# Total drag questions

Join Date: Oct 2000

Location: Bristol

Posts: 461

Karl Popper has pointed out that the more information a statement contains the less likely is it to be true. The statement " If the aircraft is in motion and pitches there is..an acceleration" is a sweeping generalisation and therefore unlikely to be true. A lot of qualification is needed here.

For example, if the accelerometer (or aircraft vertical axis acceleration sensor) is mounted at the center of rotation of the pitching movement there will be no record of the angular change in pitch. If the accelerometer is a simple spring and bob mounted on the instrument panel it may well indicate a small acceleration as the rotation in pitch is read as a linear vertical axis movement at the instrument location.

If the change of pitch occurs at some point in the flight regime where a change in pitch generates a change in lift then the pitch change will be followed by a linear acceleration on the vertical axis, read by either type of measurement system. However, if the pitch change does not result in a change of lift then there wil be no follow-up acceleration on the vertical axis.

So there is at least one set of factors that would result in a pitch change of an aircraft in motion recording no acceleration at all.

Where in the flight envelope would a change in pitch not reult in a change in lift? How about the flat bit of the CL/alpha curve around CLmax?

Going back, we never did fully explore the various linear and rotational forces that have to be brought to zero to sustain a true vetical flightpath. But that is for all you youngsters

Dick

Edited to correct "Z axis", which is an earth axis, to aircraft vertical axis.

For example, if the accelerometer (or aircraft vertical axis acceleration sensor) is mounted at the center of rotation of the pitching movement there will be no record of the angular change in pitch. If the accelerometer is a simple spring and bob mounted on the instrument panel it may well indicate a small acceleration as the rotation in pitch is read as a linear vertical axis movement at the instrument location.

If the change of pitch occurs at some point in the flight regime where a change in pitch generates a change in lift then the pitch change will be followed by a linear acceleration on the vertical axis, read by either type of measurement system. However, if the pitch change does not result in a change of lift then there wil be no follow-up acceleration on the vertical axis.

So there is at least one set of factors that would result in a pitch change of an aircraft in motion recording no acceleration at all.

Where in the flight envelope would a change in pitch not reult in a change in lift? How about the flat bit of the CL/alpha curve around CLmax?

Going back, we never did fully explore the various linear and rotational forces that have to be brought to zero to sustain a true vetical flightpath. But that is for all you youngsters

Dick

Edited to correct "Z axis", which is an earth axis, to aircraft vertical axis.

*Last edited by Dick Whittingham; 21st Oct 2008 at 16:55.*

Join Date: Jul 2006

Location: Norway

Age: 36

Posts: 122

This was an interesting subject..

Initially my hunch was with ChristiaanJ, but as of now I am not sure if I completely agree.

Since there is much discussion about definitions in this thread I will state my own. I am certainly no aeronautical engineer, but I see these definitions as a prerequisite for the discussion.

I will use thrust (T), drag (D), lift (L) and weight (W). We have a vertical and horizontal axis on our aircraft through the center of gravity. Weight is always directed towards the center of the earth. Thrust and drag are for simplification aligned with the flight path, and lift is directed perpendicular to the flight path. I will also call the angle between the flight path and the horizontal through the center of gravity the climb angle (a).

To then sum up the positive vertical vector during climb we have:

T*sin (a) - D*sin (a) + L*cos (a) - W

This would give a fixed lift for any given speed regardless of pitch-angle and thus the induced drag will be the same as for that specific speed in horizontal flight.

The negative component of drag ( D*sin (a) ) and reduced vertical component of lift ( L*cos (a) ) must from what I see be countered by the increased positive component of thrust ( T*sin (a) ).

So.. neither increased nor reduced induced drag during climb.

As stated earlier I am only a layman on this subject, so any corrections making me wiser are very welcome.

Initially my hunch was with ChristiaanJ, but as of now I am not sure if I completely agree.

Since there is much discussion about definitions in this thread I will state my own. I am certainly no aeronautical engineer, but I see these definitions as a prerequisite for the discussion.

I will use thrust (T), drag (D), lift (L) and weight (W). We have a vertical and horizontal axis on our aircraft through the center of gravity. Weight is always directed towards the center of the earth. Thrust and drag are for simplification aligned with the flight path, and lift is directed perpendicular to the flight path. I will also call the angle between the flight path and the horizontal through the center of gravity the climb angle (a).

