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# Total drag questions

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# Total drag questions

17th Oct 2008, 20:19

Join Date: Jan 2005
Location: France
Posts: 2,319
Hi all,
As an ancient, I'm enjoying this, so I'll have to do some scribbling and calculating and posting this weekend to compose a proper answer!

Rainboe, welcome!
"total lift balances total mass"
Sure, but you forget again to define "total lift".
So your wing will have to produce an additional force perpendicular to the airstream.... it doesn't "know" it's climbing.
And with the wing providing typically ten times as much lift as the engine thrust (ballpark figure, obviously) it's the wing which will provide the additional force to balance the aircraft, rather than the T*sin(climbangle).

Part of the problem of this discussion is that there is no such thing as "lift" and "drag".
Strictly speaking there are only three forces acting on an aircraft (in the context of this discussion).
Weight (or mass if you like), always pointing down.
Thrust, acting pretty well along the airspeed vector (other effects being second-order).
Aerodynamic forces, acting mostly on the wings, which we like to split out as "lift" and "drag", but which again is not a scalar but a vector tilted backwards from the airspeed vector.

Thanks to all for joining in!
I'll try to come up with something 'clear' this weekend!

CJ
17th Oct 2008, 20:58

Join Date: Jan 2005
Location: France
Posts: 2,319
Originally Posted by balsa model
"you climb with power; accelerate with pitch"
Yeah, and I'm supposed to be an aeronautical engineer and I still had some problems to get my mind around that....
So don't worry, you're not the only one!

Since we have shown (some disagree) that in climb, required lift is lower...
Nope....
The "lift", in the sense of the vertical force needed to counter gravity is exactly the same as the aircraft weight (we're talking a steady-state climb).
So the force provided by the wing, which force is now tilted back by the climb angle, now has to be larger (the upwards force provided by the thrust is an order of magnitude smaller).
To increase that force, the AoA has to be larger.

CJ
17th Oct 2008, 23:11

Join Date: Oct 2000
Location: N. Europe
Posts: 436
Allright, let's do the full math in your preferred choice of coordinate system then.

ca = climb angle

Vertical force balance gives

W + D*sin(ca) = T*sin(ca) + L*cos(ca) (1)

Horizontal force balance gives

T*cos(ca) = D*cos(ca) + L*sin(ca) (2)

(2) gives

(T-D) = L*sin(ca)/cos(ca) (3)

(1) gives

(T-D) = (W-L*cos(ca)/sin(ca) (4)

(3) + (4)

L*sin(ca)/cos(ca) = (W-L*cos(ca))/sin(ca)

L*sin2(ca)/cos(ca) = W-L*cos(ca)

L*(sin2(ca)/cos(ca)+cos(ca)) = W

L = W/(sin2(ca)/cos(ca)+cos(ca)) = W/(1/cos(ca)*(sin2(ca)+cos2(ca))) =
= W/(1/cos(ca)*1) = W*cos(ca)

L = W*cos(ca)

Now, where did I hear that before... ?

The point where you go wrong is when you mistakenly assume that the thrust can safely be ignored in the vertical force balance. Take a look at the relative rates of change of the vertical component of thrust and the vertical component of lift with increasing climb angle. The longitudinal component of the weight of the aircraft increases with the sine of the climb angle and thrust has to increase with it to keep the aircraft from decelerating.

The vertical component of thrust also increases with the sine of the climb angle. The vertical component of lift, however, will only decrease with the cosine of the climb angle. In order to maintain equilibrium, the magnitude of the lift force has to decrease.
18th Oct 2008, 02:40

Join Date: Sep 2005
Age: 51
Posts: 73
for discussion purposes: Lift, Weight, Thrust, and Drag vectors in climb

Geometrically, it would appear that in equilibrium:
L = W*cos(AoC)

Last edited by balsa model; 18th Oct 2008 at 02:52.
18th Oct 2008, 08:28

Join Date: Aug 2007
Location: UK
Posts: 64
I can't even spell aerodinamics, but I'm with the lift reduces camp.

