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]

yo

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..

[Pugilistic Animus]

Performance – Class 1 Airplane Motions

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.

[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/Cha...s/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.



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]

@fly_antonov

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


Following table extracts some data from above graphs.

Code:

    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

[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.



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.