To then sum up the positive vertical vector during climb we have:

T*sin (a) - D*sin (a) + L*cos (a) - W

This would give a fixed lift for any given speed regardless of pitch-angle and thus the induced drag will be the same as for that specific speed in horizontal flight.

The negative component of drag ( D*sin (a) ) and reduced vertical component of lift ( L*cos (a) ) must from what I see be countered by the increased positive component of thrust ( T*sin (a) ).

So.. neither increased nor reduced induced drag during climb.

As stated earlier I am only a layman on this subject, so any corrections making me wiser are very welcome.

Join Date: Oct 2000

Location: N. Europe

Posts: 436

Kristian,

welcome to the discussion!

"T*sin (a) - D*sin (a) + L*cos (a) - W"

Almost correct. You included all the factors but forgot the condition, as specified by the requirement for a steady-state climb:

T*sin (a) - D*sin (a) + L*cos (a) - W

"This would give a fixed lift for any given speed regardless of pitch-angle"

Sorry, it will not. T will increase while D will remain largely constant for small changes of a, especially at high airspeeds (see the answer to OP Q2!). W will, obviously, remain constant. L will reduce, and the condition of vertical force balance as specified by the equation will remain satisfied.

BTW, you defined

welcome to the discussion!

"T*sin (a) - D*sin (a) + L*cos (a) - W"

Almost correct. You included all the factors but forgot the condition, as specified by the requirement for a steady-state climb:

T*sin (a) - D*sin (a) + L*cos (a) - W

*= 0*"This would give a fixed lift for any given speed regardless of pitch-angle"

Sorry, it will not. T will increase while D will remain largely constant for small changes of a, especially at high airspeeds (see the answer to OP Q2!). W will, obviously, remain constant. L will reduce, and the condition of vertical force balance as specified by the equation will remain satisfied.

BTW, you defined

*a*as*climb angle*and then proceeded to make conclusions about the effect of*pitch angle*from the (half... ) equation including*a*but no reference to pitch angle. Easy to mess up in this, innit?Join Date: Jan 2005

Location: France

Posts: 2,319

OK, hands up to whoever said the

Everything else is secondary, except that the thrust angle is turned up over the same angle.

If you feel like doing some scribbling, try with an aircraft of 2000 kg, a CL/CD of 10, and a climb angle of about 6° (sine = 0.1, cosine = 0.995, so practically =1).

I'll try and clean up my scribbles and post them, but here goes in the meantime.

In level flight, drag and thrust are 200 kgf.

Start a steady climb at abt 6° climb angle (nothing excessive, that's 2000 ft/min at 200 kts).

Now you have to drag 2000 kgf upstairs at W*sin(ca), i.e., 200 kgf more.

So T now becomes 400 kgf, and the vertical component = T*sin(ca) = 40 kgf.

Since we're talking steady state, the component from the lift will have to be reduced by the same 40 kgf. But.... L=W*cos(ca) is only about 4kgf less than W. So we have to

Since 2000 kg, CL/CD=10, and a climb at 6° are hardly excessive, it would seem Rainboe's intuition about looking at a steep climb was right after all....

Now to question 2.....

__real issue__is the lift vector being turned back over (climbangle) degrees.Everything else is secondary, except that the thrust angle is turned up over the same angle.

If you feel like doing some scribbling, try with an aircraft of 2000 kg, a CL/CD of 10, and a climb angle of about 6° (sine = 0.1, cosine = 0.995, so practically =1).

I'll try and clean up my scribbles and post them, but here goes in the meantime.

In level flight, drag and thrust are 200 kgf.

Start a steady climb at abt 6° climb angle (nothing excessive, that's 2000 ft/min at 200 kts).

Now you have to drag 2000 kgf upstairs at W*sin(ca), i.e., 200 kgf more.

*Aye, there's the rub....*.So T now becomes 400 kgf, and the vertical component = T*sin(ca) = 40 kgf.

Since we're talking steady state, the component from the lift will have to be reduced by the same 40 kgf. But.... L=W*cos(ca) is only about 4kgf less than W. So we have to

__reduce__the lift by another 36kgf to get back to a steady state.Since 2000 kg, CL/CD=10, and a climb at 6° are hardly excessive, it would seem Rainboe's intuition about looking at a steep climb was right after all....

Now to question 2.....

Join Date: Oct 2000

Location: N. Europe

Posts: 436

ChristiaanJ,

welcome to the dark side!

