Why heavier aircrafts take longer to slow down in the air?

[extricate]

Pardon me but..

1)Why do heavier aircrafts take longer to slow down in the air compared to lighter aircrafts?

2)Why on high profile, it's recommended to increase speed?

Trying to learn.

Thanks

[SMOC]

Newton's second law.

F=MA

[RunSick]

If you select a higher speed descending the aircraft will "obtain" this speed by PITCHING DOWN.
That way you get a faster speed but as a side effect also a higher VERTICAL speed, which will help you come back into profile.
(Oversimplified, yes, but hopefully you´ll get the idea)

[hetfield]

1) More energy
2) More drag

[extricate]

F=MA.

A=F/M. If M increases, acceleration falls. This is why heavier aircrafts travel slower? Does it have the same meaning as why heavier aircrafts will take longer to slow down?

[yotty]

Inertia. Related to mass. The same anywhere in the universe!:cool:

[AdamFrisch]

Kinetic energy. It's exactly why a truck/lorry will coast much further if you get off the gas than a car.

[DaveReidUK]

Kinetic energy.

Related to mass. The same anywhere in the universe!

[Okivan]

İntertia , momentum.

[Microburst2002]

Extricate has a point.

Newton's second law (F=m.a) is not only about Inertia. It is also about Force.

Bigger airplanes have more mass, that is obvious. But they have more Drag, too, right?

Same with energy (E=F.d). Mass has an effect in the distance, but so does force.

Therefore, if it takes more distance to slow down a big airplane than a small one, it means that the mass increase is greater than the drag increase when airplane size increases. Somehow, increase size is more efficient aerodynamically...

As for the secons question: increasing speed has two effects. One is that you exchange altitude energy for speed energy. However you will have to get rid of the extra speed energy before landing, which in turn takes distance... But exchanging both energies is not free. The air takes a "price" in the form of friction and heating an other forms enrgy loss, so you dissipare some energy when you dive, and also when you level off. You will need less miles to descend if you dive and increase speed 100 kt and later level off to decrease them again that if you just keep constant speed.

The second effect is the change in lift to drag ratio, which is what determines the descent angle. In the normal speed regime (speed above min drag speed) an increase in speed decreases this ratio, which increases the glide angle. For instance, at low speed you are more efficient and glide at 270 fpnm. Then at high speed you make 300 fpnm. And at vey high speed, 330 fpnm.

[HazelNuts39]

I'm assuming you're comparing two aircraft of the same type at different weights. I'm also assuming that the deceleration is in level flight and that engine thrust at idle is negligible.

Then it depends. The retarding force F is the drag. At high speed the drag is mainly parasitic (i.e. the drag due to lift is small in comparison) and the drag for both aircraft at the same speed is about the same. Then from A=F/M follows that the deceleration of the heavier aircraft will be less than that of the lighter aircraft.

However, if both aircraft are flying at their 'green dot' speed (where drag due to lift is of the same order as the parasitic drag), both aircraft have the same lift/drag ratio and decelerate at the same rate.

[Lyman]

But is not greendot AoA dependent? It is a stablized value, no?

Should we channel Galileo?

[HazelNuts39]

But is not greendot AoA dependent?The green dot speed varies with weight, the green dot AoA does not.

[Stuck_in_an_ATR]

Another explanation is that heavier aircraft operates at better L/D ratio for a given speed, and therefore is more efficient => ie. there's less drag to slow it down. That's why high performance gliders take water ballast - to increase performance at high speeds

[Bye]

you really need to use the formula for kinetic energy rather than F=MA

KE = 1/2 x mass x Velocity Squared.

So with 2 identical aircraft with the same drag and at the same velocity the aircraft with the higher mass will have higher kinetic energy.

Or the same aircraft both at the same mass with one travelling faster than the other, the faster one will have higher kinetic energy and thus take longer to slow down as well, But bear in mind that drag is also squared with velocity so the faster aircraft will have higher drag until it slows down to the same speed as the other one was at.

so we can't use the F=MA formula as it doesn't allow us to take account of the velocity of the aircraft at the point of forward thrust being replaced with momentum / inertia.

With 2 falling objects you initially ignore mass as the both accelerate towards the ground at the same rate of 1G, until terminal velocity is reached then if the objects have different mass as the drag of each object equals the mass of the object acceleration will become zero and terminal velocity is achieved.

so for a faster terminal velocity with the same drag, you need a heavier aircraft.

water sufficiently muddy i hope.

GB

[Microburst2002]

OK, we are talking about same model at the same descent IAS? Like a heavy "full house" A330 and an empty A330 flying at 300 kt?

Then it is all about L/D ratio.

The distance for altitude is equal to the L/D ratio.

L/D ratio depends on AoA. L/D ratio increases with AoA up to a maximum (min Drag AoA) and then decreases again. Normally we fly at AoAs below max L/D AoA.

Imagine those 330s. The heavy one, flying at max L/D (green dot speed) would make, say 240 kt, whereas the light one would be making only 210 kt. But they are both flying at 300 kt, so the heavy one is 30 kt closer to green dot that the light one. Its L/D ratio is better. Its AoA is closer to min Drag, to green dot. That is why the heavier airplane flies more miles for the same altitude.

Another way to look at it. Imagine two blocks on skis on two icy downslopes, sliding at 120 kmh. The heavy one will have enough forward force to overcome the drag of those 120 kmh, with a shallower slope than the light one, due to its weight.

As for the second one: speed increase is the same as reducing AoA, which means getting farther from min drag AoA, and decreasing L/D.

[framer]

It is worth remembering that at a given weight you will always have the same inertia regardless of whether you are doing 340kts or standing still. The same cannot be said for momentum.

[Chu Chu]

In general, drag is proportional to area, while mass is proportional to volume. Very roughly, if you scaled an airplane up by a factor of two, you'd increase the drag by a factor of four, and the mass by a factor of eight. Since the larger airplane would lower drag relative to its inertia, it would decelerate at a lower rate. Of course it isn't this simple, especially when you bring induced drag into the equation, but I think the general point holds.

[misd-agin]

"2)Why on high profile, it's recommended to increase speed?"



L/D (better gliding performance) is almost always slower than descent speed. Speeding up generates more drag which results in your energy being closer to the profile.

Also increasing speed results in a faster rate of descent getting you closer to your profile.

The reserve applies if you get low on profile. Slow up closer to L/D speed. You should get closer the VNAV descent profile as your gliding performance at L/D is shallower than it is using the VNAV descent speed..

Adding power isn't the most efficient way to correct a small deviation below profile.

[DaveReidUK]

It is worth remembering that at a given weight you will always have the same inertia

Inertia is a function of mass, not weight.

On its way to the Moon, Apollo had a lot of inertia but weighed nothing at one point. :O

[veloo maniam]

Why is it then some airplanes, after t/off and when given high speed
climb reply by saying " ok on passing 5000ft high speed"
Any reason for this? :ok:

[Microburst2002]

Company speed limit to 250 below 5,000 ft

[extricate]

Thanks everyone for the replies.

Ok i understand that a heavier aircraft has more energy but how does it take a longer time to slow down?

Is there a better analogy?

If i were to compare a lorry to a car, you would say the lorry will coast further than a car. But can i not argue that gravity will pull the heavier lorry down more than the car and effectively slowing down the lorry faster?

Can anyone counter this? Pardon me, i don't possess any physics knowledge, just logic.

Pls advise.

Cheers

[HazelNuts39]

But can i not argue that gravity will pull the heavier lorry down more than the car and effectively slowing down the lorry faster?Yes you can, up to a point. The retarding force that can be derived from the friction between tires and road increases proportionally to the load that the weight exerts on the road. Provided the brakes of car and lorry are sufficiently powerful and equally efficient, car and lorry will slow down at the same rate, even though their masses are different. If the lorry has a higher aerodynamic drag than the car, it may even slow down faster. In practice that is unlikely, however, because the tires and braking systems of modern cars are more effective/efficient than those of lorries.

But your original question was about aircraft, not cars and lorries, and the physics are somewhat different.

[Bye]

simple analogy.

i throw a ping pong ball at your head at 20 MPH.

then i throw a snooker / Billiards / pool ball at the same place at the same speed (20 MPH )

which one hurts the most ??

Or if i roll a bowling ball down the alley will it knock over the pins.

now if i roll a balloon down at the same speed what happens ?????

sorry but you have to accept that the physics for now is correct and that the formula's and explanations you have are enough for you to understand why a heavier aircraft takes more slowing down than a lighter one.

Hope this helps. But if you really want to get into the physics, you really need to start at the beginning and build your knowledge up, rather than jumping in at some random point.

GB

[HazelNuts39]

you have are enough for you to understand why a heavier aircraft takes more slowing down than a lighter one.I'm wondering if your simple analogy explains that the heavier aircraft slows down more rapidly than the lighter aircraft below the minimum drag speed.

Regards,
HN

[Bye]

I'm wondering if your simple analogy explains that the heavier aircraft slows down more rapidly than the lighter aircraft below the minimum drag speed.

thats aerodynamics as opposed to basic Physics,so no it doesn't and wasn't really intended to, so failing that have a look here.

http://home.anadolu.edu.tr/~mcavcar/htk224/MinDrag.pdf

GB

[HazelNuts39]

Bye,

Post #2 states the 'basic physics'.
F/M is 'basic' aerodynamics.
Your analogy doesn't address either one.

[Yankee Whisky]

DaveReidUK (http://www.pprune.org/members/214092-davereiduk)

Join Date: Jan 2008
Location: Reading, UK
Posts: 856


Quote:
It is worth remembering that at a given weight you will always have the same inertia
Inertia is a function of mass, not weight.

On its way to the Moon, Apollo had a lot of inertia but weighed nothing at one point

If that is true, would inertia in zero gravity space be equivalent to Vsquared ?:bored: ?:rolleyes: It seems to me that there is not much attention paid to the cleanliness of aerodynamic profiles in modern heavies and that is also a contributing factor to it taking longer to slow down. My high performance glider, when loaded with water has a longer slowing down time when decellerating in level flight (vs a Schweitzer 222, which is a big drag queen in the gliding world !);)

[Bye]

post #2 states "some" basic physics but is not the correct formula to explain why a heavier object has more kinetic energy than a lighter one at a given speed and drag.

