Maneuvering Speed

11 Answers

1. 
   

   
   

   Gary Moore on Apr 06, 2011

   “The maneuvering speed decreases as the aircraft’s weight decreases from maximum takeoff weight because the effects of the aerodynamic forces become more pronounced as its weight decreases” – there is a pretty good explanation over at http://en.wikipedia.org/wiki/Maneuvering_speed

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2. 
   

   
   

   John A Lindholm on Apr 06, 2011

   Simple answer is that Va varies with weight because it’s predicated on design and ultimate “g” loading…..  therefore, the lighter the aircraft is, the lower Va is.  Basically, it’s the airspeed an aircraft will stall at before it breaks.  

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3. 
   

   
   

   Micah on Apr 06, 2011

   
   Why? Lift available. That’s how I describe it. This is the explanation I use.
    
   L = AOA * AS (simplified version)
    
   Because greater flying weights require greater lift, these require higher angles of attack at a constant airspeed. 
    
   L = W (unaccelerated flight)
   W = 2,000 ; L = 2,000
   (2,000) = AOA * 100 kts ; (Values are placeholders and not actual values; these values represent real values but are not real) In this case the AOA = 20
    
   In this airplane you add 1 pax and bags. Now:

   W = 2,500 ; L = 2,500
   (2,500) = AOA * 100 kts ; In this case the AOA = 25. 
   The point is that greater weight requires higher angle of attack to maintain the same airspeed. As John said, you’re closer to stalling AOA in the same conditions at higher gross weight. But we don’t fly AOA and instead fly AS, so we know that the maximum safe value (in terms of structural failure vs. lift failure) of airspeed increases as weight increases. At a lower airspeed you have more “available lift” because you require less angle of attack, so your safe Va is necessarily lower.

    
   

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4. 
   

   
   

   Brian on Apr 06, 2011

   “Simple answer is that Va varies with weight because it’s predicated on design and ultimate “g” loading…..  therefore, the lighter the aircraft is, the lower Va is”
    
   Except it doesn’t answer anything. Example: What causes lift? Lift is a force predicated by aircraft design…. therefore, it allows us to remain aloft. Substitude lift for any other aerodynamic characteristic and we could completely cut aerodynamics out of a curriculum. 
    
   John, I don’t mean to attack you. Though, in my opinion, sending a student off to a checkride with these kinds of answers does a diservice to the student. 
    

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5. 
   

   
   

   Brian on Apr 06, 2011

   Micah, have you by chance flipped through my sylabus or read forum posts with that presentation? I ask because it is similar to my method of presentation, except I add a short discussion on ‘throw’ during the brief, along with a picture, to complete the topic.
    
   For instance, if we take your example:
    
   Aircraft 1 – 2000 pounds; AOA 20
   Aircraft 2 – 2500 pounds; AOA 25
    
   Recalling that each aircraft is the same, I’d make up a critical AOA; let’s say 30 in this case. Exceeding G-load has to do with the amount of time the aircraft has to accelerate. How much time you have to accelerate has to do with the distance between your flying AOA (20 or 25 in our example) and the critical AOA; 30 in this example. 
    
   The lighter aircraft, when flying at the same speed as the heavier aircraft, will fly at a lower AOA. Therefore, the distance between flying AOA and critical will be greater than that of the heavier plane. Greater distance means more time to accelerate and greater g-forces created. Just remember, time to accelerate from flying AOA to critical AOA directly translates to how many g-forces you can potentially get before stall.
    
   On a final note, a picture, depicting flying AOA versus critical for various aircraft weights, makes this portion of the explanation far easier to grasp. Unfortunately, I was unable to locate ones I know I’ve seen before online. Would anyone be able to link one?

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6. 
   

   
   

   John A Lindholm on Apr 06, 2011

   Brian…..  no offense taken, but you quoted the wrong sentence.  The one simple answer that will satisfy most examiners is: 
    
   Basically, it’s the airspeed an aircraft will stall at before it breaks.  
    
   Obviously, I would  give a student a more detailed explanation of how to get to that conclusion, but when understood, that simple statement sums it all up.  Usually I use the example of two identical boats, one heavily loaded, the other lightly loaded.  When they encounter rough water at the same speed, the light boat will bounce off the waves and possibly break while the heavy boat will plow thru the waves.  Same goes for the effect of weight on an aircraft that encounters rough air.

