Lift and Weight in Straight and Level.

[VDEE7]

If the airplane is flying straight and level, lift = weight right?
Say the airplane weighs 3000lbs, would it be correct to say the wings are producing 3000lbs of lift? Which means Load factor = 1G? LIFT/WEIGHT

What about the Tail down force? Does that require the wings to generate slightly more lift ? And if so… how are we still getting 1G in straight and level?

 

[tgrayson]

If the airplane is flying straight and level, lift = weight right?

Essentially yes, and essentially yes in climbs and descents as well.

Say the airplane weighs 3000lbs, would it be correct to say the wings are producing 3000lbs of lift?

Correct within a good approximation.

Which means Load factor = 1G? LIFT/WEIGHT

Yes, it means that.

What about the Tail down force?

It exists, although it might be a tail up force in some situations.

Does that require the wings to generate slightly more lift ?

Yes, if it's a tail down force.

And if so… how are we still getting 1G in straight and level?

(Weight + Taildown Force)/Lift = 1 G

 

[Minuteman]

In level, unaccelerated flight, the net lifting force is equal to the weight of the aircraft. What comprises that net force will vary a whole bunch:

The wings do produce the majority of lifting force, but the engines and fuselage can also have some contribution, and the horizontal stabilizer may be lifting against or with the net force, depending on the CG location and aerodynamic configuration.

So saying in "lift equals weight", you're talking about the net result of combining several different components. If the tailplane is generating a downward force, then yeah, the wings/nacelles/fuselage will have to produce more lift than just the gross weight of the aircraft to stay in level flight.

A major component of drag is a function of how much lift is being produced by different parts of the aircraft. So if you have a forward CG, your stabilizer will have to produce some downward "lift," (to counter the moment about the CG) and the wings will have to produce more upward lift to counter that downforce. Instead, if the CG is moved aft, there's less "lift" needed from that tailplane so it produces less drag, and the wings have to produce less lift to make up for the stab's downforce so there's less drag from them as well.

 

[fish314]

I think you may have trouble with what "1 G" means. Basically it means that you are falling into your seat at the same rate as the acceleration due to gravity would cause on earth.

Let me put it another way. Right now you are sitting at your computer reading this, and you are FALLING into your seat at 32.2 ft/sec^2. Of course you don't think of yourself as falling, but you really are. It's just that your seat prevents you from moving any farther down, by providing a support force. It supports your weight.

So that's "1 G". It means that you are experiencing an acceleration towards your "frame of reference" of 32.2 feet/sec^2.


Now imagine you are in one of those comfortable rolling business chairs like executives get to have in their office.... But that chair is at the top of the empire state building. Still at 1 G, right?

NOW imagine someone pushes you AND the chair off the top of the building. You are still falling at 32.2 feet/sec^2, but NOW SO IS THE CHAIR. So compared to the chair, you AREN'T accelerating. Compared to the chair you aren't "falling" (of course compared to the EARTH you are, but that doesn't make any difference until the messy sudden stop at the end).

So what do you feel. Well, you feel no acceleration compared to your immediate surroundings (your frame of reference), and hence you feel weightless, like you are floating. That's 0 G.

Ok, well you're in luck. You bought that chair from Wyle E. Coyote, and it has rockets on the bottom. After feeling weightless for a bit, the rockets kick in and begin to lift the chair skyward. Now you begin to accelerate towards the chair at 32.2 feet per second squared again, (1 g), but in addition imagine the chair is ALSO accelerating upwards compared to the earth at 32.2 feet/sec^2. So now, it's as though you were accelerating towards the chair twice as fast (64.4 feet/sec^2). Half of that is due to the gravity of earth pulling you down, and half of that is due to the rockets at the bottom of the chair moving both you and the chair UP. So compared to the chair, it's the same as if you were accelerating toward it at twice the normal gravitiational accleration. Guess what you feel? 2G. Twice as heavy as normal.

Does that clear anything up? (Or did I just make it worse?)

 

[VDEE7]

Thanks guys. it sure helps.
Now one more question.. what exactly happens to the forces when we pitch down the airplane and why does the airspeed increase without increasing power? I know it has to do with forward component of weight but how did that add to the thrust!

 

[tgrayson]

what exactly happens to the forces when we pitch down the airplane and why does the airspeed increase without increasing power?

Here are the steps:

1. When you push forward on the yoke, you decrease your angle of attack.
2. Briefly, you have an angle of attack that's suitable for a higher airspeed, but you don't have that higher airspeed yet. This results in lift < weight.
3. The aircraft becomes a projectile and starts following a parabolic arc towards the earth.
4. Up until now, the aircraft has been in rotational equilibrium, because the nose down moments were equal to the nose up moments produced by the tail. However, the sinking of the aircraft produces a component of the relative wind from below; this increases the AOA on the wing and reduces the downward lift on the horizontal stabilizer, producing a nose down pitching moment
5. The aircraft will continue to nose down as long as lift < weight. At some point on the arc, there will be enough of a component of weight along the flight path that will provide enough “thrust” to maintain the trimmed airspeed, and Lift will once again equal weight. The aircraft is in equilibrium.