F/M is that some code to describe movies of a certain type. certainly not a formula or basic aerodynamics of any kind.

I look forwards to your explanations and analogies of why a heavier aircraft below Vmd slows down quicker than a lighter aircraft.

GB

[HazelNuts39]

F/M is that some codeSMOC states Newton's second law as: F=MA. It relates force F, mass M and acceleration A. Energy doesn't come into it.

Extricate's questions in posts #1 and #5 are about A, which equals F/M. F is aerodynamic drag, and it is related to mass M by the aerodynamic lift which equals weight. So a question about A=F/M is about aerodynamics.

I look forwards to your explanations and analogies of why a heavier aircraft below Vmd slows down quicker than a lighter aircraft.It's essentially explained in post #17:
The heavy one, flying at max L/D (green dot speed) would make, say 240 kt, whereas the light one would be making only 210 kt. But they are both flying at 300 kt, so the heavy one is 30 kt closer to green dot that the light one. Its L/D ratio is better. Its AoA is closer to min Drag, to green dot.What he writes for speeds above green dot, is reversed below green dot.

Regards,
HN

[awblain]

If you load up the same truck with more weight, keeping the same brakes, then it takes longer to stop. If there's the same braking force, energy to dissipate scales with mass.

It's a little bit more complex with an aircraft, as if it's heavier, it needs more lift, which means a different speed, height or attitude to fly level. More lift usually means more drag, both in approximate proportion to weight, and so more power is required to cruise heavier.

When losing energy descending though, if you try to shed speed deliberately with speedbrakes, that amount of braking isn't affected by weight, just by the speed and spoiler setting, and so a heavier aircraft will need more energy throwing away, and a longer deployment. The energy required to drop from 10,000m would make you travel at 440m/s in addition to your 300m/s cruise speed, so you need to lose a lot of energy on descent to land at 50m/s. That scales with mass, whereas the speedbraking force doesn't.

[HazelNuts39]

If you load up the same truck with more weight, keeping the same brakes, then it takes longer to stop.Not necessarily, it depends on the brakes and the friction available between tires and road surface. That's why I added the caveat 'brakes sufficiently powerful'. The maximum retarding force from a braked, rolling wheel is obtained when one of the following conditions is reached, whichever comes first when increasing the pressure applied to the brakes:
(a) the brake pressure equals the maximum brake pressure, or
(b) the tires start skidding.
In this context 'brakes sufficiently powerful' means that condition (b) applies. The braking force is then proportional to the load on the wheel.

[Bye]

Energy doesn't come into it.

i think Mr Einstein might be turning in his grave. :\

i give up :ugh::ugh:

[smith]

We're not talking forces or accelerations here, we are talking inertia which is velocity and mass related.

Remember the equation

M1V1=M2V2 (numbers should be in subscript)

Therefore if M1 > M2 then V1 < V2

[Bye]

(a) the brake pressure equals the maximum brake pressure, or
(b) the tires start skidding.

(a) not true, under extreme hard braking there are small particles of metal embedded into the compound which get hot and "weld" themselves to the brake disc, this forms a mechanical bond which has to be physically broken rather than just relying on friction.

The friction of Carbon Carbon brakes increases as they get hotter, so for the same brake pressure as they heat up they get grippier.

(b) not true, Depending on the Mu and the surface texture the maximum retardation is just before the skid onset.

[HazelNuts39]

Bye,

How do your remarks make untrue what I wrote, that the maximum braking force is determined by either the maximum that the brake can produce, or the maximum that the tire/runway friction can produce, and that the latter is proportional to the load on the wheel?

[Bye]

because that is not what you wrote.

What you did actually write is not true.

My remarks do not make what you wrote untrue, what you wrote that is untrue makes it untrue. I have merely pointed out that what you wrote is untrue.

GB

[awblain]

Ms 39,

You're right - I was assuming progressive braking: dumping of heat in the truck brakes at some acceptable level without locking the wheels. If you keep maximum braking force applied, limited by rubber-road friction, then the braking force would rise linearly with weight (until you ripped the tread off the wheel or set fire to the brakes).

[HazelNuts39]

I have merely pointed out that what you wrote is untrue.
If you have, I don't get it. Must be my fault.

[DaveReidUK]

If that is true, would inertia in zero gravity space be equivalent to Vsquared ?

Inertia is zero gravity space is no different from inertia anywhere else. It's measured in kg. Nothing to do with V or V².

[HazelNuts39]

According to Wikipedia:
In common usage the term "inertia" may refer to an object's "amount of resistance to change in velocity" (which is quantified by its mass), or sometimes to its momentum (http://en.wikipedia.org/wiki/Momentum), depending on the context. The term "inertia" is more properly understood as shorthand for "the principle of inertia" as described by Newton in his First Law of Motion (http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion); that an object not subject to any net external force moves at a constant velocity. Thus an object will continue moving at its current velocity (http://en.wikipedia.org/wiki/Velocity) until some force causes its speed or direction to change.

[Bye]

inertia is mass and vice versa, momentum is mass or inertia x velocity

A moving body has momentum and kinetic energy, under the laws of conservation of energy, for the body to lose momentum, or velocity you have to move some of the kinetic energy into another form of energy.

Drag converts kinetic energy into other forms of energy such as heat, noise and kinetic energy in the air.

when the thrust equals the drag the body remains at a constant velocity.

Reduce thrust and the body will lose kinetic energy thru drag as heat etc, and will slow down until once again the thrust equals the drag. and we are again stable.

Now if you are in a glider the only thrust you have is in fact the momentum you have which is your mass x velocity in straight and level flight, or you can fall to earth and use G to give you acceleration which will again increase your velocity and hence your momentum. Or you can use a thermal say and raise your height increasing your potential energy which you can then exploit to convert it to kinetic energy once again, only to lose it as drag in the form of heat etc.

The key to understanding this is the conservation of energy and managing the energy available in whatever form you have it, it could be chemical energy in your fuel, or potential energy in your altitude or kinetic energy in your velocity and mass. BUT once you let it out of your system into another system in some form no use to you such as hot air or noise its very very hard to get it back to do work for you.

I'm not a pilot, i just design the hot bits that push you guys along, but i am pretty sure i'd rather have some altitude and speed in reserve as energy stores, just in case my bits let you down at the critical moment.

GB

[yippy ki yay]

Again, like HazleNut, assuming level flight

http://upload.wikimedia.org/wikipedia/commons/0/04/Drag_Curve_2.jpg

Looking at the graph above...

Let's put a value of the minimum drag speed depicted on the diagram (courtesy of wikipedia) as 230kts for the heavier aircraft and 210kts for the lighter one.

So a speed of say 300kts will be a long way to the right of the minimum drag speed. As speed decreases AoA will increase. As AoA is increasing, initially its getting more to its optimum AoA. Form drag will be decreasing at a faster rate than induced drag is increasing ie. total drag is decreasing. As stated by Microburst the heavier aircraft, initially, is continuously flying closer to its optimum AoA than the lighter aircraft. Therefore it will be experiencing less total drag than the other aircraft and thus decelerating at a slower rate.

However once airspeed goes below Vmd for the heavier aircraft induced drag starts increasing at a faster rate than form drag is decreasing and so there is a rise in total drag. In the meantime however the lighter aircraft is now getting closer to its Vmd and so this aircraft will now start decelerating at a slower rate. Once both aircraft are below Vmd, I could well be wrong here though, I guess induced drag becomes the major player, the heavier the weight the more lift required. More lift = more induced drag = an increased rate of deceleration for the heavier aircraft.

[Sunamer]

Gee...People, are you serious to discuss this subject for 3 pages?
P = mv, where P - momentum, m - inertial mass, v - speed.

1.
Putting aside drag and all aerodynamics, and using the formula above we have:
The faster you go the more momentum you gain, OR
More mass you have, more momentum you obtain when moving with the same speed.


2. If we also have aerodynamics involved here, then we should separately do part 1, and then figure out drag. After that we can decide whether bigger AC can slow down faster than smaller one or not.

But drag doesn't change P given constant V and m.


Is not it that obvious? :ugh:

[Microburst2002]

If I had my books and my scanner, I would post here the L/D curve, and a graph of an airplane gliding angle.

If an airplane has a Lift of 10 tons, and a drag of 1 ton, it has a L/D of 10 and it glides at an gradient of 10%. That is a fact. It takes some graphics and formulas, though.

The same airplane weighing 20 tons and with a drag of 2 tons, has the same L/D ratio, and therefore glides exactly the same, at a 10% gradient.

What's more. A 1 ton cessna with a 100 kg drag will have the same L/D ratio of 10 and glide at 10% gradient.

L/D ratio determines glide angle.

The L/D ratio is equivalent to the CL/CD ratio. CL and CD only depend on AoA, therefore L/D only depends on AoA, for a given configuration. Not on weight, not on speed. Only on AoA.

For 1g flight at a given weight, there is a relationship between speed and AoA, so that to each speed corresponds an AoA. Therefore, to each speed during a 1g glide at a given weight corresponds a L/D ratio.

Normally airplanes fly at AoAs smaller than max L/D AoA. When flying at AoAs higher than that, we are said to be in the "reverse command region", or the "back side of the curve". So an increase in speed is always decreasing the AoA and the L/D ratio, and therefore increases the glide angle (reduces the gliding distance).

The AoA-Speed relationship at 1g flight varies with weight. At increased weights, increased AoA are necessary to increase Lift in the same amount for a constant speed.

If you glide at the same speed but higher weight, your AoA is higher, and your L/D ratio increases, and so your glide angle reduces (gliding distance increases).

Now think of two big football balls (football balls are spheric, ok? I don't know nor I want to know what "soccer" is...) One filled with lead, the other with air. Which one will travel faster on a given slope? And Which one will need less slope to maintain a given speed? Well, the heavier needs a shallower slope because its weight can overcome more drag than the heavy one, right?.

