c/l 1/2RhoV2S

[t'aint natural]

I’m just a helicopter instructor so I don’t know nuthin and I wondered if somebody could put me straight on a matter. Is an aircraft producing the same amount of lift when it is climbing as when it is descending at the same speed? (Presuming all considerations and configurations equal). If (as I suspect) so, why is it going up/coming down?

[chris barlow]

Intriguing question. Yes it's producing the same amount of Lift in both cases assuming that Thrust and Drag cancel each other. It's an initial condition problem. It's going up because that's the way it's pointed and there's no net force to rotate the velocity vector downwards. Similarly for the descent. A net force was supplied initially to rotate the velocity vector either up or down. Now it's there it'll take another force to rotate it back the other way. With drag and thrust cancelled, and the vertical component of lift equal to the weight in both instances, both cases are in perfect equilibrium.

[Pub User]

As a helicopter instructor, you must have passed rotary-wing aerodynamics exams, which are gernerally much more complex than their fixed-wing counterparts.

Unfortunately a picture paints a thousand words, and I'm not clever enough to draw a picture here. Just sketch yourself a diagram of lift/thrust/weight/drag vectors and alter them for different flight profiles. You will soon see how each state is actually an equilibrium of forces, although the thrust one is obviously larger in a climb than a descent because it opposes some weight (always vertical) when climbing, and is helped by it when descending.

[EchoTango]

Thats rather basic.
Vertical lift component = weight. You maintain altitude.
Vertical lift component > weight. You go up.
Vertical lift component < weight. You go down.
Thats why you fiddle with those cyclic and collective gadgets.

ET

[humpty]

When flying S&L, thrust and drag cancel each other out, as do weight and lift, hence no acceleration.

Now, if I point the nose up, a part of the thrust vector points upwards, and (once all the other forces have settled down and we're in a steady climb) there's again no acceleration because I'm in a constant climb. Therefore since the plane hasn't got any heavier, there must be less lift generated from the wings than in S&L since the total upward force is lift + upward portion of thrust.

Now, I point the nose down and, (if I'm not idling the engine totally) after everything settles down, I'm again in a zero acceleration situation. Since part of the thrust vector is down (assuming I am actually pointing the nose below the horizon) then there must therefore be more lift generated from the wings than in S&L.

I understand it, I can explain it (well, you tell me...) but it still seems wrong!

[Checkboard]

EchoTango the situations you illustrate involve unbalanced forces around the aircraft (if the other forces don't change), and will result in acceleration - a change in speed or direction (i.e. transitioning to the climb or decent) so they don't address a steady climb or decent.

humpty - spot on with the climb, and a good explaination, but a bit of an error with the decent.

The Lift is that part of the force exerted by the wings perpendicular to the flight path.

The load factor in straight and level flight is equal to one, as the Lift must oppose the Weight. Any aircraft in a steady climb or decent requires less lift than the straight and level case. (Sounds strange, I know.)

Consider this: as you sit in your chair at your computer, you are experiencing "1 g", all of which is through your seat on the chair. If you now tilt your chair back onto its rear legs, then you are still experiencing "1 g", however, the force is now shared by the seat of the chair, and the back.

It's the same with aircraft, tilt the aircraft back (in a steady climb) and the force is now shared between the thrust and the Lift. Tilt it forward (in a steady descent) and the force is shared by the Lift and the Drag. In either of these cases Lift required is reduced.

Same applies in a descending turn or climb.

As an aside, your stall speed is reduced in a climb or descent for the same reason, as the stal speed is dependant on the lift required from the wings - less lift = less stall speed. The stall speed in a verticle climb (no lift required) is zero - something all aerobatic pilots (performing "Stall Turns" or "Hammerheads") know.

[humpty]

Oh well, serves me right for getting cocky and thinking I understood some of this stuff :rolleyes:

But, having thought about this a bit more, one thing I don't understand: if the lift vector is tilted off-vertical, then surely there's less of if going straight up (sin of the angle off centre). Since thrust and drag are assumed to be along the aeroplane centre-line, then this must mean that if they're equal (and therefore cancel each other out exactly) then there must be more lift, not less....

Assuming that this is wrong, then the only other explanation I can think of is that the thrust does not equal the opposite of the drag when climbing or descending. Is that right???? Would explain the thing about 'excess thust causes climbing', which I've never really understood either.

[t'aint natural]

Doh! The lift vector is tilted ‘backwards’ in a steady climb and its vertical component is foreshortened. The lengthened thrust vector (to maintain the steady climb) has a vertical component which outweighs the foreshortening of the vertical component of lift. Ergo: Up?
I’m not a fixed-wing pilot… is it possible to descend in a perfectly horizontal attitude, in which lift and weight remain the same as they were in level flight, and only thrust and drag are altered?
Pub User and ET: I think I was gazing out of the classroom window the day we did lift. But the reason we rotary chaps wiggle the collective thingy is to increase/reduce the pitch of the blades, thus increasing/reducing lift and going up or down. It’s easy for us, you see… personally I remain to be convinced that fixed-wing aircraft can descend at all.

[ 06 September 2001: Message edited by: t'aint natural ]

[SOHCAHTOA]

Humpty
When flying S&L, thrust and drag cancel each other out

Is thrust not in excess of drag ?
Not a dig mate, im just trying to keep up with you boffs

Im probably going to show up my ignorance here.
Lift is a function of dynamic pressure, AoA and surface area.
So in a descent at a constant speed (IAS) and with a constant surface area (ie no flap) if you lower your AoA, then surely you reduce the amount of lift produced ??

Good discussion, coz Im studying principles of flight at the moment and this forum is really helping. :) :)

[chris barlow]

Assuming no acceleration, then Thrust equals Drag and the magnitude of either is irrelevent since they completely cancel each other out. What you are left with is a vertical weight and a tilted lift force. If the angle of climb equals the angle of descent ( my assumption in my earlier post ) then the lift force must be identical in both cases ( Lift*sin(flightpath) = Weight ).

[chris barlow]

Make that Lift*cos(flightpath)=Weight, assuming flightpath is measured from the horizontal.

[Checkboard]

Honestly! - what is all this rubbish about thrust equalling drag in either a climb or a decent :confused:

Doesn't anyone here push the power up to begin (and continue) climbing, and reduce it decending! :rolleyes:

In short, the loss of the vertical component of lift is made up for by the vertical component of an increased thrust vector during a steady climb.

The loss of the vertical component of lift in a decent is made up from the vertical component of drag in a steady decent.

[4dogs]

Checkers,

In my helicopter, if I set my power (rotor thrust) to maintain S&L at a speed say halfway between Vmp and Vmo then I can happily climb and descend at constant power simply by changing my speed. This is a critical teaching point for instrument flying (visual flight involves much more TLAR or WMIW).

I know that you use the same technique in descending your aeroplane (say 250kt @ idle isn't good enough - solution: increase to 290kt) and that you may have used it during fuel critical drift up.

Sorry, just being picky....