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# Total drag questions

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# Total drag questions

16th Oct 2008, 07:08

Join Date: Dec 2002
Location: Chester
Posts: 68
Total drag questions

I have two drag related questions.

1) I have a vauge recollection that total drag in the climb is higher than in level flight due to the increase in lift associated induced wing tip drag. Anyone confirm this and / or got any idea as to what the percentage increase in drag is in a low rate (say 500 ft/min) climb approaching say 35,000 feet vs level flight at 35,000 feet.

The only figures I have are for a 737 - 800 @ 70 Tonnes FL 350 at minimum drag speed (0.74 mach).

For this condition the total drag is 8,750 pounds.

2) The same Boieng 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?

16th Oct 2008, 13:46

Join Date: May 2003
Posts: 398
I'll take a stab at this...RMC, I think you may have it a bit backwards...

Lets consider the aircraft at FL 350, at 220 kcas. If it's climbing then one can expect the thrust vector to be canted up slightly, hence providing some force in the vertical direction. Therefore, for a steady climb, the wing will be producing less lift than if the aircraft were in level flight at the same speed, and thus the climbing aircraft is at a lower Coeff of lift, Cl

Hence, since a component of total drag includes induced drag (Di), and Di is roughly porportional to Cl squared, the aircraft described above will have less drag in the climb than in level flight.

So, this is in fact the exact opposite to what you were wondering. Further, a portion of induced drag is in fact from wing tip vortices, and thus the decrease in induced drag in the climb, (versus the same CAS at that FL in cruise), means that the wing tip vortices drag will also be less for the climb situation.

Caveats apply.

Is there any way to post a picture of this graph you have of the drag of the 737? Perhaps upload it to a site such as photobucket (Image hosting, free photo sharing & video sharing at Photobucket), then link to there from a reply to this post?

Perhaps others (Mr Enicalyth?) will take glee in shooting down my analysis, or can take a stab at your second question.

TDV
16th Oct 2008, 14:12

Join Date: Jul 2006
Location: 'tween posts
Posts: 195
my 2 cents

1.in level flight thrust=drag & lift= wt
2.any increase in pitch/angle of A will increase lift thus climb
3.increase in pitch/angle of A will also increase drag thus we loose speed
4.we compensate speed loss with thrust.
to sum up in climb lift+thrust> drag+wt
waiting to be shot down

Last edited by gearpins; 16th Oct 2008 at 14:14. Reason: spell check
16th Oct 2008, 15:07

Join Date: Oct 2000
Location: Bristol
Posts: 461
Dunno what happens in real life but thr theory says that in a steady climb lift is less than in level flight - Wcos(climb angle) instead of W. Thrust is up, at Wsin(Climb angle) plus drag. Drag is reducing as climb angle increases until in a vertical climb it is zero lift drag only, as lift is zero.

Dick
16th Oct 2008, 16:46

Join Date: Jan 2005
Location: France
Posts: 2,319
Let me have a go...

Let me define lift and drag as follows:
Lift is the force acting on the aircraft at right angles to the air speed vector.
Drag is the force acting on the aircraft at right angles to the lift, hence along the air speed vector.

In steady and horizontal flight, that means the lift points straight up, and is equal but opposite to the weight, which points straight down.
The drag points horizontally backwards along the line of flight, and is equal but opposite to the thrust, which is pointing forwards *)

In other words, the classic diagram we all know.

Now, what happens in a steady climb at the same speed?
The weight still points straight down.
But lift, drag and thrust now all have rotated over the climb angle.

Since we are talking about a steady climb, the weight is now balanced, not by the lift L (as defined before), but by Lcos(climb angle).
And that means that L has increased, which in practice is achieved by an increase in angle of attack (AoA, alpha).
And an increase in L means an increase in induced drag.

CJ

*) I've ignored all the second-order effects, such as the thrust line not being exactly along the line of flight, etc. which are not part of the basic question.
16th Oct 2008, 17:50

Join Date: Oct 2000
Location: N. Europe
Posts: 436
W = L*cos(climb angle) + T*sin(climb angle) - D*sin(climb angle)

Climbing means thrust will no longer equal drag in a steady state. Hence, part of the weight is being carried by the thrust of the engines and lift will in fact decrease.

