rates of descent

[b777pilot]

hi all,

two jet a/c's of the same type, descending at the same speed, throttles are at idle, one significantly lighter than the other, which will have a higher rate of descent?

or will both descend at the same rate? thks.

[Keith.Williams.]

The simple answer is that the heavier aircraft will (probably) have a lesser glide endurance and hence a greater rate of descent.

The rather more complicated answer is it also depends on the speed chosen. Best glide endurance, which means lowest rate of descent, is achieved by flying at Vmp. Vmp is proportional to the square root of weight, so Vmp for the heavier aircraft will be greater than that for the lighter aircraft. But both aircraft are flying at the same speed so one will be closer to its Vmp than the other.

If the speed chosen is at or slightly more than Vmp for the lighter aircraft, then that will be the one closest to its Vmp. This will tend to give the lighter aircraft a reduced rate of descent compared to that of the heavier aircraft. If the speed chosen is at or slightly less than Vmp for the heavier aircraft, this effect will be reversed, with the heavy aircraft having the advantage.


But at the end of the day the actual rates of descent wil depend upon the combined effects of the different weights and the speed chosen.

If this is a JAR exam question they are probably looking for the heavier aircraft having the greatest rate of descent. The examiners' argument will probably be something along the lines of "jet aircraft usually descend at or close to Vmd in order to get maximum range. So neither aircraft will be very close to their respective Vmp. So the weight difference will be the dominant factor, giving the heavier aircraft the greatest rate of descent."

[square leg]

Rate is a question of time.

The heavier A/C will be flying faster than the lighter one in order to create enough lift.

They will BOTH travel the same distance.

Thus, the heavier one needs a higher ROD to reach the same geographical point in less time (higher speed).

[BEagle]

For glide angles of less than about 15 deg (and most transport jets glide at far less than that!), max glide range occurs at the best L/D value; this corresponds to a specific angle of attack independent of weight. If the ac has been designed sensibly, ideally it will land at a relatively high weight to yield a large payload, hence the descent IAS chosen should result in an angle of attack close to the ideal Cl/Cd max value. But if the ac is at a much lighter weight, at the same angle of attack value it would need a lower glide IAS to descend for max range at Cl/Cd max. This isn't usually possible as most ac use a fixed IAS for the descent, not AoA. So the lighter ac may well be flying at an AoA which is other than optimum; it will therefore not 'glide as well' as it will be descending at a higher rate than the heavier ac.

I did some sums on this once with simple Cl and Cd curves - and Cl/Cd all plotted against AoA. Let's say that your 'heavy' ac decends with Cl of 0.4 and Cd of 0.016 at a velocity of Vh at the AoA corresponding to best glide value. A lighter ac at 60% of the other's weight would require a Cl of 60% of 0.4, i.e. 0.24 (all other factors being the same), but from the fixed Cl/Cd curve the Cd would be 0.011, leading to a descent rate 15% higher than for its heavier brother.

The significant item in the question is the requirement for both ac to fly at the same speed, not at the same AoA!

[gordonsmall]

The following describes how I understand it (i.e how it was explained to me):

You have 2 767's flying in formation side by side, one very heavy and the other very light. The engines on both aircraft are shut-down at the same time, and the aircraft are both glided to a dead stick landing at their respective best glide speeds.

Both aircraft will glide exactly the same distance, and should land right next to each other. The heavier one will land first however due to it's higher groundspeed (due to higher glide speed) and higher descent rate, the lighter one will take a bit longer but won't get any further.

Regards,
Gordon.

[BEagle]

Gordon, RTFQ! The question specified that the ac fly at the same speed!

[Right Way Up]

Ready to get flamed for this, but....
I think the lighter one will descend at a higher RoD. Certainly on jets that I have flown (747,A320,737) descent profiles are much steeper for light aircraft.

[Intruder]

I think the lighter one will descend at a higher RoD.
I think you're right!

In basic aero we learned that an airplane will have the same glide ratio at its best L/D Angle of Attack, regardless of weight (let's discount residual thrust at idle). That means that even though the airspeed is higher for a heavier airplane at that AOA, the glide profile will be the same, and rate of descent will be directly proportional to airspeed.

Let's assume the heavier airplane is at its best glide speed. That means the lighter airplane will be at a speed higher than its best glide speed, and therefore at a non-optimum AOA. The lighter airplane will therefore have a worse glide ratio, and will descent faster at that airspeed.

