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airborneforever
4th Oct 2009, 17:00
I fly the 737 NG series.
Recently my captain showed me something which completely foxed me and forced me to come here and share my qaundry with you folks.
My captains theory was that taking the same set of conditions for two aircraft, a heavier one will start its descent earlier!!:confused:
We checked this on the FMC and a difference of 10 tonnes to the ZFW on the Perf init caused to ToD to shift approximately 30 miles out!!!
Could somebody please explain how this works!:O
Regards,
A

Slasher
4th Oct 2009, 18:03
Of course it descends earlier! For a given IAS a light-weighing aircraft will fall out of the sky. Reverse for heavier wieghts.

Compare now to a case of a given AoA for descent. The clue is in the green dot speed which varys with GW ( or in nonglass-speak "best lift/drag speed clean" ).

Muse over those 2 lines above while dusting off your Student Pilot theory book and the penny'll drop (without a FMC box!) ;)

Microburst2002
4th Oct 2009, 18:33
Hi airborne

You are puzzled because typically the test questions are about airplanes gliding at best gliding speed (best gliding AoA, actually). In that case, weight makes no difference in gliding distance, only in gliding time (heavier, faster).

However, as said in the previous post, when two airplanes are gliding at a given airspeed, say 300 kt, the heavier glides more miles. Why? Because the heavier one is gliding at a more efficient angle of attack.

Gliding distance is proportional to the Lift-Drag ratio. The mor L/D, the more distance. Lift-Drag ratio is equal to Lift Coefficient-Drag Coefficient ratio. These depend solely on the angle of attack (well, mach number also has an influence...). Remember how weight, AoA and speed are related and you will see that heavier at constant speed means less AoA, and therefore (in the stable speed region, of course) more L/D.

I hope it helps.

MarkerInbound
4th Oct 2009, 22:03
The same reason gliders load up with water to go cross country.

35hPA28
4th Oct 2009, 22:35
For a given IAS a light-wieghing aircraft will fall out of the sky like a bucket of sh!t. Reverse for heavyer wieghts.

when two airplanes are gliding at a given airspeed, say 300 kt, the heavier glides more miles.

I thought I knew the answer to this but you two guys really got me confused.

Considering that two identical airplanes can glide the same distance regardless of different weight, but at different speeds, I understand there must be (lower) speeds at which a lighter airplane will glide further if both glide at the same speed.

What am I missing? Thanks in advance! :ok:

Wizofoz
4th Oct 2009, 22:38
All true,

HOWEVER a consequence of just that is that if you fly the lighter aircraft at its best L/D speed, and fly the heavier aircraft at that same speed, the HEAVIER aircraft will now descend more steeply.

Jumbo Driver
4th Oct 2009, 22:47
Surely, the simplest way of explaining it is to say that the heavier aircraft has more Kinetic Energy (1/2mv²), which will carry it further before it reaches the ground ... ?

Similarly, a heavier aircraft will take longer for a given speed reduction than a lighter one ...


JD
:)

Wizofoz
4th Oct 2009, 22:55
Jumbo,

No, nothing to do with it. As I said, fly the light aircraft at best L/D, and the heavier one at that speed, and the heavier aircraft will descend more steeply. Fly them BOTH at best L/D and glide distance will be identical.

It's a little complicated by the fact that they are NOT gliding, and the residual thrust will have a greater effect on the lighter aircraft than the heavier one, but the principle is as outlined

The heavier one may have more energy to dissipate, but it will also produce more drag, so it kinda evens out.

35hrPa28,

Our answers crossed, but yes, you are right. HOWEVER the usual descent speed for jets are well in excess of their best L/D speeds, so in any usual situation, the heavier aircraft will glide further.

QJB
5th Oct 2009, 03:42
This got me a little confused at first because I was thinking that a heavier aircraft at the same IAS would require a greater value of Cl ie greater angle of attack, more drag. But I guess if normal descent speeds are above best L/D speed, then this would make sense as the drag curve shifts to the right with increasing weight. Does this mean that if the normal speed for descent was the best L/D speed, that a heavier aircraft would glide less distance for the same IAS?