It is easier looking at in in a coordinate system fixed to the velocity vector. Assuming thrust to remain parallell to the direction of flight, the steady state equation perpendicular to the direction of flight:

L = W*cos(climb angle)

And again, we see that lift decrases with increased climb angle.

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.

Last edited by ft; 16th Oct 2008 at 19:18. Reason: Grammatical blooper
16th Oct 2008, 18:53

Join Date: May 2008
Location: Windsor CA 95492
Age: 92
Posts: 49
I think ft is right, and the vertical flight case illustrates the point simply. It is called VTOL, trouble is you need thrust a bit larger than weight
16th Oct 2008, 20:05

Join Date: Jan 2005
Location: France
Posts: 2,319
I see I'll have to do the rest of the diagrams and maths before I can convince anybody...

T*sin(climb angle) - D*sin(climb angle) pretty well cancel out, and are an order of magnitude smaller than L and W.

ft,
Stick with the original coordinate system.
Otherwise your "L = W*cos(climb angle)" is wrong for a steady-state climb condition.

If you convert to a coordinate system locked to the velocity vector, the weight of the aircraft is no longer W, but W / cos(climb angle).

CJ

PS: VTOL is not really relevant here, no?
No lift, no drag. Just T=W.
16th Oct 2008, 21:02

Join Date: Oct 2000
Location: N. Europe
Posts: 436
ChristiaanJ,
I do believe you will have to show all the diagrams and math. I note with interest how changing the coordinate system will change the weight of the aircraft. Will the local gravitational acceleration be altered through changing the coordinate system? If not, I regret not having discovered this while I was still doing mass properties!

As for the relevance of VTOL, I trust it you do not contest the fact that lift will be zero when climbing vertically? If we assume you are right and that lift does indeed initially decrease with increasing climb angle, at which climb angle will lift start decreasing with increasing climb angle?

Cheers,
/Fred
16th Oct 2008, 23:53

Join Date: May 2003
Posts: 398
ChristiaanJ, you have defined lift to be different from the rest of us. Here is what you said:

"Lift is the force acting on the aircraft at right angles to the air speed vector."

And since the airspeed vector is not level for a climb, your definition of lift has a backward cant that is equal to the climb angle. However, in fact, lift is typically defined as the vertical component of the force produced by the wing, ie in a direction straight up, regardless of the airspeed vector. The force produced by the wing will normally have a rearward component, which is equal to the induced drag, and which is included as part of the total Drag.

ft is correct, and Keith's case of taking the situation to it's extreme also provides a simple example that supports it.

Lift force is vertical, opposite to weight. And since we're talking about an equilibrium situation, and thrust has a vertical component, then lift must be less that weight by the vertical contribution of the thrust.
17th Oct 2008, 00:09

Join Date: Sep 2002
Location: La Belle Province
Posts: 2,113
I'm with ChristianJ on the lift vector direction. Lift is normal to the velocity vector, drag along the velocity vector. It's not fixed in earth axes.
17th Oct 2008, 01:00

Join Date: May 2003
Posts: 398
Yes, now I think about it a little more, lift is generally portrayed as perpendicular to the velocity vector, not to the Earth's axis. Disregard my post above with respect to that. Thanks, MFS, for not flaming me on that one.
17th Oct 2008, 01:02

Join Date: May 2006
Location: Londonish
Posts: 780
Ditto. (MFS) It's far easier to build and understand the picture that way, but I don't think it matters what you take as your frame of reference, so long as you treat all the forces in a consistent manner:

If you work parallel and perpendicular to gravity, weight is unfactored, thrust, drag and lift all have angular terms (parallel and perpendicular components). if you work parallel and perpendicular to flight path, the thrust, drag, lift are unfactored and weight has an angular term.

Using vertical flight as a thought experiment, the lift requirement reduces as we pitch to vertical flight, which should reduce the induced drag (but not parasite/form drag). However, as much as that is happening, you are also introducing a new retarding component. It's easiset working parallel and perpendicular to the aircraft path to see this as some component of gravity working normal to the path, and some component working paralell (aft) to the path - it's not drag per-se, but it requires additional power to overcome it, just like drag (or at the same power setting you slow down, reducing drag until some point where everything reaches equilibrium).

