#### What is Maneuvering Speed?

A common FAA knowledge test question is “Which of these V speeds is not found on the airspeed indicator?” The answer is usually design maneuvering speed, also known as V_{A}. Many learners might then wonder *why* this speed is not found on the airspeed indicator along with other important V speeds such as V_{NE} and V_{FE}. Perhaps it is really not that important? Well, the FAA certainly believes that it is important enough to warrant a test question! We will learn exactly what maneuvering speed is and why, despite not being present on the airspeed indicator, it is still a very relevant V Speed.

One of the reasons V_{A} isn’t found on the airspeed indicator is probably because, like V_{X} and V_{Y}, **maneuvering speed changes with weight**. The speed found in the POH or placard is applicable for a particular weight, usually maximum gross weight. We often do not fly our airplanes at its maximum certified weight, so what happens to maneuvering speed as we fly at these lighter weights? It decreases! The next few sections offer both a simple and thorough explanation as to why that is, as well as provide ways to calculate a “new” maneuvering speed if flying below maximum gross weight.

##### The Simple Explanation

Turbulence results in vertical gusts that momentarily increase or decrease the angle of attack that an airplane is flying at. There exists a direct relationship between the amount of lift produced and the angle of attack, so if angle of attack is doubled, then the amount of lift produced is also doubled. Normally, the amount of lift produced by an airplane is equal to weight – we call this 1G. “G’s” really just describe the ratio of lift to weight. If the angle of attack is doubled, thereby doubling lift, the ratio of lift to weight is now 2 to 1 – we call this 2G’s. Airplanes are certified to structurally handle a certain amount of G’s – this is called the “limit-load factor” and is 3.8 G’s for the normal category. Maneuvering speed helps us ensure that we never cross the limit-load factor.

How does this work? For every cruising airspeed, the airplane is flying at a particular angle of attack. During cruise, the airplane is flying at a relatively low angle of attack. During slow flight, the airplane is flying at a high angle of attack. For primary flight training, we know that the airplane will stall if it exceeds the “critical angle of attack,” which happens to be about 18 degrees or so. Let’s say the airplane is cruising at a speed that has it flying at 9 degrees of angle of attack. Because the airplane stalls at 18 degrees angle of attack, the angle of attack can at most be doubled before stall. In other words, the airplane will stall at 2G, which is less than 3.8G. Now, if that same airplane was instead cruising at a speed that has it flying at 2 degrees of angle of attack, the angle of attack can be increased 9-fold before stall. That means the stall will occur at 9G! This is much more than what the airplane is structurally certified for.

Because a stall will reduce the G’s experienced, we would rather have the airplane stall momentarily than experience lasting structural damage. If we fly above maneuvering speed, the airplane can *aerodynamically *support the G’s, but cannot *structurally* support the G’s (greater than limit-load factor). If we fly below maneuvering speed, the airplane can *structurally* support more G’s, but cannot *aerodynamically* support more G’s (stall occurs at a G loading below limit-load factor). The maneuvering speed is the boundary between these two categories – it is the exact speed where the airplane will stall at the limit-load factor.

So how does weight play a role? If the airplane is lighter, it needs to produce less lift to counteract weight, which means that the angle of attack for any given cruise speed will be lower than what it was before. Previously, the cruise speed may have put the airplane at 6 degrees angle of attack, which means that it would have stalled at 3G’s. But now, that same cruise speed may put the airplane at only 4.5 degrees angle of attack, which means that it will experience 4G’s before stalling! Flying at this cruise speed is now unacceptable considering that the limit-load factor is 3.8 G’s. For this new weight, we must now fly at a different speed: one that results in a higher angle of attack. This will ensure that the airplane will stall before exceeding the limit-load factor. To increase the angle of attack that an airplane cruises at, speed must be decreased (think again to cruise vs. slow flight). Thus, when weight is reduced, maneuvering speed is also reduced.

How much should maneuvering speed be reduced for a given decrease in weight? There is a formula to calculate it! All we need is maneuvering speed at a given weight, along with a new weight. **Divide the new weight by the published weight, take the square root, and multiply by the published maneuvering speed.** Or, in math speak:

v_{A, new} = v_{A, old} * \sqrt{\frac{W_{new}}{W_{old}}}

There is also a rule of thumb, if you find square roots inconvenient or scary. **Take the percent decrease in weight and decrease maneuvering speed by half of that.** Which is better? It is important to note that the rule of thumb will always result in a higher value for maneuvering speed than the formula. This means that flying at the speed given by the rule of thumb may result in exceeding limit-load factor before stall. However, the difference between what the rule of thumb yields and what the formula yields is about 1 knot – insignificant for practical purposes. So, it’s hard to go wrong with using either. In the next section, we’ll dive a bit deeper with the derivation of the formula, the derivation of the rule of thumb, and a comparison of both. Caution: this involves some Calculus.