Thridle Op Des
5th Oct 2009, 05:49
Wiz,

I must admit that I though Jumbo's idea was closer except the heavier aircraft possesses more POTENTIAL energy (ht above ground x mass) which is then converted during the descent as Jumbo says into Kinetic energy via 1/2mV2, and therefore has more energy to glide further. I can see that the polar diagram for LD changes with mass but I would have thought that the energy the aircraft possesses at the start of descent will be the bigger influence.

I am standing to be corrected!

TOD

Wizofoz
5th Oct 2009, 05:53
QJB,

Not quite sure what you mean.

If Best L/D were the usual descent speed, a heavier aircraft would descend at a higher speed than a lighter one.

I guess the best way to put it is this- Best L/D speed increases as a function of weight. Fly faster OR slower than best L/D, and you will not glide as far.

I would think that a descent with an FMC cost index of zero would have the aircraft descending at very near best L/D, perhaps modified slightly for residual thrust. Would anyone like to clarify that?

Intruder
5th Oct 2009, 06:22
Normal descent speeds are 290-310 KIAS. I don't know of any current transport airplane (get rid of Concorde and SR-71) whose L/Dmax is at 300 KIAS at landing weight. Therefore, a lighter airplane descending at 300 KIAS will not go as far as the heavier airplane, because the heavier one is closer to L/Dmax.

mansah
5th Oct 2009, 08:13
it has to do with the momentum folks..if they start their desent at the same point the heavier aircraft has the chance to overspeed to keep up the profile.. that not being allowed makes it start descent ealier.:ok:

enicalyth
5th Oct 2009, 10:54
Assume still air, ISA, M0.78 and FL360. Aircraft a) = 60000kgf; b) = 50000kgf. The L/D for a) is 16.98 and for b) is 15.84 so that all things being equal aircraft a) has the advantage of cleaving through the air more efficiently at ToD and therefore must commence its descent earlier. QED?

Capn Bloggs
5th Oct 2009, 11:11
At the same speed, a lighter aircraft must descend more steeply so that the reduced weight vector provides enough effective thrust to counter the effectively-constant drag:

http://i521.photobucket.com/albums/w334/capnbloggs/descending.jpg

Brian Abraham
5th Oct 2009, 14:04
Recall reading that an emergency descent in a light weight 747 is something to behold, very steep nose down attitude. Similar principles at play.

Microburst2002
5th Oct 2009, 14:50
How many posts!

It is all about L/D ratio. Of course we can look at it from the point of view of Energy or Momentum if we take into account all the factors.
However, the L/D is the best explanation.

1-The gliding ratio is equal to the L/D ratio (By geometry)
2-The L/D ratio is equal to de CL/CD ratio (Dividing L formula by D formula)
3-Both CL and CD depend on the AoA alone (and Mach Number, ok...)
3-For a constant IAS and straight flight path, the heavier the airplane, the greater the angle of attack (From Lift formula)
4-flying in the stable speed region (AoAs above best finesse) the heavier has a more efficient AoA. (see any CL/CD versus AoA graph)
5-Therefore, the heavier airplane has a better L/D. It glides more miles for altitude.

If the descent speed was very very low so that the heavier was flying in the reverse command region, wich would be nonsense, it would glide less than the lighter one, because it would have a less efficient AoA.

XPMorten
5th Oct 2009, 15:23
You got to look at the wing airfoil to understand this.

Modern commercial transports use laminar flow airfoils which
utilize a drag polar 'bucket'. Where you are in the bucket
depends directly on your AoA. As the planes burn fuel during
cruise, step climbs into thinner air are performed to maintain
that optimal lowest-drag AoA.

Since these airplanes are designed for efficient cruise at higher weights
and high altitudes, when they get light and low, the AoA is too
shallow to keep the wing inside the low Cd area and drag actually increases.

The lighter the plane, the shallower the AoA at a given altitude.
-> the higher the drag from the wing.

So, at the same speed, the light plane will have more drag than
the heavy one and needs to do a steeper descend to maintain airspeed.

