Why does an aircraft descend quicker when it is lighter?

[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]

Glider Performance Airspeeds

I hope this picture is sufficient.

Few things to notice.

1. 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).
2. The angle between the tangent and the horizontal axis is same as glide path angle.
3. 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.
4. If you chose to descend at 70 Knots TAS, the heavier one will descend slower and shallower than the lighter one.
5. 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.
6. 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.

Code:

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

1. 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.
   
   
   
   
2. 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
   
   
3. 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]

It is about Efficiency

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?

[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?