Now, from the energy point of view you might have trouble understanding, because here there is a lot of energy dissipation. Actually, descent is all about energy dissipation (you glide at quite a constant speed and yet lose altitude, right?). You can't use energy conservation here...

finally, two airplanes of same model flying at maximum glide speed (min drag speed or max L/D speed) will glide the same, irrespective of the weight. Only the heavier will fly faster.

[HazelNuts39]

yippy ki yay,

If I understand you correctly, I think you agree with Microburst and me. Just a small correction:
As stated by Microburst the heavier aircraft, initially, is continuously flying closer to its optimum AoA than the lighter aircraft. Therefore it will be experiencing less total drag than the other aircraft and thus decelerating at a slower rate.Actually, above the minimum drag speed the heavier aircraft will have more total drag than the lighter one, but its drag-to-weight ratio will be lower and therefore it decelerates at a slower rate.

Sunamer,

1. Agreed
2. That's what we do

The reason for discussing it for so long is that it is not generally true to say that the heavier aircraft decelerates more slowly than the lighter one.

Microburst2002,
Normally airplanes fly at AoAs smaller than max L/D AoA. True, for transport aircraft, except in a few minutes after takeoff and prior to landing. Gliders fly below minimum drag speed when circling in thermals to gain height.

Regards,
HN

[Owain Glyndwr]

Microburst

If I had my books and my scanner, I would post here the L/D curve, and a graph of an airplane gliding angle.

finally, two airplanes of same model flying at maximum glide speed (min drag speed or max L/D speed) will glide the same, irrespective of the weight. Only the heavier will fly faster.
This what you want?
http://i1081.photobucket.com/albums/j351/OwainGlyndwr/LiftDragvsSpeed.jpg

At anything above 230 kts on this aircraft the Lift/Drag ratio will be better at the higher weight and the aircraft will take longer to slow down;below that speed the opposite is true.

[DaveReidUK]

Now if you are in a glider the only thrust you have is in fact the momentum you have

You were doing OK, up until that part :*

[Crabman]

I've seen the name Newton come up. Where is Bernoulli? Also, where does the conveyor belt come into this?

[Microburst2002]

Bernoulli has absolutely nothing to do here. Like I said, gliding at constant speed is about energy dissipation. Bernouilli is a particular case of the energy conservation theorem, so they are incompatible things...

As for the conveyor belt, I have no idea! what is that?

[Oldaircrew]

The formula for drag is: Cd=1/2Rho x V(sq) x S

Rho = Air density
V= velocity
S= area

By increasing your speed, you increase your drag exponentially and therefore increase your rate of descent.

[loop swing]

I don't mean to be fussy, but Cd = Measured drag / Theoretical drag
Drag = Cd x 1/2 rho x V squared x S

Higher weight means a higher wing load allowing a further glide distance / time to slow down?

[Sunamer]

To HazelNuts39,
The reason for discussing it for so long is that it is not generally true to say that the heavier aircraft decelerates more slowly than the lighter one.

Agreed. When considering AoA of empty AC, AoA may be less, thus producing more efficient aerodynamic profile, than when the same AC weighs more, so it needs more AoA to compensate for more mass = more lift is needed.
That's my understanding.

:}

[HazelNuts39]

When considering AoA of empty AC, AoA may be less, thus producing more efficient aerodynamic profile, than when the same AC weighs more, so it needs more AoA to compensate for more mass = more lift is needed. That is true at any speed. For the variation of L/D in Owain Glyndwr's graph the wing's aspect ratio (roughly the ratio between wing span and average chord length) determines the efficiency of producing lift. As such it is perhaps as important as the 'efficient aerodynamic profile'.

Regards,
HN

[Bye]

Now if you are in a glider the only thrust you have is in fact the momentum you have
You were doing OK, up until that part

Please explain why that is wrong Dave.

GB

[DaveReidUK]

Thrust and momentum are completely different quantities, with different units/dimensions.

A glider usually has momentum, but has no thrust (unless it has a pop-up sustainer or SL engine).

Good illustration here

L/D Ratio (http://www.grc.nasa.gov/WWW/k-12/airplane/ldrat.html)

of the three forces (lift, drag, weight) acting on a glider.

[Bye]

We have 2 aircraft
plane A weighs 300 Tonnes and plane B weighs 200 Tonnes.
Both are identical and have the same wing with the same Lift drag ratio of 17 similar to a 747.
Both aircraft are cruising straight and level at 400 Km/h.
Plane A at 300 tonnes lift has a drag of 300 / 17 which equals 17.65 units of drag.
Plane B at 200 tonnes lift has a drag of 200 / 17 which equals 11.76 units of drag.
So plane A has 1.5 times more drag than Plane B. at this speed.
Plane A has a momentum of Mass x Velocity which is 300 x 400 = 120,000.
Plane B has a momentum of 200 x 400 = 80,000
So again we have plane A with 1.5 times the momentum of plane B.
So all is nicely in equilibrium with both planes nicely in straight and level flight.
Now we want to reduce the speed of both aircraft to 300 kmh.
Plane A has a kinetic energy of ½ x 300 x (400 x 400) = 24,000,000
At 300 kmh it will have KE of ½ x 300 x (300 x 300) = 13,500,000.
This means plane A has to lose 10,500,000 units of KE.
Plane B has a KE of ½ x 200 x (400 x 400) = 16,000,000
At 300 Kmh it will be ½ x 200 x (300 x 300) = 9,000,000
So plane B has to lose 7,000,000 units of KE.
Plane A has to lose 1.5 times the energy as plane B.
So as long as the lift drag ratio stays the same between the 2 velocities and the amount of lift needed, both aircraft will slow down at exactly the same rate.
If the lift drag ratio alters due to plane configuration it will affect the rate of deceleration between the 2 aircraft.
BUT maximum lift drag ratio is independent of the weight of the aircraft, the area of the wing, or the wing loading.

i‘ve tried to keep this simple so forgive the rubbish math’s, but I thought I’d give you all something to pull apart for a while.
GB

[Bye]

Thrust and momentum are completely different quantities, with different units/dimensions.

A glider usually has momentum, but has no thrust (unless it has a pop-up sustainer or SL engine).

Good illustration here

L/D Ratio

of the three forces (lift, drag, weight) acting on a glider.

Sorry Dave but you're wrong here.

There are 4 forces acting on a gliding some of the time.

Your nasa illustration shows a gliding falling from the sky, BUT if he builds up some speed ( momentum ) and pulls back on the stick good n hard what happens, well he will go back up again against gravity untill he runs out of velocity at Agogee and then gravity will take over again and he will fall to earth again

During this upwards flight bit he is using the momentum built up as THRUST, and yes its called thrust becuase the momentum is PUSHING the glider upwards against gravity with the kinetic energy being converted to potential energy that is known as height, as is the tension on a kite string that stops it fluttering away in the wind.

so during this upwards flight mode using the momentum as thrust 4 forces are acting on the glider, not 3.

GB

[Owain Glyndwr]

@Bye
Thrust and momentum are completely different quantities, with different units/dimensions.Sorry Dave but you're wrong here.
Dave is right and you are wrong I'm afraid.

Thrust (like any force) is a rate of change of momentum not momentum. Different units and different dimensions.

We have 2 aircraft
plane A weighs 300 Tonnes and plane B weighs 200 Tonnes.
Both are identical and have the same wing with the same Lift drag ratio of 17 similar to a 747.
Both aircraft are cruising straight and level at 400 Km/h.
Plane A at 300 tonnes lift has a drag of 300 / 17 which equals 17.65 units of drag.
Plane B at 200 tonnes lift has a drag of 200 / 17 which equals 11.76 units of drag.Wrong again. The 300 tonne aircraft will have a lift coefficient 1.5 times greater than the 200 tonne if they are flying at the same speed. They cannot have the same drag coefficient or the same lift/drag ratio so the rest of the argument falls by the wayside - even without perpetuating the thrust = momentum error.

And just for the record, the 'heavy' aircraft in the picture I posted earlier was 1.3 times the weight of the 'light' one

[Bye]

THRUST
Pronunciation (Uk):
Dictionary entry overview: What does thrust mean?
• THRUST (noun)
The noun THRUST has 5 senses:
1. the force used in pushing
2. a strong blow with a knife or other sharp pointed instrument
3. the act of applying force to propel something
4. verbal criticism
5. a sharp hand gesture (resembling a blow)
Familiarity information: THRUST used as a noun is common.

• THRUST (verb)
The verb THRUST has 8 senses:
1. push forcefully
2. press or force
3. make a thrusting forward movement
4. impose or thrust urgently, importunately, or inexorably
5. penetrate or cut through with a sharp instrument
6. force (molten rock) into pre-existing rock
7. push upward
8. place or put with great energy
Familiarity information: THRUST used as a verb is common.

[Bye]

They cannot have the same drag coefficient or the same lift/drag ratio so the rest of

Read it properly before jumping in.

I did NOT say they have the same drag coefficient did i ????

in fact i seem to think i said that plane A had 1.5 times the drag of plane B which is NOT saying the drag is the same is it ???:ugh::ugh::ugh:

i said they DID have the same lift / drag ratio, purely for the maths, as a theoretical calculation.

Why can't they have the same lift/drag ratio. ?????

[RunSick]

In the context of this discussion:

3. (Engineering / Aeronautics)a. a propulsive force produced by the fluid pressure or the change of momentum of the fluid in a jet engine, rocket engine, etc.

Source:
thrust - definition of thrust by the Free Online Dictionary, Thesaurus and Encyclopedia. (http://www.thefreedictionary.com/thrust)

[DaveReidUK]

Sorry Dave but you're wrong here.

There are 4 forces acting on a gliding some of the time.

Your nasa illustration shows a gliding falling from the sky, BUT if he builds up some speed ( momentum ) and pulls back on the stick good n hard what happens, well he will go back up again against gravity untill he runs out of velocity at Agogee and then gravity will take over again and he will fall to earth again

During this upwards flight bit he is using the momentum built up as THRUST, and yes its called thrust becuase the momentum is PUSHING the glider upwards against gravity with the kinetic energy being converted to potential energy that is known as height, as is the tension on a kite string that stops it fluttering away in the wind.

so during this upwards flight mode using the momentum as thrust 4 forces are acting on the glider, not 3.

