Maneuvering Speed

:yeahthat:

And flying at a heavier weight means for any given airspeed you are closer to the critical angle of attack than you are when flying at a lighter weight for the same airspeed.
 
To overstress an aircraft, your control input or something else (updraft for example) has to displace the airframe in pitch at a rate/acceleration that exceeds the acceleration/G limits of the aircraft. Simple definition right there, if you can get past the physics'y wording. In order to generate this rapid movemement about pitch, you have to do it with lift. The most lift you can instantaneously generate at a given airspeed is seen when you make a rapid, full deflection of the control stick/yoke into your lap. Assuming you were flying in level 1 G flight prior to making this control movement, then you were already generating enough lift to exactly oppose the weight of the aircraft. In order to accelerate the aircraft in pitch enough to overstress it, you will now have add a certain amount of excess lift (above the amount needed for level flight) to create this movement/acceleration (see first sentence definition). We will call this amount of excess lift X. This means that the total lift required to cause the overstress is the sum of X and the weight of the aircraft (which again equates to the amount of lift needed to just sustain level 1G flight). As you can see, with a heavier aircraft, this sum total increases to a greater total amount of lift required to overstress. As most pilots know, in very general terms, you can generate more lift at a higher airspeed than you can at a lower one. So putting it all together, the more the weight goes up, the more total lift you need to overstress, and thus the greater airspeed you need to generate this amount of lift. So (using made up numbers here) in a generic light single, if I am flying at 120 kts at a weight of 3000 lbs, I can generate enough lift to throw 3 G's (the load limit in this example aircraft) on the airplane and not stall the wing. If I am flying still at 120 kts, but now I weigh 5000 lbs, my wing can no longer generate enough total lift to both oppose my extra weight, AND generate enough excess lift to reach 3 G's......I hit 2 G's before the wing stalls perhaps. In other words, I will need to fly faster at 5k #'s to generate enough total lift to cause the same acceleration of 3 G's because my wing can't do it at 120 knots. For this aircraft, the manuevering speed at 3k #'s would then be 120 kts, and V_a would be some airspeed higher than that at 5k #'s. Maybe that makes sense, maybe it doesn't. The other easy way to say it is F = ma. Or more specifically a = F/m. Weight being a function of m. Lift being another way of saying F, and a being G force. With a fixed F (maximum lift possible at a given airspeed), a increases as m decreases.
 
Yeah sorry I think I blacked out a little while typing that. Just tell him/her to push the "I believe button" :D
 
///AMG said:
Yeah sorry I think I blacked out a little while typing that. Just tell him/her to push the "I believe button" :D
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staples-easy-button.jpg



In seriousness, as already stated, it is all about the 'throw' so to speak. Flight at any given speed requires a greater AOA with greater weight. If that concept is understood, make sure it is, explaining Va changes with weight is easy. I like to explain it using a simplified lift formula: lift = speed * AOA.
 
I like to think of it this way:

A heavier aircraft requires a greater angle of attack than a lighter aircraft at any given airspeed in order to maintain level flight. Because of this, the heavier aircraft has less reserve angle of attack before it stalls and could possibly reach the limit load factor. The heavier aircraft can be flown faster without exceeding the limit load factor because it will stall much sooner than it would if it were lighter.
 
There are really two lines you can take to explain this. The first is the angle of attack explanation. Saying basically, that as the aircraft weighs more it is operating closer to the critical angle of attack.

The other explanation is the load factor explanation. As the plane flys faster the wing produces more lift and is able to apply more force to the airframe. Your job is to keep the amount of lift generated below an amount that would exceed the rated load factor the the aircraft, if full control throw were applied.

Pick one that works for you and have fun.
 
FDX8891 said:
I like to think of it this way:

A heavier aircraft requires a greater angle of attack than a lighter aircraft at any given airspeed in order to maintain level flight. Because of this, the heavier aircraft has less reserve angle of attack before it stalls and could possibly reach the limit load factor. The heavier aircraft can be flown faster without exceeding the limit load factor because it will stall much sooner than it would if it were lighter.
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That is the correct answer.

On my last 135 checkride, The FSDO guy who used to teach aerodynamics said this topic was the least understood among flight instructors. He stated that "Heavy = harder to throw it around" is partially correct, but not the correct answer.

Rod Malacho explains it very well if you can look up his article.
 
azaviator08 said:
Can anyone tell me a simple way to explain to my student how maneuvering speed decreases with a decrease in weight??
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You know, I just print this off:
http://www.aopa.org/members/ftmag/article.cfm?article=829

And have them read it. Every single student. Examiners will be more impressed when your student fully understands the subject then just giving a statement like, "harder to push a heavier object around" yet they still dont understand it.

Also its easy to teach because all you gotta do is print it off and have them read it.

