krujje,
"...how do you change your AoA without adjusting your climb angle?"

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...
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

"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. :)

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

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.

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 :O )

D*mmit! Bang to rights there... :)

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?

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!>

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):D

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... ?

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. :)

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... ?

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

.. 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.

john,
Welcome to this thread!

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

.. 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

This discussion would be so much easier with a blackboard and chalk...

I like drag, anything in red or with sequins....

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.

ChristiaanJ, you quoted me...

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.





.

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.

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.

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 = 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? :)

OK, hands up to whoever said the 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.....

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. ;)

LOL.... But I don't mind admitting I got it wrong.... Nothing beats pen and paper and thinking about it, rather than "off-the-cuff".
Right. My mistaek was not looking up cos(ca) for something like 6° climb angle, which is pretty well negligeable compared to sin(ca), hence negligeable in the vertical force balance relative to T*sin(ca), even with T ten times less than W.
Oops. Yes, I goofed there. But even in my quick-and-dirty example, T-D is still 200 kgf, so the vertical component is still 20 kgf, so the lift will still have to decrease by about 16 kgf.

Thanks all! I'm enjoying this.

CJ

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.

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.

And there's where you went wrong :) Humour me - consider the problem with the aircraft deck angle as your frame of reference. Forget factoring L, D, T, and look at what happens to W...

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.

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:

http://img127.imageshack.us/img127/5...malluv4.th.gifhttp://img127.imageshack.us/images/thpix.gif

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:

http://img255.imageshack.us/img255/6...mallju1.th.gifhttp://img255.imageshack.us/images/thpix.gif

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!

That's where you are trying to disguise an assumption as a fact. On what do you base the 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?

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.

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.

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

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.
http://img.photobucket.com/albums/v3...Total_drag.gif
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

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

Rainboe,
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. ;)

ChristiaanJ,
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?

ft, increasing headwind will do this. Wouldn't most test aircraft have inertials that would record this?
Windshear recognition comes to mind as a similar result of changing air mass velocity.

ft,
Re the quiz:
My guess would be either thermals, or a standing wave flying parallel to a mountain ridge.

A cunimb might also do the trick, but I doubt you would do perf test flying near a cunimb.

CJ

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.