Why does an aircraft descend quicker when it is lighter?

[Pontchertrain]

Hello guys,
I got this question from Ace The Pilot Technical Interview and am bit confused about the answer, can anyone elaborate.


Why does an aircraft descend quicker when it is lighter?
Because an aircraft is restricted to a maximum speed during descent, the heavier aircraft has to maintain a lower rate of descent than a lighter aircraft; otherwise, it would overspeed. Remember, heavier aircraft have a greater momentum and this weight driven momentum will produce a greater speed in vertical dive. Therefore a heavier aircraft has to start its descent earlier than a lighter aircraft because it has to maintain a shallower descent.
In other words a lighter aircraft can descend later and quicker than a heavier aircraft because it can maintain a greater vertical descent profile without overspeeding.

[fly_antonov]

It depends alot on the properties of your aircraft and what speed you are flying at.

An aircraft has what is called a drag curve.
An aircraft requires an upward force named lift to fly.
Your wings generate lift and by this process occurs induced drag, which is highest at low speeds and decreases with increasing speed.

The heavier your airplane, the more lift you need to produce to balance the added weight, so you produce more induced drag.
By increasing drag, you dissipate more kinetic energy, which means that depending on descent profiles, aircraft, speed and air density the given assumption may not be correct.

Too many factors, it should be studied case by case, which makes the assumption totally irrelevant.

[dragqueen120]

f = ma xxxxxxxxxxxxxxxxxxxxxxxxx

[TowerDog]

You can tell the diff in a B-747:

Empty it comes down like a rocket, heavy it takes some planning as the momentum is money in the bank and will keep ya going..:cool:

[Pugilistic Animus]

Performance – Class 1 Airplane Motions (http://www.centennialofflight.gov/essay/Theories_of_Flight/performance/TH24.htm)

I checked it over and it accurate:)

[Andy_20]

Funnily enough after reading this post i just came across a question exactly like this in the Bristol QB. Confused me also. :ouch:

[justasmallfire]

it all comes down to the lift formula.if your heavier you need more lift to balance the weight in descent(in descent lift should equal weight so you fly down rather than drop(stall) so you fly at a higher angle of attack to create more lift and therefore a slower speed over the ground.increasing lift (decreasing speed) or increasing speed and keeping the same angle of attack would also be an option but then the descent would have to begin earlier as the time to descend would take longer and the descent gradient would be a lot shallower.

[kaptene]

It does make sense, however, I'm assuming that the heavier airplane will have to travel a a higher speed than the light one, which means the ground speed is greater, which means also that the pilot of the heavier airplane will increase the descent rate to reach its destination. To prevent this, he will need to begin the descent earlier.

[mad_jock]

Just think of it as energy.

The higher the mass the more energy you have to get rid of. Ep=mgh

At max limiting IAS you have a fixed max drag which won't change with weight.

So per min you will be able to get rid of xxx amount of energy call it Ed

If you look at Ed=mg(delta)h, g is fixed so the only thing that can vary is the (delta)h which is your rate of decent.

(delta)h = Ed/mg so if you increase m (weight of the aircraft) you decrease (delta)h.

Basic physics

[Keith.Williams.]

We can also look at the situation in terms of aerodynamics.

Glide endurance is maximum (and so rate of descent is minimum) at Vmp.

Glide range is maximum at Vmd.

Typical descent speeds are greater than both Vmp and Vmd.

Both Vmp and Vmd increase with increasing aircraft weight.

So if two aircraft descend at the same speed (higher than Vmp and Vmd) the heavier aircraft is closer to its vmp (so its rate of descent is less) and closer to its Vmd (so its descent range is greater). So the heavier aircraft must start its descent sooner and further from the planned level off position.

[fly_antonov]

At max limiting IAS you have a fixed max drag which won't change with weight.


D = 1/2 Cd Rho Vsquare S
Cd increases with weight.

I came to the following conclusion that is best illustrated by an example.

Take an imaginary aircraft and name it the B797 or whatever you like.
Loaded to the following 3 different all up weights it behaves as follows:

100 tons, AOA 5 degrees, L/D ratio 20
200 tons, AOA 10 degrees, L/D ratio 25
300 tons, AOA 15 degrees, L/D ratio 15

The aircraft will produce induced drag as follows:
100 tons, a drag force equivalent to 5 tons (100:20)
200 tons, 8 tons (200:25)
300 tons, 20 tons (300:5)

Analysis:
Compared to the 100 tons aircraft, the 300 tons aircraft will have 3 times more potential energy to dissipate but has 4 times more induced drag to do so.
If we suppose that we' re at at the low-speed end of the drag curves and that total drag is 80% induced drag + 20% parasite drag, then the 300 tons aircraft will be able to descend at a larger angle and faster than the 100 tons aircraft because it will have 200% more potential energy but 240% more total drag (induced drag will then be 94,1% of total drag because parasite drag stays the same 1,25 tons).

However this is not true for the 200 tons aircraft vs 100 tons aircraft comparison because the 200 tons has 100% more potential energy to dissipate and only 60% more induced drag or 54% more total drag at the low-speed end of the drag curves.

It is also not true in the 300 tons vs 100 tons comparison if we consider flight at a higher speed where for instance total drag is 20% induced + 80% parasite. In that case, the aircraft will have 200% more energy to dissipate but only 80% more total drag to do so.

The point of equilibrium would be at the minimum drag speed, which may or may not be achieved during the descent depending again on several factors.

