and I just pitched for 55 knots to see what would happen. The airplane started descending (sinking), which is really what I expected it to do.

My limited understanding of why this happens is that total drag (mostly induced drag at high AOA) exceeds thrust available and the airplane develops a sink to try to regain equilibrium. Is this correct? Can someone with more knowledge explain why this happens? Maybe I'm overthinking it... I just know that high drag at low airspeeds typically results in a high sink rate. I just can't explain exactly why. Some people have said that weight exceeds lift, but I'm pretty sure this isn't the case as the airplane is not stalled.

  • $\begingroup$ at higher sink plane is just using more potential energy (altitude) to maintain speed. $\endgroup$ Dec 17, 2018 at 3:18
  • $\begingroup$ Please elaborate on the "this" in "My limited understanding of why this happens is that . . . . " $\endgroup$
    – skipper44
    Jul 28, 2021 at 12:41

2 Answers 2


total drag (mostly induced drag at high AOA) exceeds thrust available and the airplane develops a sink to try to regain equilibrium.

That is exactly right. At best L/D, induced drag is half of total drag and at lower speeds it becomes dominant, growing with the inverse of airspeed squared.

Some people have said that weight exceeds lift, but I'm pretty sure this isn't the case.

You did well not to listen to those people. If the aircraft doesn't accelerate downwards, lift still equals weight. At a constant sink rate the aircraft does not accelerate.

As you say, the aircraft is in an equilibrium. Because you choose to add less power than what was necessary for level flight, the aircraft behaved similar to a glider. Allow me to use a glider for my explanation. Below is one with the dominant forces added as arrows in level flight. Clearly, there is nothing to compensate drag (red):

Glider in level flight

This airplane is not in equilibrium and does what you did in order to slow down: It flies level while drag is exceeding thrust (which is per definition zero in a glider).

So what to do? Just like when flying a turn, the pilot will now tilt the lift vector. That is the biggest force at his command, and he does so by pitching down. Now the airplane is in a glide, the direction of flight is slightly down and the lift vector (which is per definition perpendicular to the direction of the airspeed, which again is equal to the direction of flight if there is no wind) is equally tilted forward. This is what the next sketch shows:

Glider in a glide

Now we have a component of lift pointing forward, and that component is equal and opposite to drag. The rest of the lift is needed to compensate the weight, so the lift force is a tiny amount larger than weight. The sink rate is speed times the tilt angle, which in turn is what is needed to have the sine of lift $L$ equal drag $D$. Mathematically speaking, this is: $$D = L\cdot sin(\gamma)\;\;\;\text{and}\;\;\;\gamma = arcsin\left(\frac{D}{L}\right)$$ with $\gamma$ as the angle of the direction of flight relative to the horizon. The sink rate $v_z$ is $$v_z = v\cdot sin(\gamma) = v\cdot\frac{D}{L}$$ That is the direct answer to your question: The sink rate is the product of speed and the inverse of the lift-to-drag ratio. At small angles the ratio is approximately equal to its sine when expressed in radians, so I left the trigonometry out of the last equation.


First let's add another point so we can compare all 3 at 55 knots "backside of the power curve".

  1. 55 knots straight and level
  2. 55 knots descending at partial power
  3. 55 knots descending no power.

1 is horizontal to the horizon. 2 is at a descending angle to the horizon. 3 is descending more rapidly. This is sort of what you do when you approach for landing and modulate your throttle. Remember PITCH CONTROLS AIRSPEED. When you cut power, the nose will drop and speed will increase ("the plane develops a sink to try to regain equilibrium"), until the elevator raises the nose, lowering airspeed.

This relationship (equilibrium) of forward set CG and elevator pitch is fundamental in understanding how pitch controls speed. Now let's add number 4, more power than 1. Your plane will CLIMB at 55 knots!

Now "high drag at low airspeed typically results in a high sink rate". This is what flaps are for. The more drag, the steeper the angle of descent.

Notice you can change your angle of descent (sink rate) at CONSTANT AIRSPEED by either reducing power or adding flaps. Both techniques can be used for landings.


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