Full Scale Load Factor

[Stainless Skills]

I asked my instructor what the relationship was between Load Factor which is basically + or - g forces, the weight of the plane at the CG and how far one would have to pull the stick to turn. As well as how weight affects the safe maneuvering speed.

Suppose you have a 100kg plane. At a constant bank and speed, you make a 180 degree turn pulling back on 10 degrees on the elevator. Hypothetically!-->, if the plane was 200kg could you pull back 20 degrees on the elevator and keep the same circumference of the previous circle in another 180 degree turn? Is the relationship linear, as in the example? Is it exponential? Is it neither and does it change for different airplanes?

If you have an airplane and then stick an elephant in it, does the safe maneuvering speed for turbulence (where you can do no damage to the airframe by pushing the stick to its limits) does this maneuvering speed increase or decrease when the airplane gets heavier.

I know the wing has to be at a higher AOA when the plane is heavier for the same airspeed but does the control surface effectiveness decrease dramatically? Would the elevators effectiveness decrease if lead was added to the airplane at the CG? Or does the elevator effectiveness around the longitudinal axis stay the same (discounting inertia). If this was the case would it just take more elevator throw to put the plane and wing at a greater AOA to accomplish the same turn because the plane is heavier and it tends to ‘slide’ through the turn more?

Thanks allot, ground school manual does not really cover this to any depth.

[Stainless Skills]

What I really want to know is WHY does maneuvering speed increase with weight?
And how does elevator or any controls effectiveness change with the increase in weight?

[sweetpea]

you also have to factor into each equation thrust......doesn't matter where your CG is......more weight = more thrust to push around any given axis

[Stainless Skills]

Forget about thrust, lets use a glider, it has nothing to do with what I want an answere to.



Get this: Supposedly maneuvering speed decreases when weight decreases.

BUT:

A 100LB plane pulling 2G's produces 200lbs of force on the wing and attachment, in my case on a 172

A 50LB plane pulling 2G's only produces 100lbs of force on the wing and attachment.

Maneuvering Speed wise the 100lb plane can travel faster than the 50lb plane, safely.

Does this mean that it is harder to pull more G's (load factor) when you are heavier at a given velocity? To me this does not really make sense because the only thing effecting G pulling is the elevator.

Lets just use the elevator and forget about gravity in our discussion because the elevator is most representative and gravity plays no role.

[Stainless Skills]

The heavier the airplane, the less effective the elevator?

Is there a region of reverse g pulling effect when it comes to weight? like when you go slower in a 172 considering induced drag and parasite drag? Because the elevator decreasing in effectiveness as much as one might imagine it in the real world when you increase the weight seems alarming.

[Stainless Skills]

I have been reading into the wee hours of the morning on this and just get more and more confused, some of the stuff that pops up on google is astoundingly assanine though.

[Black Bird]

The reason why manuvering speed increases with weight is because a heavier plane will dampen oscillations quicker than a lighter plane. I could go into it more, but your flight instructor should have went into more detail in their explanation.

The easiest way that I can explain it is imagine two balls. One obviously lighter than the other. The ligher ball will bouce more and higher than the heavier ball when dropped from the same height. Now instead of the ball hitting the ground, it is the wind hitting the bottom of the wing. When the wind hits the bottom of the wing, if the airplane is going to fast for a certain weight, more load will be put on the wing and it will stress before it stalls. I think that you need to look up divergent and convergent oscillations. Once you have an understanding of stability, than you will understand the purpose of manuvering speed.

Control Surfaces

Control thro effectiveness decreases as weight increases. Read Newton's second law again. Control throw effectiveness increases as airflow over the surfaces increase. The heavier the aircraft the more control thro is need to impart change in direction.

In revewing your questions, your instructor is to blame. Find another instructor with a better grasp on aerodynamics. You will not find a simple explanation here. Stick to your AIM/FAR.

[Black Bird]

G pulling also happens without using the elavator. Ask your instructor about wind shear.

[OverTemp]

Your instructor is lacking.