Conclusion, it' s a case by case problem of different L/D values for different angles of attack and different values of density (density can affect AOA) and speeds and drag curve profiles.

To illustrate, see how L/D ratio can decrease with increasing AOA, while CL is increasing:

http://www.faatest.com/books/FLT/Chapter17/Drag_files/imageQDT.jpg


(http://www.faatest.com/books/FLT/Chapter17/Drag_files/imageQDT.jpg)

[Keith.Williams.]

Fly antonov,

What you have demonstrated is that the effect described in the question is true only at certain speeds.

We can see this if we go back to my aerodynamics approach.

Let's imagine two aircraft of the same type and configuration, one heavy and one considerably lighter, descending at the same speed. This speed is the Vmp for the lighter aircraft.

The lighter aircraft will be achieveing its maximum glide endurance and hence its minimun sink rate. So it will descend more slowly than the heavy aircraft. We have now reversed the scenario posed in the initial question.

But if we now have both aircraft descend at the Vmp for the heavier aircraft, we will once again find that the heavy aircraft descends more slowly than the lighter one. Provided of course that the weight difference is sufficiently large.

[fly_antonov]

Yes that should be correct.
What may be true for a B747 or a B737 may not be true for an F-22 or a Spitfire or even an E190, it all depends on where their VMP/VMD/VMT are located in their speed envelope.

Though I don' t know, it may also not be true for a B747 that initiates a descent from its operational ceiling because you flirt alot with VMD at those altitudes (IAS/CAS-wise).

So generally speaking the initial statement that refers to any aircraft is not correct unless we add some conditions and precisions.

[jimmygill]

The above question is an interesting one, and the explanations offered here all have some correctness and some hypothesis.

I will begin with first limiting the scope of the question.

Why does an airplane (not all aircrafts are airplanes) descend at a higher rate of descent (quicker) when it is lighter and descending at a max limited descent speed?


Lets start with two airplanes of same make airplane A (lighter one) airplane B (heavier one). Lets assume they both chose to descend at same speed of 250 Knots IAS.


If both of these airplanes are flying at the same IAS, it is natural that the heavier one must produce more lift than the lighter one to balance its weight.

Since B cannot increase its speed, the only way for B to generate more lift as compared to A, is by flying at a higher angle of attack (AOA) then A. Lift(L) and Drag(D) both are dependent on AOA. Lift to drag ratio determines the slope of a gliding flight. In fact D/L is same as the glide path gradient.

Here is L/D vs AOA curve from FAA PPL test supplement.

http://upload.wikimedia.org/wikipedia/commons/9/9e/FAA_Lift_Drag.JPG

I have marked on the curve the operating points of the airplanes A and B. It can be seen that there is a considerable improvement in L/D for the heavier aircraft. Hence aircraft B will descend via a shallower path when compared to aircraft B which will go through a steeper path. Since the lighter airplane has a steeper path at same airspeed as the heavier airplane, the lighter one will also descend quicker.


A lot depends on the speed selected for descent, if the selected speed corresponded to a particular AOA on the right side of the L/Dmax (i.e. low speed regime) the heavier airplane will descend quicker and steeper than the lighter one. This will be because now, an increased angle of attack will move the aircraft to a lower L/D situation. This proves that any explanation based on the weight alone without resorting to the underlying aerodynamics is wrong.

I am sure there will be some glider pilots on the forum which can throw more light on this.

[jimmygill]

I made the post without having read yours, content is almost identical. :)


Following table extracts some data from above graphs.


Airplane A Airplane B Change
(100 tons) (150 tons) in Result
AOA L/D AOA L/D L/D
2.0 7.6 3.0 9.8 Positive Lighter Descends Quicker
5.0 12.2 7.5 12.2 No Change Both Descend The Same Way
8.0 12.0 12.0 10.0 Negative Heavier Descends Quicker

[mad_jock]

Quite interesting how people think to solve this problem

Those curves only do lift related drag they don't do form drag.

The limiting factor of an aircrafts speed is either a structural limit (something will fall off) or an aero one (critical mach number exceeded) At that point for an aircraft the drag is going to be at a max for the aircraft . This won't be weight related.

My instant thought was that max rate of decent is when you have zero lift or even neg lift and max drag. This would be likened to the old physics experiment of 2 balls in exactly the same aero profile but deferrent weights being dropped and there terminal velocity. Now in real life with an aircraft you won't get this you have to keep the thing going forwards. But the max speed is always limited by the aircraft limits so there forefore by definition for anyweight the aircraft is always at max drag (presuming the aircraft is in the same config). Because you want as little lift as possible I discarded the variation of the lift related drag because you will be running at the very bottom of the curve and in the grand scope of things it will vary slightly but it will be under 5% in the grand scale of things.

Drag is energy removal you have zero (in know residual thrust but its going to be the same for any weight) so the max rate of decent is going to be a function of the energy removal of potential energy.

Thats the engineers way of thinking about a system.

To be honest its like a blinding light after 4 years at uni doing Mech eng. That it doesn't matter what system you are looking at, be it Thermo,static structure ,dynamic,fluid. If you look at the energy of a system if you can work out what its got, whats getting put in and whats coming out. The resultant effect can usually be explained by some fairly simple energy conversion. Now whats actually happening in the energy conversion can get very brain intensive. Which is whats happen here. People have focused on the mechanics of the the conversion instead of boxing off the system and applying basic physics laws. Energy in (which is aero) Energy out (which is drag which is taken as constant) and a start energy state of what ever the potential energy of the aircraft is.

