G and its affect on drag

[stiknruda]

Took a chum flying last night and did some gentle rolls and loops.

Later on the ground he asked me how the aircraft didn't stall as the ASI was reading well below Vs.

I explained that drag is directly proportional to G and at zero G - drag is zero.

My problem is that he didn't believe me! Can any of you shed any light on this and perhaps point me to a formula like the lift formula where I can show him that when 'g' = 0, Drag as a product must be zero?

He quickly understood that when S is reduced by the wings being knife-edge that L in turn must reduce from the Lift equation

Or have I been kidding myself for the last 30 years!!

Thanks

Stik

[mad_jock]

Its a change in concept which helps

If you explain instead of using airspeed but explain it in relation to angle of attack and critical angle of attack.

So for 0g there must be zero lift coming from the wing therefore the wing must be at zero angle of attack which is less than the critical angle of attack therefore the wing can't be stalled.

[keith williams]

I'm not sure why you were concentrating on drag to explain why Vs is zero in zero g flight.

But if the problem was how to explain the relationship between stalling speed and load factor the following may help.


L = CL ½ Rho Vsquared S


At the stall in 1g flight L = 1 x W = CLMAX½ Rho V1gstall squared S……equation 1

At the stall in 2g flight L = 2 x W = CLMAX ½ Rho V2gstall squared S……equation 2

At the stall in 0g flight L = 0 x W = CLMAX ½ Rho V0gstall squared S……equation 3


The general equation for stalling speed at any load factor is

Vs at any load factor = Vs at 1g x the square root of the load factor.

[DB6]

Bloody hell! I am now deeply traumatised by being transported back to my first year university exams.
Drag isn't zero at 0 AoA. Lift induced drag is, but form drag isn't.
The wing isn't stalling because the AoA isn't above the critical AoA. That remains constant at about 16-17 degrees, the g you have to pull to reach/exceed it varies with speed. Below Vs (1g stall speed) if you are not stalling you must be pulling less than 1g.

[mad_jock]

Correct on the drag bit DB6 although I presumed he was on about lift and had a typing fart.

[stiknruda]

Thank you!

The power of Pprune! I have exactly what I wanted - thank you. Keith W supplied the exact bit I was looking for with his equation 3!

Cheers

Stik

[EK4457]

Wow. Lots of equations that lost me. My degree in Chemistry and Physics was obviously a waste of time.

Is it not;

g force = load factor = L/W

If g force = 0, then L must = 0. Therefore induced drag = 0

Simples?

Of course profile drag will never be zero unless you are stationary. And at zero g on Earth Mr Einstein says this is impossible as you are in freefall.

In theory, you will dropping like a stone and not flying, so of course you will not have a stall speed.

In reality you will always have air flowing over the wings creating lift so will hardly ever be truly at zero g. This will only be for a brief moment as you go from a positive g to a negative g.

EK

[BEagle]

He quickly understood that when S is reduced by the wings being knife-edge that L in turn must reduce from the Lift equation.

That's a new one on me - wing area being dependent on angle of bank!

It is, of course, bolleaux! Sorry Stik....

[Oktas8]

In theory, you will dropping like a stone and not flying, so of course you will not have a stall speed. In reality you will always have air flowing over the wings creating lift so will hardly ever be truly at zero g. This will only be for a brief moment as you go from a positive g to a negative g.

Perhaps you might replace "dropping like a stone" with "accelerating at 1g". If you start by going up very fast, you can accelerate toward the earth at an acceleration rate of 1g (that is, with zero lift) for quite a while before you end up going down very fast. This would be zero-g flight sustained for long enough for a passenger to feel quite unwell. Considering airflow over the wings, the whole manoeuvre might be flown at -4° angle of attack, giving zero coefficient of lift and therefore zero lift at any speed.

Cheers, O8

[hihi]

Drag isn't zero at zero G. At any time the aircraft is moving through the air, there will be drag (parasite), and at any time the wings are producing lift, there will be drag (induced).

Flying below VS will not make the aircraft stall. A stall is caused by the wing exceeding the critical angle of attack - the wings are now no longer producing enough lift to counteract weight of the aircraft. VS is therefore a reference speed as to when you can expect to exceed the critical angle of attack in normal flight conditions. Explain to your student that VS changes with the weight of the aircraft - an increase in weight means an increase in VS, and vise versa. When flying at unusual attitudes, the weight of the aircraft will not always be supported entirely by the wings, and the load factor will not always be equivalent to 1G. At any time the load factor drops below 1G, the apparent weight of the aircraft is also reduced, and therefore the lift requirement and VS are reduced accordingly.

Hope that helps.

[Pilot DAR]

At any time the load factor drops below 1G, the apparent weight of the aircraft is also reduced, and therefore the lift requirement and VS are reduced accordingly.

Well, considered slightly differently, when "the pilot reduces the load factor below 1G".... With all the appropriate cautions and preparations for conducting practice stalls, when just at the point of the stall breaking, and with an eye on the indicated airspeed, push a bit more than you would normally to recover the stall. You will, of couse, experience less than 1G. You may also notice an indicated airspeed somewhat lower than you were expecting to see. There are several factors which will affect the accuracy of the airspeed indication, but you should see the effect. If not, try it in a bit of a climb with some power on, and leave the power on, as you go over.

[bingofuel]

I think Mad_jock wins the coconut for the plain English answer that can be understood by the layman!

The 'chum' asked a general question, no need to get all involved with equations, just give him a simple explanation that is generally correct.

[keith williams]

That's fine but the original post actually asked

Can any of you shed any light on this and perhaps point me to a formula like the lift formula

Sound to me like he was asking for some kind of formula or equation!

And

So for 0g there must be zero lift coming from the wing therefore the wing must be at zero angle of attack

Isn't true except for a symmetrical aerofoil.

[hihi]

Quite rightly. A general purpose aerofoil produces zero lift at approx -4 degrees AoA.

[mad_jock]

Fair cop but as stick is flyin an aero's machine its likely it won't be a general purpose aerofoil

[Rick777]

The simple answer is that Vs only applies in 1g level flight. In fact stall has nothing to do with speed. It is (as several others have said) only about angle of attack. When you exceed critical angle of attack you stall.