[BEagle]

Agreed. As I posted earlier - but you've explained it much more concisely!!

[jtr]

Assuming the lighter jet is further away from its max glide range speed (which at say 300kt it most likely will be) then yes it (the lighter) will have a higher rate of descent.

Fly at anything other than max glide range speed, and you will not glide as far, therefore if both a/c are doing the same speed, the lighter a/c must descend faster.

Good question.

[BEagle]

I understand that this used to be a Cathay Pacific interview question, incidentally!

[oldebloke]

The heavier aircraft ,due to weight has higher kinetic energy,gets to descent speed(best l/D ratio)early.Light a/craft stuffs the nose down to reach it's L/D speed,once there meanders down some time later.Both reach the same spot(about 75 miles from FL350).
Used to talk of this Circa DC8 in the 70's..
Cheers...needless to say been done in the Sim':O

[faq]

Why do gliders carry water ballast?

[Keith.Williams.]

when the engines fail, or are set to idle the aircraft has a limited store of energy. This stored energy is made up of potential energy (weight x height) and kinetic energy 1/2 mass x velocity squared. From the start of the glide the stored energy is depleted in pushing the aircraft forward against the drag force.

When the energy store is completely exhausted the aircraft will be standing still on (or in) the ground. Unless the engines are restarted, this energy cannot be replenished. So for maximum glide endurance, which means minimum rate of descent, the glide must start at the maximum possible height (giving max potential energy), then it must be flown at the speed at which energy depletion rate is least. But energy depletion rate is power required, so maximum glide endurance from any given height is achieved by flying at Vmp.

Power required is equal to drag x TAS, so with both aircraft at the same speed (as specified in the question), the heavier aircraft which will be generating more induced drag, will require more power. So the heavier aircraft will deplete its energy faster than the lighter aircraft. This means that the heavier aircraft will be descending faster than the lighter one.

As stated in my original post, the overall effect depends upon what identical speed both aircraft are flown at. High speeds will tend to favour the heavier aircraft and low speeds will tend to favour the lighter one.

Why do gliders carry water?...... Because this enables them to glide faster. For maximum glide range, you must fly at Vmd and Vmd is proportional to the square root of the weight. So if you are in a glider race and want to get to a fixed point before the competition, it is useful to increase your weight by carrying water. But if you want to glide for the longest time you must dump the water to miniimise your energy consumption rate.

Think about it..........If increasing aircraft weight will increase glide endurance then something like an A380 will almost be able to glide forever. Do we really believe this??????

[square leg]

... to fly faster. Gliders are under time pressure when it comes to competitions. They have a set course that has to be flown and documented by GPS loggers or, in the older days, by camera shots of the turning points (quite a complicated affair considering that you have to fly and do a semi chandelle and then press the camera take button).

My previous post:rolleyes: (upstairs) refers more to gliders, apart from the fact that I did NOT read b777pilot's question properly.

So, the glider pilot wants to fly fast around a given course (distance) in order to be home first and to be able to score the most points. You try and keep your average speed up as high as possible, depending on how strong the lift is (using a McCreedy Ring). Thus, for a GIVEN SPEED, the heavier glider will fly faster (for the same distance). The horizontal component of the lift vector is greater on a heavier glider than on a light one (speaking about two identical gliders, one with, the other without water ballast).

Then there is penetration of a strong headwind for example. Take a ping-pong ball and throw it as far as you can. Now take a stone of similar size to the ping-pong ball and throw it as far as you can. The stone will travel much further.

[Bellerophon]

square leg

...Thus, for a GIVEN SPEED, the heavier glider will fly faster...

:D:D

Regards

Bellerophon

[Intruder]

...Thus, for a GIVEN SPEED, the heavier glider will fly faster...
No...

For a GIVEN SPEED, the heavier glider is flying the same speed!

For a GIVEN SPEED, the heavier glider has a better glide ratio, and will have to circle less in the "lifties"!
:ok:

[square leg]

:O I was just checking...;)

Intruder is right though (thanks also to Bellerophon). My mistake:p

What I meant is:\ the heavier glider is given a speed to fly and the other one is also given a speed to fly. So they both have given speeds:D

the same difference...

Nah, it's about glide ratio at the same speed and about time if they fly at their OPTIMUM speeds for their individual weights (masses).