We are looking at a rate so we have to bring time into it as well.

I have disregarded some effects that do vary but you can do that as they have little or no influence on the out come. They are under 5% (more likely 1%) and your variation of weight is 50%-100% its going to be by far the controlling variable.

Anyway thats my thought as a Pilot and ex Mechanical Engineer

[jimmygill]

Those curves only do lift related drag they don't do form drag.

Your assertion is wrong.

If those curve consider only the drag due to lift, i.e. the induced drag the L/D curve will be a monotonically decreasing hyperbola with infinite value at near zero AOA.

Lift induced drag coefficient Cdi ∝ Cl*Cl (Lift Coeff squared), hence lift by drag Cl/Cd ∝ 1/Cl.

My instant thought was that max rate of decent is when you have zero lift or even neg lift and max drag.

Zero lift and all drag is not a flying object (like kite, firsbee, airplane, glider), Zero lift and all drag is more like a falling stone, hail, rain, round parachute, such an object will fall ultimately with its terminal velocity.


Because you want as little lift as possible I discarded the variation of the lift related drag because you will be running at the very bottom of the curve and in the grand scope of things it will vary slightly but it will be under 5% in the grand scale of things.


This is where you are committing the fundamental mistake. It is true that for a descent lift required by an airplane is less than the lift required for cruise. But not near zero lift.

If the airplane were descending through a glide path of ϑ,
the lift required to sustain weight W of the airplane will be

L=W*cos(ϑ) or W = L/cos(ϑ)

For a glide path of -3 degrees L = W*0.998, or just 0.2% less than cruise lift, descend doesn't take you to the bottom of the curve, a free fall may.

What if the airplane were climbing at 3 degree, we put three degrees again and find the
L = W*0.998

Surprise, even for climb we need less lift than level cruise.


Drag is energy removal you have zero (in know residual thrust but its going to be the same for any weight) so the max rate of decent is going to be a function of the energy removal of potential energy.

Drag is a means of energy removal, in fact when you multiply Drag by TAS, what you get is Force*Distance/Time = Power


Hence
D*TAS = Weight*(Descent Rate)

Energy Dissipated = Change in Potential Energy

D*TAS = [ L / cos(ϑ) ] * TAS*sin(ϑ)

TAS cancels out from each side,

D = L * tan(ϑ)

or D/L = tan(ϑ) = (ϑ) .... ( For small thetas)

This is how you write the energy conservation from an engineer's point of view.

D/L = Flight Path Angle ( expressed in radians)

or Flight Path Angle = 60* D/L ( Expressed in degrees)


I hope this helps... if you have nay doubt just post here...



But I like you thought process, it is somewhat akin to detective work.

[Pugilistic Animus]

Love it :D

[mad_jock]

If you look at Ed=mg(delta)h, g is fixed so the only thing that can vary is the (delta)h which is your rate of decent

And

Energy Dissipated = Change in Potential Energy

Its exactly the same thing except I have used very basic theory where as you have resorted to using the greek alphabet.

At Vmo what is the drag going to be at various weight say 50% 75% and 100% of MTOW. In relation to each other?

Or put another way when are you going to have maximum energy removal from the system?

Right tfound the graph I was looking for.

http://www.dynamicflight.com/aerodynamics/drag/avd.gif

As you can see the max drag is at the highest airspeed which is limit by Vmo. Therefore to get max rate of decent you need to go as fast as possible. Which is where my very simple equation comes into play.

That graph you have is for induced drag only. There is a whole family of graphs to do the performance for total drag.

[Keith.Williams.]

It's interesting that this debate is all about drag.

I say interesting because it is not drag that matters in determining the sink rate in a glide. It is the power required.

When the engines are running they provide energy to the airframe.

When the engines stop working the airfarme has a certain amount of stored energy (Kinetic + Potential)

Throughout the subsequent glide it uses up this energy to move forward through the air.

The glide endurance is the time it takes to use up all of the stored energy.

The rate at which energy is used is power.

So for maximum glide endurance (which will give us minimum sink rate) we require minium power required.

The greater the difference between our arspeed and Vmp, the greater will be the rate of descent. If we start at Vmp, then accelerating or decelerating will increase the rate of descent

It is of course true that because power required = Drag x TAS, the drag is relevant to the debate. But it not the only relevant factor.

[mad_jock]

Fair point Keith.

I am proberly focusing to much on the max rate because thats what in real life causes you most problems with decent managment. High and need to get the height off can be a major issue where as below profile might be fuel hungry but it cause alot less problems.

O aye and if you use my theory in a tech interview you get the job :p

[flapassym]

Heavy car, top of hill, release brakes, will be going faster at bottom of hill than light car due to stored (potential) energy!

It's no more complicated than that!

[Keith.Williams.]

The original question was essentially "Why is the rate of descent less and the distance required to descend more, when an aircraft is heavy, compared to when it is light.

I'm not at all sure how your heavy and light car analogy explain this phenomenon.

[jimmygill]

Its exactly the same thing except I have used very basic theory where as you have resorted to using the greek alphabet.


Well you wrote the same as I did as far as the energy conservation is concerned, but we drew different inferences out of this, pardon my greek letters, they were just for the symbology.


At Vmo what is the drag going to be at various weight say 50% 75% and 100% of MTOW. In relation to each other?


