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flybubba
11th Jun 2005, 03:39
In the typical descent profile for a jet from the enroute cruise altitude, one starts with a constant mach and then transitions to constant IAS. Maybe I should already know this, but I started thinking about what the glide angle is throughout the descent: ie, is it constant? I have verified that for a constant IAS descent from altitude, the angle is constant (I also verified this with a home pc flight sim profile). The case of constant initial mach and transition to constant IAS is more complicated. As one descends initially in constant mach, the IAS is continually increasing, and so is the glide angle. As one transitions to constant IAS, the glide angle then stays constant. I am just posting this for re-affirmation of what I believe to be true. This then relates to another question: Do FMS systems take this into account (the change in glide angle as a result of the constant mach/constant IAS transition) in predicting top of descent points?

FullWings
11th Jun 2005, 05:28
I think you're pretty spot on with your logic.

As for FMC predictions, I reason they must take the Mach/IAS transition into account otherwise we would always end up 'off profile' not long after the TOD. Another clue is that they ask for Mach no. & IAS as input...

Old Smokey
11th Jun 2005, 05:30
I wouldn't agree with everything you've said there flybubba, in both the Mach and the IAS phase, descent angle constantly increases.

(1) During the Mach phase, IAS is constantly increasing, with increasing drag, thus, both the Angle and Rate of descent constantly increase.

(2) During the IAS phase, drag is ALMOST constant, whilst TAS decreases. Thus the Rate of Descent is almost constant against a decreasing TAS, thus the descent angle constantly increases. Rate of Descent, whilst almost constant, will decrease slightly as drag reduces as EAS reduces for the same IAS with decreasing altitude. This latter factor slightly reduces the rate of increasing descent angle, but only slightly.

Regards,

Old Smokey

FullWings
11th Jun 2005, 05:41
Oooh, I'm not sure I agree with:
(2) During the IAS phase, drag is ALMOST constant, whilst TAS decreases. Thus the Rate of Descent is almost constant against a decreasing TAS, thus the descent angle constantly increases. Rate of Descent, whilst almost constant, will decrease slightly as drag reduces as EAS reduces for the same IAS with decreasing altitude. This latter factor slightly reduces the rate of increasing descent angle, but only slightly.For a given airframe weight the glide angle is pretty much purely dependent on IAS. TAS is a bit of a red herring in this discussion.

If you descend at constant IAS, your glide angle will stay the same throughout (neglecting things like change of wind/ISA deviation/fuel burn/changing idle thrust, etc.). As your TAS reduces, so will your ROD to compensate but the glide angle stays constant.

Try it sometime in a real aircraft. Start a fixed IAS descent at, say 250Kts @ 40,000'; note idle ROD: it will be around twice what you'd expect at sea level because your TAS is about twice your IAS at that altitude. Your glide angle is fundamentally the same.

flybubba
11th Jun 2005, 20:43
I would have to agree with Fullwings on this (constant IAS = constant glide angle). I see the logic that you are trying to use, Old Smokey. I think you are trying to equate vertical descent rate with drag, thus believing that the vertical descent rate doesn't change. I don't think this is a valid assumption. There are different ways to think about this: Here is one way which may help: IAS is really the dynamic pressure. Lift is a product of dynamic pressure and angle of attack. The glide angle depends on the ratio of lift to drag. At constant IAS, the lift and the drag don't change, thus the glide angle remain constant. For what it is worth, I have run these profiles on the microsoft flight sim. The descent angle is constant with the constant IAS descent. Now this assumes that they have used valid code in the simulation, but since these codes are widely available I don't expect it's a problem.

Regards

Old Smokey
12th Jun 2005, 12:54
Oh my Gawd FullWings and flybubba, I think that I crashed and burned on that one. I think that the two of you are very much closer to the truth than I was. Some re-thought needed on my part. There's a few mitigating factors which will lead to a reducing ROD and Descent Angle during the IAS phase, needs a few days thought for me to put it properly into words.

Thanks for keeping me honest,

Regards,

Old Smokey

flybubba
12th Jun 2005, 16:29
Hey Old Smokey,
Thanks for your honesty. I look at this message board as a way to exchange ideas and learn. I made the original post just to bounce my ideas off others, not being quite sure if my logic was correct. If we leave our ego's behind, and are willing to think things over carefully, and admit when we are wrong, much can be learned. I guess I would call that "intellectual honesty".

Regards,
flybubba

FlyingTom
13th Jun 2005, 22:12
The answer I was once given to this question was that the glidepath angle is non linear. The formula for a M0.73/280IAS descent is

: (((FL-10) * weight) / 17) + 40 = distance from airport. +/- 2nm for each 10kt of wind !!! Weight in tonnes.