However,

L = W*cos(ca)

does hold true. You forgot to factor in D*sin(ca) in the world-vertical force balance. Have a look at balsa model's diagram in post #26.

Thanks for the credit for post #6 and #58.

welcome to the dark side!

However,

L = W*cos(ca)

does hold true. You forgot to factor in D*sin(ca) in the world-vertical force balance. Have a look at balsa model's diagram in post #26.

Thanks for the credit for post #6 and #58.

Join Date: Jan 2005

Location: France

Posts: 2,319

Originally Posted by

**ft**ChristiaanJ,

welcome to the dark side!

welcome to the dark side!

However, L = W*cos(ca) does hold true.

You forgot to factor in D*sin(ca) in the world-vertical force balance. Have a look at balsa model's diagram in post #26.

__decrease__by about 16 kgf.

Thanks all! I'm enjoying this.

CJ

Join Date: May 2006

Location: Londonish

Posts: 780

In all fairness Rainboe, not everyones mind is wired to think that way / can do that - it's a visual 'I see' thing. Sometimes to get the message across you have to speak *their* language, not your own. Bit like someone trying to convince you or I with trigonometry when all they need is a decent picture.

Join Date: Jul 2006

Location: Norway

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Posts: 122

Hello ft, glad to be a part of the discussion!

You are quite right:

T*sin (a) - D*sin (a) + L*cos (a) - W = 0

That's the total equation.

And my intention was of course to refer to (a) as 'climb angle' throughout the post.

But still my point stands. If one climbs with the same speed and the same weight as a similar plane would for level flight the angle of attack would be the same in both situations. Therefore I haven't included it in the equation.

Assuming then the airspeed is the same the force perpendicular to the thrust is the same. So the lift vector is the same.

The vertical component of lift - L*cos (a) - will on the other hand be less. By increasing thrust and hence the vertical component of thrust - T*sin (a) - one compensates for the reduced vertical component of lift.

You are quite right:

T*sin (a) - D*sin (a) + L*cos (a) - W = 0

That's the total equation.

And my intention was of course to refer to (a) as 'climb angle' throughout the post.

But still my point stands. If one climbs with the same speed and the same weight as a similar plane would for level flight the angle of attack would be the same in both situations. Therefore I haven't included it in the equation.

Assuming then the airspeed is the same the force perpendicular to the thrust is the same. So the lift vector is the same.

The vertical component of lift - L*cos (a) - will on the other hand be less. By increasing thrust and hence the vertical component of thrust - T*sin (a) - one compensates for the reduced vertical component of lift.

Join Date: May 2006

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If one climbs with the same speed and the same weight as a similar plane would for level flight the angle of attack would be the same in both situations. Therefore I haven't included it in the equation.

weight on the wings reduces, so the AOA will reduce. Still works in your frame of reference, but you need to plug in real numbers to see how it comes to pass.

Join Date: Oct 2000

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Posts: 436

Allright, it is Friday. Time for some entertainment! I was stuck in a dull meeting and ended up scribbling a few illustrations for this thread. Here’s how most people seem to think about it at first:

Same old reference system we grew up in (the antipodeans excepted), so I don’t suppose that is too surprising. The problem is that intuitive understanding of the issue at hand is not obvious and if you try to go the analytical route, you end up with the trigonometrical mess I posted a while ago (and yes, Rainboe, some of us do both the nice piccies and the analytical part behind closed eyelids while holding a nice cuppa – no promise not to spill though!).

If you instead think about the climb case in the aircraft reference frame, like this:

everything becomes much clearer. Suddenly, the lift decrease (and thrust increase) is easy to understand intuitively. Even sorting it out analytically becomes a breeze. The only real problem is that the water will seem likely to spill out of the lake, but that’s a small sacrifice to make on the altar of aerodynamical understanding.

Yes, you can all have my autograph when I become famous for my art!

Same old reference system we grew up in (the antipodeans excepted), so I don’t suppose that is too surprising. The problem is that intuitive understanding of the issue at hand is not obvious and if you try to go the analytical route, you end up with the trigonometrical mess I posted a while ago (and yes, Rainboe, some of us do both the nice piccies and the analytical part behind closed eyelids while holding a nice cuppa – no promise not to spill though!).

