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# Total drag questions

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# Total drag questions

19th Oct 2008, 17:00

Join Date: Jan 2005
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krujje,

You're confusing climb angle and pitch attitude!
The climb angle is 90°.
For a "steady-state" vertical climb, lift has to be zero or the climb wouldn't stay vertical, so you have to fly at zero lift AoA.
On most aircraft, that corresponds to a slightly nose-down pitch attitude.
So while your flight path is vertical, the nose of your aircraft is not pointing exactly to the zenith....

Oh, and that leads to a nice one....
In our theoretical perfect steady vertical climb..... the thrust line is unlikely to be perfectly aligned with the flight path. So there will be a small but noticeable component normal to the flight path.... which you then have to compensate for with a small but measurable amount of "lift".

fullyspooled, re accelerometers, I fully concur.
But you said...
You do however correctly point out (in my lesson of the day) that the accelerometer will indicate temporary accelerations caused by pitching moments.
Negative. If the accelerometer is installed in the right location, pitching moments accelerations become second-order.
Of course.... if you install it under the pilot's bottom... that's a different story.

ft, re zero lift... so sorry to contradict you slightly, above, because your reasoning is basically correct!

CJ
19th Oct 2008, 20:13

Join Date: Oct 2000
Location: N. Europe
Posts: 436
"However, unless you have an all-moving wing, how do you change your AoA without adjusting your climb angle?"

You are confusing deck angle with climb angle. The climb angle will be 90 degrees, but the pitch angle will be less.

I now see that Rainboe with some additional elegance pointed out the exact same thing in the next post.

"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."

Yes, and this compensation will be some residual lift. However, in this discussion we have been careful to specify that we assume the thrust to remain parallell to the direction of flight, a reasonable simplification to make in a discussion of basic principles of flight.

"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."

Not unless you factor in other forces in the aircraft xz plane perpendicular to the direction of travel it won't.

"Nobody is saying the wing does not produce lift in a vertical climb."

Yes, I am. As long as we stick with the basic assumption above.
19th Oct 2008, 21:37

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ft,
Each time, we're talking about what are essentially four vectors in the vertical plane: W, L, T and D.
So let's not fully neglect the fact that they are not necessarily at right angles all the time.

In steady vertical flight, L and the normal component of T are obviously second-order. So, I garee with you, to explain the basic state of affairs, T=W+D makes perfect sense.

But to get back to the original question, I'm still convinced that in a "typical" aircraft in a "typical" steady climb, the extra force needed to compensate W/W*cos(climbangle) i.e., a slightly higher vertical force, comes for 90% from a slight increase in "lift", by means of a slightly higher AoA.
Only 10% will come from the vertical component of the increased thrust - which will have increased slightly to compensate for the increased drag from the slightly higher AoA (second order effect).
It will have changed by T*sin(climbangle), which will still be far less than the "lift" effect.

Yes, I know I'm sticking my neck out... especially since I promised to look at it this weekend but had other things to do...

Just to clarify, when I talk about a "typical" aircraft and a "typical" climb, I'm talking about a Cessna, or a Boeing, where lift/weight is about 10 times thrust/drag and a climb angle of something like 5° to 10° at the most.

Talking about vertical climbs confuses the issue, since by then you need engine power/thrust sufficient to maintain steady vertical flight, and the aerdynamics are no longer the same at all.

CJ
19th Oct 2008, 23:52

Join Date: May 2006
Location: Londonish
Posts: 780
Q2:

No, the drag will not be the same at both M.8 and M.67 in the climb; The drag will be lower at .67 climbing than it will be at .8 climbing; i.e. climbing will favour the lower speed regime.
20th Oct 2008, 02:48

Join Date: Sep 2005
Age: 51
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Mark1234 said:
Nothing! I answered question 2, with explanation back on page 1
Ooops.. I see it now: post #14. Sorry, it didn't get through to me, at the time - I was too focused on question 1 controversy.
Honestly though, we might have to deduct some points for saying "you're going to move the induced drag curve a bit to the left". Under s.s. climb, induced drag curve shrinks down by a factor of cos(AoC). The portion of total drag curve that is dominated by induced drag will appear to move to the left.
(I'm really crying for attention here )
20th Oct 2008, 03:13

Join Date: May 2006
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D*mmit! Bang to rights there...
20th Oct 2008, 06:38

Join Date: Sep 1998
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I'm loving this thread! Thanks chaps.

Could I ask about a glider going by virtue of a thermal from a flight path angle of 0 to a flight plath angle of >0 at the same IAS?