I suggest you Google Newton's Second Law before you dig yourself in an even deeper hole.

[Bye]

In the context of this discussion:

3. (Engineering / Aeronautics)a. a propulsive force produced by the fluid pressure or the change of momentum of the fluid in a jet engine, rocket engine, etc.

As i used the word thrust in the context of a GLIDER and wrongly assumed that people here would know that a glider doesn't have any rockets or jet engines and therefore was simply trying to show that the thrust in a conventional jet engine is replaced by momentum in a glider in certain circumstances.

so as a GLIDER does not have any other propulsive unit i was using the word thrust in its correct context.

sorry i should of realised this is just a troll fest and not actually based on real science.

GB

[Bye]

Dave i am genuinely interested as to your belief that Newtons second law is relevant.

please explain

[Bye]

thank you hn39 WE AGREE.

[DaveReidUK]

Dave i am genuinely interested as to your belief that Newtons second law is relevant.

Newton's Second Law can be summarised as (F)orce = (M)ass x (A)cceleration.

So in your glider example, pulling up into a climb (acceleration) is the result of a force with a vertical component (lift).

I can't really put it any more simply.

[Bye]

Maybe not but you could put it correctly.

immediately you pull up into a climb you start to DECELERATE as you are now being pulled back to earth by GRAVITY. not ACCELERATE.

you only have your momentum to FORCE you up which is being taken away by gravity.

this is a VERTICAL climb and there is no contribution of lift from the wings by the way.

and even if there was, it would still be the force of momentum pushing you along against the drag and gravity 4 forces in play not 3.

[HazelNuts39]

Well ...

The two aircraft shown in post #49 have the same L/D at about 226 kts. At that speed, and only at that speed, both airplanes slow down at the same rate. As the speed reduces below 226 kts the L/D's of the two aircraft will change differently and hence the rate of slowing down will be different, as shown in the graph.

P.S. I deleted the above immediately after posting it Because I saw that while writing there had been several new posts and I wanted to read those first. Sorry for any confusion.

P.S.2 I suggest you could consider the component of weight that acts in the direction of the flight path as the glider's equivalent of thrust, but momentum is not thrust.

[Owain Glyndwr]

Read it properly before jumping in.

I did NOT say they have the same drag coefficient did i ????No you didn't - I did as part of an explanation.

in fact i seem to think i said that plane A had 1.5 times the drag of plane B which is NOT saying the drag is the same is it ???:ugh::ugh::ugh:That is what you said, but it is just as wrong. Aircraft A will NOT have 1.5 times the drag of Aircraft B if the only difference is weight.

i said they DID have the same lift / drag ratio, purely for the maths, as a theoretical calculation.And that is where you went wrong because you postulate a non-valid equivalence

Why can't they have the same lift/drag ratio. ????? If I really have to explain that I'm wasting my time. If you understood the basics of how drag varies with lift coefficient and how lift coefficient varies with speed and weight you wouldn't need to ask that question.

As to force vs momentum

Momentum = mass * velocity
Force = mass * acceleration
Acceleration = rate of change of velocity
Force = rate of change of momentum.

[DaveReidUK]

the force of momentum

OK, this is where we came in, I give up.

sorry i should of realised this is just a troll fest and not actually based on real science.

Well we can agree on something at least: the rest of us can't hope to match your grasp of aeronautics.

[Bye]

thank you HN39 we agree yet again, i was merely trying to explain that the gliders equivalent of thrust was its momentum.

But the context police jumped on me looking for a fight.

If you understood the basics of how drag varies with lift coefficient and how lift coefficient varies with speed and weight you wouldn't need to ask that question

i clearly don't and have in fact asked that question, would you now be so kind as to enlighten me please.

Dave think of a skier going down a ski jump ramp, at the end the ramp curls upwards to the sky, now as the skier gets to the turned up bit what happens to him, oh he goes upwards, is he creating lift, no then what carries upwards as he leaves the ramp, i guess it must be the momentum he built up on the way down.

and i don't really care how much you insult me about my knowledge.


the force of momentum
OK, this is where we came in, I give up.

sorry i keep forgetting i have to explain everything and leave nothing open to deliberate misintrepration

The word "of" from the dictionary.

of1    [uhv, ov; unstressed uhv or, especially before consonants, uh] Show IPA
preposition
1.
(used to indicate distance or direction from, separation, deprivation, etc.): within a mile of the church; south of Omaha; to be robbed of one's money.
2.
(used to indicate derivation, origin, or source): a man of good family; the plays of Shakespeare; a piece of cake.
3.
(used to indicate cause, motive, occasion, or reason): to die of hunger.
4.
(used to indicate material, component parts, substance, or contents): a dress of silk; an apartment of three rooms; a book of poems; a package of cheese.
5.
(used to indicate apposition or identity): Is that idiot of a salesman calling again?

so you see i was using this word to try to explain that the force is derived from the momentum or that its source was the momentum, not that the 2 words are the same.

:mad:

[Microburst2002]

We have 2 aircraft
plane A weighs 300 Tonnes and plane B weighs 200 Tonnes.
Both are identical and have the same wing with the same Lift drag ratio of 17 similar to a 747.
Both aircraft are cruising straight and level at 400 Km/h.

Sorry, that is not possible. Identical airplanes with different weights an the same cruising speed can't have the same L/D ratio, because they have different AoA. From then on, any conclusion you come to is arguable.

L=q.S.CL
D=q.S.CD

L/D = CL/CD , that is exclusively dependent on AoA. They can'y fly at the same AoA.

As for the slowing down subject:

The heavy airplane has more Drag than the light one, when speed is constant. When you cut thrust to idle the difference between thrust required and actual thrust (let's say nil for simplification) is greater: greater retarding force than in the light airplane case, which tends to decrease stopping distance. However, the higher weight airplane has more mass, which tends to increase stopping distance. Which effect prevails?

D varies because CD varies, when q and S are constant. CD varies because CL does vary too (they are childrem of the same father, AoA). AoA varies linearly with weight, and so does CL. However, CD is not linear. At lower AoAs, CD varies less than linearly with CL. At higher AoAs the opposite is true.

So, at low AoAs, D will vary less than mass for a given weight change. Inertia prevails at low AoAs or high speeds. Mass increase is greater than the drag force increase and as a result acceleration is reduced.

At high AoAs or low speeds, air viscosity prevails over inertia. Mass increase is less than the drag force increase and as a result acceleration is increased.

At least that is the conclusion I have come to thinking about this. I'm standing by for your comments.

[DaveReidUK]

sorry i keep forgetting i have to explain everything and leave nothing open to deliberate misintrepration

I said I wasn't going to argue with you any more, and I won't.

But that remark can't go unchallenged. I can't see any evidence of anyone deliberately misinterpreting your posts. We may not agree with them, in fact we don't, but that not the same thing at all.

PPRuNe is full of spirited discussions on matters of physics, science, engineering, aeronautics, etc, which we all enjoy and which illustrate why many of us chose aviation as a career.

But no poster is infallible. Most of us, if challenged by 3 or 4 people who clearly have some background knowledge of the subject matter, would at least be prepared to consider the possibility that we might be wrong, and to listen to their honest attempts to explain the theory and to provide understandable examples to illustrate it.

You might want to pause to reflect on that.

[Radu Poenaru]

Gezzs what a blood bath in this thread..

I'm assuming "longer" means dt (more time to) slow down from the same speed

Now stop killing the momentum and inertia, it has nothing to do with our problem!

First let's have Newton in here:
a) Force = Mass * Acc
Now take two airplanes with mass M1 and M2 where M2 is heavier:
b) M2 = 1.3* M1
The lift produced must be just as big aswell:
b) L2 = 1.3 * L1
Now if you want airplane nr 2 to decelerate faster, from (a) and (b), the drag needs to be bigger for the heavier airplaine in respect to it's mass:
c1) F2 = M2/a2 remember that F in our case is Drag
c2) F1 = M1/a1 and because we need
c3) a2> a1 then the drag is automatically:
c) D2 > 1.3 *D1
Now combine (c) and (b) a little bit and you end up with a simple condition
(d) L2/D2 < L1/D1

Lift/Drag ratio needs to be bigger for the second plane. We know that L2>L1 from (b), and we assumed they both start at the same speed, so it's only Alpha that changes to create more lift:
e) Alpha2>Alpha1 (note that we do not know the actual A2/A1 ratio .. it's a matter of wing profile)

So we need a better lift/drag ratio (d) for a higher alpha (e) for airplane nr 2.
Take a look at a polar Cl/Cd graph. This only happens at speeds below max lift/drag ratio, ie below best glide. Above that speed the ligher airplane will slow down faster.

[Bye]

but Dave your so called honest attempts to explain that a glider only has 3 forces acting on it was wrong, and it seems once challenged you just resort to insults.

You claimed that a glider only has 3 forces acting on it, yet in the very same article you linked it even states that the glider has 4 forces acting on it.

So to say you clearly have a background in the subject when all you rely on is google is a bit rich and highly insulting to those of that rely do have a background in aeronautical engineering.

When you fail to win the argument on substance you result to syntax then finally to insults - classic troll behaviour pattern.

i have repeatedly stated that my aerodynamics is flawed, i am a thermodynamic engineer who designs jet engines for living, yet when i ask to be enlightened by the "experts" there is no reply.

It seems a trait here is that people just want to pull every poster down and show the world they are cleverer than the poster rather than actually discussing a subject, as you have shown without resorting to insults when you can't win.

Maybe you would also like to reflect on your posts, re read them and realise that maybe just for this one in a million possibility that you were not correct in stating a glider has only 3 forces acting on it. Oh and i didn't see 3 or 4 other posters supporting your claim as to only 3 forces acting on it, just yours.

Kind regards.