XPM

Wizofoz
5th Oct 2009, 15:53
XP,

It is exactley the same for DC3, C172, Tiger moths and gliders- nothing to do with modern aerofoils.

mansah- nothing to do with momentum either.

XPMorten
5th Oct 2009, 17:45
It is exactley the same for DC3, C172, Tiger moths and gliders- nothing to do with modern aerofoils.


I didnt say that, but since you brought it up, modern airliners use
supercritical airfoils which usually have a "flatter" bottom drag bucket.
They also have a higher L/D ratio for the same Cl than conventional "old" airfoils
so I wouldn't say "exactley the same".. .

XPM

framer
5th Oct 2009, 18:37
Ok, this is how I've always looked at it, and it's how I want to keep looking at it, so without going into "lift buckets" and "super-critical airfoils",
am I basically right?

Aircraft A =60 ton and best L/D speed of 230kts
Aircraft B = 50 ton and best L/D speed of 215 kts
Both descend at 300kts
Aircraft A is closer to it's best L/D speed and therefore will be at a more efficient AOA and will need to start descent earlier.
Correct?

Intruder
5th Oct 2009, 18:38
it has to do with the momentum folks..if they start their desent at the same point the heavier aircraft has the chance to overspeed to keep up the profile.. that not being allowed makes it start descent ealier.
No momentum at work here...

Exactly the opposite. The LIGHTER aircraft can "overspeed" its L/Dmax speed by a larger margin, so it can descend more rapidly.

Wizofoz
6th Oct 2009, 09:04
Thridle Ops,

Sorry, missed your post.

Yes, the heavier aircraft has more energy, but it also has more mass that needs accelerating, and the two exactley cancel each other out.

That's why big rocks fall at the same rate as little rocks as Gallelao postulated.

framer- exactley so.

Thridle Op Des
6th Oct 2009, 10:05
Hi Wiz,

At the serious risk of appearing dim (I do a good impression on a regular basis), can I chase this energy thing one more time? I agree that Galileo was correct if a B747 and a C152 were dropped from 10,000' with no forward velocity and in a vacuum that they would both make holes in the earth at the same time. However we have a different situation here; two aircraft a 300KIAS and FL300 but 20 tonnes different in mass. The drag and lift which were not considered in Galileo's proposal now come into play. During the descent both aircraft remain at 300 KIAS (until 10,000' of course so as not to upset anyone). There is no acceleration involved, in fact there is a deceleration. The heavier aircraft would have used a lot more fuel to achieve the climb to FL300 which is stored latently as Potential Energy and therefore I would have thought would fly the furthest.

In helicopter terms with which I am more familiar, the heavier helicopter in autorotation flies further and descends slower than the lighter one with the same collective blade AOA (generally minimum collective pitch), similarly the Auto Revs with a lighter helicopter is lower since the potential-kinetic trade is not as great.

I look forward to your thoughts.

Regards

TOD

Capt Pit Bull
6th Oct 2009, 10:26
That's why big rocks fall at the same rate as little rocks as Gallelao postulated.

Only in an airless environment. Dave Scott could only demo this at Hadley Rille after all.

There are two issues at work here. Firstly the difference between the speed schedule actually being flown and the L/D ratio, as has been discussed previously.

The other important thing to remember is that we are not actually talking about gliding. Unless you've actually shut the engines down there will be residual thrust; if you have the rpm floor raised for any reason (e.g. anti ice on) then flight idle thrust might be quite substantial. The residual thrust is proportionately more significant for lighter aircraft.

pb

Jumbo Driver
6th Oct 2009, 10:26
It is a known fact that a heavier aircraft will take longer for a given descent at the same speed than a lighter aircraft. Our rule-of-thumb clipboard data for the early 747s indicated that for every extra 10 tonnes, the descent distance would increase by approximately 2nm - and this was consistently borne out in practice.

The question is, why? I still think it must be to do with the increased energy (P.E. - not K.E. as I erroneously said earlier :O) retained in the heavier aircraft at altitude.

Perhaps we should be analysing the converse to reach a conclusion and consider why a heavier aircraft takes longer to climb to a given altitude than a lighter one ... ? That must be to do with greater energy required, surely ...? If that is so, then there is more P.E. available in the heavier aircraft at altitude to be released in the form of work on the descent (as TOD most eloquently says), is there not ... ?