If you instead think about the climb case in the aircraft reference frame, like this:

everything becomes much clearer. Suddenly, the lift decrease (and thrust increase) is easy to understand intuitively. Even sorting it out analytically becomes a breeze. The only real problem is that the water will seem likely to spill out of the lake, but that’s a small sacrifice to make on the altar of aerodynamical understanding.

Yes, you can all have my autograph when I become famous for my art!

Join Date: Oct 2000

Location: N. Europe

Posts: 436

Originally Posted by

**KristianNorway**But still my point stands. If one climbs with the same speed and the same weight as a similar plane would for level flight the angle of attack would be the same in both situations.

*assumption*that AoA would stay the same after entering a steady climb?

AoA will, under the specified conditions, have to be changed to whatever it needs to be to maintain equilibrium. And that

*does*mean a decrease...

Again, doublecheck with the extremes and if it doesn't match up, you're probably doing something wrong. Will the AoA be the same in a vertical climb as in level flight?

Join Date: Jul 2006

Location: Norway

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Posts: 122

Indeed.. sounds right. Do you have any way of proving this? I would very much like to understand the fact better.

It would be interesting to see how much the climb angle would affect the AoA.

I would estimate the effect to be fairly small, but still.

In any other case the AoA would be directly connected to airspeed.

It's interesting because it tells us how much of an effect that will have on lift during climb.

It would be interesting to see how much the climb angle would affect the AoA.

I would estimate the effect to be fairly small, but still.

In any other case the AoA would be directly connected to airspeed.

It's interesting because it tells us how much of an effect that will have on lift during climb.

Join Date: Feb 2001

Location: Milkway Galaxy

Posts: 218

Hi

I am on leave now.

At my first flight soon (A 330), I will check the AOA from ACMS page for same weight and IAS: Initially at level flight and immediately after this, at stabilized climb.

I hope this will put a full stop to this discussion...

Regards.

I am on leave now.

At my first flight soon (A 330), I will check the AOA from ACMS page for same weight and IAS: Initially at level flight and immediately after this, at stabilized climb.

I hope this will put a full stop to this discussion...

Regards.

Join Date: Jan 2005

Location: France

Posts: 2,319

**Rainboe**,

Been there, done that.

Initially I expected a something squared to jump out of the woodwork and bite...

I've now come over to the dark side

But....

**JABBARA**,

Can you give us a typical figure for the climb angle (or vertical speed and TAS) of your stabilized climb?

For instance, for a climb at 200 kts TAS and 2000 ft/min, the climb angle is about 6°. Since the cosine of 6° is about 0.995, the lift (normal to the flight path) at that sort of climb angle is reduced only by a very small amount (see all the earlier formulae...), and so is the AoA (we're talking a percent or so).

So unless you do some careful flying and 'measuring', you may not see the effect at all.

CJ

Join Date: Jan 2005

Location: France

Posts: 2,319

**Rainboe**,

Don't over-simplify... the increase in alpha is not the same as the cosine(climbangle). But your basic intuition of decreasing alpha with climb angle is right (except possibly for something with L/D less than one.... I'll have to look into that....)..

**To all**,

OK, promises, promises...

Here's my scribble.

Not quite so artistic, so let me explain.

What we are looking at, are the horizontal and vertical force balance (steady state, remember?).

I've been looking at the other extreme from Rainboe's example, i.e., the typical climb rather than the vertical climb.

With a climb at 6° (my earlier example) or even 15° (cos = 0.97, quoted by Rainboe)...

Horizontally:

- the drag component barely changes,

- thrust changes a lot, because suddenly the thrust has to "drag" L*sin(ac), the horizontal component of L, "upstairs".

Vertically:

- the weight doesn't change,

- the drag component is now tilted downward,

- the thrust component is tilted upwards, and thrust is now larger than the drag, so the net result is an upwards force.

To get back to a steady state, we'll have to reduce L*cos(ac).

After plugging in the rest the maths... that comes down to reducing L, hence the AoA.

So, conclusions?

- At the typical climb angles of most aircraft, be them Cessna or Airbus, you will barely notice the change in angle of attack (it's all cosine something, and the cosine is almost 1 (one) ).

- You will certainly notice the difference in thrust (or engine power - which is the same in this context) to maintain a steady climb under the same conditions as in horizontal flight.

- If you're a fighter jock with more puff behind you than your aeroplane weighs... have a go at all means to try the other extreme of our discussion. But remember what we said. You may need to pitch down a bit for that perfect vertical climb.