More and more of my ground-school now days is me doing less teaching, but giving them the resources to learn. I got a whole folder of files that I created one page at a time that I print off and give to every student at the appropriate time. Works well :)
 
///AMG said:
To overstress an aircraft, your control input or something else (updraft for example) has to displace the airframe in pitch at a rate/acceleration that exceeds the acceleration/G limits of the aircraft. Simple definition right there, if you can get past the physics'y wording. In order to generate this rapid movemement about pitch, you have to do it with lift. The most lift you can instantaneously generate at a given airspeed is seen when you make a rapid, full deflection of the control stick/yoke into your lap. Assuming you were flying in level 1 G flight prior to making this control movement, then you were already generating enough lift to exactly oppose the weight of the aircraft. In order to accelerate the aircraft in pitch enough to overstress it, you will now have add a certain amount of excess lift (above the amount needed for level flight) to create this movement/acceleration (see first sentence definition). We will call this amount of excess lift X. This means that the total lift required to cause the overstress is the sum of X and the weight of the aircraft (which again equates to the amount of lift needed to just sustain level 1G flight). As you can see, with a heavier aircraft, this sum total increases to a greater total amount of lift required to overstress. As most pilots know, in very general terms, you can generate more lift at a higher airspeed than you can at a lower one. So putting it all together, the more the weight goes up, the more total lift you need to overstress, and thus the greater airspeed you need to generate this amount of lift. So (using made up numbers here) in a generic light single, if I am flying at 120 kts at a weight of 3000 lbs, I can generate enough lift to throw 3 G's (the load limit in this example aircraft) on the airplane and not stall the wing. If I am flying still at 120 kts, but now I weigh 5000 lbs, my wing can no longer generate enough total lift to both oppose my extra weight, AND generate enough excess lift to reach 3 G's......I hit 2 G's before the wing stalls perhaps. In other words, I will need to fly faster at 5k #'s to generate enough total lift to cause the same acceleration of 3 G's because my wing can't do it at 120 knots. For this aircraft, the manuevering speed at 3k #'s would then be 120 kts, and V_a would be some airspeed higher than that at 5k #'s. Maybe that makes sense, maybe it doesn't. The other easy way to say it is F = ma. Or more specifically a = F/m. Weight being a function of m. Lift being another way of saying F, and a being G force. With a fixed F (maximum lift possible at a given airspeed), a increases as m decreases.
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Whoa :drool: Thanks for the physics lesson - I get that, but I think many students would get lost at F=ma.

"You must be an economist. Your answer, while technically correct, is useless." ;) :laff:

A heavier aircraft requires a greater angle of attack than a lighter aircraft at any given airspeed in order to maintain level flight. Because of this, the heavier aircraft has less reserve angle of attack before it stalls and could possibly reach the limit load factor. The heavier aircraft can be flown faster without exceeding the limit load factor because it will stall much sooner than it would if it were lighter.
Click to expand...

That's the explanation that's parse-able and verbally teachable without a white board. The tl;dr factor applies to verbal explanations, too, and you don't have to be a physicist to fly an airplane.
 
Autothrust Blue said:
Whoa :drool: Thanks for the physics lesson - I get that, but I think many students would get lost at F=ma.

"You must be an economist. Your answer, while technically correct, is useless." ;) :laff:
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I kinda liked AMG's explaination - I sorta always looked at it like Va = the "Velocity of Acceleration" in relation to F=ma and all of it's implications.

One thing though that I would add is that in some aircraft the category changes (ie limiting load factor is higher) with weight/cg changes so inspite of weighing less, the structural integrity can be higher resulting in a miminal change in speed for a large difference in weight....something to think about I guess.

A lot of it probably just depends on the aircraft and what exactly is the limiting factor (physical object speaking) - engine mounts, wings, trim tabs, etc
 
ryan1234 said:
I kinda liked AMG's explaination - I sorta always looked at it like Va = the "Velocity of Acceleration" in relation to F=ma and all of it's implications.
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True - AMG - please don't read my snarky comment as anything other than snarky - I understand what you said, but I've also had physics. :cool: I think it would be difficult to explain to someone who hadn't had it, was the point.

:beer:
 
Autothrust Blue said:
True - AMG - please don't read my snarky comment as anything other than snarky - I understand what you said, but I've also had physics. :cool: I think it would be difficult to explain to someone who hadn't had it, was the point.

:beer:
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No worries. Just threw the mathematical stuff in there as an alternative, in case said student was more of a technical thinker. I think most of us know that this route would probably result in eye glazing in 90% of flight students
 
It seems to me that maneuvering speed isn't just about overstressing a plane, but also about control. A 160-square-foot wing, at its critical angle, at a given speed will have the same stress in the spars whether the plane has a skinny pilot and one gallon of fuel or full of passengers and fuel. Imagine yourself in an ultralight at 100 knots in turbulent air. Even if the plane could withstand the stress, it would be a very wild ride.
 
amjon said:
Yeah they tend to add humor to the lesson. (At least when I draw them.) ;)
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My lead the other day tried to draw an AA-10 Alamo (eastern bloc air-to-air missile) and it ended up looking like a penis with a bowtie
 
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