JD
:)

Sir Donald
6th Oct 2009, 11:23
The topic question, ''wow'' why the varying ToD with weight in Airline flying.
Replies to topic=aerodynamic theory.

The two are separate subjects.

In airline flying the descent assumes a constant speed schedule.

Glide descent is a constant Angle of attack descent.

Now what happens in each and why I shall leave it up to you.

Wizofoz
6th Oct 2009, 11:47
It is a known fact that a heavier aircraft will take longer for a given descent at the same speed than a lighter aircraft.

No, that is not definitivley true. It is true if both aircraft are descending at a speed which is greater than the heavier aircrafts best L/D, which is the case in almost all normal ops, but it is not a "Known Fact" in all cases.

Most efficient descent speed is very close to Max L/D. If the lighter aircraft is flying AT that speed, the heavier aircraft will be flying BELOW IT'S most efficient speed. It will be on the back side of the drag curve and therefore descend more steeply.

Dusthog
6th Oct 2009, 12:36
Potential energy is energy that is stored within an system.
Think of it as the heavier aircraft has more stored energy.
In the descent a light and a heavy airplane has the same drag, except for the induced drag.
When it comes to helicopters you can see that a heavy loaded helicopter often
autorotates at a lower vertical speed.

Cheers

Jumbo Driver
6th Oct 2009, 13:42
Wizofoz, I accept your correction. My earlier comment was based on 747 operations, where clean descent speeds are almost always above the best L/D speed for the weight. I also agree that the best descent speed is close to the best L/D speed for the weight, furthermore that the best L/D speed increases with increased weight.

However, is it not the descent angle we should be looking at when comparing descent performance between different weights? Are you saying that, if two otherwise identical aircraft descend at their best L/D speed (i.e. producing their best angle), the descent angle is the same for both aircraft? If this is so, why does the greater P.E. stored in the heavier aircraft not have any effect on the angle? If it does affect the angle, surely it will be shallower because of the faster forward speed ... and therefore Q.E.D. ...?


JD
:)

QJB
6th Oct 2009, 13:49
It seems simple to me. At the same speed the only thing that matters in determining the ToD is the glide angle. Glide angle depends on the lift to drag ratio. Better lift to drag ratio = Shallower glide angle.

A greater weight will require a greater angle of attack for any given airspeed: different lift drag ratio for the same airspeed.

If you are on the left side of the drag curve then an increase in AoA will reduce the aircrafts L/D max, more weight = steeper glide. If you are on the right side of the drag curve an increase in AoA will increase L/D max, more weight = shallower glide.

Therefore whether a heavier or lighter aircraft has to start descending earlier is entirely dependent on what speed it is to be maintained during the descent, and what side of the drag curve this speed is on.

Capt Pit Bull
6th Oct 2009, 13:54
Are you saying that, if two otherwise identical aircraft descend at their best L/D speed (i.e. producing their best angle), the descent angle is the same for both aircraft?

Yes, he is.

In a glide the descent geometry is entirely determined by the L/D ratio, which in turn is entirely determined by Alpha.

(assumptions: zero thrust, still air)

Yes, the heavier aircraft has more GPE to burn off. This can be expressed in terms of energy=forceXdistance. In this case the force is Drag

If the aircraft has 10% more mass, it has 10% more lift... and 10% more drag (IF it flies the optimum L/D ratio, i.e. the same Alpha, i.e. a faster speed).

So if the energy to be burnt off is +10%, and the drag is +10%, we can see that the distance is going to be the same. We just fly faster and get there quicker.

pb

Microburst2002
6th Oct 2009, 18:02
From the point of view of Energy:
A descent is about dissipating energy.
The airplane's energy has to be reduced from cruising level at cruising speed to 50 ft ATL at Vapp.
Where does all this energy (potential and kinetic) goes? It has to go out of the airplane. It is transferred to the air. Basically the airplane moves air (and heats it, as well). The more the drag, the more the energy dissipated. Gliding at minimum drag speed (same as best L/D) means dissipating less energy, which means a longer glide.