I will get the data for this from a real aircraft, B737-700


L/D Data for 737-700 @ 0.78/280/250 KIAS
UNITS
Weight 40.0 50.0 60.0 70.0 (1000 kgf)
L/D 15.4 17.8 19.6 20.6
Drag 2.6 2.8 3.1 3.4 (1000 kgf)
ERR 2.6 2.8 3.1 3.4 (250kts)*(1000kgf)
ERR' 0.65 0.56 0.51 0.49 (25kts)
ERR(fpm)1621 1405 1277 1213 (fpm)

ERR = Energy Removal Rate
ERR' = Energy Removal Rate Per Unit Weight
ERR'(fpm) = Energy Removal Rate Per Unit Weight (expressed in feet per minute)








Or put another way when are you going to have maximum energy removal from the system?


Maximum energy removal rate is for the heaviest aircraft.
But maximum per unit weight energy removal rate is highest for the lightest aircraft. Interestingly enough it has the same units as speed and can be conveniently expressed in fpm.





Your graph is fine and my graph is fine but there are differences.

1. Your graph has speed on its horizontal axis, my graph has angle of attack on horizontal axis
2. Your graph doesn't depict L/D directly, my graph does.


The way to infer L/D curve from your graph is by assuming that lift is constand and speed is varying, that way L/D curve from your graph will be reciprocal of total drag curve.


Now a few arguments why my graph includes the parasite drag as well as induced drag just like your graph. My graph sure doesn't break them down in two components.

At very high speeds where angle of attack approaches zero, lift induced drag must be zero, your graph says that. My graph at zero angle of attacks shows a positibe coefficient of drag, this is parsite drag coefficient.

Please note that my graph doesn't depict lift or drag, instead it depicts the lift and drag coefficients, thats why the shape of the curve is different from yours.


As you can see the max drag is at the highest airspeed which is limit by Vmo. Therefore to get max rate of decent you need to go as fast as possible. Which is where my very simple equation comes into play.


Its not the max drag that matters, its the least L/D which will give fastest and steepest descent. Your graph hence cannot be used to describe the phenomenon under consideration.

[jimmygill]

It's interesting that this debate is all about drag.

I say interesting because it is not drag that matters in determining the sink rate in a glide. It is the power required.


Its not the drag that matters.
Its not the power required that matters.

Its the power required per unit weight of the aircraft that matters here, this quantity has a direct relation with the all so familiar Lift to Drag ratio.


Heavy car, top of hill, release brakes, will be going faster at bottom of hill than light car due to stored (potential) energy!
It's no more complicated than that!


My instructor always stressed on Keep It Simple and Stupid. But this is analogy way too simplistic.

You have controls in aircraft using which you can set its speed for a descent.
I haven't seen any car with down the hill cruise control while in neutral gear.

[mad_jock]

At very high speeds where angle of attack approaches zero, lift induced drag must be zero, your graph says that. My graph at zero angle of attacks shows a positibe coefficient of drag, this is parsite drag coefficient.


which is exactly what I said I was talking about.


Free body diagram of the two cases of weight.

Energy in arrow forward
Energy out Drag arrow to the back.
Energy in the system potential inside the box.
Energy out arrow pointing down (gravity)
Energy in pointing up (lift)

Max airspeed equals max drag which is the same for both as the Vmo doesn't change with the weight. As per the graph I showed if you look at the total drag.

As soon as you go away from Max Drag you are creating more variables that you have to account for and also will affect the result. Which is why I took the situation of Vmo because it zero's out or decreases a multiple of variables to insignificants. Induced drag tends towards insignificant so your variation in L/D ratios can be ignored the AoA of the wing can be ignored. By taking any other speed other than maximum you are into calculus to work out what all the different variables will actually mean and how they relate to each other.

Your 737 data proves the point, the only variable that is constant is the speed. Drag is different for each weight, the energy removed is different for each weight. You have 4 variables which can't prove a relationship between two of the variables.

Where as my solution the drag is the same for all weights, the energy removed is the same and the speed is the same. We only have two variables the weight and the rate of decent. Which you can prove the relationship.

It is a very simple problem which is given to second year Mech Eng students to teach them how use freebody diagrams for pratical problem solving and manipulations of variables to form simplified systems from complex ones and to also learn how to choose boundary conditions to find a relationship.

Yes you could go into into L/D ratios and all that but they would fail because they wouldn't have proved the relationship.

Now with your 737 data get a set of weights where the drags the same and the energy removal rate is the same and the speed is the same. Then we will be talking about the same thing. I will save you some time it won't happen but if you try Vmo it will be quite close.

[mad_jock]

Actually I have worked out how you could do it.

You would have to do some none dimensionless analysis so that you could normalise the rate of decent against the other variables.

A bit like using reynolds number in the wind tunnel.

Its 15 years since I have done any of that in anger so your on your tod.

[Capt Pit Bull]

Holy Moly.

Although this one may seem counter-intuitive when you first think about it, most students are happy enough with Rate Of Climb = Excess Power / Weight.

The easiest way to deal with descending is just to think of it as a rate of climb that happens to be negative.

Rate Of Descent = Power Deficit / Weight.

Immediately you can see that increasing the weight makes the descent rate less.

pb

[jfkjohan]

Dear all,

I have actually had the same question on my mind last week when I was reading through Gary Bristow's "Ace" book as well. I remember in school, we learned that "the heavier aircraft reaches the ground faster than a lighter aircraft".