In any steady state climb the accelerometer indicates less than 1 G. The steeper the (steady) angle of climb, the less the G force, until we reach the vertical where the G force (and stalling speed) reach zero.

I can't explain why, but surely if the acceleration force is reduced it must follow that lift has reduced also!

Standing by to learn something new today.

Last edited by fullyspooled; 18th Oct 2008 at 08:44.
18th Oct 2008, 09:33

Join Date: Oct 2000
Location: Bristol
Posts: 461
Here's one more misconception. An aircraft G meter measures CN, the coefficient of normal acceleration. "Normal" means on the aircraft normal axis. In a steady climb although the aircraft is experiencing 1g the element on the normal axis is less than 1g. In a vertical climb the G meter will register zero

Dick
18th Oct 2008, 11:14

Join Date: Aug 2005
Location: Estonia
Posts: 834
An aircraft G meter measures CN, the coefficient of normal acceleration. "Normal" means on the aircraft normal axis.
So, does it mean that on takeoff acceleration and landing braking deceleration, the G meter will register 1,0, because the aircraft vertical speed and acceleration are zero and the accelerations along fuselage are not counted? And that during steady horizontal cruise in a slightly nose-up attitude, the g meter will register less than 1,0?
18th Oct 2008, 12:14

Join Date: Oct 2000
Location: N. Europe
Posts: 436
"So, does it mean that on takeoff acceleration and landing braking deceleration, the G meter will register 1,0, because the aircraft vertical speed and acceleration are zero and the accelerations along fuselage are not counted?"

Yes.

"And that during steady horizontal cruise in a slightly nose-up attitude, the g meter will register less than 1,0?"

Yes.

"Yes Dick, you can calibrate your g meter as you like. For convenience, subject to normal surface gravity, you can make it 1.0. Out floating in free space subject to no thrust, you can make it zero. Then you can take that meter and put it on the surface of any planet to get a good reading of gravity relative to the Earth's surface, or stick it on an aeroplane in Earth's atmosphere in a steady climb, cruise or descent, with no acceleration effects in which case it would read 1.0!"

If the local gravitational acceleration is 1g and, assuming as is the norm a one-axis g meter, if the g meters sensing axis is aligned with the local gravitational field.

In the case of an aircraft which is not level, the g meter will not be aligned with the local gravitational field and will thus only register the component of local gravity parallell with the g meters sensing axis. Hence it will read less than 1 g.
18th Oct 2008, 16:02

Join Date: Oct 2000
Location: N. Europe
Posts: 436
Rainboe,
if that is indeed the case then your accelerometer

a) does not measure the normal load factor, which is the load factor we are usually interested in in aviation

b) differs from most other such accelerometers employed in aviation.

I'd say you are the one throwing in unnecessary complications by turning away from the industry norms without clearly specifying so.
18th Oct 2008, 16:07

Join Date: Oct 2000
Location: Bristol
Posts: 461
Rainboe, does your dangly g meter read 1 positve g in all steady flight angles up to the vertical? And does it read 1 positive g in inverted flight? And what does it read on acceleration on takeoff or deceleration on landing?

Dick
18th Oct 2008, 16:58

Join Date: Jan 2005
Location: France
Posts: 2,319
Rainboe,
Is your "g meter" just a thought experiment, or does it exists, and if so, is it used in aircraft?

I can't quote recent aircraft, but in the big white pointy one, the 'g' indicator ran off a one-axis accelerometer installed with the sensitive axis along the z-axis of the aircraft, hence it indicated normal acceleration relative to the aircraft coordinate system.
The accelerometer was installed on frame 56, so as close to the "average" CG as possible, to minimise pitch angular acceleration effects.
As it was, there was nothing specific about the accelerometer... identical accelerometers were used to sense longitudinal and lateral acceleration (for other systems). They were just mounted differently.

Your "g meter", as you describe it, would be "contaminated" by longitudinal acceleration or deceleration.