From the point of view of forces:

The heavier, the shallower you have to descend to achieve a given forward force. For a given airspeed you have more Drag in the heavy airplane (because of the increased AoA required to maintain flight path) but this effect is negligible, compared with the effect of the weight component along path. So the result is that the heavier, the shallower the angle of descent for a given airspeed

Wizofoz
6th Oct 2009, 18:56
However, is it not the descent angle we should be looking at when comparing descent performance between different weights? Are you saying that, if two otherwise identical aircraft descend at their best L/D speed (i.e. producing their best angle), the descent angle is the same for both aircraft? If this is so, why does the greater P.E. stored in the heavier aircraft not have any effect on the angle? If it does affect the angle, surely it will be shallower because of the faster forward speed ... and therefore Q.E.D. ...?


If I'm getting the question right Jumbo, consider this- If two aircraft are descending at the same speed, their angle of descent will be DIRECTLY related to their rate of desecent- More angle at the same speed=more rate and Visa-versa.

It is also a curious product of vectors that a particular aircrafts best L/D is independent of wieght. It can ALWAYS achieve the same still air distance regadless of weight- that's why Gliders have the facility to carry ballast- They can glide just as far at a heavier weight, but faster and with a higher descent rate, which can be optimal ig there are plenty of strong thermals around.

XPMorten
7th Oct 2009, 08:02
To get an idea of the numbers..

XPM

http://www.xplanefreeware.net/morten/DOCS/763.jpg

mansah
7th Oct 2009, 08:36
A heavier aircraft needs a shallow descent relative to a lighter aircraft , because it needs to check its rate of descent at given speed, due to its weight or more precise the product of mass and the given speed which means momentum...I am willing to be corrected :cool:

Wizofoz
7th Oct 2009, 14:57
I here-by correct you!!

read the posts above..

HarryMann
8th Oct 2009, 02:16
At the same speeds (true in general terms)

1)
At the same speed, a lighter aircraft must descend more steeply so that the reduced weight vector provides enough effective thrust to counter the effectively-constant drag:What he said ^ The force equation ^

2) Equating potential energy at ToD with energy burn during descent (Drag x speed)
You need more drag to dissipatethe greater PE, so it has to act for a longer period

^ The energy equation ^

Provisos
a) Residual thrust ignored (we're talking round terms)
b) Higher induced drag of heavier a/c ignored* (we're flying considerably faster than Vmd, thus Cdi becomes much less than half the total)

* In practice the difference between the ToD distances is reduced by b), and the nearer Vmd we fly, the smaller the difference will be**

** Proof : Imagine flying below Vmd, the heavier a/c will have a very steep descent due to very high Induced Drag (in absolute terms).

1) & 2) show that both force and energy equations can be equally valid solutions. Both, and sometimes momentum balances as well are frequently used in combination (they should all agree)... momentum differences don't 't apply being insignificant in this situation.

The general theory of descent profile used to be (1970' s from memory), to delay as long as possible and descend as steep as possible, cabin pressure change considerations being limiting (the human ear being more sensitive to rate of pressure increase than rate of decrease).
The theory being, to stay in cold thin air for as long as possible, for efficiency whilst minimising flight times; it seems thinking has been adjusted somewhat, to further reduce fuel burn.

Sciolistes
8th Oct 2009, 09:43
My captains theory was that taking the same set of conditions for two aircraft, a heavier one will start its descent earlier!! We checked this on the FMC and a difference of 10 tonnes to the ZFW on the Perf init caused to ToD to shift approximately 30 miles out!!!
Could somebody please explain how this works.
No way would the FMC would calculate a difference of 30 miles for the T/D based on a 10 tonne difference of ZFW! There is some other factor or simple finger trouble at play.

Anyway, a heavier 737 would descend later as a heavier aircraft will descend faster (at least initially). In 737 normal operations, faster means more drag and more drag means a steeper descent. 737 pilots know that adding speed steepens the descent gradient more than it takes to loose said speed during the decelleration phase - the effect of parasite drag at operationally normal hight speeds.