GB

[HazelNuts39]

Urbandictionary.com:
Troll: One who posts a deliberately provocative message to a newsgroup or message board with the intention of causing maximum disruption and argument

[DaveReidUK]

Oh well, it's a damp Saturday and I have nothing else to do ...

You claimed that a glider only has 3 forces acting on it

I did, and it does.

yet in the very same article you linked it even states that the glider has 4 forces acting on it.

It doesn't. Read it again.

"The thrust (http://www.grc.nasa.gov/WWW/k-12/airplane/thrust1.html) is determined by the size and type of propulsion system (http://www.grc.nasa.gov/WWW/k-12/airplane/bgp.html) used on the airplane and on the throttle setting selected by the pilot." (those emphases aren't mine, they appear in the original article).

The US convention is to use the term "airplane" to mean a powered, fixed-wing aircraft - I used to argue, once upon a time, that gliders were also "airplanes", but I was wrong about that, it's their word and they get to define it. :O

The terms "propulsion system" and "throttle" also give a pretty good clue as to what we're talking about here.

So please, one more time, tell us all what this mysterious 4th force (in addition to L/D/W) is that acts on a glider in flight ?

[HazelNuts39]

There are only two forces acting on a glider: an aerodynamic force and a gravitational force. The gravitational force is directed towards the center of the earth and is equal to the product of mass m and gravitational acceleration g.

The aerodynamic force is usually represented by its two components: the drag force D acting opposite to the direction of movement, and the lift force L perpendicular to it.

For a glider moving at constant altitude the lift force is equal to the gravitational force: L = m * g ... (equation 1)

There is no force opposing the drag force so according to Newton's second law there results an acceleration a resulting from the equation: - D = m * a ... (equation 2)

(The sign convention is positive for forces and accelerations in the direction of motion, negative in the opposite direction)

Combining equations (1) and (2) we get: a / g = - D / L

No need to discuss momentum or energy.

P.S. If you measure accelerations in kts/second, then g is about 19.05 kts/second

[DaveReidUK]

No need to discuss momentum or energy.

Whew!

Thanks for that - I was beginning to worry that when I did my aero engineering degree I must have missed a lecture where they explained that momentum was a force. :O

[HazelNuts39]

Dave,

We're debating with a person who claims to be an engineer designing aero engines yet writes such nonsense as (see his first post #15 on this thread):

... we can't use the F=MA formula as it doesn't allow us to take account of the velocity of the aircraft at the point of forward thrust being replaced with momentum / inertia.

... if the objects have different mass as the drag of each object equals the mass of the object acceleration will become zeroThen states his objective as:
water sufficiently muddy i hope.Regards,
HN

[Microburst2002]

Using Newton (forces, inertia, acceleration, momentum) is complicated here, because airplanes move a lot of air. It is not just ann ideal body moving in outer space.

For F=m.a we need to know the force, which is D, and it is a complex dorce to work with. It doesn't behave linearly.

For KE= 1/2 m v2, we don't know how much energy is put in the air, so it is difficult to calculate decceleration rate based on that.

The best approach, I deem, is the empirical formulae of L and D, which are based on the empirically obtained values of CL and CD. And then Newton.

Mass increases, Weight increases proportionally, AoA increases proportionally, CL increases proportionally, CD increases but less than proportionally to CL (at typicall AoAs), so D increases less than mass does. Newton's second and Galileo's equations then take us to the conclusion: The heavier a/c will use more time and distance to slow down.

[Bye]

hen I did my aero engineering degree I must have missed a lecture where they explained that momentum was a force.

and i missed the bit where i claimed that it was well.

But hey you can't let the facts get in the way of an attack now can you. :D

what i said and repeated was that a force is derived from the momentum to make the glider go up against gravity.

In this circumstance there are 4 forces acting on the glider not 3.

The momentum is not a force it creates a force on the mass to push it in a direction momentum has a quantity and a vector.

[DaveReidUK]

i assume you can prove this claim that i am not an engineer designing aero engines

For the avoidance of doubt, I'd like to make it clear that it wasn't me who made that claim about you.

But you certainly made this one, when addressing me:

So to say you clearly have a background in the subject when all you rely on is google is a bit rich and highly insulting to those of that rely do have a background in aeronautical engineering.

Remind me again, who exactly is being "highly insulting" ?

Pot ? Kettle ? Black ?

[HazelNuts39]

Bye,

I'm just quoting the facts as you wrote them in your posts. It is up to the reader to form his/her opinion about your sincerity in this debate.

If you're not playing the game of trolling, then take the sound advice offered to you in good faith in post #75.

Regards,
HN

[Owain Glyndwr]

Bye,

You have complained that nobody will explain the basic aerodynamics to you but you yourself gave a reference to a document that does just that in post #27. Is it unreasonable to expect that you have read this?

You are a design engineer and own a company that designs aero. engines so you ought to be able to put some numbers in the equations given in your reference, or employ someone who could do it for you, so if I give you some realistic values you can check what others are saying for yourself.

For an aircraft of the B737/A320 class:

Cdo (profile drag coefficient) 0.018
K (induced drag factor) 0.0425
Wing area 1300 sq.ft
Weights between 180,000 and 120,000 lb

Over to you

[Lyman]

Adding any energy to the system (thrust) was mooted when it was eliminated in the original question. The question had everything to do with reducing velocity, "slowing down". This involves dissipating energy, not accreting it.

I think calling momentum 'thrust' is not appropriate. Momentum is the result of thrust, not its own precursor. The identified heavy aircraft is unpowered by definition.

all respect.

[Owain Glyndwr]

Bye
Owain, sorry but someone posted that the link i posted contained references that were wrong so i expected them to be corrected before i could use them

I don't know what that means exactly, but if I knew that references in one of my posts were wrong I would feel duty bound to issue a correction. As it stands the reference you quoted DOES contain the explanation you are looking for and you can access it and put in the numbers I gave, so you can, if you wish, make an independent verification of what others have been saying here.

Once again, over to you.

[Bye]

HN39 said the material in the link had errors in it.

He didn't say what they were. and i couldn't see any errors but then i know nothing.

so i was hoping he would point out where the material was in error.

i have worked through with your numbers and thanks for them, and i am wiser for it thanks to you.

oh and forgot to state that FORCE = MOMENTUM / TIME

GB

Edit, having gone thru again, i still can't get why the 2 aircraft at different weights can't have the same L/D ratio. the graph shows they do at 226 kts, and my sums show they do as well with the data you gave.

I can get them to have the same L/d ratio at a certain V every time, so i would be grateful if you would explain again why its not possible and where the graphs are wrong.

GB

[HazelNuts39]

Dear mr Bye,

I'm not aware of anything in my writing that calls for the apology you ask for. I've pointed to an incongruence that I perceive between your claim of being an engineer and the understanding of elementary physics that you display in your posts. That leaves two possibilities:

Either you are an engineer and deliberately pretend to misunderstand the physical realities you write about, or -

You really don't understand physics and refuse to learn more about it, which in my mind is not compatible with having had an engineering education.

I do apologize for a mistaken remark that I made regarding the paper referenced in your post #27. That paper discusses speed stability at two arbitrily chosen speeds V1 and V2. When writing my post just before going to bed I mistakenly associated those with the V1 and V2 speeds that are used in the operation of civil transport airplanes. Thinking about it later I realized my error and deleted that remark first thing the next morning.

[DaveReidUK]

the rest of us can't hope to match your grasp of aeronautics

You're right, I did indeed say that. I've seen nothing in your subsequent posts to make me change my view.

and i missed the bit where i claimed that it [momentum] was [a force]

Happy to help you recall:

Your post #43:
in a glider the only thrust you have is in fact the momentum you have

Your post #59:
yes its called thrust because the momentum is PUSHING the glider upwards against gravity

I rest my case.

[Owain Glyndwr]

Bye

Because you specified that they were both flying at 400 km/hr (250 kts).

226 kts is the ONLY speed at which they can have the same L/D.

[DaveReidUK]

Yet you claim i said that momentum was a force, sorry i don't see it, i see the word thrust, but not the word force

OK, I understand now - you don't consider thrust a force.

So presumably the thrust that's produced by the engines you design isn't capable of making the aircraft move.

Words fail me. :ugh:

[Lyman]

hi bye,

I see no libel. I also see an ephemeral chance for truce.

IMO. Thrust is a force that is not contained in the object to which it is applied.

So, to apply the word Thrust to the system under discussion, we look for a separate object, divorced from the airframe, such that it is subject to, and gains from, the force contained in the system.

The Airmass is acted upon, and reacts to, the entry of the aircraft into its separate region. The aircraft imparts energy to the airmass, the airmass reacts, the airplane slows. Accelerating the airmass produces momentum therein, the aircraft accelerates negatively, etc.

As I say, ephemeral. But not wrong, and in the proper spirit, etc.

bye, bye

[yippy ki yay]

We're debating with a person who claims to be an engineer designing aero engines yet writes such nonsense

I read this in the context that an engineer is writing something that, to the author of the above quote, didn't make a lot of sense for someone who is suppose to be an engineer...not the way you have seen it which I assume is the author saying that because what you wrote didn't make sense to him you cannot be an engineer.

Much like someone saying to me that they can't believe I claim to be pilot if I did a landing like that - They're not claiming I'm lying about being a pilot, they're just claiming that as a pilot they can't believe I actually did a landing like that!

And off the record I nail every landing - on the record though... :eek:

[Sunamer]

Already have one engineering degree, and working toward computer science master's degree.

all this thread is .....mm....weird! I initially (silly me) thought that there is only momentum envolved. Then, ok, if you need to account for different attitude (=different AoA + different profile, obviously) for different weights you can do it too. But after you calculate initial momentum which doesn't change (I don't consider burning fuel here, since that wasn't the initial question and there is no point in making task over-complicated) it shouldn't be too hard to account for different AoA + profile drag too. induced drag will be almost the same, I assume.

I think the thread needs to be closed since there is no point, I am afraid, to continued!
One side tries to stick to physics, others try to develop their own...ehmm...understanding which is a bit strange to me! :(

All sloppiness that are presented here can't be a sign of good knowledge in physics in any way.
Everything that was said here about legal claims whatever it might be, simply beyond anything logical...don't want to comment on that.
:ooh:

[mm43]

I don't have an engineering degree, but have been interested in the points raised.