From TOD the angle starts above average and by ED it is less than average. Divide by 3 works quite well I find.

I will look at the pitch angle next time I descend as this formula doesn't answer that question.

Pack2
14th Jun 2005, 07:56
Old Smokey
I am doing the JAA performance exam in a few weeks and I spend a lot of my time in this forum as it is a great place to learn.

In your experience do you have a quick method of mentally working out the required ROC to meet a given missed approach climb gradient. Lets say I have a 120 kt ground speed in the missed approach and need to make good a gradient of 3.5%
Thanks

FullWings
14th Jun 2005, 12:34
...a quick method of mentally working out the required ROC to meet a given missed approach climb gradientTry 1 in 60 rule? 120/60 = 2, x 3.5degs = 7, so roughly 700' per minute...

One knot is close to 100fpm.

square leg
14th Jun 2005, 13:45
Are u talking about 3.5% or 3.5°?

If 3.5° then yes 750' per minute is correct.

If 3.5% then I think that 420 ' per minute is correct. (Rule of thumb: 3.5%x120kts=420' per minute)

FullWings
14th Jun 2005, 14:14
Yes, you're right, for a %age gradient, climb rate = climb gradient x groundspeed.

Old Smokey
14th Jun 2005, 17:55
Last things first, Pack2, the formula offered by square leg is absolutely correct for practical purposes, but if you’re doing the maths, forget the percentage sign,
(Rule of thumb: 3.5%x120kts=420' per minute)
Just multiply the Groundspeed in Knots by the percentage gradient, i.e. multiply 120 by 3.5 in the example, DON’T take 3.5% of the G/S. 3.5% of the G/S gives you the Vertical Speed in Knots, and as the conversion from Knots to Feet per minute is as near as dammit to 100 (101.27 to split hairs), the ‘built in’ factor of 100 eliminates the need to work percentages.


With regard to my earlier discussions with flybubba and FullWings, my few days ‘think’ time is up, and I should have listened to my own words in a post which I made on another thread (difference between Vy and Vx), when I said –
Essentially all aircraft performance depends upon the 4 basic vectors, Weight, Drag, Lift, and Thrust. These may all be resolved geometrically, which, when there is an excess of thrust, yield a climb angle. Following on from this, all climb performance, when related to the 4 basic vectors indicate that all climb performance is at a particular angle, thus, the amount of excess thrust determines the climb angle.
Similarly, all descent performance, when related to the 4 basic vectors indicate that descent performance is at a particular angle, with the amount of excess DRAG Vs Thrust (Idle) determining the descent angle. The descent rate will then be determined as a function of descent angle and TAS. Variations in descent angle will then depend directly upon drag for the speed schedule flown (ignoring variations in Net Idle thrust during descent). To examine total drag, we must consider the Low speed Polar, due to Equivalent Airspeed (EAS), and the High speed Polar due to wave drag when flight is above Mcrit.

Take, as an example, an aircraft with a Mcrit of M0.73, and a descent profile speed schedule of M0.78 / 300 KIAS (These are ‘ball-park speeds for B737, A320, DC9, Learjet aircraft). Mach/CAS changeover Pressure Height for this speed schedule would be 29323 feet.

(1) For the descent to 29323 ft, High speed (wave) drag would be constant, due to the constant Mach number, but Low speed (dynamic) drag will be increasing due to increasing CAS and EAS. Thus, total drag and TAS would be increasing during descent, resulting in increasing descent angle and rate.

(2) At Mach/CAS changeover Pressure Height (29323 ft) the EAS is 286 knots whilst the Mach number is M0.78. Further descent at a constant CAS of 300 knots will be occasioned by decreasing wave drag as Mach number now reduces, until 25943 feet, when Mach = 0.73 (Mcrit) and wave drag is now zero. In this same phase of descent, EAS increases slightly to 288 knots. Thus, in this phase, wave drag decreases significantly, whilst dynamic drag increases slightly. The net result is that, in this phase, descent angle, TAS, and descent rate all reduce.

(3) Below the level where Mcrit was passed (25943 ft), at a constant CAS of 300 knots, EAS increases from 288 knots towards equality with CAS at Sea level. At 10000 feet EAS is 297 knots, and dynamic drag has increased by 6.3% between 25943 ft and 10000 ft. This leads to a steepening descent angle, but, as TAS decreases at a faster rate, Rate of Descent steadily decreases.

The summary is that initially, at the ‘Mach stage’, descent angle and rate constantly increase. At the CAS phase, descent angle and rate initially decrease, but then descent angle increases to a small degree, whilst rate of descent steadily decreases. In the ‘win a little, lose a little’ CAS descent phase, the average is fairly close to a constant angle, and may be considered as thus for PRACTICAL purposes.