With ECON speed, if you add 10 tonnes to the ZFW in a 737 I bet the T/D wouldn't change by more than a mile or two and then the distance from ToD to BoD would be reduced, not increased, due to the steeper path due to additional parasite drag. I reckon it would make little difference if you have a speed restriction, descend at ECON or a hard speed.

To get an idea of the numbers..
The 737 doesn't carry the same momentum as a 767. The 767 performance table (for a fixed speed schedule) above suggests that decelerating an additional 40 tonnes overrides additional inefficiencies of a said heavier weight to steepen the initial part of the descent.

It seems simple to me. At the same speed the only thing that matters in determining the ToD is the glide angle. Glide angle depends on the lift to drag ratio. Better lift to drag ratio = Shallower glide angle.
All the discussion of descending at best L/D is probably a bit of a tangent, as medium jets don't descent at best L/D (green dot speed - about 200 to 230 kts), not even with a cost index of zero, at which the 737 will have a descent speed of something like .72/250 for descent. But if you were to fly at green dot speed, as Wizofoz says, the distance would be pretty much the same, as would the glide angle, just the rate of descent would increase with the increased speed (until weight increases so much that parasite drag becomes the inhibitor).

Capt Pit Bull
8th Oct 2009, 10:05
Hate to be picky, but:

2) Equating potential energy at ToD with energy burn during descent (Drag x speed)


Force x Speed = power (not energy)

You need more drag to dissipatethe greater PE, so it has to act for a longer period

Force x Distance = Energy transferred (i.e. work done)

Talking about a "longer period" implies you mean time; if you are talking about forces and time you are into the realms of accounting for momentum, not energy.

If there is greater PE, then Force X Distance needs to increase in order to get rid of it. So you can have more drag for the same distance (the best L/D scenario), or the same drag for a greater distance, or some other combination (e.g. a little bit more drag for a greater distance; the 'high speed descent' scenario).

pb

XPMorten
8th Oct 2009, 10:26
1)
Quote:
At the same speed, a lighter aircraft must descend more steeply so that the reduced weight vector provides enough effective thrust to counter the effectively-constant drag:
What he said ^ The force equation ^

The light aircraft will have MORE drag than the heavy in a, same speed scenario (like in the 767 case above).
The reduced weight will give the light aircraft a shallower AoA moving
it to the left side of the optimum airfoil drag bucket which means increased
drag and having to do a steeper descend to maintain airspeed.
If we do some calculations on the 767 chart, we find that the
Light acf has an average descend angle of -3,03 deg and the heavy acf -2,69 deg.
So the difference is only a 0,34 deg average descend angle.

http://www.xplanefreeware.net/morten/DOCS/dragpolar.jpg

Energy calculations don't really work well in an environment where there is energy loss. In cruise, the two B767 engines produce around 20.000 lbf of thrust which means thats the amount of drag you got. Thats alot
of friction (energy loss). It's like having a B737 engine at TO thrust sealevel pointing in the wrong direction... .

XPM

Capt Pit Bull
8th Oct 2009, 11:50
Energy calculations don't really work well in an environment where there is energy loss.

Have to disagree with that statement. In fact it would be pretty difficult to do an energy based analysis if the wasn't energy lost!

Everything else being equal, energy based calculations are always easier because you're dealing in scalars not vectors.

Never mind the L/D ratio, CD, CL, which way the forces are pointing.... if you know the cruise thrust, speed and mass of the aircraft the rate of descent is a one line calculation. And the descent angle not much more.

pb

concordino
8th Oct 2009, 17:43
Humbly put in simple words :

The heavier plane has a higher potential energy due its weight.

Thus it requires extra miles to dissipate that extra potential energy.

And i think applying the same principle to the Kintetic energy will have the same results if the two planes are travelling at the same descent speed.

XPMorten
8th Oct 2009, 20:18
EDIT: some error in the calc

HarryMann
9th Oct 2009, 01:37
The light aircraft will have MORE drag than the heavy in a, same speed scenario (like in the 767 case above).
The reduced weight will give the light aircraft a shallower AoA moving
it to the left side of the optimum airfoil drag bucket which means increased
drag and having to do a steeper descend to maintain airspeed.
If we do some calculations on the 767 chart, we find that the
Light acf has an average descend angle of -3,03 deg and the heavy acf -2,69 deg.
So the difference is only a 0,34 deg average descend angle.I'd like to see your calculations..

