question about stall speed and weight

[shiver]

ok so i was wondering if someone could explain this to me.

why does your stall speed increase with increased weight?

[Runaway Gun]

Because Lift opposes weight. And when lift is less than weight, it needs to increase. This is done by increasing Angle of Attack, or speed (or a few other factors). So to get MORE lift to counter the INCREASED weight, we need to fly Faster and provide MORE V squareds....

[eckhard]

As Runaway Gun said, if the weight increases, the lift also has to increase.

The lift formula is <Lift = 1/2 rho x Vsquared x wing area x CL>

Therefore if we keep the air density (rho) and the wing area the same, the only things we can adjust are the speed (V) and the angle of attack (which influences CL).

So if we need more lift due to increased weight we have to either increase the speed (V) or the angle of attack (CL).

Given that a particular wing will (nearly) always stall at the same angle of attack, the only way we can get more lift at that angle of attack is to increase the speed.

[Ocampo]

Now, how about a little spin on the topic to shed a bit more light to our matey (and to several others of course, including myself).

Why does increased G force also increases stall speed?

Reading some of my notes, I found/concluded that, since you are maintaining more than 1G, you maintain that G force by pitching up i.e., high angle of attack; so you are already limited by means of angle of attack to provide more lift, given that if you increase the angle of attack even further you would approach the CLmax quicker than if you were flying at unaccelerated flight (1G).

Am I right?

[FE Hoppy]

First separate pitch and AOA. I could have a negative pitch angle and still be at MAX AOA.

Now think about what weight is. If I increase the effect of gravity by accelerating away from the earth I am effectively increasing the weigh of the aeroplane and therefore need more lift to oppose the increased weight. As CLmax remains the same, the only way to increase lift is to increase V^2

[Ocampo]

Gee, that makes perfect sense, I wonder why they don't break it down that easily on textbooks!

Thank you very much, sir

[eckhard]

And as an extra refinement to the above excellent posts, the increase in stall speed will be equal to

the square root of the 'load factor' in the case of 'g'; or

the square root of the weight increase, in the case of 'weight'.

For example, if you are turning at 60 degrees of bank in level, balanced flight, you will be pulling 2g. The load factor is 2. The increase in the stall speed will be the square root of 2 which is approx 1.4. So if your stall speed in level flight was 60kts, at 60 degrees of bank it will be 60 x 1.4 = 84kts.

Similarly, if you double the weight of the aircraft, you will increase the stall speed in level flight by a factor of about 1.4.

[FE Hoppy]

I'm a simple man. I need simple explinations.

[john_tullamarine]

Some observations to herd the previous posts into one paddock -

(a) increased mass has the same outcome as increased g so far as the wing looks at weight. Our unfortunate and relaxed use of "weight" when we really mean "mass" is a source of great confusion for many folk.

Recall from basic principles of flight classes (a long time ago) that W = mg, so that, if the net m, and/or g value varies, then W varies.

(b) at the stall (Vs)

CLmax = lift / (½ ρ Vs² S)

which can be simplified to

CLmax = lift / (k x Vs²)

if we lump the other factors sensibly together as some constant (k) for the one aircraft and altitude.

We can get rid of the k constant by talking in terms of proportionalities and simplify the equation to

CLmax ≈ lift / Vs²

If we further constrain altitude to be constant so that lift = weight (KISS principle), then

CLmax ≈ weight / Vs²

If we presume CLmax (ie stalling CL) to be a constant, which is reasonable for a given aircraft (and configuration) at a given altitude, then

Vs² ≈ weight

as we can lump the CLmax value (just another number) in with the k and get rid of it as well for proportionality considerations. So now, if we vary weight (ie mass, or g),

Vs ≈ √ W


We use this relationship by comparing two different cases -

Vs2 = Vs1 x √ (W2/W1)


In the real world we need to apply PECs and figure the sums in CAS rather than IAS otherwise all bets are off. This presumes nil instrument error and we can ignore compressibility effects for typical situations.

So, for example, if level stall (presumed at 1g) occurs at 52 KCAS for 1200 kg, we would expect the stall at 1500 kg to occur somewhere near

Vs = 52√(1500/1200) = 58 KCAS

As an aside, if you think back to principles of flight, you probably looked at n-g diagrams which have curved stall lines at the low speed end for positive and negative g loads. The curve is just the equation above.


If you are uncomfortable with the sums, we can describe what happens in the following way

(a) weight (strictly mass) variation is similar to g variation, so far as the end "weight" is concerned

(b) for all intents and purposes, stall speed varies with the square root of the weight variation


Hopefully I haven't just confused everyone more than they were at the start ?


It is important to keep in mind that PECs can go a bit strange post stall.

As a consequence, when we play test pilot at the flying school with a small Cessna and the IAS goes off the clock (indicates something approaching zero ?) in the stall, that "observation" is more about PEC than the actual speed (which is definitely not zero) of the aircraft through the air.