If you insist on working with reference to the gravity plane, in the climb you have:
Vertically, positive up:
weight (-ve), a component of thrust (+), a component of drag (-) a component of lift (+ve)

Horizontally, positive left:
a component of thrust (+), a component of lift (-ve), a component of drag (-ve)

That's much more complex to understand (draw it!), has the same resultant - you're increasing the -ve components, and have to increast the thrust, (or allow the aircraft to slow until drag reduces to match thrust), in order to have everything balance.

, post appeared while I was posting!
17th Oct 2008, 01:20

Join Date: May 2006
Location: Londonish
Posts: 780
I'll also have a stab at 2), though I'm guessing somewhat here - happy to be set straight!

The only way you can have the same drag at 2 differing speeds is if they're either side of the minimum drag point, recall a L/D curve is some sort of U shape with: On the left high drag dominated by induced drag courtesy of excessive AOA. On the right high drag dominated by parasite drag due to excessive speed. The L/D curve is the result of combining the parasite drag curve, and the induced drag curve. Somewhere in the middle there's a low point where the combination hits a minimum.

given that, and what we already decided about lift in the climb being slightly reduced: you're going to move the induced drag curve a bit to the left (the wing is compelled to generate less lift at any given speed - effectively you have a lighter aircraft). that will move the min l/d left, and closer to the .67M figure.

So, I believe that transitioning from level flight to climb will favour the lower speed from a drag point of view. I have no idea of the magnitude of the change, or if it's significant. I would also expect it would only really figure at high altitude, as at any lower levels, the speed at which the low point occurs (best glide effectively) will be well to the left of where I'd expect anyone to be operating.

Does that seem reasonable?
17th Oct 2008, 03:16

Join Date: Sep 2005
Age: 51
Posts: 72
my turn

CJ: I think that you might have made a manipulation mistake with your vectors or the trig.
With your definitions (and references), I get:
L = W*cos(angleofclimb),
D = T - W*sin(angleofclimb)
This happens to satisfy all extreme cases of horizontal flight and vertical flight (if power setting allows such a steady state).

Some assumptions are missing in the question.
If he doesn't touch the throttles as he levels off, the aircraft will accelerate and D will progressively increase to match T.
If he keeps his airspeed as he levels off, he will have to reduce the throttles. In this maneuver, D will be constant throughout.

(2) The second question is even more open.
You can't maintain speed and transition to climb without touching the throttles.
If you don't touch the throttles, we still need to know something about the vertical speed. Are they supposed to be the same? You will need less angle of climb for the same VS the faster you go.

I'm not waiting for the return fire - I'm going to bed - bm.
17th Oct 2008, 06:54

Join Date: Dec 2002
Location: Chester
Posts: 68
Hawk 37, You asked for a link to the graph...can't send you there directly but SmartCockpit - Airline training guides, Aviation, Operations, Safety type in search 737 & drag and it will come up with the Boeing presentation on high altitude flight. I think it is slide 13.
17th Oct 2008, 17:29

Join Date: Oct 2000
Location: Bristol
Posts: 461
Answer 1: Lift and drag in a climb are less than in level flight at the same speed.

Answer 2: Yes, with qualifications. The normal drag curves are for level flight and therefore based on one specific weight/lift. In a climb lift is less than in level flight so the speeds taken from the level flight curves are no longer accurate. At small angles of climb the error is small, but as climb angle increases from zero Vimd and best rate of climb speeds reduce, reaching zero in a vertical climb

Dick

Edit for typo
17th Oct 2008, 18:17

Join Date: Aug 2001
Location: Dorset
Posts: 775
Now here's a really counter intuitive one:

As an aircraft goes from climbing flight to level flight at constant IAS what happens to the angle of attack?
17th Oct 2008, 19:20

Join Date: Oct 2000
Location: Bristol
Posts: 461
Hi Keith,

Dick
17th Oct 2008, 19:53

Join Date: Sep 2005
Age: 51
Posts: 72
Now here's a really counter intuitive one:

As an aircraft goes from climbing flight to level flight at constant IAS what happens to the angle of attack?
Ignoring changes in air density, at constant IAS, AoA is a fuction of required lift. Since we have shown (some disagree) that in climb, required lift is lower, the answer must be:
AoA will increase in level flight.

And yes, it is counter intuitive.
Personnaly, I started suspecting that aerodynamics will never be intuitive to me when I realized that people who say "you climb with power; accelerate with pitch" are right. Now if I'm in doubt, I draw them vector balance diagrams, and write power balance equations. If I can't prove it, then I must say that I don't know the answer.