 

[VDEE7]

Nice and Brief!! Thanks.
P.S I loved your Va change with weight explaination in another thread!
Nice to have guys like you around and willing to help

 

[nosehair]

You bought that chair from Wyle E. Coyote

The only problem with that is the rockets are guaranteed to not fire, or mis-fire.

 

[fish314]

The only problem with that is the rockets are guaranteed to not fire, or mis-fire.

Yeah, but using Wyle E. Coyote physics, you also wouldn't start to fall until you actually looked down and then put out a sign that said "Yikes!!!".

 

[tgrayson]

Nice and Brief!! Thanks.
P.S I loved your Va change with weight explaination in another thread!
Nice to have guys like you around and willing to help

Thank you. A few things:

1. You might think your question basic, but it's not. You won't find an aerodynamics book anywhere that directly addresses the steps I listed. They do have math that shows the end result, but not the mechanism by which it is accomplished.
2. The addition of thrust doesn't really change anything, other than the resulting climb angle after the angle of attack change.
3. You mentioned using "pitch", but it's really an angle of attack change. I sense a virtual roll of the eyes when I make that distinction, but confusing the two is the basis for a great number of misunderstandings. If you were behind the thrust curve when you push on the yoke, you would indeed increase your airspeed, and experience a brief downward pitch, but the end result would be a higher pitch, due to the reduced drag on the airplane. So a reduced angle of attack produces an increase in pitch.

Using the same reasoning I used in my previous post, you can deduce a lot of aircraft behavior. Can you come up with the steps that result from an increase in thrust in level flight, with no angle of attack change?

 

[VDEE7]

Back to the load factor..

How come the load-factor doesn't increase in a constant bank level turn when we increase the airspeed? I dont understand how the decreased rate of turn helps maintain the load-factor!!

 

[tgrayson]

Back to the load factor..

How come the load-factor doesn't increase in a constant bank level turn when we increase the airspeed? I dont understand how the decreased rate of turn helps maintain the load-factor!!

In order to increase airspeed, you must lower your AOA. Lift is proportional to AOA * V^2, so as the AOA goes down, the V is going up. Thus the total quantity of lift in a steady, constant banked turn doesn't change. And since n = L/W, the load factor doesn't change.

The decreased rate of turn isn't particularly relevant.

 

[fish314]

Back to the load factor..

How come the load-factor doesn't increase in a constant bank level turn when we increase the airspeed? I dont understand how the decreased rate of turn helps maintain the load-factor!!

The short answer to your question is because of the word LEVEL. The bank angle that you pick in the turn determines what load factor will be needed, regardless of your airspeed. All that your airspeed determines is how much load factor you CAN produce, not how much you need to produce to maintain the turn at a particular altitude.

Another way to think of this in addition to what Tgrayson already posted is this:

You go into a bank, and consequently some of your lift is now pointing sideways, and some of your lift is pointing up (at least until you hit 90 degrees of bank).

But in order for your turn to be a LEVEL turn (remember we were talking about level turns, right?) the amount of your lift that is pointing up must equal weight.

This is exactly the same as what happens straight and level, but when you are straight and level ALL of your lift is UP (opposite weight) and therefore the TOTAL of your lift is acting opposite your weight. (Disregard the tail down force, which is basically negligible in comparison).

Same thing in a turn, but now the TOTAL lift force must be bigger so that the FRACTION of it that is acting up still equals weight. The more bank angle, the more the total lift force must increase, so that the fraction of the lift acting upwards (opposite weight) is still equal to the weight.

Therefore, load factor NEEDED in a level turn is a function of bank angle ONLY.

Speed and angle of attack come into how much lift you can PRODUCE, not how much lift you NEED. Now how do we get that lift to increase? Well there are two ways (if you discount changing the shape of the wing using flaps or slats).

Lift = Cl * rho* V^2*S/2 where "Cl" is the coefficient of lift, V is the velocity (in true airspeed, and notice that it is squared), S is the plan form area of the wing, and rho is the density of the air. Assume that we can't change S or rho, and what's left to change?

Only Cl and V. V is velocity, so we could produce more lift by going faster. But we don't have to. We could also change the "Cl" or coefficient of lift of the wing. Cl is related to a lot of different things, like camber, the shape of the wing, etc. One thing that it is related to is the angle of attack, and that relationship is linear throughout almost all of the range of angles of attack that you would be flying the wing. (It does, however, have a peak near the critical angle of attack and after that the wing's Cl and therefore the lift produced by the wing drop off sharply). What that means is that you can increase lift by increasing velocity (and keeping angle of attack the same), or you can increase lift by increasing angle of attack and keeping velocity the same. (And of course you could also increase BOTH if you wanted to).

So now with that in mind, back to your original question, which was "how come load factor doesn't increase in a level turn when we increase airspeed".