Are you perhaps ignoring:

a) Heavy a/c has more induced drag
b) The drag bucket only applies to wing profile drag, so isn't the whole aircraft Cdo component (watering bucket effect down)
c) Are you sure where the lighter a/c is on the bucket at ToD? Looks to me (without the chore of checking) it may well be right in the middle of bucket!

Microburst2002
9th Oct 2009, 01:43
Indeel all the energy has to be lost during the descent, otherwise the airplane would accelerate to supersonic speeds. We cannot apply Energy conservation law.
I think it is just geometry. The gliding ratio is the same as the L/D ratio.
It comes directly from a graph.

Capn Bloggs
9th Oct 2009, 01:56
Pitbull,
Never mind the L/D ratio, CD, CL, which way the forces are pointing.... if you know the cruise thrust, speed and mass of the aircraft the rate of descent is a one line calculation.
How about sharing it with us. :confused:

XPMorten
9th Oct 2009, 09:30
HarryMann,

a) Heavy a/c has more induced drag
b) The drag bucket only applies to wing profile drag, so isn't the whole aircraft Cdo component (watering bucket effect down)
c) Are you sure where the lighter a/c is on the bucket at ToD? Looks to me (without the chore of checking) it may well be right in the middle of bucket!

a) Right, but it (B767 example) also has a 4000 lbf of extra gravity component working as thrust in the opposite direction which also plays a part in this.
b) The fuselage is offcourse also designed to have lowest drag at optimum
cruise lvl (2,5 deg AoA or so). Thats why it isn't symetrical top/bottom.
So changes in AoA away from optimum will give higher fuse drag as well.
c) Right, see attached derived from the B767 diagram.
The difference in RoD don't happen before the low FL300's.

http://www.xplanefreeware.net/morten/DOCS/RoD.jpg

http://www.xplanefreeware.net/morten/DOCS/RoD2.jpg

Microburst2002
9th Oct 2009, 10:51
Hi XPMortem

Regarding those CD vs CL graphs of NACA airfoils:

I thought that (for positive CLs) CD was always increasing with AoA, wether with or drag bucket or not, since CD is basically the sum of CD parasite (constant with AoA) and CD induced (increasing with AoA).
I am not sure now but the graphs I remember comparing conventional and laminar NACA airfoils had its minimum drag coefficient at CL=0.
However, in the 767 graphs you show, both airfoils have the minimum drag at a positive CL.
What is the reason for that effect?
¿Do you also have the L/D graph versus AoA for the 767?

Thank you

As for the energy calculations ¿What is your opinion about the following?

Let's consider a descent from cruising altitude to a given altitude at a constant IAS for two identical airplanes, light and heavy.

Their potential energy is to be dissipated, transferred to the air. The heavier, the more potential energy. For a similar rate of dissipating energy (for a similar drag) the heavier needs more time to dissipate its potential energy. Same speed, longer time, more distance.
At a given airspeed, the heavier has more drag (or so I used to believe) which means a faster rate of spoiling energy. But this effect does not overcome the effect of the greater potential energy of the heavy one.

We should consider kinetic energy too, since at constant IAS there is a decrease in TAS as we descend. This decceleration is identical in both heavy and light. However, the decrease in kinetic energy is greater in the case of the heavy, because of its mass. So, not only the heavier has to transfer more Potential energy but it also has to transfer more Kinetic energy. At a similar rate of loosing energy, or Drag, the heavier will need more time and hence more distance.
Again, the heavy has more drag for the same airspeed, but the mass effect is stronger.

Capt Pit Bull
9th Oct 2009, 12:38
Pitbull,

Quote:
Never mind the L/D ratio, CD, CL, which way the forces are pointing.... if you know the cruise thrust, speed and mass of the aircraft the rate of descent is a one line calculation.

How about sharing it with us.