It stood out to me when this question came out, and I can now SOMEWHAT understand/appreciate why --- drag?

Please help me (and others) understand this. So, to keep it simple, the heavier the aircraft, the more lift it requires hence the higher AoA thus the more induced drag it generates.

So if the lighter & heavier aircraft had the SAME AoA, the heavier aircraft would descend faster due to momentum/kinetic energy?

This has been bugging me since last week! Please help, anybody! Thanks!

@jfkjohan

[jimmygill]

@jfkjohan
"the heavier aircraft reaches the ground faster than a lighter aircraft"

What you learnt at the school is correct. The actual context of the learning was this:

Airplane B is heavier than Airplane A. Both are same make. Which one will descend faster for best range glide.


Same make means same L/D characteristics.
In order for these aircrafts to maximise the glide distance each of them should glide at maximum L/D. L/D is purely an aerodynamic variable, function of airplane shape and angle of attack only.

Maximum L/D is achieved at a fixed angle of attack for a fixed shape of the aircraft.

Now the problem is constraining the Angle of Attack. If A and B fly at the same angle of attack, and if B has to produce more lift the only way B can do this is by flying at a higher speed.

Yes the heavier aircraft glides at a higher speed for maximum range glide.

But for a speed restricted descent, the heavier aircraft may have a higher or lower sink rate depending on the L/D ratio at which the heavy aircraft is operating.



Please help me (and others) understand this. So, to keep it simple, the heavier the aircraft, the more lift it requires hence the higher AoA thus the more induced drag it generates.


I will try to build up on the simple sentence you started.




The heavier the aircraft the more lift it requires.
If the airspeed is constrained only way to increase the lift is by increasing angle of attack.
An increase in angle of attack will always increase the coefficient* of drag and coefficient of Lift.
But an increase in AOA may increase the L/D ratio, or it may decrease the L/D ratio.
The variation in L/D ratio depends on which region of the L/D curve the aircrafts are operating.
As you can see in the following curve for Angle of Attack between 0-6 L/D increases with increase in AOA. For AOA beyond 6 degrees, L/D reduces with AOA.
Hence a heavier aircraft is a better glider than a lighter on in low AOA regime. And a lighter aircraft is a better glider in high angle of attack regime.

http://www.faatest.com/books/FLT/Chapter17/Drag_files/imageQDT.jpg

Thanks to kieth for pointing out the mistake.

@Capt Pitt Bull
Rate Of Descent = Power Deficit / Weight
Your hypothesis work for low angle of attack regime and fails in high angle of attack regime. Even though it is easy to explain, it is wrong.

[mad_jock]

Which isn't the question that has been asked.

It is nothing to do with glide distances or speeds in the glide

It is purely Why does an aircraft descend quicker when it is lighter?

Its purely asking why the rate of decent is higher when you are lighter

Generally aircraft don't operate on the dirty side of the drag curve unless your on approach. Pit Bulls method works for anything above minimum drag. I know these days of advanced fuel performance the wisdom dictates we don't decend at Vmo but its always at a speed which is faster than minimum drag.

[Dick Whittingham]

The question was "why does an aircraft descend quicker when it is lighter?" and the answer is that it doesn't - except in the special case where you required to descend at a fixed airspeed above your minimum drag speed. As you all know if you fly at Vimp or Vimd the heavier aircraft will have a higher rate of descent. If you choose a particular descent angle then again the heavier aircraft, going faster, will have a higher rate of descent. If you can manage a terminal velocity vertical dive the heavier aircraft will go straight down faster.

If you are ordered by ATC to descend at a given airspeed or take it to the limit at Vmo/Vne then to hold the required airspeed the heavier aircraft will have to adopt a shallower descent angle and therefore will have a lower rate of descent.

So, in summary, if you are free to choose your airspeed the lighter aircraft will stay up longer. If you must maintain a fixed airspeed in a cruise descent the heavier aircraft will give you the demanded airspeed in a shallower descent

The reason why the question pops up in interviews and in the CQB is because the aircraft behaviour in a cruise descent - well known in practice - appears paradoxical in theory. Anyway, from Keith and Mad Jock you have all the aerodynamics you could wish for to deploy as your answer

Dick

[jimmygill]

I apologise Pitt Bull, your formula is correct

Rate Of Descent = Power Deficit / Weight.
= (Drag*TAS) / Weight
= TAS * (Drag/Lift) Lift = Weight (approx)


In airplane A and B TAS is same for both,

hence Rate of descent is determined by the Ratio of Drag to the Lift in the case.


Just putting the weight in the denominator is robbing us of accuracy amd lending credence to inappropriate analogies such as a car rolling down the hill.

[mad_jock]

The equation Capt Pit Bull stated is exactly the same as mine but uses different terms. I used a constant of energy lose per min so you could get a rate of decent per min.

Putting the weight on the bottom does nothing to affect the solution of the relationship. It gives us an inverse relationship. Accuracy is an issue with expicit solutions not implicit.

Why do you insist on bring L/D none dimensional ratio into it and TAS its not required and adds nothing to the relationship of rate of decent to weight.

People have focused on the mechanics of the the conversion instead of boxing off the system and applying basic physics laws

As I said very very common mistake for engineers early on in their academic education and for some it never does click. Thankfully they make very good accountants

[Farrell]

Well.... if he was confused before he read this thread.... :E

[mad_jock]

LMAO Farrell

Or one from my past which I always forget the real reason for the correct answer.