Hopefully now you see that the answer is, "because load factor REQUIRED to keep the airplane level is based only on BANK angle, but load factor AVAILABLE is based on airspeed (and angle of attack... sort of)".

 

[tgrayson]

But in order for your turn to be a LEVEL turn (remember we were talking about level turns, right?) the amount of your lift that is pointing up must equal weight.

It's easy to get the wrong idea from this. Regardless of whether the turn is level, or descending, or climbing, the vertical forces must be equal; otherwise, there would be a constant vertical acceleration. A steady climb or descent is not an acceleration.

Now, it's true that lift is mathematically equal to weight only in a level turn, but in cilmbs and descents, the vertical forces are still equal, it's just that drag and thrust provide the missing component of lift. But this contribution is small, unless the climb or descent angle is large. Consider: you can calculate the load factor of a constant banked turn in a descent by the following formula:

n = cos(descent angle)/cos(bank angle)​

So even if we were descending at a 6 degree angle with the horizon, the load factor would be

cos(6)/cos(60) = 1.989​

which is 99.4% of the load factor in level flight.

So the reader would be drawing the wrong inference to say that the descent was caused by the lack of lift, since the up forces are still equal to the down forces; rather, the quantity of lift needed is reduced, because the drag of the airplane is supplying part of it. So the reduced lift is essentially caused by the descent, rather than the descent caused by lack of lift. The descent itself is produced by an insufficient quantity of thrust.

 

[VDEE7]

In order to increase airspeed, you must lower your AOA. Lift is proportional to AOA * V^2, so as the AOA goes down, the V is going up. Thus the total quantity of lift in a steady, constant banked turn doesn't change. And since n = L/W, the load factor doesn't change.

The decreased rate of turn isn't particularly relevant.

I like how all your answers are based on your "Core concepts". They seem simple but they're really involved in pretty much everything.
The reason I asked about the airspeed effect on load-factor because I read this:

FAA-H-8083-25 Says
"For any given bank angle, the rate of turn varies with the
airspeed; the higher the speed, the slower the rate of​

turn. This compensates for added centrifugal force
allowing the load factor to remain the same"

This confused me a a little bit, I liked your explanation better.

Fish.. thanks for your input too. good info!!

 

[tgrayson]

FAA-H-8083-25 Says
"For any given bank angle, the rate of turn varies with the
airspeed; the higher the speed, the slower the rate of
turn. This compensates for added centrifugal force
allowing the load factor to remain the same"

Yuck. FAA publications uniformly garble the ideas of centrifugal/centripetal force, load factor, rate of turn, slip/skids, etc. I recommend that you not read those sections or you'll be crippled for life.

 

[B767Driver]

Now, it's true that lift is mathematically equal to weight only in a level turn, .

Hey TG....great posts...very well written. But wouldn't lift be greater than weight in a level turn? A portion of the wing lift produced goes to a horizontal vector...and a portion of the wing lift produced goes to vertical vector. Hence you feel a few more g's in your seat.

That also reminds me of a piece of trivia. Q: In an unaccelerated, steady state condition when is the only time that the wing produces more lift than the airplane weighs? A: A level turn.

Like you stated...I don't like the whole "lift = weight, thrust = drag" instruction. I believe we're much better suited for understanding the forces applied when dealing with "upward, downward, forward and rearward" forces.

Keep up the good work.

 

[tgrayson]

But wouldn't lift be greater than weight in a level turn? A portion of the wing lift produced goes to a horizontal vector...and a portion of the wing lift produced goes to vertical vector. Hence you feel a few more g's in your seat.

Thank you, sir. Yes, you're right. But it's funny, my sloppiness is repeated in lots of aerodynamics book. Lift is most properly used for the aerodynamic force generated by the wing perpendicular to the relative wind. But when banked, only a portion of it opposes gravity. I sloppily used the word "lift" for that portion.

<<That also reminds me of a piece of trivia. Q: In an unaccelerated, steady state condition when is the only time that the wing produces more lift than the airplane weighs? A: A level turn.>>

Except that a turn is anything but unaccelerated.

<<I don't like the whole "lift = weight, thrust = drag" instruction. >>

Yeah, we see those in FAA publications and I wonder that anyone is able to derive any insight from them.

 

[B767Driver]

Except that a turn is anything but unaccelerated.

.

Right...I should have left out unaccelerated.

 

[B767Driver]

Yeah, we see those in FAA publications and I wonder that anyone is able to derive any insight from them.

This is where the "Gleim" type of learning really hurts us. We all know that the horizontal component of lift makes the airplane turn. But how many pilots know how that occurs?

Do ailerons cause the airplane to turn? No. They just re-orient the lift vector...and the wing actually provides the turning force. Once the wing lift has been re-oriented sideways...we actually put the ailerons away (neutralize them for the most part) and let the wing do the work.

Many pilots would object to the fact that the wing causes the airplane to turn...yet all of them would agree that the horizontal comp of lift does cause the turn.