Top of Descent Versus Gross Weight
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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 ^
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 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.
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
Last edited by XPMorten; 8th Oct 2009 at 10:03.
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Energy calculations don't really work well in an environment where there is energy loss.
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
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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.
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.
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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.
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.
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!
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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.
I think it is just geometry. The gliding ratio is the same as the L/D ratio.
It comes directly from a graph.
Pitbull,
How about sharing it with us. 
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.
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HarryMann,
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.
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!
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!
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.
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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.
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.
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Pitbull,
How about sharing it with us.
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.
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.
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.
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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
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
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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
XPM
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
XPM
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when very light your AOA is high so is your drag,
in level flight yes, but not in descent. May be I am wrong
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...
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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.
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.
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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
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
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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.
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.
Last edited by KristianNorway; 14th Oct 2009 at 14:11.
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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 .