The OP's request related to two similar aircraft, one heavier than the other, and why the heavier aircraft took longer to slow down than the lighter one. Assuming that the initial speeds are the same, the mass of each aircraft along with the L/D ratio (associated with AoA) will become the variable. Discount the flight idle fuel usage and thrust which are equal for both aircraft, then we are essentially dealing with kinetic energy.

EK = ½mv²

The L/D ratio along with AOA will change in a similar (though different) fashion for both aircraft as they slow, so it can be assumed that the Mass and V² are the only real factors we are dealing with. SLOWING is the operate word, and it follows without too much argument that the aircraft with the higher mv² product will always take longer to slow than the lighter one.

My 3 year old grandson proved it to me when he set up two parallel lines of Thomas wooden railway and placed one engine and a carriage on one line, then one engine and a wagon loaded with sand on the other. He pushed them off together, and you know the answer - he observed it also.

[Sunamer]

yes. all that is true, but

the thing that needs to be accounted for also is:
unlike in the case with two different-weighted railroad engines, that have different mass but identical railways, identical speed and identical friction (drag is the same here)
two identical planes with different masses will have different AoA due to the fact that once needs more lift to keep AC at the same altitude if AC has more mass, than derivation of AoA = aerodynamic profile will be (probably) less efficient and this will create more profile drag (not induced, though). Since it creates more drag, it probably in some cases will be enough to counteract more momentum of heavier AC and thus the result will be the same = two aircraft will slowdown the same way.

everything depends upon particular coefficients and values. There is no general solution here to say that yes, it will slow down slower (or faster), I am assuming.

[DaveReidUK]

What my solicitor has said is that the intent is clear that his intention was to discredit my standing as a professional engineer on a public forum.

Out of interest, is your solicitor also recommending that you maintain your novel "thrust is momentum, it isn't a force" proposition that, frankly, wouldn't get you a pass in GCSE mechanics ?

[HazelNuts39]

My 3 year old grandson proved it to me when he set up two parallel lines of Thomas wooden railway and placed one engine and a carriage on one line, then one engine and a wagon loaded with sand on the other. He pushed them off together, and you know the answer - he observed it also. Let's study that in some detail. Presumably the speeds achieved by your 3 year old grandson were low enough for the aerodynamic drag to be insignificant. So the retarding force is the result of the friction between the wheels and the track, and the friction in the wheel bearings. Normally one would expect those frictions to be proportional to the loads on the wheels, i.e. the masses of the two trains. In that case the two trains would slow down at the same rate. If that didn't happen, there must be other differences besides the difference in mass. It would be interesting to extend the experiment to find out what these differences were. If you loaded both trains to have the same mass, would they decelerate at the same rate?

[DaveReidUK]

is your solicitor also recommending that you maintain your novel "thrust is momentum, it isn't a force" proposition that, frankly, wouldn't get you a pass in GCSE mechanics ?

No

Good - then you should take his advice.

Some might think that displaying a woeful grasp of basic mechanics would be enough, on its own, to discredit anyone's standing as a professional engineer, without needing any external assistance.

But I couldn't possibly comment on that ...

[john_tullamarine]

Folks,

Please be aware that the basic RoE in Tech Log include -

(a) posts and statements therein do not have to be correct but they do have to be reasonably polite and civil lest they be censored.

Note that this does not proscribe spirited and vigorous discussion including pointed jousting. However, the essential thing is to play the ball and not the player.

(b) one of Tech Log's underlying values is in education which is why we are happy to let discussions go around in circles for a while .. if they don't sort themselves out, folks tend to lose interest and such threads sink to the bottom of the abyss.

A good way to underpin one's technical knowledge is to be forced to explanation and justification and, in the process, those whose knowledge could be a little shakey may end up with a benefit from the discussion

(c) now, I know the identities of only a few folks on PPRuNe but I do have to say that we are VERY honoured to count amongst our number some very erudite, well qualified, and experienced folk in a range of technical and other disciplines

(d) I note that this discussion appears to be centring around engineering and aerodynamic matters. I would point out that (at least) one of the participants in the discussion is an aerodynamicist whose standing in the Industry is unquestioned. In my view, his counsel ought to be heeded in matters aerodynamic.


Unfortunately I hadn't been monitoring this thread until certain matters were drawn to my attention. However, I will do so from here on in. Hopefully, my infrequent forays into censorship will not be needed henceforth.

[HazelNuts39]

Hi John,

Thank you for your intervention. I wasn't prepared for the emotional response to my remark. I shouldn't have made it and regret that I did.

Regards,
HN

[john_tullamarine]

When I get to be perfect (SWMBO indicates that such an outcome is not likely in my lifetime) I might expect others to be so.

Occasionally we all are at risk of shooting from the hip .. fact of life, I guess.

[Sunamer]

A good way to underpin one's technical knowledge is to be forced to explanation and justification and, in the process, those whose knowledge could be a little shakey may end up with a benefit from the discussion

yes, but not in the case where person doesn't consider facts and vigorously defend his PoV. :ugh:
Going personal isn't a very good option though, but, honestly, I can see why some of respectable PPRuNers did just that. :\

Such level of misunderstanding of physics can be insulting by itself, and if a person having that "knowledge" claims to be a person who relies on that knowledge PROFESSIONALLY, than this makes it twice more insulting. :}

[Lyman]

Only ducks are safe. If one is a professional, and not a duck, one will be carrying about a load of water on one's back.....

[Bye]

Sunamer.

Such level of misunderstanding of physics.

i would be grateful if you would point out where i demonstrated this to you.

i have already explaned that i ( wrongly but innocently ) used the word thrust not to describe the thrust from an engine but to describe the effect of momentum in a glider or ski jumper on a ramp that will allow him to overcome gravity without lift for a period of time dependent on the amount of momentum acquired.

So i admit i used the word thrust ( and did so previously but it was ignored rather conveniently ) when i should of used another more appropriate word in this instance, i would like to see where else my Physics was so far off.

GB

[john_tullamarine]

A few more comments, if I may ..

yes, but not in the case where person doesn't consider facts and vigorously defend his PoV.

I have no problem with folks defending their position. If they are wrong, fine, then one hopes that the discussion can work through to the point where that is accepted and the learning process progresses.

On some occasions, brick walls are maintained and the thread is subject to general loss of interest and dies a natural death.

It doesn't necessarily follow that one who holds whatever qualification or accreditation is the font of all wisdom. Folks should be wary of falling into the trap of presuming such. I'm a chartered engineer, for instance, and know a little about a few things .. however some will attest to my lack of knowledge in a great many areas ... the secret is knowing one's areas of strength and weakness .. and acting accordingly.

Going personal isn't a very good option though

and, generally, is the main reason for the very occasional censoring of posts in this Forum.

Such level of misunderstanding of physics can be insulting by itself

such a comment is a tad precious, I would opine ? Then again, I suspect that you are speaking with a smile so all probably is OK

if a person having that "knowledge" claims to be a person who relies on that knowledge PROFESSIONALLY

in a nominally anonymous website, unless one actually knows the person behind the post, there is an inherent danger in presuming too much about the quality of a given post.

[DaveReidUK]

So i admit i used the word thrust ( and did so previously but it was ignored rather conveniently ) when i should of used another more appropriate word in this instance

That's gratifying to hear.

And, given that the offending posts have all been removed, we'll forget about the equally inappropriate term that you directed at those who had the temerity to point out that your usage was incorrect. :)

[mm43]

If you loaded both trains to have the same mass, would they decelerate at the same rate?As usual, you've raised a very valid point.;)

Well, I took both locos/engines and tried the experiment again with just them. One was Thomas and the other was Billy, with identical wheel base / bearings etc.., and yet Thomas came to a stop earlier. So, I got out some scales and weighed each loco; Thomas = 61 grams and Billy = 54 grams, and to equalize the mass I taped a NZ 10c coin (3.3g) + 20c coin (4.0g) to Billy.

To ensure that variable track friction wasn't a factor, I placed both locos on the kitchen bench top and pushed both in the same direction together. They both came to a stop with Thomas slightly ahead by 2mm after traveling about 80cm from the point where the propelling force ceased.

Now these bench top tests with kids toys are far from the ideal, yet they are capable of demonstrating the not so obvious.:uhoh: :ok:

[HazelNuts39]

Hi mm43,

Thank you for sharing your exciting experiment. Thomas, the heavier engine in the first test, stopped in the shorter distance?

Bye,

The problem lies not in the use of the word thrust. You are wrong when you equate momentum to a force. Momentum and force are like apples and oranges. An apple can equate to another apple, it cannot equate to an orange.

Force = rate of change of momentum (post #71).

You yourself wrote in post #90: Force = Momentum/Time

If that is correct, then Force = Momentum cannot be correct.

Let's examine the ski jumper. If we ignore the aerodynamic forces acting on him, then after leaving the ramp the only force he is subjected to is gravity. Because there is no force opposing gravity, his vertical motion slows down at the rate of 19.05 kts/second. If there were no force acting on him he would continue to move at constant speed in a constant direction.

[aerobat77]

somebody claims a question and you all start to fight ,throwing wild formulas, in the very urgent try to demostrate the own internet superiority without thinking about the question by itself.

we have to ask if the statement"why heavier aircaft take longer to slow down in the air" is in any case true at all .

as an extreme example lets take the space shuttle which is (was) at approach basicly an aerodynamic glider airplane. and for sure significantly heavier than a small one-seater glider.

did it take longer for the space shuttle to slow down in the air than for the small glider ?

cheers

[yippy ki yay]

From reading the original question I believe the OP, and all subsequent replies, were referring to the same aircraft.

[Microburst2002]

Gennlemen

One thing that amazes me about physics is that we don't even know what a force actually is... For us it jIs ust whatever makes a mass to accelerate or decelerate, whatever the hell it might be. There is something that makes that happen, and they called it Force.

But we don't really know what is the nature of forces, or their origin. Again our fate is to recur to the empirical method.

