Aircraft certified by the FAA under 14 CFR Part 23 have many V speeds, one of them which manufacturers are required to determine and publish in the POH is maneuvering speed, or Va.

In aviation, the maneuvering speed (Va) of an aircraft is an airspeed limitation selected by the designer of the aircraft. At speeds close to, and faster than, the maneuvering speed, full deflection of any flight control surface should not be attempted because of the risk of damage to the aircraft structure.

This speed varies with the weight of the aircraft. Why does this speed vary with weight? Additionally since manufacturers often only provide that value at max gross weight, how does one figure out what that Va airspeed is when not at max gross weight?


4 Answers 4


Compute maneuvering speed below max gross using the formula $V_A\sqrt{\frac{W_2}{W_1}}$, where $V_A$ is the maneuvering speed at max gross, $W_2$ is actual weight, and $W_1$ is max gross.

We can derive this relationship — or for any other V-speed such as stall speed of landing speed that varies with weight — from the lift equation. In steady-state flight, weight equals lift so

$$W_1 = \frac{1}{2} C_L \rho v_1^2 S$$

and likewise for $W_2$ and $v_2$. Dividing the first by the second cancels the coefficients and leaves

$$\frac{W_1}{W_2} = \frac{v_1^2}{v_2^2}$$

Take the square root of both sides and solve for $v_2$ to arrive at the general formula

$$v_2 = v_1 \sqrt{\frac{W_2}{W_1}}$$

John Denker provides an intuition for why the relationship works the way it does.

Unlike $V_{NO}$, the maneuvering speed varies in proportion to the square root of the mass of the airplane. The reason for this is a bit tricky. The trick is that $V_A$ is not a force limit but rather an acceleration limit. When the manufacturers determine a value for $V_A$, they are not worried about breaking the wing, but are worried about breaking other important parts of the airplane, such as the engine mounts. These items don’t directly care how much force the wing is producing; they just care about the acceleration they are undergoing.

By increasing the mass of the airplane, you decrease the overall acceleration that results from any overall force. (Of course, if you increase the mass of cargo, it increases the stress on the cargo-compartment floor — but it decreases the stress on unrelated components such as engine mounts, because the acceleration is less.)

Later in the same section, Denker clarifies.

Finally, we should note that there are two different concepts that, loosely speaking, are called maneuvering speeds.

  • The design maneuvering speed, which we can denote $V_{A(D)}$, is primarily of interest to aircraft designers, not pilots. The designer must choose a value for $V_{A(D)}$ and then build an aircraft strong enough to withstand certain stressful maneuvers at that speed. Higher values of $V_{A(D)}$ promote safety, by forcing the design to be stronger.
  • The maneuvering speed limitation, which we can denote $V_{A(L)}$, is of interest to pilots. It is an operating limitation. It appears on a placard in the cockpit. Lower values of $V_{A(L)}$ promote safety, by restricting certain operations to lower, less-stressful airspeeds.

Denker, John S., See How It Flies, §2.14.2 “Maneuvering Speed,” accessed 16 Aug 2015.

The discussion above pertains to the maneuvering speed limitation, i.e., values a pilot would find in a POH or on placards. For example, the humble Cessna 152 POH shows $V_A$ decreasing with decreasing weight: 104 KIAS at the max gross of 1,670 pounds, 98 KIAS at 1,500 pounds, and 93 KIAS at 1,350 pounds. The reader will also note that these values fit the general relationship given at the beginning of this answer.