CJ
18th Oct 2008, 17:27

Join Date: Oct 2000
Location: Bristol
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Rainboe - you're blathering

Dick
18th Oct 2008, 18:20

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Post #7
Or, for a more intuitive approach, analyse the extremes. Assume the climb angle to approach the vertical. Lift will go to zero. It thus makes sense to assume that lift will in fact decrease with increasing angle of climb.
In a straight vertical climb, wing lift will definitely not got to zero. If the aircraft is climbing vertically, nose pointing up, then the wings see a relative airspeed, therefore they produce a force perpendicular to the relative velocity vector. In this case the velocity vector is perpendicular to the ground, as the aircraft is moving straight up. So the force produced by the wings is acting to push the aircraft away from the vertical. It's kind of counterintuitive to call it "lift", since it's not acting to lift the aircraft, but it is wing lift nonetheless. So, in a straight vertical climb, lift is not zero.

Furthermore, if the climb velocity is not that great, then there is a good chance that the wings will actually be stalled, which means there will be airflow separation over the wing surface, and that means there will be lots of drag, not no drag. Therefore, the thrust from the engines not only has to act against the weight of the aircraft, but also against a good amount of drag.

Just a thought. Sorry for the digression.
18th Oct 2008, 19:32

Join Date: Jun 2008
Location: Nouvion
Posts: 42
If something is in steady state , it is in balence i.e. all accelerations acting on it cancel out . So surely it's a constant climb rate , as opposed to a steady state climb ? For the aircraft too climb , the total upward vertical acceleration i.e the total lift made up of wing lift componant and thrust lift componant , must be greater then the total downward acceleration due to gravity i.e. weight , therefore it cannot be steady state ! Steady state can only be applied to cruise where the aircraft neither climbs or descends , speeds up or slows down.

Krujje , quote:

Furthermore, if the climb velocity is not that great, then there is a good chance that the wings will actually be stalled, which means there will be airflow separation over the wing surface, and that means there will be lots of drag, not no drag. Therefore, the thrust from the engines not only has to act against the weight of the aircraft, but also against a good amount of drag.

Sorry have to disagree , I think you are confusing a stalled wing due to excessive AoA with one that is not producing lift due to insufficient airspeed . In vertical flight , AoA will be such that the wing produces no lift and therefore induced drag will fall to zero , of course all other forms of drag will remain. All upward acceleration will be provided by engine thrust alone acting against weight and the remaining drag....as for a rocket .
18th Oct 2008, 19:47

Join Date: Oct 2000
Location: N. Europe
Posts: 436
krujje,
sorry, no joy.

In a steady state vertical climb, the lift will most definitely be zero. As you say, the lift is perpendicular to the trajectory through the air. With no opposing force, the trajectory would not remain vertical very long if the wing indeed generated lift. With lift, a vertical climb is not steady state.

You will have to reduce AoA to the zero-lift AoA for the climb to remain vertical. At the zero-lift AoA, the lift coefficient is zero and... and here it comes... there's zero lift.

With no lift being generated, critical AoA cannot be reached (as this would mean generating lift) and a stall is impossible. The wing will not stall until you go into a tail slide, at which point the climb will not be a climb nor vertical. Of course, steady state means no deceleration and no risk of going into a tail slide either, so stalling is in fact not possible given those boundary conditions.

As long as there's forward airspeed, there will be drag.

T = W+D in a vertical climb.
18th Oct 2008, 19:51

Join Date: Oct 2000
Location: N. Europe
Posts: 436
steady state effectively means "no acceleration".

For an aircraft to climb in a steady state, the forces striving to accelerate the aircraft upward have to equal the forces striving to accelerate the aircraft downwards. Force balance means no acceleration means steady state.

If the upward forces were in fact greater than the downward forces, as you suggest, the aircraft would be accelerating upwards and not in a steady state climb at all.
18th Oct 2008, 22:21

Join Date: Aug 2007
Location: UK
Posts: 64
I can state with complete certainty that all military and civilian aircraft that I have flown and displayed in bygone days used accelerometers of a type that Rainboe is not familiar with, and they do indeed read less than 1G while in any steady state of climb, and precisely zero G while climbing (or descending) in the vertical.

