Top of Descent Versus Gross Weight
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?
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?
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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.
Exactly the opposite. The LIGHTER aircraft can "overspeed" its L/Dmax speed by a larger margin, so it can descend more rapidly.
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.
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.
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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
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
Last edited by Thridle Op Des; 6th Oct 2009 at 09:06. Reason: properly define helicopter blade angle in auto
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That's why big rocks fall at the same rate as little rocks as Gallelao postulated.
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
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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
) 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
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
) 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
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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.
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.
It is a known fact that a heavier aircraft will take longer for a given descent at the same speed than a lighter aircraft.
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.
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Law of physics
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
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
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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
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
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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.
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.
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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?
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
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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
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
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. ...?
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.
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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 
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At the same speeds (true in general terms)
1)
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.
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:
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.
Last edited by HarryMann; 8th Oct 2009 at 01:26.
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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.
Could somebody please explain how this works.
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..
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.
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Hate to be picky, but:
Force x Speed = power (not energy)
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
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
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