Stall Speed Decreases With Decreasing Weight, A student stated..... Options

[BMeister]

So a discussion came up

a student stated at private level: the aircraft stalls at a slower airspeed when the aircraft is less heavier...

which in a way is ultimately true, but it is important to stress that an aircraft stalls at any power setting and any airspeed...


but it's true that an aircraft will stall at a slower speed when less heavier than max gross weight, and if I'm understanding correctly why this is, i'd be quite happy ...


so this is why I think it does what the student states:

At max gross weight a 172 will stall in the clean configuration at 48Kias (based on max gross weight)

The aircraft would stall at (for example/and simple terms) at 40Kias (in the clean configuration) if the airplane were 500lb lighter in weight

this is because: the lighter airplane wings are producing more lift than when it's heavier therefore the critical angle of attack is less until it ultimately slows down in airspeed where by applying more elevator back pressure raises the induced drag and critical angle of attack slowing the airplane more until it exceeds this AoA and stalls... 18-20' degee's

so am I correct in thinking.... a lighter plane will have a lower AoA at the same power settings than an aircraft which is heavier will need to apply more back pressure to sustain flight at a specific altitude therefore exceeding AoA?

your opinions rock...
Cheers!

[c150student]

Firstly, a plane can be less heavy, not less heavier. BUt I digress...

As we all (I hope) know, we can get more lift from an aerofoil by increasing its angle of attack (AOA). This works up till a point, beyond which the airflow can no longer stick to the upper surface of the aerofoil, so it will become turbulent, and the aerofoil will stall.

To keep an aircraft flying level, lift must exactly oppose weight. Therefore, a [more] heavier aircraft will need more lift to prevent it from descending.

Weight = lift

Lift = Coefficient of lift x 1/2 x Area of wing x Air density x Velocity squared

For an increased weight, we need to increase lift. To increase lift, we can increase any of the values on the other side of the equation. 1/2 is a constant and air density is usually constant. We can change the area of the wing (can anyone say 'flaps'?) or the velocity. However, as we are discussing stall speeds here, we wouldnt be increasing velocity, and we would want to leave the plane in one certain configuration, therefore we would concentrate on the Coefficient of lift.

To be honest, I dont know how this is calculated. All I do know is that it takes AOA into account. Cl is higher as AOA increases, up to a point (the stall). To maintain a certain value of lift, we can decrease velocity as long as we increase another value. For this, we will keep 1/2 x air density (rho) and the area of the wing the same, only changing Cl.

I now refer you to this basic graph: http://www.aviation-history.com/theory/lift_files/fig9.jpg

As you can see, the Coefficient has a maximum value that it can take. To keep the lift generated constant, we can decrease the velocity, but would have to increase the Cl. This works fine, up until this peak. When the critical AOA is met, the aerofoil will stall, and the Cl will decrease (this is where the graph peaks). Now, both the velocity and Cl are low, meaning that lift generated is low. This means the plane descends (stalls).

A heavier aircraft needs more lift. Therefore, for any value of Cl (which doesnt change depending on the aircraft weight.... only the shape/AOA of the aerofoil) the velocity must be higher. Therefore, you can reduce the velocity to a point where Cl must take its maximum value in order to balance out weight. This is the stall speed. A lighter aircraftt, however, doesnt need as much lift to remain aloft, therefore the velocity/Cl (AOA) can be reduced further. The velocity at which Cl must take its maximum value is therefore lower.

Stall speeds are determined by the speed at whic the aircraft stalls when maintaining level flight. A heavier aircraft will need to be travelling faster than a lighter one at a certain AOA in order to stay aloft (or have a higher AOA for a certain speed).

Stall speed is also related to wing loading. An aircraft in a certain configuration has a fixed wing size. If you increase the aircrafts weight, the wing loading (amount of weight each area of wing has to support) increases, increasing stall speed. This is also what happens in a turn (a 60 degree bank at a level altitude will generate 2Gs, therefore each unit area of wing will be supporting twice what it normally does. This increases stall speed, and is why you have to be particularly cautious with your airspeed in a steep turn).

Anyway, I hope this answers your question (moreover, I hope what I just wrote is correct!) though if Im honest, it wasnt really clear what you were asking.

[Kilrah]

Ouch - I didn't have the motivation to read c150student's post seeing its length - it's not that complicated!

the aircraft stalls at a slower airspeed when the aircraft is lighter...

Yes.

the lighter airplane wings are producing more lift than when it's heavier

No! Assuming constant vertical speed or more precisely 1G flight, lift = weight, so lighter plane = less lift. I suspect you got it but just got the words wrong, but as it's the key point to the rest... better make sure.

Taking the above, a heavier plane needs more lift to stay up. You have 2 ways to get more lift: increase speed, or increase AoA.
The plane will stall when you exceed the critical AoA, so let's make that a fixed value, the light plane is flying just at its critical AoA. Now make it heavier. We said the heavier plane needs more lift - you can't increase AoA or you'll stall - but you can increase speed to get that lift. So there we are, in the same plane (same critical AoA), if you increase weight you need to fly faster not to stall in 1G flight.

but it is important to stress that an aircraft stalls at any power setting and any airspeed...

Yes, but still the stall speed is called as such and we need to deal with that... what it doesn't explicitly state and needs to be remebered is that it is valid for 1G flight.