[FlightDetent]

This reminds me of the thread on max drag/thrust thread some time ago. All contributors know the stuff pretty well but no one answers the question. Eventhough it is the precise wording of it that only allow us to answer properly. :sad: Typical JAR ATPL catch-up.

So, two aircraft (or one a/c on two occasions), same speed, different weights. ROD?

Can not be answered. It is possibe that the lighter will have lesser ROD. It is also possible that the heavier will have a higher ROD. It is also possible that they will have the same rate of descent. :{

The real perfomance is L/D, Height/Distance or glide ratio whatever it is - airframe characteristic. No matter the weight, as described in previous postings. Yet, for every given weight the optimal speed must be selected in order to achieve this. Not the case of our example.

To assess this properly, one would need to know the optimal speed for the two weights and then be wise enough to judge how close each of the planes would be. (I am not!)

Now, imagine a side-view of the two, descending.
Once again, these two are able to glide at same angle, but need different speeds for it. The question in place does not allow them.

Let's assume the selected speed is the optimal glide speed of the lighter aircraft computed as to include residual thrust/weight effect. This light aircraft would then fly the best glide, longest distance. The heavier one tough, is off its speed to achieve this optimal ratio, so it performs worse. They still fly side-by side, but the heavier one loses more height. Higer ROD. In pure theory it also falls behind some. Do the vectors. :cool:

Let's assume the selected speed is the optimal glide speed of the heavier aircraft. ... blah, blah, blah ... It is higher ROD for the lighter one now.

There's really no need to elaborate on the third option I've put forward. :ok:

Cheers,
FD.

PS: For those still wondering, yes "aircraft flying at max speed demonstrates both max thrust and max drag at the same time".

[Panama Jack]

Interesting question.

I wonder if we are getting the distinction between maximum gliding distance and minimum descent rate confused?

Drawing from my glider experience, there are two important speeds-- the so-called "Best L/D" (Vmp) and "minimum sink speed". Best L/D, as mentioned, is proportional to weight-- a higher weight will require a higher speed to be able to fly the maximum distance. It is the best compromise between forward penetration and altitude loss.

However, when flying in areas of updrafts the speed would be reduced-- to minimum sink speed. Minimum sink speed would provide the lowest descent rate (in feet per minute, knots, or whatever unit you prefer tu use).

If two gliders, at the same weight were flown in still air, the one maintaining Best L/D would have a higher descent rate than the one maintaining Minimum Sink speed. Because of this, the glider flying at Best L/D would land before the one flying at Min Sink. However, the glider flying at Best L/D would cover a much greater distance than the one flying at minimum sink speed.

[Scallywag]

Everyone is getting too deep here. let's RTFQ and give a simple answer.... IT'S THE LIGHTER ONE.

"two jet a/c's of the same type, descending at the same speed, throttles are at idle, one significantly lighter than the other, which will have a higher rate of descent?"

In order to maintain the same speed with idle thrust the lighter aircraft will have to increase pitch down resulting in a higher rate of descent. Those of you who have operated empty sectors in jets compared to a full load will have already experienced this many times, and some very high descent rates can be achieved in a light jet aircraft of same type at same speed as a heavy one provided thrust remains at idle.

KISS

[ft]

Take a DC-9-41. In a speedbook I have handy I find a maximum listed weight of 52t and a minimum weight of 30 t.

Doing a bit of algebra, I find that the rate of descent is proportional to the drag (D) through the mass (m) of the aircraft. The drag will of course in turn depend on the mass, D(m).

Thus, the difference in descent rate, heavy aircraft descending faster being the positive, can be described by

V/S difference = D(52)/52 - D(30)/30 = ( 30*D(52)-52*D(30) ) / (52*30)

In other words, if

30*D(52)-52*D(30) > 0, the heavy aircraft will descend faster.

30*D(52) > 52 * D(30)

30/52 * D(52) > D(30) <=> 57% > D(30) / D(52)

Then the question boils down to if the drag at 30 tons is less than 57% of the drag at 52 tons at the given airspeed. If that is true, the heavy aircraft will descend faster.

I find it highly unlikely.

Cheers,
Fred

[Clickclick]

Answer from a lazy pilot:

Heavy or light have same best glide distance except the lighter one fly slower IAS. To maintain same speed with heavier one, pitch down increase rate of descent.

answer: lighter a/c higher rate of descent.