More or less drag? Lift?
20th Oct 2008, 06:50

Join Date: May 2006
Location: Londonish
Posts: 780
Sure

1) Transient increase in AOA while it's being accelerated, which corresponds to a nice feeling in the pilot's butt.. (probably a continued increase in AOA too due to beardy glider pilot slowing and cranking it onto a wingtip - but we'll ignore that)
2) ASI will surge (I have no idea why, that's empirical)
3) Once established in the theoretical climb, without circling, All will be as before. The glider is moving in exactly the same manner realtive to the airmass that it was, the 'deck angle', AOA, drag and lift etc., will all be as before, just that the whole lot's going up.

<also a glider pilot, without beard... and yes, I'm having a very bored day at work!>

Last edited by Mark1234; 20th Oct 2008 at 07:01.
20th Oct 2008, 10:17

Join Date: Dec 2002
Location: Chester
Posts: 68
Guys...thanks so much for the replies to this thread. Have learnt a lot. I hear a lot of non believers knocking PPRuNe as a source of technical info...the quality of your replies has been excellent.
Note to mod (please can I have that free personal title now)
20th Oct 2008, 18:33

Join Date: Oct 2000
Location: N. Europe
Posts: 436
Originally Posted by ChristiaanJ
But to get back to the original question, I'm still convinced that in a "typical" aircraft in a "typical" steady climb, the extra force needed to compensate W/W*cos(climbangle) i.e., a slightly higher vertical force, comes for 90% from a slight increase in "lift", by means of a slightly higher AoA.
a) The required net vertical force generated by T, D and L will still be the same as in level flight: W.

b) I don't know where you get W/W*cos(climb angle) from. It's the same thing as cos(climb angle) BTW... and why would you ever need to compensate a plain cosine?

c) You saw all the math above, giving you L = W*cos(climb angle). If you still think lift needs to increase, perhaps you should point out the error in the calculations... ?

Last edited by ft; 20th Oct 2008 at 18:59.
20th Oct 2008, 18:58

Join Date: Oct 2000
Location: N. Europe
Posts: 436
As for #2, I'll have a more analytical stab at it:

In both cases, total lift generated will decrease by the same amount as a climb is initiated. L = W*cos(climb angle), no speed dependency.

Parasite drag will essentially remain the same and can thus be ignored in this context.

As for induced drag,

L = S * rho/2 * V^2 * C_L

or, solving for the lift coefficient,

C_L = L/(S * rho/2 * V^2) = k1*L/V^2 (k1 is a constant)

The induced drag coefficient is proportional to the lift coefficient C_L squared,

CDi = k2 * C_L^2 = k2 * k1^2 * L^2 /V^4 = k3* L^2/V^4

(Induced drag is proportional to the induced drag coefficient)

The rate of change of the induced drag coefficient as the lift changes is calculated,

dCDi/dL = 2 * k3 * L / V^4

We see that V^4 remains. Conclusion: The rate of change of the induced speed as the amount of lift generated changes is highly dependant upon the true airspeed. The amount of change of total drag as you enter a climb is very unlikely to be the same at M.67 and M.8. At a high speed, the change in induced drag will be a lot smaller than at low speed.

It's late(ish). If I messed something up I'm sure I will be corrected.
20th Oct 2008, 21:04

Join Date: Jan 2005
Location: France
Posts: 2,319
Rainboe,
Don't tell me....
I should print out this thread, sit down with pen and paper and a calculator, and try to get my ideas sorted out, rather than answering all these posts off-the-cuff.

By the time you talk about a really steep climb, you're talking about hanging on your prop, or sitting on your exhaust... i.e., a thrust-to-weight close to 1, and your wing normal force is becoming pretty well irrelevant in your steady climb.

CJ

And... oh shoot....
Suddenly saw I missed half a dozen posts, and it's getting late at UTC+2 (I think). See you tomorrow... ?
20th Oct 2008, 21:15
Moderator

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Note to mod (please can I have that free personal title now)

.. too far down the food chain to be of much use to you there ..

... than answering all these posts off-the-cuff.

A lesson which just about all of us learn early in our careers .. in my case, being a bear of very little brain (as Milne would have it) .. I think it took me a tad longer than the average teddy bear to learn that lesson as a young engineer ... with several monumental clangers along the learning path.
20th Oct 2008, 22:03

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john,

I think you'll find two kinds of bears of little brain here....
The young withersnappers (sp?) that know it all, and the ancients (of which I am one) that still remember something, and don't mind getting it wrong, because they no longer have a boss looking over their shoulder.

Please, let's keep this one going, if only between us, until we all agree?