"Which wing section is more prone to Icing a thick section or a thin section?"

MJ's Answer with fingers crossed "a thin section is more prone to icing"

"why is that?"

MJ's answer "I don't have a clue to be honest I just know DC3's don't have any de-icing boots on the wing and they have big thick sections and they have flown all over the world without dropping out the sky due to icing"

"well thats the most orginal answer we have had to that question, and can't fault your logic. Tell me what you know about type IV de-ice fluid?"

[Keith.Williams.]

An increase in angle of attack will always increase the total drag.

If this is correct then why does the total drag decrease when we decelerate towards Vmd in level flight?

[mad_jock]

Actually I know its a bit of a thread diversion.

Why is a thin section more prone to icing than a thick one?

Just in case the interviewer doesn't have a sense of humour like the last one.

[jimmygill]

Why do you insist on bring L/D none dimensional ratio into it and TAS its not required and adds nothing to the relationship of rate of decent to weight.

TAS comes into picture because both the aircrafts are flying at a common limiting descent speed.

And rate of descent comes out to be
ROD = TAS / (D/L)

How can one ignore this important non dimensional characteristics?


@keith
I have made the edit in the original post. Thanks for pinting out.

[mad_jock]

Because its irrelvant to forming the solution. It looks sexy and try's to make the solution look aero dynamic, but that is the sum total of its usefulness.

All you do by bring TAS into it is introducing a heap of enviromental variables. Which then cancel each other out because your comparing aircraft in exactly the same situation bar the fact that there weight is different. Again its irrelevent to the solution apart from it makes it look aero dynamic.

You are trying to give an expicit answer to something that doesn't require one.

The point I am making is that if you start bring L/D ratio's and other such factors into play to answer the question. You will just come across as using techno waffle to answer a simple question.

So my answer would be

"The heavier aircraft has more potential energy than the lighter one and at a constant airspeed they both remove the same amount of energy via drag in the decent. Therefore for over a period of time the lighter aircraft would have lost more altitude than the heavy one ie the lighter one has a higher rate of decent"

Thats the answer they are looking for which for wannabies is the answer they should give if they get to a oral tech interview.

[jimmygill]

"The heavier aircraft has more potential energy than the lighter one and at a constant airspeed they both remove the same amount of energy via drag in the decent. Therefore for over a period of time the lighter aircraft would have lost more altitude than the heavy one ie the lighter one has a higher rate of decent"


How can they remove same amount of energy via drag when they have different magnitude of drag? Did the initial question state that they have same drag?

Isn't it from the very basic physics that force (Drag) multiplied by displacement in the direction of force is energy, and rate of change of energy is power.

Since the mistake lies in the very first sentence, we should not move on to the second sentence.

When Galileo threw the 1 pound and 10 pound weights from the Leaning tower of Pisa he found both traveled at the same speed. If you do not include the aerodynamics, them you can replace the airplanes with stones and get results in variance with Galileo's experiment.

In the following quote I am replacing the word "aircraft" with stone. Do we see something wrong here?


"The heavier stone has more potential energy than the lighter one and at a constant airspeed they both remove the same amount of energy via drag in the decent. Therefore for over a period of time the lighter stone would have lost more altitude than the heavy one ie the lighter one has a higher rate of decent"


Role of L/D in descent seems sexy thats fine, and thats the only correct and consistent way of explaining the phenomenon.


Thats the answer they are looking for which for wannabies is the answer they should give if they get to a oral tech interview.
They are looking for these answers because the so called "experienced aviators" get most of their knowledge from folklore in and around the airport bar.



The only environmental I introduced is same TAS, to make sure that we are comparing apples to apples. Please note the word "constant airspeed" which you too have used in your explanation, I have emphasized it above.


And one more thing, I know it may again sound sexy.

When I limit myself to the ratio L/D, in deriving the Rate of descent relation, I am doing basic Kinematics and not aerodynamics.

Aerodynamics is not considered until I explain the relation between L/D and TAS, Weight and Angle of Attack.

[mad_jock]

I don't see a problem at all with that statement.

The internal system of the free body diagram could be an elephant, stone or what ever you like.

The droping of anything is not an equilibrium system its accelerating(dynamic). After x amount of time the system energy isn't equal to each other, where as with our boundary conditions the light aircraft will have the same system energy as the heavy aircraft after a period of time. I know the Kinetic energy will be different but the Ek will cancel out because the speed of the aircraft will be the same at the start and end of the time period which is one of the the boundary conditions "constant speed"

If you can't see that an accelerating object is a dynamic system and our discussion is a static system in equilibrium your missing the basic foundations of my argument.

You can sex up what ever you want but it doesn't change the fact its techno waffle of "if you can't dazzle them with brillance baffle them with bollocks". And take it from an expert of "baffle them with bollocks" thats what you are doing.

[jimmygill]

How can they remove same amount of energy via drag when they have different magnitude of drag? Did the initial question state that they have same drag?


You didn't answer this.

[Keith.Williams.]

If the interview board accept an answer that is concerned only with the fact that the heavier aircraft has more energy, this will relieve the candidate of the burden of demonstrating even the most rudimentary knowledge of aerodynamics.

This is exactly the wrong way to eliminate the large number of recent/current JAR ATPL students who pass their exams by simply memorising answers from online databases. Mony of these candidates have only the most rudimentary konwledge, so a little bit of probing will quickly separate the wheat from the chaff.