[Fast Jet]

There is a formula for the stalled condition in it is the Weight as W - if you can get the formula for the stall - then play about with the W then you will find the answer to your question. if you are looking for it there is a square root in it - 2WL or something I forgot, I will go and look for it. You will find it in your books you study for the PPL. And here we enter the stalled condition. . . .

[BMeister]

There is a formula for the stalled condition in it is the Weight as W - if you can get the formula for the stall - then play about with the W then you will find the answer to your question. if you are looking for it there is a square root in it - 2WL or something I forgot, I will go and look for it. You will find it in your books you study for the PPL. And here we enter the stalled condition. . . .

VA decreases by 5% for every 10% reduction in weight, is this the same with the stall speed, because this is not dicussed in the private instrument or commercial...

or flight instructor for that matter

[Fast Jet]

VA decreases by 5% for every 10% reduction in weight, is this the same with the stall speed, because this is not dicussed in the private instrument or commercial...

or flight instructor for that matter

AWWW MAN !!!!! I just wrote an epic post on this the size of war and peace and the b----y thing just disappeared - GRRRRR!!!! I am NOT happy shopper!!!!!"£$%^&*()_+ B----R!!!!¬! G-- D----t!!!!!! ANYWAY! (God!!)

Anyway, ahem, There is this (G-----n!) formula which is Vs = Sq rt (then you write in the square root box) 2W over rho times CL max times S (thats a big S) where S is the wing area. rho is the density of the air CL is the co-effecient of Lift the max bit is you trying to extract more Lift by pitching upwards, this only works to the point of stall - then - you stall.

Anyway, (I am still brassed of) Weight is inversely proportional to Lift. The heavier you are, the more Lift you need.
If you fly at constant altitude in my Panavia Tornado and without doing anything you receive a load of fuel (more weight) from the Tri-Star then the Tornado would descend sending fuel splashing all over the place - so you have to slightly increase you r lift = increase your angle of attack to the relative airflow or relative wind over the wings by pitching up slightly to maintain (LIFT) altitude. Also, if you are flying along and you open the door and throw out 2 tonnes in weight the the aircraft will sail upwards at an alarming rate due to the weight loss wheeeeeeee!! So you have to trim forward and fly at a lesser angle of attack OK?

If you are REAL heavy you have to fly at a higher angle of attack (pitch up) for the SAME airspeed - therefore you are nearer to the stall angle the whole time. therefore pull back alittle bit more and you will stall.

If you reduce the lift by reducing the air density Rho then you can have real fun at 45,000 feet where the air is so thin that all you need to do to stall is apply back pressure on the stick- and hear this, if you apply forward pressure on the stick the stall warner will go off also!!!!! Figure that one out. Its a lodda fun!!!!

Hope this helps.

Also, watch it with heavy weights!!!! You have to keep the aircraft in balance because all up weight and C.G. location must be considered1 An increase in weight will equal an increase in Lift required hence an increase in stall Speed Vs 20% increase in weight gives a 10% increase in Vs. Forward C.G. gives high balancing tail load (Down) and therefore an increase in LIFT is required. from the wings. Hence the increase in the stall speed. With an AFT c.g. this would reduce the tail load to balance so therefore Lift required, so it reduces the stall speed.

Enjoy !

[The Airbuser]

You don't need fancy-pants formula for stall speed!

It's easy, the square root of your weight is your stall speed, then, on a PA28-181 at its max T/O weight (2550lbs), the stall speed is 50KIAS = Sq root of 2550.

Also, as a side note, remember BMeister that, AOA is referred to the relative wind, not the horizon line; then less IAS, higher angle of attack, and lower "horizontal flow", if you will, speed; air is flowing towards the wing too horizontal, and since we are nose-up, the wing is too vertical, thus, creating turbulent airflow upon contact with the wing surface. So you would need a way for that flow to get more "vertical", so you increase speed. That's why some fighter jets can climb vertically, their pitch is high, but they have massive engines producing enough thrust to keep the airflow parallel to the wing (even though climbing vertically is just a raw display of powerplant engineering! )

Hope it clears more clouds than it created!

Cheers,
Eddie

[effte]

You don't need fancy-pants formula for stall speed!

It's easy, the square root of your weight is your stall speed, then, on a PA28-181 at its max T/O weight (2550lbs), the stall speed is 50KIAS = Sq root of 2550.

Stall is when the lift you are trying to generate exceeds the lift generated at the critical AoA. At a fixed AoA,

L = k * V^2, with k being a constant.

At a given amount of lift, the stall speed is thus

V = sqrt(L/k) = sqrt(1/k)*sqrt(L) = k2 * sqrt(L)

The stall speed is proportional to the square root of the lift required, or in unaccelerated level flight the weight of the aircraft.

If the constant k2 for a given model of PA28 happens to be right for giving the stall speed as the square root of the weight in pounds, so be it. You're in luck, if that's all you're gonna fly and it's always gonna be in imperial unit world. It is unlikely to hold true for any other aircraft though. Ponder what the stall speed of a 50t DC-9 would be otherwise... not to mention the real heavies.

So yes, you do need the fancy pants formulae. k will in actual practise involve air density and maximum lift coefficient.