  • $\begingroup$ @RyanBurnette: No, you have it backwards. $v_A$ is defined with stall speed and maximum load factor, and then all parts need to be made strong enough that they don't break off when maneuvering at this speed. $v_A$ is only a design speed - some speed needs to be defined, and that is the definition for one load case of the vertical. What is not covered is building up of sideslip with several rudder inputs and then adding the load of sideslip and full deflection, however, a good design will also cover this case (it would be easier to calculate, anyway). $\endgroup$ Aug 24, 2015 at 22:41
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    $\begingroup$ "So you say that Va goes up when the aircraft mass goes up? This is certainly not true." Quoting from Cessna 152 1980 POH: Va: 1350 Pounds -> 93 KIAS, 1500 Pounds -> 98 KIAS, 1670 Pounds -> 104 KIAS. $\endgroup$
    – E4z9
    Mar 25, 2018 at 18:38
  • $\begingroup$ I have often wondered whether in some a/c like gliders, the concept of Va going up as aircraft weight increases is not really appropriate. I understand why that makes sense in some a/c where Va is an acceleratn limit, but if the extra weight is being added exclusively to the fuselge in the form of pilots sitting in strong seats, and there's not a lot of stuff bolted on like engines and batteries, it seems in some aircraft perhaps Va ought to go down as weight goes up, because the mass of the wings absorbs proportionately less of the total lift force, so there is more strain on the wing bolts. $\endgroup$ Aug 6, 2019 at 2:30
  • $\begingroup$ @PeterKämpf I believe you are thinking of the design side and are far more qualified to discuss that aspect. This answer shows how to determine from more-or-less design value the different limitation values at varying weights. $\endgroup$
    – Greg Bacon
    Aug 6, 2019 at 10:46
  • $\begingroup$ @GregBacon: I get what John Denker says, but in order to limit accelerations, you limit accelerations (maximum and minimum gs), not speed. Again, this is getting it backwards. What he means is the speed for maximum control deflections, and - given a high enough v$_A$ during design - this will be lower in order to limit accelerations. But the better procedure is to limit those accelerations directly. $\endgroup$ Aug 6, 2019 at 21:06

At the normal stall speeds given by $V_s$, the load on the aircraft is 1g, and the lift is equal to weight. i.e $L = W$.

In case of maneuvering, the load factor is greater than one, and we have $L = nW$, with $n$ being the load factor.

We have, $L = nW = \frac{1}{2} C_L \rho V^2 S$.

This gives maneuvering speed , $V_a$ = $V_s \sqrt{n}$

It can also be written as $V_s$ = $\sqrt{\frac{2 n W}{\rho C_{Lmax} S}}$

At maximum weight, this gives, $V_A$ = $\sqrt{\frac{2 n W_{max}}{\rho C_Lmax S}}$

For other weights, we have, $V_a$ = $\sqrt{\frac{2 n W_{a}}{\rho C_Lmax S}}$

For the same load factor, we then have, $V_{a} = V_{A} \sqrt{\frac{W_{a}}{W_{max}}}$, where $V_{a}$ is the maneuvering speed at weight $W_{a}$ and $V_{A}$ is the maneuvering speed at maximum weight $W_{max}$

As the maneuvering speed depends on the load factor and the stall speed, it depends on the weight of the aircraft (which decides the stall speed). It is basically a structural limit.

Another thing to note is that the aircraft can sustain structural failure even below maneuvering speed when multiple large control inputs are given.

  • $\begingroup$ This is generally right - but note that $V_A$ can be be defined as anything from $V_S\sqrt{n}$ to $V_C$ - cruise speed. $\endgroup$
    – NathanG
    Aug 15, 2015 at 13:38

The maneuvering airspeed $v_A$ ensures that the maximum structural load is not exceeded even with maximum control surface deflection. Wikipedia says this is valid only when a single control surface is maximally deflected, but really the regulations try to make sure that any combination of single control inputs is safe. To determine the structural load, one must know the mass of all non-lift-creating parts. Fuel in the wing tanks does not count, as it is carried by the lift created right around it and does not increase the wing root bending moment.

Given: A maximum wing root bending moment $M_{b_{max}}$, a maximum load factor $n_{z_{max}}$, a maximum lift coefficient $c_{L_{max}}$ and a gross mass of all parts supported by the wing (fuselage, payload, fuselage-mounted engine, …) of $m_{nlc}$. Further, let's assume that the center of lift of one wing with the area $½S$ is out at a wing station $y_L$ (to be precise, use the center of lift with maximum down aileron deflection, measured from the attachment point of the wing at the fuselage), and the mass of the wings (plus fuel in the wing tanks and engines on the wing) is $m_{total} - m_{nlc}$. The lift of one wing is $$L = ½S\cdot\rho\cdot ½v^2\cdot c_{L_{max}}$$ and the root bending moment is $$M_b = \left(L - ½(m_{total} - m_{nlc})\cdot n_z\right)\cdot y_L$$ Note that the bending moment ist not just lift times lever arm, but is reduced by the part of the airplane's mass which is contained in the lift-creating parts, all calculated for one side! The maximum allowable speed at which this bending moment reaches the maximum allowed value is $$v_A = \sqrt{\frac{4\cdot\frac{M_{b_{max}}}{y_L}+2\cdot(m_{total} - m_{nlc})\cdot n_{z_{max}}}{S\cdot\rho\cdot c_{L_{max}}}}$$ Now you need only to know how the total flight mass $m_{total}$ of your aircraft is divided between what is carried at the wing root and what is the wing and attached to it. Increasing the payload carried in the fuselage, or adding fancy equipment in the cockpit will reduce $v_A$, and increasing fuel in the wing tanks will have no effect. If you add mass outside of $y_L$ (like with tip tanks), the bending load is decreased and $v_A$ goes up.