Its not a state secret.

Rate of Climb = Excess Power / Weight

Hopefully that'll ring a bell.

This is just a case of a negative value for Excess Power that will result in a negative value for Rate of Climb. i.e.

Rate of Descent = Power Deficit / Weight.

In this case the power deficit is Drag x Speed, and since in the cruise thrust = drag, if we knew the cruise thrust we also know the drag at the start of the descent.

i.e.

Rate of Decent = Thrust Reduction (from that required to maintain level flight at the TAS in question) x TAS / Weight

Stick it all in the same units and you're in business.

kijangnim
9th Oct 2009, 18:25
Greetings,
As far as I am concern, very light requires as much as very heavy, very heavy because of inertia, very light because of the amount of speed to loose to transition to flap maneuvering speed.
when very light your AOA is high so is your drag, in level flight yes, but not in descent. May be I am wrong

XPMorten
9th Oct 2009, 20:25
Looking at some A320 FDR data.

In cruise optimum altitude FL 370, 65T, pitch is +2,5 deg, AoA around 6 deg.
Descending in the FL300's, pitch -2 deg, AoA 4,5 deg
Descending in the FL200's pitch -1,5 deg AoA 3,5 deg

This clearly shows the acf moving significantly to the left
in the drag bucket on descend, but I guess we already knew that :p

XPM

HarryMann
9th Oct 2009, 22:00
when very light your AOA is high so is your drag,
NO, AoA is LOW!

in level flight yes, but not in descent. May be I am wrong

You are

Microbusrt

XPMortens wing Cdo (wing only, profile drag only) plot looks fine to me...

It is quite possible ( & common even) for a section to have lower drag at a small +ve Cl than at a lower one, or even zero (take the case of heavy simple camber for instance, that won't like a low alpha!)

I'm just questioning the relevance of laminar flow low drag buckets in this context...

HarryMann
9th Oct 2009, 22:27
Simply put:

The heavier aircraft would overspeed if it tried to descend at the same angle as the lighter one!

Because it has more power at its disposal, all the way down, by surrendering potential energy

Along the same lines of Capt Pit Bull's

Rate of Descent = Power Deficit / Weight.

This also requires the assumption that the drag of an aircraft does not go up in proportion to it weight - and of course it doesn't.

FlightDirector7
14th Oct 2009, 08:05
Guys lets keep it simple....lets not get into L/D ratios and coefficients of lift....it has absolutely nothing to do with that.

The answer is : a heavier aircraft needs to descend earlier because as any aircraft is restricted to a certain VMO/MMO for descent, a heavier aircraft has a higher momentum and so needs to start descent earlier and maintain a shallower rate of descent or else it will overspeed as a result of the weight driven momentum.

The shallower the rate of descent the greater the ground speed. You cover more ground distance per every 100 feet vertically being heavier. So simply, a heavier aircraft needs to start descent earlier than a lighter aircraft, as a lighter aircraft still being restricted by VMO/MMO for descent, is less likely to overspeed as a result of its weight driven momentum.

Hope this helps

cheers

KristianNorway
14th Oct 2009, 10:49
What a discussion. I think Intruder was right at the 5'th of october already.

Drag curve shifts right with an increase in weight. Hence, best L/D increases, and therefore best glide speed increases with weight.

With a set speed of 300 kts in the FMC, which is well over best glide speed, that speed is closer to L/D for a higher weight. Remember, the speed was fixed. Therefore that speed is comparatively more efficient, and you glide longer.

If the descent speed was set to ECON, the distance shouldn't shift with an increase in weight. Only the speed should differ. That's why we operate a low cost index and ECON descent, giving us a speed of about .77M/255kts during the descent. Then T/D usually is around 120-140 miles, non-dependent of weight.

Try selecting ECON descent speed and then changing the weight. TOD should stay where it is.

Flyman35
14th Oct 2009, 16:06
Higher weights increase the descent distance (early TOD) because of the reduction of descent gradient and as rate of descent = VD-VT / W where V=velocity D= drag and T= thrust in stabilized flight if Weight increases the rate of descent will increase for a given IAS and vice versa .