   +14 Votes 16 Votes  2 Votes

7. 
   

   
   

   Brian on Apr 08, 2011

   “Basically, it’s the airspeed an aircraft will stall at before it breaks.” 
    
   This tells us what maneuvering speed is. How does it answer: “Why does maneuvering speed varie with A/C weight.”? 
    

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8. 
   

   
   

   Andy Neumann on Jul 11, 2011

   That wikipedia article is spot on, but is that really what the examiner wants to hear from you as you pretend to be teaching him like a student pilot?  It’s going to go way over nearly everyone’s head. 
   Here’s my best attempt to answer this question succinctly and clearly as on a CFI oral exam. I’m not sure I succeeded.
   1. Airplanes are designed to withstand loads up to a certain level.  We measure these loads in terms of g’s, and that’s why in the limitations section of the AFM you have load limits in terms of g’s. 
   2. When you increase the AOA, you increase the load factor until you reach the stalling AOA at which point the load factor reduces rapidly.  However, if you reach the breaking load factor before you reach the stalling AOA, the airplane breaks and the party’s over.  Conversely, if the stalling AOA is reached before the breaking limit is reached, then you can’t break the airplane because it will stall first (at which point the load factor reduces rapidly). 
   3. The question then becomes, which airplane is flying closer to it’s stalling AOA–and therefore more likely to stall first before it breaks?  The heavier airplane.  Thus, the heavier airplane can be flown safely at a faster speed than the lighter airplane without worry of exceeding it’s limiting designed load factor. 
   ugh

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9. 
   

   
   

   Armando Vazquez on Jul 23, 2011

   First of all we need to know the meaning of Va : ” is the maximum speed where full, abrupt control movement can be used without overstressing the airframe”, so this speed is calculated to avoid stress in the structure of the airplane, is a structural limitation. If we recall newton 2nd law that states : Force= (mass)x (accelaration), then we have the answer. Weight is the product of (mass) x ( gravity of the earth), so, If our airplane is carrying less weight in it wings…. the aircraft is capable of flying slower because we have “extra capacity” in the airframe to cope with the imposed forces to the airframe. In case we are flying the aircraft at it maximum gross weight, the speed is higher beacuse other wise if we exceed that speed the forced imposed to the airframe can produce structural damage.

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10. 
    

    
    

    Jordan on Jun 23, 2012

    Just to add a little bit!
    Maneuvering speed remains the same regardless of weight… when considering structrual limits. The airplane doesn’t feel acceleration, it feels forces. It will break apart because of forces, not accelerations. Consider two airplanes, a 2,000 lb and a 4,000 lb airplane. The engineers have determined that at 20,000 lb of lift force, the wings will separate from the fuselage. So for the 4,000 lb airplane that would require a 5G acceleration. For the 2,000 lb airplane that would require a 10G acceleration. This shows that the limit load factor actually increases with a decrease in weight.
    Regarding the speed remaining the same: if you want to analyze an airfoil’s lift capability, it doesn’t matter if it’s attached to a 2,000 lb fuselage or a 4,000 lb fuselage. All that matters is: L=0.5*rho*V^2*S*Cl. Consider the density and surface area of the wing to be equal to each other (they’re identical wings). So all that remains to affect lift are velocity and Cl. Well since they’re identical wings, the Cl would also be identical and they’d both be at a maximum. So now it’s just down to velocity. If you want to make lift the same in both cases you have no choice but to make the velocity the same as well. Therefore, maneuvering speed remains the same with a decrease in weight.
    Can anyone guess why we instead consider maneuvering speed with regard to a limit load factor instead of forces? 

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11. 
    

    
    

    Francisco on Dec 10, 2014

    Mr. Jordan’s argument is fine for the Wing structure, but not the whole aircraft structure. The easiest to remember is the engine mount.
    Regardless of load, the engine still has the same mass in at max gross and at below max gross weights, therefore more G’s (in the lighter aircraft) will result in higher forces or stress on the engine mounts, and in a single engine also the firewall and the fuselage structure leading to the wing root.
    The lighter airplane will not bend the wing at Va, but you may bend the fuselage!

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