I think I demonstrated to you what is the effect of weight in drag for normal flight AoAs: less than proportional than the mass increase. It coincides with the intuitive notion of higher momentum or higher KE will leas to longer stopping distance. However, at high AoAs drag force will increase more than proportionally to mass, and the heavy airplane will actually slow down faster than the light one, which is against anyones' instinctive answer.

So Newtons second law is the right choice, after the above analisis.

a=F/m.

Sometimes F increases more than mass, sometimes it does increase less than mass. sometimes acceleration is slower, sometimes it is faster.

It all depends on the AoA.

Owain said above the two airplanes can fly at the same speed, 226 kt, with the same L/D, which is amazing to me, but since I tend to believe him, I thought a bit about it and I deem that the heavy at 226kt is above max L/D AoA while the lighter one, at that speed is below it, and they happen to be the same L/D. Is that right?

[Owain Glyndwr]

Microburst

I thought a bit about it and I deem that the heavy at 226kt is above max L/D AoA while the lighter one, at that speed is below it, and they happen to be the same L/D. Is that right? Yes, that's right. If you look at the graphs I posted (#48) you will see that at 226 kts the high weight aircraft is flying below minimum drag speed (AoA above that for max L/D) and the light weight above its minimum drag speed (AoA below that for max L/D). I guess that if one had the time and the inclination one might be able to derive some sort of formula to tell you what that speed might be for any given set of coefficients, but to be honest I can't be bothered ;)

Of course if one is thinking about how long it takes for aircraft with different weights to slow down then the drag/weight will vary with airspeed differently for each aircraft so the fact that at one particular speed the values of D/W i.e.1/(L/D) are the same is unimportant, particularly since that speed is going to be pretty close to the lower limit of authorised airspeeds (1.25~1.3 Vs?)

[HazelNuts39]

one might be able to derive some sort of formulaEquation (9) in the paper referenced in post#27 gives the minimum drag speed Vmd. It corresponds to a lift coefficient equal to √(Cd0/K).

If Vmd1 and Vmd2 are the minimum drag speeds at different weights, then the speed at which the airplane has the same L/D at both weights is given by:

V = √(Vmd1*Vmd2)

P.S.
A pretty useless formula, isn't it? It took me a whole evening to work it out. Just for the fun of doing it. But the result is a beauty, for those open to it.

[Owain Glyndwr]

Hi HN39

Equation (9) in the paper referenced in post#27. It corresponds to a lift coefficient equal to √(Cd0/K)

I read that as an equation to calculate minimum drag speed. Maybe the way I phrased things, but I meant an equation to tell you at what lift coefficient two aircraft of differing weight could have the same L/D. Or am I missing something?

[Capt Pit Bull]

Somewhat as a johny come lately, a few observations:

Quite a lot of dodgy Physics in this thread.

You can go at the OPs question using momentum if you like. You can also go at it using KE. Momentum is handy if you want to know how long a time a velocity change requires. KE is handy if you want to know how long a distance a velocity change requires.

(Generally, energy changes are easier to deal with because they are scalars.)

Bye, and gliding etc. Sorry Bye, but your Physics is full of holes.

In my experience people with a dodgy grasp of mechanics usually fail at Newton one. This is evidenced by trying to find some 'Thrust' to explain why the glider keeps moving. If you resolve forces on axes along/across the flight path then a component of weight is equal and opposite to drag, ergo there is no resultant force ergo the velocity is constant.

So, the gliders equivalent to thrust is a component of its weight, not it's momentum.

Pb

[Lyman]

Capt bull...

If frame of reference can be exchanged, how is thrust not acceleration due gravity, when descending? How is dissipation of momentum not due to the thrust provided by drag? If the calculus is interchangeable due the acceptance of any increase in acceleration as due to thrust, positive or negative, then any loss of velocity can be expressed in terms of only two forces, gravity and drag, both of which accelerate the airframe.

Owain. If the heavy a/c is at AoA producing maximum lift, is it not About to Stall? There can be only an instant of maximum lift without power?

How can one equation serve to express two variable results, when it cannot contain a non common variable?

Sorry, it's late.

[Owain Glyndwr]

Lyman

If the heavy a/c is at AoA producing maximum lift, is it not About to Stall? There can be only an instant of maximum lift without power? Any aircraft at the AoA for maximum lift is about to stall no matter what the weight. But who is talking about such conditions? Not I for sure! We have been discussing the point at which L/D is the same for two different weights - 226 kts in the example I gave. For those aircraft stall would have been at about 170 kts or 194 kts - close to the LH ends of the lines.

How can one equation serve to express two variable results, when it cannot contain a non common variable? Allow me to give a Lymanesque answer - How can a bus with #29 on the front go to a different destination from the one with #41?

[Its early] ;)

[BOAC]

Then again, Owain, is it not true that a #29, full of passengers, would take longer to slow down than an identical but empty bus #41?

[Owain Glyndwr]

Then again, Owain, is it not true that a #29, full of passengers, would take longer to slow down than an identical but empty bus #41?

Sure its true, which is a good safety feature because the empty #29 just behind will be able to stop in time if the first #29 does an accelerate/stop :ok:

[HazelNuts39]

... if the brakes are too weak to lock the wheels or to activate the ABS.

[Microburst2002]

I hope I manage to paste here this graph...

http://www.lets-go-fly.com/Lift%20and%20Drag%20curves%20for%20the%20Wing.pdf

On the slope of the CD curve is the answer of the effect of weight o a flying airplane slowing down rate.

Absolutely no momentum or Energy analysis is valid without taking this into account

edit, I dont know how to paste the graph alone without the text...

[Capt Pit Bull]

Lyman

If frame of reference can be exchanged, how is thrust not acceleration due gravity, when descending? How is dissipation of momentum not due to the thrust provided by drag? If the calculus is interchangeable due the acceptance of any increase in acceleration as due to thrust, positive or negative, then any loss of velocity can be expressed in terms of only two forces, gravity and drag, both of which accelerate the airframe.

Too many variables, too many questions to understand what you are getting at.

Thrust provided by drag? Conventionally we talk about thrust as provided by engines.

I was briefly visiting gliding because it got a brief mention in the thread, although the OP was talking about deceleration in level flight.

how is thrust not acceleration due gravity, when descending?

BZZZTTT! Dimensions clash!!!!

How is dissipation of momentum not due to the thrust provided by drag?

BZZZTTT! In level flight unbalanced drag will dissipate momentum, yes, but drag does not 'provide' thrust.

Notice I said

gliders equivalent to thrust is a component of its weight

It's equivalent in being (1) a force that is (2) equal and (3) opposite to the drag, thereby (4) keeping the aircraft at a constant speed. But is not actually thrust in terms of being a powerplant squirting air around.

(Personally, as an aside, I find gliding a lot easier to explain by forgetting about drag as a stand alone force. It's only a method for conveniently resolving the total aerodynamic reaction for times when we want to show 4 forces. For a steady glide it's actually a lot easier to say total reaction is equal and opposite weight. Therefore no unbalanced force therefore steady glide.

Energy burnt of as force x distance, power as force x TAS, therefore KE dissipated as drag x TAS. Easy, no vectors because Energy is a scalar.)

pb

[Lyman]

Captain Pit Bull

It's equivalent in being (1) a force that is (2) equal and (3) opposite to the drag, thereby (4) keeping the aircraft at a constant speed. But is not actually thrust in terms of being a powerplant squirting air around.

Agreed. However, relative to the airframe, there is no difference in drag v. Thrust, simply because no energy is being consumed in the form of fuel. With the standard four forces model, cruise flight is balanced among four assumed forces. When one of the required forces is diminished (or enhanced),
a change in E occurs. To slow, Thrust is removed, and the other three are left to establish some form of attitude that will result in arrival.

Simplistically, which of the two aircraft under discussion can remain aloft the longest, the heavier? The lighter? Because that introduces Galileo, no?

Thanks for your response, lyman

[Lonewolf_50]

What an interesting thread and discussion.

When I go back to the original post, I see a problem statement that asks us to consider deceleration between two aircraft of different mass. (I assume the frame of reference is Earth, where a difference in weight can be equated to a difference in mass). ;)

One is asked to explain the time it takes to change velocity from one value to another, with the only difference in initial condition being difference in mass. (Note that I had to make a number of assumptions to parse that problem statement).

I have enjoyed the seven pages of discussion into the esoterica of deceleration, rates, inertia, momentum, and enery in an aerodynamic context. Good stuff! :ok:

From the discussion, I find myself aligned with those people who approach the problem from the change in energy state. Were I asked that question with no help from 20 pound brains, that is the direction from which I would answer it.

Question asked:

Pardon me but..

1)Why do heavier aircrafts {sic} take longer to slow down in the air compared to lighter aircrafts? {sic}
This appears to me to be an example of begging the question, and indeed, assuming the answer to a question in the asking. :E

A more correctly phrased question would be:
Do heavier aircraft take longer to slow down in the air compared to lighter aircrafts? {sic} If so, why is that?

[QUOTE] 2)Why on high profile, it's recommended to increase speed? [QUOTE]
Is it? By whom?

Depending on how you are flying, you may wish to maintain approach speed and reduce power to increase rate of descent, or you may wish to increase drag by doing things like adjusting flaps, or dropping the landing gear to increase drag. (Or, if in a glider, descending in a slip to also increase drag ... )

Once again, I'd say we have an example of begging the question. :E

Provisional answer:

1) Given all other things being equal (which some of the above conversation suggests they might not be) and
assuming the "I need to slow down" action taken will be identical
in terms of what change to the aircraft is executed to change energy state (example, change a flap setting or reduce engine power an identical amount)
then the aircraft with the greater energy at the beginning would more likely take longer to slow down ... On The Assumption that they are both attempting to get to the same "V" when slowing down in the first place. (Can we assume that? )

I have been cheered to learn that "it's more complicated than that" during the discussion, because of aerodynamic unique considerations. Again, my thanks to the 20 pound brains. :ok:

I am not certain the person framing the question was clear on that or not.