The only way in which the technical interview can serve any useful purpose is if the board ask questions which cannot be memorised prior to the event. This will of course require more effort on the part of the board.

If the slower rate of descent of the heavier aircraft were due entirely to its greater stored energy, then this would be true at all airspeeds. But it is not true at all airspeeds. Given the right combination of weights and airspeeds, the lighter aircraft will sink more slowly.

The relationships between the two weights, the speed chosen and the relationship between that speed and the Vmp for each aircraft all play a major part in determining the rates of descent.

If the board wish to make the process worthwhile I suggest they use a series of questions such as:

a. What effect does increasing aircraft weight have on maximum glide range? Answer none.
b. What effect does increasing aircraft weight have on maximum glide endurance? Answer increasing aircraft weight decreases maximum glide endurance.
c. So why does my 777 require more time and distance to descend from cruise level when it is heavy than when it is light?........

This series of questions might also shed some light on how the candidate will deal with unexpected emergencies.

[jimmygill]

c. So why does my 777 require more time and distance to descend from cruise level when it is heavy than when it is light?........



Seems to me that our heavier 777 is a better glider than its lighter friend for the given descent airspeed.

Could it be that the heavier one is operating at a higher L/D than the lighter one? I believe so.

Why does the heavier one has higher L/D?
Because at the same airspeed heavier one needs to fly at higher angle of attack, the L/D ratio at this higher angle of attack is more than the one for the lighter 777.


How do you know the L/D ratio is higher at higher AOA?
For that we must take a look at the L/D vs angle of attack curve, and plot the operating points for these two cases.

Is the L/D ratio higher for all angle of attacks higher than the one at which the lighter 777 descends?
No, with increase of angle of attack the L/D ratio increases first, then levels off and then gradually decreases.

[mad_jock]

How can they remove same amount of energy via drag when they have different magnitude of drag? Did the initial question state that they have same drag?

It doesn't, its an implied boundary condition

But i can produce exactly the same question using fluids, thermo, dynamics and statics as the base line which is why we used this "problem" as an example for undergrads. I had to make them up to teach the fact that system means nothing when looking at a problem. I have exactly the same problem for each perversion including electrical using exactly the same energy conversions. I.e potential to some system which has a disapation factor.

It doesn't matter your perversion the energy freebody diagram will give the solution to the relationship.

Before I go any futher do you actually know what a free body diagram is and how engineers use it to simulate systems? Or for that matter what a boundary condition is?

[jimmygill]

Before I go any futher do you actually know what a free body diagram is and how engineers use it to simulate systems? Or for that matter what a boundary condition is?


No sir I don't know a free body diagram. I am a college drop out. I have heard of boundary layer but never came across boundary condition.

But I will request you to explain me step by step your argument, and considering my background I hope you will elaborate a little more.

[jimmygill]

http://www.jdburch.com/ballast.gif
Glider Performance Airspeeds (http://www.jdburch.com/polar.htm)

I hope this picture is sufficient.

Few things to notice.


The heavy glider and the light glider have the same best L/D ratio, note the two polars share the same tangent passing through (0,0).
The angle between the tangent and the horizontal axis is same as glide path angle.
If you fly both the gliders for maximum range, the heavier one will have higher TAS and higher sink rate, but will have same glide path angle as the lighter one.
If you chose to descend at 70 Knots TAS, the heavier one will descend slower and shallower than the lighter one.
If you chose to descend at 48 Knots TAS, the heavier one and the lighter one will both descend at same sink rate and hence same flight path angle.
If you chose to descend at 40 Knots TAS the heavier one will descend faster and steeper than the slower one.



@mad_jock
Please note that picture looks very sexy to me, but the picture is an observation not an explanation. Can you explain the above picture with your weight energy hypothesis?

[Keith.Williams.]

Mad Jock

The need to recite your CV suggests that you have lost the argument (or at least your composure).

The validity or otherwise of a person's comments is determined by what they say, not by who they are or were.

You have not answered the fundamental question:

"If the rate of descent is determined by the amount of energy that an aircraft has, why is it that?

a. When the descent is carried out at the speed that is Vmp for the lighter aircraft, the heavier aircraft has the greater rate of descent.

b. But at typical airliner descent speeds the lighter aircraft has the greater rate of descent."

The heavier aircraft has the greater energy in both cases.


Jimmygill

Seems to me that our heavier 777 is a better glider than its lighter friend for the given descent airspeed.

Could it be that the heavier one is operating at a higher L/D than the lighter one? I believe so.

OK, here's the follow-on question.

We take an aircraft and carry out a glide at Vmp. We note the Power required, L/D ratio and the rate of descent.

We repeat the glide at the same weight but now at Vmd. The L/D ratio has increased, but so as the rate of descent.

We repeat the process at a speed greater than Vmd. The L/D has decreased but the rate of descent has again increased.

It appears that the link between L/D ratio and rate of descent has broken down when we accelerated beyond Vmd.

But in each case the power required and the rate of descent both increased.

Does this not suggest that the link between power required and rate of descent is more direct than the link between L/D ratio and rate of descent?

[jimmygill]

For an equilibrium flight

RoD = TAS / (L/D) (For idle power glide)

This is derived not from aerodynamics but from simple consideration of equilibrium.


RoD@Vmp = Vmp / (L/D@Vmp)

Rod@Vmd = Vmd / (L/D@Vmd)

RoD@Vxx = Vxx / (L/D@Vxx) where Vxx is a speed greater than Vmd



Vmd > Vmp also L/D@Vmd > L/D@Vmp
The increase in the numerator (TAS goes up from Vmp to Vmd) more than compensates for increase in the denominator (L/D)

Hence we get RoD@Vmd > RoD@Vmp.