##### The In-Depth Explanation

Before we get into the math, here are definitions of the variables used in the formulas:

L – Lift

W – Weight

α – Angle of Attack

C_{L} – Coefficient of Lift (proportional to α)

ρ – Air Density

v – Velocity

S – Wing Planform Area

k – An Arbitrary Constant

c – Another Arbitrary Constant

d – Yet Another Arbitrary Constant

v_{A} – Maneuvering Speed

Load \ Factor = \frac{L}{W}

Equation 1: Definition of load factor. As discussed previously, it is simply the ratio of lift to weight.

C_L = k\alpha

Equation 2: Coefficient of lift is proportional to angle of attack. So, if angle of attack is doubled, coefficient of lift also doubles.

\begin{aligned} L = \frac{1}{2}C_L\rho v^2S = \frac{1}{2}k\alpha\rho v^2S \end{aligned} = c\alpha

Equation 3: Lift equation. Assuming that density, velocity, and wing planform area are constant, lift is directly proportional to angle of attack.

v_{A} = \sqrt{\frac{2L}{C_L\rho S} } = \sqrt{\frac{2(W * Limit \ Load\ Factor)}{k\alpha_{critical}\rho S} }

v_{A} = d\sqrt{W}

Equation 4: Equation 1 is substituted in for lift. Since limit-load factor, k, critical angle of attack, density, and wing planform area are all constants, it is all combined into d. Thus, maneuvering speed is proportional to the square root of weight

\frac{v_{A, new}}{v_{A, old}} = \frac{d\sqrt{W_{new}}}{d\sqrt{W_{old}}} = \sqrt{\frac{W_{new}}{W_{old}}}

v_{A, new} = v_{A, old} * \sqrt{\frac{W_{new}}{W_{old}}}

Equation 5: To eliminate d, we take two combinations of weight and maneuvering speed. The “old” combination is what is given in the POH or placard. The new weight is also known. So, the unknown new maneuvering speed is isolated and the formula is derived.

##### A Rule of Thumb

Now let’s derive the popular rule of thumb: **Take the percent decrease in weight and decrease maneuvering speed by half that amount.** To do this, we will use Calculus – specifically, a first-order Taylor Series expansion.

f(x) \approx f(a) + f'(a)(x-a)

f(W) \approx f(W_{old}) + f'(W_{old})(W-W_{old})

Equation 6: First order Taylor Series expansion for a function centered around point *x=a*. In other words, we can determine a good approximation for our function (formula) for weights approximately equal to the given weight.

f(W)= v_{A, old} * \sqrt{\frac{W}{W_{old}}}

f(W_{old}) = v_{A, old}

f'(W) = \frac{v_{A, old}}{2W_{old} *\sqrt{\frac{W}{W_{old}}}}

f'(W_{old}) = \frac{v_{A, old}}{2W_{old} }

f(W_{new}) \approx v_{A, old} \ + \frac{v_{A, old}}{2W_{old}} (W_{new} - W_{old})

= \frac{2v_{A, old}*W_{old}}{2W_{old}} + \frac{v_{A, old}* W_{new}}{2W_{old}} - \frac{v_{A, old}*W_{old}}{2W_{old}}

= \frac{v_{A, old}*W_{old}}{2W_{old}} + \frac{v_{A, old}* W_{new}}{2W_{old}}

v_{A, new}= v_{A, old}*\frac{1}{2}[\frac{W_{old} + W_{new}}{W_{old}}]

Equation 9: Mathematical representation of the rule of thumb

The steps above calculate the required terms for equation 8 and then perform algebraic simplification. Finally, we arrive at equation 9. To confirm that this is indeed the mathematical representation of the rule of thumb, let’s plug in some numbers. If old weight is 100, and the new weight is 80, that is a 20% reduction in weight. According to this formula, old maneuvering speed will be multiplied by 0.9, which is a 10% reduction in maneuvering speed.

##### Comparison

How good is the rule of thumb, and is it OK to use? As mentioned before, the rule of thumb is only a good approximation at weights close to gross weight. This is clear when we plot the two relations (figure 2), where the two curves predict essentially the same maneuvering speed towards the right end. To highlight an egregious error: if weight is 0% – in other words, a 100% decrease in weight – then maneuvering speed should be 0. However, the rule of thumb says that maneuvering speed should be decreased by only 50% (half of the 100% decrease in weight)!

This is an exaggerated example, however, because empty weight constitutes a large part of the total weight. Let’s take a case study: a 1979 Piper Warrior II. This airplane has a gross weight of 2325 lbs. Empty weight is 1500 lbs, which is already almost 65% of the total weight. After adding a 150 lb pilot and a quarter tank of fuel, the realistic minimum weight that this airplane should be flying at is around 1725 lbs. This is 74% of gross weight. Looking at figure 3, the percent difference in predicted maneuvering speeds when loaded to 74% of gross weight is 0.98%. For this airplane, where maneuvering speed at gross weight is 111 knots, the rule of thumb yields a result 1 knot higher than the actual formula – a relatively insignificant error. Also, as per figure 3, this is the worst possible error, because the error decreases as weight is increased. So we conclude that the rule of thumb is good for practical purposes.