CJ

Join Date: Feb 2001

Location: Milkway Galaxy

Posts: 218

Chritiaanj,

I will try to record as many as paramaters when I do this test. But of course to be able to read a remarkable AOA difference in two cases, I will try do it as low altitude as possible (to get a significiant difference between cruise and climb thrusts) and as close as possible to Green Dot speed (to get a noticable body attitude during climb). Obviously I should comply with ATC requirement as well. I hope ATC instructions coincide with my goaled flight conditions.

I will try to record V/S and TAS as well, but in this issue, I do not consider they are so important as body attitude.

Anyway, I will record and you will decide.

Regards

I will try to record as many as paramaters when I do this test. But of course to be able to read a remarkable AOA difference in two cases, I will try do it as low altitude as possible (to get a significiant difference between cruise and climb thrusts) and as close as possible to Green Dot speed (to get a noticable body attitude during climb). Obviously I should comply with ATC requirement as well. I hope ATC instructions coincide with my goaled flight conditions.

I will try to record V/S and TAS as well, but in this issue, I do not consider they are so important as body attitude.

Anyway, I will record and you will decide.

Regards

Join Date: Oct 2000

Location: N. Europe

Posts: 436

Originally Posted by

**Rainboe**In a vertical climb, AoA for all intents and purposes would be zero,meaning the fuselage would have negative pitch (ie be pointing at about 86 degrees pitch whilst plane is climbing vertically)

todays nitpicker comment: In a vertical climb most aircraft (which have cambered wings) will in fact have a zero-lift AoA which is negative rather than zero, giving you a few additional degrees nose down relative to vertical to add to those caused by the angle of incidence.

Originally Posted by

**ChristiaanJ**Don't over-simplify... the increase in alpha is not the same as the cosine(climbangle). But your basic intuition of decreasing alpha with climb angle is right (except possibly for something with L/D less than one.... I'll have to look into that....)..

Rainboe is spot on with his assumption which is not at all an oversimplification. In the normal flight region, lift for most wings is more or less linearly connected with the AoA. Hence it is perfectly valid to assume AoA = cos(climb angle) when L = L_level_flight*cos(climb angle).

If the lift coefficient curve is highly non-linear, Rainboe's assumption won't be correct. However, for your typical wing in normal speed flight, that's not the case.

Tried tilting your diagram so that the lift vector is poing straight up and redoing the calculations yet, as in my very artistic sketches before?

Horizontally

Thrust equals drag plus the longitudinal component of weight

Vertically

Lift equals the flight path normal component of weight.

Much easier, methinks.

L/D isn't a factor at all in how much the lift will decrease. We know exactly how much lift will decrease. The thing L/D in the current flight conditions will tell us how much drag will decrease as a result.

Jabbara,

thanks for volunteering. However, as has been pointed out, it will be surprising if your AoA sensor can sense an AoA decrease of a few hundreths of a degree. Don't worry about the body attitude, it doesn't matter. Only climb angle does. However, slower will give a larger absolute change in angle of attack, everything else remaining equal.

Will the A330 tell you approximately how much thrust you are generating? That's a more interesting parameter to compare, as the change should be close to W*sin(climb angle).

TAS and V/S are interesting, as they will tell us the climb angle. Of course, any change of AoA will cause ASI errors... which again might or might not be compensated for by the ADC based on if it can sense the reduction in AoA or not. Not that I think these factors will be significant at the

*huge*AoA changes we are considering here.

Oh, the joys of flight testing. Todays quiz: We once (well, more than once but bear with me) experienced climb performance data from a performance flight test which was way outside of the normal parameters. The aircraft just climbed as a bat out of h_ll for a segment of the recorded climb. What do you think happened, and how were we able to determine the cause and compensate?

Join Date: Oct 2000

Location: N. Europe

Posts: 436

hawk37 nailed it, the aircraft climbed through a layer of gradual wind shear.

No inertial sensor worth its salt nor any external tracking as it was a line aircraft performance flight test.

The solution to these problems is to always fly the same climb on the reciprocal course along the same track. Looking at the data from the two climbs together, it is then obvious what happened.

No inertial sensor worth its salt nor any external tracking as it was a line aircraft performance flight test.

The solution to these problems is to always fly the same climb on the reciprocal course along the same track. Looking at the data from the two climbs together, it is then obvious what happened.