:}

[catpinsan]

a perusal of kermode (pg192) reminds us that keith has got it nailed

having said that, a landing wt of 120 tons gives us a dist required to descend of about 10 nm more than at 100 tons at the SOP speed of .80/300 - that's where the chosen speed comes into play i guess with these speeds being well above best L/D

[Checkboard]

Actually, Keith got it a bit wrong! :)

The simple answer is that the heavier aircraft will (probably) have a lesser glide endurance and hence a greater rate of descent.

In all descents at line flying speeds, the lighter aircraft will have a steeper angle, which at the same speed will mean a higher ROD.

[Keith.Williams.]

Checkboard,

I think that you will find that this is because "line flying speeds" are close to Vmd and therefore higher than Vmp for both the light and heavy aircraft. This means that if they are both flying at such speeds, the heavy aircraft will be closer to its best glide endurance speed (its own Vmp) than the lighter aircraft will. So at such speeds the light aircraft will have the lesser endurance and hence higher ROD.

But if you were to glide both aircraft at the Vmp for the light aircraft it would (probably) have the greatest endurance.


FT,

Your equations are interesting but I am not convinced. If I may quote you,

"Then the question boils down to if the drag at 30 tons is less than 57% of the drag at 52 tons at the given airspeed. If that is true, the heavy aircraft will descend faster....I find it highly unlikely".

Well Cdi is proportinal to Cl squared and at a common speed the Cl at 52 tons will be 52/30 = 1.733 times that at 30 tons. So the Cdi at 52 tons will be 1.733 squared = 3.00 times sthat at 30 tons. This means that the induced drag at 53 tons will 200% greater than that at 52 tons. or to put it another way the induced drag at 30 tons is only 33% of that at 52 tons.

[ft]

Keith,
you're forgetting that we are talking about the total drag, not only the induced drag.

[Keith.Williams.]

FT,

No I did not forget it, I just did not discuss it.

At Vmd the induced drag is half of the total drag. Only the induced drag is affected by weight changes. So at Vmd the increase in total drag for any given weight increase will be half of the increase in Induced drag. In effect this means that for small % weight changes, there will be a 1 to 1 ratio between weight increase and total drag increase.

So using the figures in my previous post (Induced drag increased to 300% of its previous value). If this happens at Vmd then total drag increases to 150% of its previous value. So the drag of the 30 tonne jet is about 2/3 that of the 52 tonne jet.

But at lower speeds the induced drag is a much greater proportion of the total drag. So the increase in total drag is a much greater proportion of the increase in induced drag. The relationship between weight increases and total drag increases become smuch greater than 1 to 1. This means that at low speeds such as Vmp it is entirely possible that the drag on the 30 tonne jet will be less than 57% of that of the 52 tonne jet.

The overall effect of this drag-weight-speed relationship is that the heavy jet will have the greatest ROD at low speeds and the lighter jet will have the greatest ROD at high speeds.

Readers who prefer the "nose further down to maintain speed at low weights - therefore greater ROD" argument, should consider what happens at low speeds. The heavier aircraft will have an inefficiently high angle of attack, so it will produce a great deal of drag. This high drag will use up its energy more quickly.

The fact that gliding at or close to Vmd causes light aircraft to sink faster, does not mean that this is true at all speeds. It will just appear to be inevitable if you do most of your gliding at or close to Vmd.

[ft]

Keith,
good post. The fact that it is certainly not a black/white scenario is very clear now IMO!

Cheers,
Fred

[unruly]

In my experience:

95,000lbs, 250 kts = 1,500 fpm
78,000lbs, 250 kts = 1,800 fpm

my answer? light aircraft, faster descent, same forward velocity.

But I do appreciate this discussion! Very nice!

I only wish the question should have been more specific. ;)

[Capt Fathom]

Phew...too many cobwebs to come to grips with the theory here.

What about this then...
How long is a piece of string?

Just kidding...:E

[b777pilot]

i am glad to have brought this up again cuz' i know zilch.

yes, it was hashed some time back and KE=1/2 mv(squared) was used to illustrate this.

it was proven (altough with much debate) that a lighter a/c would have a higher ROD than a heavier a/c.

sorry i have not put in my response, blasted thing went dead on me.

yes, it is about rate of descent, two perfectly fine aeroplanes descending at the same speed. with thrust at idle.