Rainboe, think of it this way. Imagine you have a set of scales under your bum while seated in whatever airplane you fly. While in straight and level flight the scales would read exactly your weight, ie 1G, but while climbing in the vertical plane all of your weight would be acting towards your seat back, and not on the scales. The scales would therefore read zero. While in a steady state climb, they would read somewhere between the two extremes. The accelerometer we use in aircraft act in a plane aligned exactly as the scales I have just described, and they are in no way used for measuring the acceleration forces one may observe while braking after landing, or while accelerating along the runway during take off roll. For the gague to do this it would have to be laid flat on the aircraft floor!

You do however correctly point out (in my lesson of the day) that the accelerometer will indicate temporary accelerations caused by pitching moments. It will also indicate the constant acceleration (greater than 1G indication) throughout any duration of a balanced erect turn - or less than -1G while in a balanced inverted turn. For display pilots the accelerometer is used not only to measure the stresses on an aircraft, but also as a guide to determine when the aircraft will stall.

That much I did not have any misconception about, but I am confused when it comes to the relationship of wing loading and lift. Surely if the G force is reduced, so must the wing loading. As we have not in anyway reduced the wing area, how is it possible that the lift cannot reduce also?

Over to the scientists....

Last edited by fullyspooled; 18th Oct 2008 at 23:19.
19th Oct 2008, 03:23

Join Date: Sep 2005
Age: 51
Posts: 73
original question:
2) The same Boeing chart shows the total drag curve at all speed ranges. The level of drag at mach 0.8 is the same as mach 0.67 for the level flight 35,000 feet 70 Tonne 737).

The second question is: can I assume the total drag in a climb at mach 0.8 is the same as the total drag in the climb at 0.67?
My answer - assuming that we are comparing climbs of equal RoC - nope.
The parasitic drag components will remain unchanged (no change in speed). The induced drag of the slower case will go down farther, since the AoC will have to be greater to achieve the same RoC, and from our previous discussion re question 1, the greater is AoC, the smaller is required lift and thus the smaller is induced drag. On top of that, in the slower case, the induced drag component is a greater proportion of total drag to begin with, making the above reduction even more poignant.

What do I win?
bm
19th Oct 2008, 07:44

Join Date: May 2006
Location: Londonish
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Nothing! I answered question 2, with explanation back on page 1
19th Oct 2008, 15:47

Join Date: Jan 2007
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Sorry have to disagree , I think you are confusing a stalled wing due to excessive AoA with one that is not producing lift due to insufficient airspeed
Yes. My bad. There will be no wing stall. Stall comes as a result of too-high AoA, not insufficient velocity. In straight-and-level flight, the two are linked, as minimum amount of lift is required to balance weight. However, in the theorized straight vertical ascent, thrust is balancing weight, so wing-lift is what it is. When one examines non-traditional scenarios, one must also re-examine assumptions taken for granted.

ft;

In a steady state vertical climb, the lift will most definitely be zero. As you say, the lift is perpendicular to the trajectory through the air. With no opposing force, the trajectory would not remain vertical very long if the wing indeed generated lift. With lift, a vertical climb is not steady state.

You will have to reduce AoA to the zero-lift AoA for the climb to remain vertical. At the zero-lift AoA, the lift coefficient is zero and... and here it comes... there's zero lift.
Sorry. I have to disagree. It is true that the trajectory would not remain vertical very long if the wing generated lift. And it is true that you will have to reduce AoA to the zero-lift AoA in order to achieve this. However, unless you have an all-moving wing, how do you change your AoA without adjusting your climb angle? Remember the constraint on this theoretical problem: straight vertical climb. If you adjust your climb angle to other than 90 degrees, you then have the probem of a thrust component not in the vertical direction, which has to be balanced by a force perpendicular to your trajectory.

All I'm trying to say here is that for most applications, this straight vertical climb we're talking about does produce some wing lift. You could have zero-lift in the case where the wing zero-lift angle is parallel to the aircraft longitudinal axis, but very few aircraft are designed this way.