Christian
20th Oct 2008, 22:13
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.. somewhere in the past I recall it being put in terms of "age and treachery will always win out over youth and enthusiasm .. ?"
20th Oct 2008, 22:28

Join Date: Jan 2005
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Originally Posted by john_tullamarine
.. somewhere in the past I recall it being put in terms of "age and treachery will always win out over youth and enthusiasm .. ?"
John, PLEASE? This is a fun topic, even if we don't all agree...
Keep American politics out of it.....
We would like to keep this in "Tech Log", but this way it'll be relegated to "Jet Blast" in no time!

I think the current participants would like to continue kicking the subject around until we can all agree.

Christian
21st Oct 2008, 00:52

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This discussion would be so much easier with a blackboard and chalk...
21st Oct 2008, 01:03

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I like drag, anything in red or with sequins....
21st Oct 2008, 01:22

Join Date: May 2006
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Christiaan,

I can see where you're coming from, and at face value it's kinda compelling, but I believe your simplifying assumptions are flat wrong, hence the problem - you've stated a couple of times you're ignoring '2nd order effects of thrust vector etc', yet you do pay attention to the 2nd order effect of the tilting lift vector. That's not cricket - if the 2nd order effect of 1 is significant, so is the other.

Staying well away from the vertical, I think/hope we all agree that in level flight, T=D, L=W (and we'll leave it at that).

Rotate to a 10degree climb, stabilise so we're unaccelerated.

Now, you rightly point out that the W vector is unchanged (I'm trying to stay in your earth parallel frame of reference), and the L vector is tilted slightly back. But the T vector is also tilted slightly up, and the D vector down; sure, it's small, but so is the backward tilt of L. Don't ignore it

So we now get that (vertical component of L)+(vertical component of T)-(vertical component of D)=W Nothing too radical so far.

Now the question is, 'does the component of T' make up for the loss of L, and vertical component of D?

If we climb at the same speed as we were in level flight, T must increase - we all know that intuitively, but the maths is harder in your frame of reference:
(H component of D)+(H component of L) = (H component of T).

I don't have any numbers, but I'd argue that that increase in T is what makes the (vertical component of T) term larger than the reduction in L due to the tilt of the lift vector.

As previously stated you're free to pick any frame of reference you choose, so long as all terms are treated equally to that frame - it's common practice in maths to choose a frame that simplifies the calculations:

It's much easier to visualise if you take the aircraft as the frame of reference, then the only thing that factors is W; Pitch the aircraft and whole frame of reference; T must increase to combat the component of W that is now acting rearward along the deck plane, and L must decrease because it is combating a reduced component of W.

And no, I don't know it all.. I'm working the maths as I go along. Quite happy to be proven wrong, and learn something.
21st Oct 2008, 01:41

Join Date: Aug 2007
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ChristiaanJ, you quoted me...

fullyspooled, re accelerometers, I fully concur.
But you said...
Quote:
You do however correctly point out (in my lesson of the day) that the accelerometer will indicate temporary accelerations caused by pitching moments.
Negative. If the accelerometer is installed in the right location, pitching moments accelerations become second-order.
Of course.... if you install it under the pilot's bottom... that's a different story.
And I have pleasure in responding......

My Dear Sir, and I mean that most sincerely, but I think you are now guilty of splitting hairs, or dare I say it, having gotten slightly confused! (Should I say ooops now, I wonder?)

If the accelerometer is installed "in the right location," and by that I mean vertically on the instrument panel, it will indeed reflect accelerations occasioned by pitching moments! By the very nature of "pitching," assuming that one has sufficient airspeed to alter the trajectory of flight in the pitching plane, an acceleration takes place - and any such acceleration will be measured, and indicated on the accelerometer mounted as I have described.

The accelerometer, as has been previously mentioned, cares not at which attitude the airplane is, it simply measures acceleration in the "pitching" plane of an aircraft that is IN MOTION. If the aircraft is in motion and pitches, there is undoubtedly an acceleration, and I promise you that it WILL be measured by an accelerometer installed on the instrument panel!

However, if the aircraft is NOT in motion, you can pitch it all day long and no accelerometer in the World will notice a jot of change. An accelerometer does not in any way detect changes of pitch - that I know, and I can further testify that when correctly flown, several auto -rotational manoeuvres that involve massive pitch changes at VERY low airspeeds result in very small accelerometer deflections - due only to the fact that "directional" motion is limited to almost nothing. The lomcevak is a classic example.

Perhaps I was guilty of not qualifying my statement as thoroughly as you seem to do so naturally, but I trust that we now agree. If we do, I have another topic of discussion that I just know you will love to debate.........once this thread has finished.

.

Last edited by fullyspooled; 21st Oct 2008 at 02:02.