Manuevering speed has nothing to do with elevator authority. That has more to do with CG than anything else... the more nose-heavy the less authority.

In light civil aviation manuevering speed is a speed which, if you stay under, you will not over-stress the wing with abrupt and excessive control inputs. Flying just below manuevering speed allows the airplane to stall just slightly before it over-Gs. Since you have no G-meter you have to have some way to control your g-loading.

So... how does a heavier plane fly differently than a lighter plane? A heavier plane flies, in level flight, at a higher angle of attack than a lighter plane. As I hope your instructor taught you, a plane/wing does NOT have a stall speed... it only has a stall angle of attack. Since it is already has a higher angle of attack when it's heavier it takes less G-loading to get it to it's stall angle of attack. If you fly a little faster, you decrease the level flight AOA to about what it was at the lighter weight (or less), then you can achieve the same G-loadings as the lighter aircraft before stall.

I haven't done much civil aviation, just to get my ATP in a seminole, but I do have 3000 hours military and this is what I understood about the reasoning for the seminole's manuevering speed. In military aviation we fly with G-meters... we don't really worry about manuevering speeds as maximums... we more look at minimums we need to be able to achieve AT LEAST the desired G-loading for a given mission and use the meter to keep from over-G'ing. Some of the "electric jets" have flight control systems that won't let you over-G.

If you want to learn more, do a search for V-G Diagram, Corner Speed, or Coffin Corner.

[OverTemp]

By the way, a 500,000 lbs 747 and a 2,000 lbs bug smasher will both pull exactly 1.1547 Gs in a 30 degree bank turn and 2 Gs in a 60 degree bank turn. Moreover, if they are both travelling at 150 knots, they will both have exactly the same turn radius.

Same with an SR-71 and a C-130... if travelling at 250 knots they will both pull the same turn radius at a given bank, regardless of weight. The only things that matter for turning performance are speed and bank angle... they alone determine the radius (and if they plane will stall and fall out of the sky) My point, weight has nothing to do with turning performance... only the point at which the plane will stall.

[tommy321]

Here's a paraphrasing of how William Kershner describes it in the Advanced Pilots Flight Manual (definitely recommend picking it up)

Imagine you're flying an aircraft that has a maximum load factor of 4Gs, and gross weight of 3000lbs. At max gross weight, while flying right at the maneuvering speed, You suddenly pull back on the stick and pitch upwards. Since you're right and maneuvering speed, when you do this sudden pull, the wing will stall, and you will feel 4Gs (the limit load factor). This is the definition of maneuvering speed - the point where the wing stall and the limit load factor occur simultaneously.

So, if you're experiencing 4Gs in a 3000lb aircraft, the wing must be producing 3000lbs*4Gs = 12000lbs. The wing produces 12000lbs at the critical angle of attack and maneuvering speed.

Now... lets lighten the plane. Your plane now weighs 1500lbs (to keep the numbers simple). Remember that the wing, and airfoil are the same though. Once in cruise at maneuvering speed, we again yank back on the stick. The wing angle of attack increases to the critical angle, and since the wing characteristics have not changed, the wing still produces 12,000lbs of lift in this condition.

What's the load factor now? 12,000lbs/1500lbs = 8Gs. We're now subjecting the plane to an 8G acceleration. While the plane is lighter, the fixed components are no lighter. The engine weighs the same, and therefore the engine mounts that were certified to 4G are now being subjected to 8G. Not a good situation.

It'd be tough to show Kershner's derivation here, but it boils down to the ratio of maneuvering speeds is equal to the square root of the ratio of the aircraft weights.
V2/V1 = sqrt(Weight2/Weight1) in our case, V2/V1 = sqrt(1500/3000) = 0.7 or the new maneuvering speed is 70% of the old one.

So, if you're operating below maximum gross weight (which you usually are), the maneuvering speed is actually less than the "book" maneuvering speed.

Tom

[Stainless Skills]

Thats totally awsome, I understand now.

But if I could only understand the logic of weight and manuevering speed in turbulance