Vxx > Vmd, but L/D@Vxx < L/D@Vmd in order for RoD@Vxx > RoD@Vmp

Here increase in the numerator and decrease in the denominator both favor an overall increase in RoD


In considering a speed limited descend the case to case variation in TAS is not permitted.
Hence for cases where TAS is kept constant RoD is purely a function of the ratio L/D.

This is the case for a speed restricted descent such as in a 777.
For different weights and same TAS, the 777 will have different angle of attacks in a descent.

The heavier 777 will always have higher AoA.
But a higher AoA doesn't always translate into a better L/D, in some cases of TAS, an increase in AoA may worsen the L/D. The later cases are the ones which can't be explained by the argument of high momentum, high energy etc.. which prompts me to reject such arguments.

Also there are several angle of attacks "or speeds" for which a certain increase in AoA will lead to no change in value of L/D.
In such cases the changed in RoD is entirely due to change in the TAS of descend.

[Capt Pit Bull]

Don't have a huge ammount of time, so a few points:

1. Thread is shooting off on tangents. Yes, if you change the speed you change the descent rate, e.g if you choose to fly at Vmp in both (heavy and light) cases then you are choosing to introduce an additional variable.

2. The question does not state "at a fixed speed", but then again it doesn't have to. Check the preamble of the LOs, it specifically states:

Normally it should be assumed that the effect of a variable under review is the only variation that needs to be addressed, unless specifically stated otherwise.

As such, the question is asking about varying weight and we can take it as read that the speed is not a variable. If you choose to frame an arguement around varying the speed you just are not ATQing.

3. I accept, that as Dick says, the high speed descent is a 'special' case in that it is not a 'general' solution, however it is the majority case in civil air transport operation. Since alpha is low, lift dependant drag is relatively unimportant and therefore the effect of changing weight on total drag is minimal. This is the case that the question refers to, and also the case that confuses student pilots. My point was that it LOOKs counter intuitive at first glance, but if the student 'gets' ROC = specific excess power (which IS intuitive) then you have a great way of breaking down the counterintuitive barrier.

edit: Looking back, see ther is no 'question' as such, I am referring to the way such things are usually phrased to provide a brain teaser!

[Keith.Williams.]

Jimmygill,

I have no problem with anything that you have just said.

But you appear to be determined to ignore the relevance of power required

As you have said:

At any speed ROD = TAS / (L/D)

Strictly speaking the lift in this equation should be the weight

So we can restate the equation as

ROD = TAS / (W/D)

Rearranging this gives

ROD = TAS x (D/W)

ROD = (TAS x D ) / W

But TAS x Drag = Power required

So ROD = Power required / Weight

Which is what pretty well every text book on this matter states.

Instead of all of the drag curves and polar diagrams, all we needed to do was look at the power required curves for the two aircraft.

My reason for feeling uneasy about excessive emphasis on the L/D ratio is that it ignores the vital part that the TAS has to play in this matter.

As we accelearte towards Vmp the rate of reduction in drag is greater than the rate of increase in TAS, so the power required decreases.

At Vmp the rate of decrease in drag exactly balances the rate of increase in TAS so the power curve is flat.

Between Vmp and Vmd the rate of decrease in drag is less than the rate of increase in TAS so power required increases.

Above Vmd both drag and TAS are increasing, so power required increases.

At each stage along the speed range it is not just the L/D ratio that is determining the shape of the power required curve (and hence the way ROD changes). It is both the TAS and the L/D ratio.


In considering a speed limited descend the case to case variation in TAS is not permitted. Hence for cases where TAS is kept constant RoD is purely a function of the ratio L/D

This is true, but it ignores the fact that the chosen TAS has a bearing on the outcome. Pick a different TAS, and you may well get the opposite outcome.

[jimmygill]

Thanks Kieth for pointing out the mistake.

[Pugilistic Animus]

Aside from this excellent discussion. I think that "Aerodynamic for Naval Aviators" by Hugh Harrison Hurt Jr. is an ideal book for these explanations after that the business of aerodynamics gets just plain nasty:\

[smecucci]

This happens mainly with jet airliner descending at fixed airspeed: The heavier the plane the closest the flight to the maximum efficiency. Hence the more distance necessary for a descent.

[Q07]

Hmm I still can't get my head around this...

Let's take the same aircraft, but with two different weight at the same altitude.
Let's put these 2 aircrafts at Vmd.
Glide range is the same, right? So from the same altitude we can all agree that we're descending along the same glide slope.
BUT the rules says that the heavier aircraft has a higher Vmd ie should cover the distance faster to reach the same point on the ground.
The aircraft descends quicker when it is heavier.
Which is the exact opposite of the statement in the question...

Am I missing something here? :ugh:

[Alex Whittingham]

Yes, you have the aircraft operating at different speeds, which was not assumed in the question. Let's put these 2 aircrafts at Vmd

[SaulGoodman]

It does not descent quicker. It descents in a shorter distance for the same vertical speed ;)

[sablatnic]

That is why we pour water into the wings of the gliders, to make them stay aloft. And to get them up in the air we fasten a rope underneath them. But don't think about it.

[angelo1977]

Would be much easier to answer this question:

WHEN does an aircraft descent quicker when is lighter?