Lifting Out of the Weeds

When we take to the sky, we make a reasonable assumption that our airplane will fly if we keep its speed above stall speed. That’s true, right?

To answer that question, we need to look at the lift formula. Yes, that means math, but it’s pretty easy math, so read on with the confidence that you don’t need to pause here for another cup of coffee.

Sure, I could write the lift formula down and detail the meanings of the variables, but that could lead to confusion. Heck, when they get down into the minutiae of the lift formula, even aerospace engineers can get sideways about it.

So, instead of going down that path, let’s address what we actually want to know about the lift formula and how we can use it. What follows should be understandable and improve your intuition about lift.

Cutting through all the variables, math, and stuff we can’t change, the lift formula tells us two really important things:

1. Lift varies more or less linearly from no lift at zero angle of attack to maximum lift at critical angle of attack.
2. The total available lift varies by the square of calibrated airspeed (CAS). For our purposes here, we’ll assume our airspeed indicator doesn’t have an installation error (i.e., IAS = CAS in this article) but don’t forget that your airplane likely does.

We only need to know one thing — stall speed — to figure out everything else. Yes, we could use the lift formula to figure everything out, but we’re going to do something a little more practical (and fun). Let’s imagine we’re flying our airplane and doing very careful power-off stalls to determine stall speed in landing configuration (Vs0) and stall speed clean (Vs1).

We perceive lift as G-force. When the lift of the wing is equal to gravity, we experience straight-line flight (unaccelerated). The slowest we can go and maintain that 1G of lift is, yep, stall speed. At that speed, we will be at the airplane’s critical angle of attack. Any attempt to increase lift by increasing AoA will be met with a stall.

As we speed up, we have more lift available with which to maneuver. We can now figure out how much speed we’ll need in order to do whatever it is we want to do.

For example, if you have an airplane with a Vs1 of 60 KCAS, and you want to do a 60˚-bank steep turn, you need to be able to pull 2 Gs at or above stall speed. But what is stall speed at 2 Gs?

You must increase the speed by the square root of 2, which is 1.41, to avoid an accelerated stall. Therefore, you must go faster than 85 KCAS to avoid stalling.

Oops, I just did math. But it was easy and extremely useful. Let me summarize.

1. √"Number of Gs You're Gonna Pull" = some number 
2. Multiply that number by your airplane’s Vs1  
3. Go that fast to avoid stalling when pulling the "Number of Gs You're Gonna Pull."

And check this out: We can calculate Va this way too. If our airplane has a load limit of 3.8G, and our Vs1 is 60 KCAS, then Va = sqrt( 3.8 ) x 60, which is approximately 117 KCAS.

Another summary for you:

1. √("Your Airplane's Load Limit in Gs")  x  Vs1 = some number
2. That number is your Va.

Okay, enough math. Maybe you don’t want the hassle of using a calculator. So, let’s look at something pilots use all the time: a diagram.

Here is a Vg diagram, sometimes called a Vn diagram, for the Cessna 172 operating in the utility category (so, Va aligns with 4.4G instead of the 3.8G Skyhawk pilots normally think about). Note that this graph expresses airspeed in miles per hour. I have no idea why they didn’t use knots, but here we are.

The Vg diagram provides the speed and load factors on its axes for a particular Skyhawk. This is the “envelope” we talk about, which tells you what your airplane can and can’t do. The outer edges of the shape on the graph depict a stall.

The graph is pretty intuitive to use. If your load factor is 2G and your indicated airspeed is 60 mph, you’re stalling. Note that this graph assumes max gross weight.

A stall is depicted on the diagram as a line rather than a point. That is because stall speed depends on G-loading. If you unload down to a zero-G load factor, notice on the far left side of the graph that stall speed drops to zero. So, those lines really represent the critical angle of attack.

Let’s use this graph to understand how easy it is. How fast do you need to go to be able to do a 60˚-banked level steep turn?

Starting on the left side of the diagram, follow the 2G load line until it intersects the stall line. From that point, follow the line straight down to the answer. For our C172 loaded to max gross weight, that minimum speed is 88 mph, which we convert to 77 KIAS.

But who flies at max gross all the time? No one who burns fuel. So, how do we correct our speeds to accommodate different gross weights?  

More math! But fear not. It’s no more complicated than the performance calculations you already know.

We can use the square or square root function of lift and airspeed to calculate new limits on the diagram. Just take the square root of the ratio of current gross weight to max gross weight. Here’s the formula.

New speed = speed x √(gross weight / max gross weight)

Admittedly, that’s starting to look too much like engineering math for my taste. Here’s an easier way of framing it (for me, at least):

1. What’s your current gross weight? Okay, take that number and…
2. …divide it by max gross, which gives you some number.
3. Now, take whatever that number is and, as I say, “square root it” to get another number.
4. Now, take that number and multiply it by the stall speed at max gross.
5. That’s your stall speed at your current gross weight.

Let’s run through an example. You want to determine your stall speed in regular ol’ 1G flight, but you weigh less than max gross.  

The Vg diagram says that the 1G stall speed is 64 mph, which is stall speed at max gross. But you are arriving home after a long cross country with only about ⅓ fuel in the tanks. You’re definitely not at max gross.

Upon figuring out how much fuel you have aboard, you calculate that your gross is now 1950 pounds. So, let’s figure out our new stall speed at 1G.

1. Current gross is 1950 pounds
2. Divided by max gross, which is 2450, we get 0.7959
3. The square root of 0.7959 is 0.8921
4. Multiply 0.8921 x 64 mph, which is max gross stall speed

…and the answer is 57.0944. So, basically, 57 mph (or approximately 50 knots).

The math-y version looks like this.

57 mph = 64 mph x √(1950lb / 2450lb)

Seriously, this stuff is a piece of cake. Just be sure to correct for installation error (Skyhawk pilots, I’m looking at you).

Know Before You Go