Note that conditions exist where the maximum tolerable structural loads can be exceeded even at $v_A$: If the pilot moves one control surface repeatedly with the eigenfrequency of a structural or a rigid-body eigenmode, the aircraft can build up angles of attack beyond those achievable with a single input. Consequently, the stresses can grow above those used for sizing the structure.

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    $\begingroup$ "Increasing the payload carried in the fuselage, or adding fancy equipment in the cockpit will reduce Va." Ah, now that's the answer the mathematically illiterate among us were looking for. $\endgroup$
    – Terry
    Aug 15, 2015 at 21:51
  • $\begingroup$ It's not true that Va ensures maximum structural load no matter what maneuver is flown. It only provides that assurance if a single control surface is fully deflected in one direction. See the other answer's reference of American Airlines 587 and the resulting CFR release. $\endgroup$
    – ryan1618
    Aug 15, 2015 at 22:24
  • $\begingroup$ @RyanBurnette: Technically, it needs two $v_A$s, one for longitudinal and one for lateral maneuvers. Also, the rate of flow angle change is important, as is the excitation of eigenfrequencies by a sequence of control inputs. This topic is much more complex than my answer shows, but I wanted to show the basic thinking behind the determination of $v_A$. $\endgroup$ Aug 16, 2015 at 20:26
  • $\begingroup$ @PeterKämpf I think you should edit the first sentence. Propagating this myth is could create a dangerous condition for pilots who operate under this misconception. $\endgroup$
    – ryan1618
    Aug 16, 2015 at 22:16
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    $\begingroup$ @RyanBurnette: Wikipedia calls it a myth, and Airbus got away with this interpretation that $v_A$ is only for single control surface deflections. But the idea is really that you can put the aircraft in any attitude with all control surfaces, and still parts don't break off. Take an accelerated stall: You fly a coordinated turn an pull until the maximum lift is reached. You use all control surfaces and still the aircraft must not break apart. You may even add sideslip - still the structure must stay intact. Wikipedia creates a myth of its own here! $\endgroup$ Aug 17, 2015 at 8:40

... So I went and watched the tutorial. Turns out Va is a case where STALLING is a built in SAFETY feature. Va is a speed limit, above where the airplane will exceed its G load limit before it stalls if abruptly manuvered.

How does weight factor in? A heavier plane flies at a higher AOA to generate adequate lift at a given speed, or the same AOA at a higher speed.

The universal AOA formula for the plane would be stall AOA/Va AOA is less than or equal to G load limit (Accounting for linearity of the lift vs AOA curve).

The heavier plane would need higher speed than a lighter plane to reach Va AOA limit in straight and level flight. Too small an AOA sets you up for exceeding G load limits in an event such as severe turbulence.

Properly working AOA sensors are valuable instruments for this application, as well as the airspeed indicator.

  • $\begingroup$ @Peter Kampf From your engineering point of view, let's design a Vstab/rudder. Should IT stall before "blow down"? Yes, here larger mass plays a role, as a full rudder deflection will stress a heavier weight more ( it's acceleration will be less) - it breaks, rather than "gives" or "rolls with the punch". Graphicly, you will see a higher peak load for a longer period of time. For placard Va, seems like they are referring more to departure from cruising AOA. $\endgroup$ Aug 6, 2019 at 15:55
  • $\begingroup$ So, for the Vstab/rudder, a brisk control application would put peak load at or near Tzero, the lighter one "gives" faster to reach "zero lift" zero load state (opposite forces rebalance). $\endgroup$ Aug 6, 2019 at 16:18

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