2) The permutations of what to do with your energy state to change profile if you are high were not, in my training, generally to speed up to change profile errors, unless one were ALSO slow and while being high on profile.

The problem statement omits what actual initial conditions are, which leaves us guessing. I'd answer question number two differently:

If you are high on profile and slow, then speeding up is a part of returning to profile.
If not, then speeding up may not be the best way to resolve your being off of profile to begin with.

For example, if you have time to change configuration early, perhaps you dive, speed up, and then dirty up early so as not to get to close to the plane ahead of you ... there's a lot to this answer left open in the question.

To answer number two, on basic principles, I'd allude to "Power plus attitude (for a given cofiguration) equals performance" in powered flight.

This answer is not provided by an aero engineer, it was provided by a pilot. Sure, I have a degree in mechanical engineering, but I can't hang with the 20 pound brains in this thread. (You all know who you are! :ok: Goodonya, all of you!)

Talking about gliders seems a red herring in this discussion.

[Lightning Mate]

Is this thread serious?

:mad::ugh::ugh::ugh::ugh::ugh::ugh::ugh::ugh:

[Microburst2002]

The energy or, more practically, the power extracted from the fuel to keep the airplane flying level at a constant speed would give us some measure of the rate of kinetic energy decrease if the airplane maintained altitude after reducing thrust to nil.

Comparing the power delivered by the engine to the airplane at the reference speed with the power the airplane would have at another reference lower speed would give us the average kinetic energy loss between that speed range, for that weight. From that we could obtain deceleration rate. But i don't now how yet...

edit the rest of the post I have quarantined for analyisisssss

[BOAC]

Is this thread serious? - it would be if you were waiiting for the #41...............................

[Lyman]

Solo Lupus 50

Talking about gliders seems a red herring in this discussion.

From the original, the question has to do with accelerating a reduction in velocity. It is a Rate question, and demands a little latitude in parameters.

So my assumption was from the beginning that the engines on both slowing a/c were 'mass only', not thrust. We could get into "windmilling" v. Flight idle, but as we see with seven pages, most here are not satisfied unless the discussion can be made more complex, not less.

To some extent it is a "Straw Thread", a vehicle for trotting out some algebra in the interest of appearing cerebrally massive.

[Owain Glyndwr]

it would be if you were waiting for the #41.....

In the rain ...

[Lightning Mate]

- it would be if you were waiiting for the #41...............................

Time to go where people know what they are talking about then.

Why is this "professional pilots forum" full of amateur idiots please?

[BOAC]

Why is this "professional pilots forum" full of amateur idiots please - I trust you are not referring to me, Carruthers - I have always prided myself on my professionalism:)

[Owain Glyndwr]

Time to go where people know what they are talking about then.

Why is this "professional pilots forum" full of amateur idiots please?

Occasionally one needs a little light relief from all the idiocy, both amateur and professional, that has plagued this thread.

There are non-piloting professionals writing here as guests, and believe it or not, some of them know what they are talking about. :8

[Lonewolf_50]

Lightning Mate: maybe lightening up would be in order, eh mate?

Here, have a pint. :ok:

Lyman:
From the original, the question has to do with accelerating a reduction in velocity.
From the original, the question has to do with deceleration. What you typed there is ... :confused:
It is a Rate question, and demands a little latitude in parameters.
Depending upon if you are solving it in the classroom, or in the cockpit. ;)
So my assumption was from the beginning that the engines on both slowing a/c were 'mass only', not thrust.
Engines are generally used on an aircraft to provide sufficient thrust to fly.
Hanging an engine on an airframe for the purpose of mass would be counter productive, would it not? :p Hanging engines on a glider seems to defeat the purpose of building a glider, does it not?
We could get into "windmilling" v. Flight idle, but as we see with seven pages, most here are not satisfied unless the discussion can be made more complex, not less.
Various means of increasing drag are yours to employ if you know your aircraft well. They become tools for deceleration, which as I noted is the core of the question in post number 1.
To some extent it is a "Straw Thread", a vehicle for trotting out some algebra in the interest of appearing cerebrally massive.

The Original Post, as I noted, begged the questions. :cool:

From there to now, it's been fun and games.
I, for one, have been most entertained.

Would you care for a pint as well? :ok:

[Lyman]

Wuff...

Slowing is acceleration with a minus, no? Nomenclature and concept.

I'll have a pint, thank you Sir! Sarseparilla....

Now slowing in a helicopter.....requires thrust. No rest for the wicket.

[bubbers44]

Maybe what seems like weight affecting descent angle has more to do with the fact we descend at about 320 knots and we are closer to our best L/D at that higher speed. If we are light our L/D speed is much lower. Once level I think the same aerodynamics apply. Being heavy always requires more miles because the wing is more efficient at the higher speed. It always worked for me.

[john_tullamarine]

Why is this "professional pilots forum" full of amateur idiots please?

One of the thread management problems we are faced with regularly relates to the basic question of what the forum is about.

(a) if commentary is restricted to a specialist group, fine, but the whole thing becomes elitist and, very possibly, we miss out on a bunch of interesting things

(b) if commentary is restricted to folks of competence above a given level, (a) probably still applies and we lose the potential of whatever inherent training/educational value may vest in a given thread.

Senior management edicts indicate that we should transfer appropriate threads to the Questions forum. I interpret that to mean threads which rapidly lose their way/provide little interest for the readership or for which requests are made from the floor for such to occur.

So far I have not had any PMs etc so requesting.

[aerobat77]

One of the thread management problems we are faced with regularly relates to the basic question of what the forum is about.


john, do you have an answer for this very interesting question ?

[Microburst2002]

Bubbers you have it right

Trying to answer the question with Energy or Momentum alone, without considering the very complex subject of aerodynamics and Drag, is a very difficult thing, if possible at all.

The answer depends on about aerodynamic efficiency, and not only on the effect of having more or less mass. L/D ratio is a measure of the aerodynamic efficiency of the airplane. Similar as the efficiency of propeller blades.

An airplane flying at a less efficient AoA is wasting a larger portion of energy in the air than an airplane flying at a more efficient one.

In normal conditions, at the starting point of the experiment, the Heavy has a higher state of energy than the Light one. Due to its mass, exclusively. After we magically remove the engines and thrust disappears, deceleration starts immediately, if both airplanes keep altitude. The heavy one has more mass and , on top of that, is aerodynamically more efficient. It will then slow down more slowly. In the back side of the curve there might be cases where it would be the other way round, though.

But if we talk just about "normal" conditions, the answer is that: increased mass plus better efficiency give a slower deceleration.

Lyman

I like this thread. If you don't, why do you lose time posting here? I don't like so many other threads that I deem quite silly and I just stop reading them, like I stop reading a bad novel.

[HazelNuts39]

The original post just says "in the air". That isn't necessarily cruise, what about other flight conditions such as holding or approach? What speeds are you at when holding?

[john_tullamarine]

john, do you have an answer for this very interesting question ?

.. hard to say.

I guess we all have a slightly differing point of view.

As for me, I have naught to prove so, consistent with what those higher up the totem pole dictate, I prefer to let a thread run its course providing that folks are reasonably polite about things and play the ball rather than the player ...

[aerobat77]

I prefer to let a thread run its course providing that folks are reasonably polite about things and play the ball rather than the player ...

i think thats the obvious job of every moderator in an open forum . but what is this forum for?

imo there is not a problem when a variety of people with a variety of personal knowledge takes place in a discussion. its finally an advantage.

the problem is imo that the very most people here , especially at some subforums , claim to be some what they never are and never were. in most cases the "dreamer" is obviously a pilot with xxxxx hours on an widebody or a high degree engineer.

you will find tons of threads where "fake" people try to sell internet knowledge and manual readings compared with computer based flight simulation as actual real facts and the debate starts until the general loss in interest ends it - until a new debate starts.

this turns nearly every thread into stupidity and the " fights" come up automatically.

the rare people who really work on a flightdeck have so no use of a forum what originally might be intended for them .

today its more and more a platform for dreamcatchers . and honestly- in nowadays i do not see many reasons for what this profession is a "dream" where the dreamers catch on it.

nevertheless -i personally would strongly vote for a subforum where you have to make a proof who you are to post ( but not sure how it should be done via internet and how anonymity is given) or a profession identification in your signature after you prooved it -with postings free for everybody.

cheers !

[Mike X]

SLF here, whom has taken years out of his life and understands the theory.

I live under the approach path in Cape Town, South Africa. Now ,being season,
Emirates & Qatar fly in a 777-300 every day. On an always straight-in approach, the engines are spooling up and down. Only the the 777s. Have A340s, A330s, 744s and the local 737s, but why only the 777s ?

[john_tullamarine]

but what is this forum for?

In a nutshell, with respect to your concerns ..

(a) inappropriate questions are moved automatically to other forums - this happens with reasonable regularity and probably is unnoticed by the majority

(b) threads which deteriorate we endeavour to get back on the rails. If that is not successful, either the thread is closed, moved or dies a rapid natural death. Occasionally, we sin bin an individual poster in the worst cases.

(c) if I get several reasonable PMs from regular and reputable posters asking for a thread to be moved, generally I go along with such requests unless there is some cogent validity in retaining the thread here

(d) the idea of restricted sub-forums comes up from time to time but has never made the grade with the folk higher up the totem pole

(e) at the end of the day, whatever we do and however we moderate, we can't please all the people all of the time ...

[aerobat77]

(d) the idea of restricted sub-forums comes up from time to time but has never made the grade with the folk higher up the totem pole

(e) at the end of the day, whatever we do and however we moderate, we can't please all the people all of the time ...

the idea of a restricted forum will surely not make the grade of the vast majority because when any serious proof of what is claimed would be introcuded the very most would lost their playground in internet dreamcatching here.

i agree that the administrators and moderators cannot satisfy everybody and will have to follow the masses for several reasons.

on the other hand these masses turn behind the anonimity of the web and free fairy taling an aviation forum in a strange direction for the last users who would like to discuss practical problems, worries , questions and suggestions of their everyday business .

nevertheless, whatever, i like pprune and enjoy the reading !

cheers