I think (I hope) that I got it right this time.

Regards,

Old Smokey

Pack2
14th Jun 2005, 22:25
Square Leg..(Has to be an Aussi)

Thanks, just what I needed.

OLD Smokey

Brilliant post ...explains a lot of theory bouncing aound in my head just now..

flybubba
19th Jun 2005, 01:39
Old Smokey,
I thought I had this figured out. Now I'm confused. I hadn't thought of wave drag, nor had I thought much about EAS, etc. But of course compressibility effects can't be ignored. When I did the descent profile simulation on the microsoft sim, I started out at say FL400 at M.73 (maybe this is too slow). In the constant mach descent, the glide angle increased as you said. I understood this as an increasing CAS, and increasing glide angle. I didn't look at the transition regime carefully. But during the constant IAS descent, the glide angle seemed fairly constant. What you have said seems to fit what I observed, but I'll need to think a bit more on this.

Regards,
flybubba

airmen
21st Jun 2005, 05:45
Try to check your FMS book, it should give you some info about how the TOD is calculated by the cptr. If I remember corectly, it is an angle that you can adjust somewhere in the PERF pages...
Try to change this value and you will see the TOD moving along the flight plan

Old Smokey, nice point;)

Old Smokey
21st Jun 2005, 09:06
flybubba,

An addendum to my previous post. I stated for the example given "a drag increase of 6.3% between 25943 ft and 10000 ft", true. I then went on to say that for practical purposes the descent gradient can be considered as fairly constant. This seems a little too simplified in view of a 6.3% increase in drag, which would lead to a noticeably steepening gradient. The addendum (fancy way to say this is what I forgot to say) is that TAS and therefore momentum is decreasing during the CAS phase, and the kinetic energy is dissipated as an effective increase in thrust, thus lessening, but NOT cancelling the overall effect of a steepening descent. This is, in fact, the opposite to climb at a constant CAS where some of the available thrust is used to accelerate the aircraft, leading to less being available for climb.

Hope that this does not add to the confusion.

Regards,

Old Smokey

flybubba
22nd Jun 2005, 02:38
Old Smokey,
I follow what you are saying about the drag change, but I try to understand this in terms of L/D. The glide angle is proportional to L/D. The drag may be changing but is not the lift also changing so as to have constant L/D? I'm only talking about the case of a constant IAS descent here (I agree about the initial constant mach descent having a changing glide angle). TAS may be changing thru all this, but I think that what really matters is L/D.
Any thoughts?

Old Smokey
23rd Jun 2005, 08:42
flybubba,

In advance, this discussion relates entirely to the 'CAS' phase below the level where Mcrit is encountered, thus all statements here are relevant to low speed drag polars only.

What you say would be (almost) absolutely true if we descended at a constant EAS, but we don't, we descend at a constant CAS.

Weight during this phase is constant, if we ignore the very small weight reduction caused by the 15 minutes or so at idle thrust. (If we did consider the weight reduction due to the fuel used, Weight, and Lift required would be LESS). Thus, Weight, and Lift, are constant, because Lift = Weight. Drag, on the other hand is increasing due to increasing EAS at a constant CAS as we descend (6.3% increase in my earlier example). So, whilst Lift remains constant (actually reducing slightly), Drag is increasing, and the L/D ratio is definately not constant. As you have correctly indicated, glide angle is directly related to the L/D ratio, and a decreasing L/D ratio will result in a steepening glide angle. This steepening glide angle is mitigated to some degree by the translation of kinetic energy into 'effective thrust' as TAS reduces.

To sum up the 3 factors involved -

(1) Weight, and therefore Lift, reduce, leading to a lower L/D ratio, and steepening glide angle,
(2) Drag increases due tincreasing EAS, leading to a lower L/D ratio, and steepening glide angle,
(3) Kinetec energy dissipation due to deceleration (decreasing TAS) translates to an effective drag reduction, and a degree of abatement of the increasing glide angle.

Regards,

Old Smokey

flybubba
23rd Jun 2005, 16:49
Old Smokey,
OK, I now understand what you are saying. As you say, Lift isn't changing, so if Drag is increasing then the glide angle is changing (inreasing). A couple of additional points: since Lift equals W times cos (glide angle), if the angle changes, then the lift will change slightly. But this is a small affect since the cos of a small angle is approx equal to one, and small changes will not be significant.
I hadn't thought much about EAS. As you point out, it is a small affect. The affect is even smaller at lower speeds (say 250 kts).
I need to think a little more about your third point relating to dcereasing kinetic energy.

Regards,
flybubba

togaroo
23rd Jun 2005, 19:46
For the descent, especially in Airbus aircraft it really depends what mode you are in as to what ROD and pitch angle you get as in a managed profile speed is the key driver, so therefore your pitch angle will vary through the various levels to meet constraints and speed targets. For least fuel, in a low aircraft congestion area pitch and speed will remain fairly constant. However in London, you rarely use a managed profile as headings and constraints will change what the FMGC has predicted. If you descend at flight idle with a selected speed, again pitch will vary to maintain the speed and roughly ROD of about 2000 - 2500fpm. I use the 3 x rule for height you need to lose plus a mile for any speed you need to lose per kt. I guess what I am trying to say is that you rarely achieve a constant descent glide angle - except for a long approach to LGW / LHR!! Best to expect the unexpected. Better to be slighty under the profile and if your going down you cant slow down!!!!

Just to complicate things further, A good technique I use to accelerate and climb, is to do so with a selected ias then you of course also increase your mach as you climb, however there will be some trade off with ROC (well for airbus aircraft anyway - fuel hungry boeings can do what ever they like!) It works for me.

AirRabbit
26th Jun 2005, 17:00
Gentlemen (and Ladies?)

All I can say is “Wow!” There is some really interesting and valuable information floating around on this thread, and I’m going to copy it and keep for my future reference. Having said that, I am hoping that I might be able to ask a question of those here, specifically Smokey and Bubba, however any of the rest of you certainly are welcome to throw your comments in as well. I’d much rather have too much information than not enough.

I have been watching the results of the Pinnacle Airlines accident unfold with great interest. A CRJ200 on a repositioning flight was dispatched using, I think, FL370. But for some reason the crew decided to climb to the service ceiling of the airplane, FL410. Records show that temperature that night was ISA+10; the crew elected to climb at a constant ROC, and upon reaching 410, they were very slow with an uncommonly high pitch attitude. They remained at 410 for about 3 minutes (with decaying airspeed and increasing pitch attitude) and lost an engine, autopilot disconnected, lost the second engine, entered stick shaker, got stick pusher, never were able to get either engine back on line (even though the ADG did deploy and did provide some electricity for some things), and wound up crashing, killing both crew members.

As you can imagine, there have been a lot of questions generated about glide angle, rate of descent, distance capability, etc., etc. I have been on a campaign for some time now, trying to spur interest in having direct reading AOA indicators in airline cockpits, and it seems that this accident may be enough of a catalyst to make that argument. I recognize that several things change throughout the descent with respect to the descent angle and descent rate; but I would like to know the following:

With a power off situation such as this, if a direct reading AOA indicator is available, is there an AOA that might be used (even if it had to be adjusted a couple of times) during the descent that might be used, instead of recalling and applying all the very good information posted in this thread, that would allow the crew in such a harrowing and difficult circumstance to get the best return from the only source of energy they have available to them, height?

Thanks in advance.

_______
AirRabbit

atyourcervix73
26th Jun 2005, 17:13
The book that comes with the shiney bird I drive (B757) gives the figure of 3.5 miles per 1000 feet of altitude loss, this is at the economy Mach/ then IAS calculated by the FMS.
Interestingly...but not surprisingly..the whole thing comes down to energy management i.e. a heavier aircraft has more potential than a lighter aircraft, so at light weights we use 3 miles per 1000 feet, and at heavy weights we use 3.5 - 4.0 miles per 1000 feet. In practical terms, trying to match the desent profile is often impractical due to ATC constraints.


:ok:

flybubba
26th Jun 2005, 21:26
AirRabbit,
I started this thread as more of a curiosity, than as something really useful. But it has made me think about things more carefully. Your suggestion of using an AOA indicator has been suggested by others, and I often wonder why we don't have them in the civil aviation fleet, but this would digress into another thread subject (as an aside Boeing had a tech report about AOA use for the airline fleet).
In the interest of simplicity, maybe I should just remember what I was first told: "when you pull back on the yoke things on the ground get smaller; if you keep pulling back they will get larger again". Also, I was once asked to explain by an examiner what keeps airplanes in the air. After launching into the usual explanations, the examiner just snickered and pulled out his wallet. He then replied "it's this green stuff that keeps airplanes in the air, don't ever forget that".

regards,
flybubba

hawk37
26th Jun 2005, 22:52
AirRabbit
As a rough approximation, 1.5 * V stall, about a normalized aoa of .44 gives best L/D for todays jets for low level ops. I'd think for above 30,000 feet, use 1.6 *V stall at sea level, .39 normalized aoa, to factor in the rise in stall CAS due to compressibility.
Hawk

bobrun
27th Jun 2005, 00:34
About the accident......the article I read mentionned that the engines might have frozen at FL410, causing them to fail!:confused: