Kinetic energy calculation

[Zoyberg]

Bookworm (who I can't get hold of) advised that the energy of a 50 tonne aircraft travelling at 140 knots (70 metres/second) equated to 125 M J

I am interested to know what the percenatge kinetic energy difference is between a
51 Tonne aircraft landing at 119 knots (fullish A319)..... and a 60 Tonne aircraft landing at 142 knots (fullish B737-800).

I know energy relates to the square of the speed but don't know how to factor in weight.

Thanks in advance

[henra]

I know energy relates to the square of the speed but don't know how to factor in weight.

Thanks in advance

Linear.
Kinetic energy = 1/2 m * v^2

[eckhard]

As KE = 1/2m x vsquared, take half of the weight (mass) for each aircraft and multiply by the square of the speed.

Make sure to use SI units: kgs for mass and metres per second for speed.

Good luck!

[bcgallacher]

This info should be in the Maint Manual - used for calculating energy of aborted t/o to determine what action to take regarding brakes.

[eckhard]

Quick calc on back of fag packet (not mine) reveals that the 737 lands with approx 67% more kinetic energy than the 319.

[hvogt]

Not the technical type myself, but I seem to remember kinetic energy is half the mass times the square of the speed.

In your example, the airbus would have a kinetic energy of

E_kin_Airbus = 0.5 * 51 t * 119 kn ≈ 0.5 * 51000 kg * (61 m/s)^2 ≈ 96 MJ

The Boeing's kinetic energy is

E_kin_Boeing = 0.5 * 60 t * 142 kn ≈ 0.5 * 60000 kg * (73 m/s)^2 ≈ 160 MJ

By my reckoning, the Boeing's kinetic energy is 67 percent higher than the Airbus':

E_kin_Boeing / E_kin_Airbus = 160 MJ / 96 MJ ≈ 1.666

(Glad you just corrected your post, eckhard)

[Zoyberg]

That was quick......thanks guys.... so have I got this right

My 51 T @119 knots has only 59.7% of the energy of the 737-800 weighing 60 T @142 knots?

(95571 MJ vs 160094MJ)

[eckhard]

(Glad you just corrected your post, eckhard)

The government health warning made my scribbles illegible!

And yes, Zoyberg; your figures seem about right.

[Bellerophon]

Zoyberg

... Bookworm (who I can't get hold of) advised that the energy of a 50 tonne aircraft travelling at 140 knots (70 metres/second) equated to 125 M ...



A 50 tonne aircraft travelling at 140 kts has a kinetic energy of: 129.68 MJ


A 50 tonne aircraft travelling at 70m/s has a kinetic energy of: 122.50 MJ


A 51 tonne aircraft travelling at 119 kts has a kinetic energy of: 95.57 MJ


A 60 tonne aircraft travelling at 142 kts has a kinetic energy of: 160.09 MJ

The answer to your question is that the B737 has almost 68% more kinetic energy at touchdown than the A319, or, alternatively, that the A319 has roughly 40% less kinetic energy at touchdown than the B737.


Kinetic Energy = Half the mass multiplied by the speed squared.

The formula: KE = ½ x M x V2

Use kilograms (as units of mass) and metres per second (as units of speed) and your answer will then be in Joules, divide by one million to get Mega Joules.


Best Regards

Bellerophon

[VH-UFO]

Nerds.......

[FCeng84]

No Nerds, No Birds

[lomapaseo]

If its a US registered aircraft use American units like pounds, and feet not this pussy stuff like meters and knots

If its German convert it to ergs :)

[5LY]

lomapaseo (http://www.pprune.org/members/48942-lomapaseo)Do you realize that you're the only ones left? You promised to change 40 some years ago and pussied out. Those of us from the cold bit north of you of a certain age speak both. Does that make us by-metric?

[fantom]

Ft-pounds is all I understand.

[kenneth house]

definitely bi-metric and stuff, momma and them even said so...

Oh yeah what is the formula for that kinetical energy, could somebody post that again?

[Bye]

i'm goin metric inch by inch

[Hydromet]

You should go metric. It's a dozen times easier.

[OATNetjets]

And do you all agree we are talking groundspeed here? I stand to be corrected...

[WeekendFlyer]

Absolutely. To calculate velocity or any of its related quantities (e.g. Kinetic Energy) you need to know the frame of reference you are using. For an aircraft the KE should be calculated with respect to the Earth's frame of reference, thus GROUNDSPEED is the correct speed to use.

Think about it: if you land in zero headwind you will need to use a longer runway (or brake harder) than if you landed in a strong headwind. This is also why landing in a tailwind is a bad idea - the KE you have to dissipate on landing increases in line with the SQUARE of the tailwind, meaning you need a much longer runway or else you will trash the brakes (or both)!

Lastly, with regards to the earlier "Nerds" comment: Engineers are good at this stuff, and some far more complex stuff too, which is why we can design, build and test some remarkable aircraft. If we didn't do this, pilots would (a) not have a job and (b) would not be able to fly around looking and feeling cool and (in some cases) superior to the rest of humanity :)

So, credit where credit is due please! :8

[kenneth house]

@ Bye and Hydromet

You guys crack me up--i spit coffee all over the keyboard...

i'm goin metric inch by inch

You should go metric. It's a dozen times easier.

[awblain]

You land at an airspeed V. If you have a tailwind deltaV, the energy you need to lose to stop depends on (V+deltaV)-squared. The fractional change in energy to lose scales approximately as (2 x deltaV / V), so it's not quite as nasty as a deltaV-squared dependence.

[Chu Chu]

Good thing, too. Otherwise, you'd have four times as much kinetic energy with a two knot tail wind as you would with a one knot wind!

[boguing]

A resident of that land to the west of the Atlantic told me that the 'old 3 4 5 right angled triangle setting out trick' didn't work in SI units. I assured him that it would, and that setting it out with Cornflake packets would work too.

If he were just a little brighter he'd have told me that Kellogg's is Murican.

[Machinbird]

Anyone care to hazard a guess as to how much energy the arresting gear on an aircraft carrier has to absorb when arresting a 40,000# F-4J flying down a 4 degree glideslope relative to the ship, flying at 143 knots with 27 knots of wind over the deck. (Standard no flare carrier landing) :8
We will assume that the ship is in a level attitude (which is not always true).

[Wizofoz]

It would seem to me that the F-4s landing gear would be absorbing any VERTICAL accelerations, leaving the cable to simple arrest the minor matter of the Aircrafts horizontal KE.

So around 65 Million Joules.

The POWER that represents? I'm guessing about three seconds to arrest the aircraft?

so around 21.5 million Watts=28 800 Horse power.

[boguing]

40,000 whats? Assume lbs, in the light of my previous? And, is the ship being sensible and steering into the wind? What is the tide speed and direction?

I'll take a shot tomorrow sometime. Std red wine excuse.

Wizofoz, don't forget that the PE absorbed on landing by suspension is restored to its' owner on rebound(s) - less any heat. E is very squirmy.

[roulishollandais]

Problem :
Energy the arresting gear has to absorb when arresting a 40,000# F-4J flying down a 4 degree glideslope relative to the ship, flying at 143 knots with 27 knots of wind over the deck. (Standard no flarecarrierlanding) We will assume that the ship is in a level attitude.

Weight M = 40 000 LBS = 40 000 x. 45359237 kg = 18 144 kg

Horizontal component of airspeed = 143 Kts x cos4° = 142.65166 # 143 Kts

EDIT PLEASE SEE MY POST #32
Horizontal component of groundspeed = 143 - 27 = 116 Kts = 59.5 m/sec to be absorbed by the arresting gear

Vertical component of groundspeed = 143 x sin4° = 9.97518 Kts # 10 Kts = 5.2 m/sec to be absorbed by the landing gear

Cinetic energy to be absorbed by the arresting gear
= 1/2 x 18144 x 59.5 x 59.5 joules
= 32 112 668 joules = 33 megajoules :)


Another ?;)

[Machinbird]

Just a thought.
The 4 degree glideslope is relative to the ship and is determined by the Fresnel lens landing system. However, if the ship is making its own wind, the Fresnel lens system is moving through space also.

I think I had better add the constraint that the ship is making its own wind of 27 knots.
That will tweak the decimal points a bit.:)

[roulishollandais]

EDIT :wrong slope correction, see my post #32: the following reckoning is wrong
The first version was correct ! ( - Hmmm so the ground-glide has that angle Å | tg Å = tg4°× 115.7/143 Å = 3° 14' ??)


:mad: I tried to imagine a very little bit changing the speed of the Ship to absorbe a part of the energy but leaved that idea, thought that I would have to absorbe the thrust at the moment of the trap, and :{ neglected !

Not only the centerline is going away, but the slope too !
Hmmm so the ground-glide has that angle Å | tg Å = tg4°× 143 / 115.7
Å = 4° 56 ' ??

[roulishollandais]

Which is the length of the wires before and after the Bird's stop ?

[henra]

Not only the centerline is going away, but the slope too !
Hmmm so the ground-glide has that angle Å | tg Å = tg4°× 143 / 115.7
Å = 4° 56 ' ??

As the lens (to which the 4° are held) is moving and therefore the 4° would be against the slower moving deck, I would say it is the inverse.
Effective glide angle would be 3° 14,4' over ground.

[roulishollandais]

:mad::mad:
before I prepared to go to church I did a wrong last minute editing... reversing the slope correction ! Agreed with my first version the piloted slope is 3°14' !
(03:46 - Hmmm so the ground-glide has that angle Å | tg Å = tg4°× 115.7/143 Å = 3° 14' ??)

During the time I was at church I did not pray a lot !:oh: I figured the Bird in the arresting gear with Ship doing its own Wind...:\ finishing with a water-speed of :E 27 kts, and discovered the slope mistake too..:E:{

The second problem I jumped over ,was I did not imagine origine of that Aircraft Carrier's "wind". So I have to edit after Machinbird's precision that the wind was the Ship's own wind / speed .

Consequently the final Bird's air-speed is 27 kt and not 0kt.

So the kinetic energy to dissipate is 1/2 M V1² - 1/2 M V2²
(V1 = 143 kts, V2 = 27kts)
= 1/2 x 18144 x (143² - 27²) x( 1852 / 3600)² joules
= 47 346 357 joules :{

"Unverified reckoning has probability near of 1 to be wrong" (Kaufmann) so ...you shall verify ! Thank you:)

[awblain]

You don't get vertical and horizontal kinetic energy, you just get kinetic energy. The same formula applies to landing on a carrier deck or a concrete runway, only the braking technique differs.

Do it all in the frame of the deck/runway, since that's where you need to end up with zero kinetic energy, unless you want a trip to the hospital.

The energy to lose in the wheel or cable braking is
0.5 x mass x (airspeed - headwind)-squared.
This assumes the headwind is in the same direction as the airspeed vector.
If not, you need to replace (airspeed-headwind) with the modulus of the vector sum of the airspeed, and the windspeed measured in the frame of the deck/runway.

If headwind=airspeed, you just latch the Harrier down without any wires, If headwind=-airspeed, you have four times the energy to lose.

Note that is not the same as Roulishollandais's
0.5 x mass x [(airspeed-squared) - (headwind-squared)].
That would give you the same braking energy advantage for a tailwind as a headwind.

[roulishollandais]

The energy to lose in the wheel or cable braking is 0.5 x mass x (airspeed - headwind)-squared. This assumes the headwind is in the same direction as the airspeed vector. If not, you need to replace (airspeed-headwind) with the modulus of the vector sum of the airspeed, and the windspeed measured in the frame of the deck/runway
So the result is 33 Megajoules from my post 27.
I changed it, thinking that energy is not a vector but a number, effectively I disliked my decomposition in horizontal and vertical. But assuming headwind is parallel to airspeed vector looks like that too : airspeed is descending the slope, headwind is horizontal !
I agree with your argument headwind/downwind in my post 32...??
Method seems still unclear with some contradictions..

Am I right with the slope?

[awblain]

For an M-tonne aircraft landing at a groundspeed of V kts (~V/2 m/s), the kinetic energy

T ~ 125.M.[V-squared] (in joules).

If V=143 kts and M=18, I reckon that T~47MJ.

Added: That's 47MJ with no headwind, and thus a deckspeed of 143kt. If the ship steams at 27kt, then the deckspeed is V=116 kts, and then T~31MJ.
The ratio is (116/143)-squared ~ 0.66.

If there's a small descent angle theta, the groundspeed is less than the total speed, and so you could perhaps justify dividing V by a factor of cos(theta).
However, for 3 degrees, cos(theta) is only about 1-0.5x(1/20-squared) or ~0.9986, so it has the same effect as adding 30kg to the landing weight, or increasing the landing speed by 0.1kt at 143kt.

If you treat the wind and airspeed as vectors, then any three-dimensional relationship between them is accounted for automatically.

[roulishollandais]

Re: 27

For an M-tonne aircraft landing at a groundspeed of V kts (~V/2 m/s), the kinetic energy

T ~ 125.M.[V-squared](in joules).

If V=143 kts and M=18, I reckon that T~47MJ.
47 or 33? (Your posts #35 #33) That is the question.
Do it all in the frame of the deck/runway, since that's where you need to end up with zero kinetic energy, unless you want a trip to the hospital.
We don’t reckon absolute kinetic energy ( we are still flying with the hospital and solar system toward Vega at 20 km/sec and kinetic energy is not zero :p) , kinetic energy changing in an other form of energy. So we have to evaluate ke at time 1 and time 2, (in any reference frame, and do the difference between ke1 and ke2.

With the Ship doing its 27 kts own wind, the Ship the arresting gear system and the traped F-4 have a GPS ref speed of 27 kts.

But the plane is stopped on the deck when the airspeed indicator shows still 27 kts should the wind be wind or Ship's speed

With a speed of zero the plane would be in the rounddown.Could one of our Aircraft Carrier Pilot confirm that thought (or deny, I am ready to accept that I should be wrong).

When the Ship does its own wind we have a balistic problem to land added to the classical relative wind problem, aswell as a parachutist landing on a truck for a movie or for fun.





And when the Ship is doing its own wind

[Wizofoz]

And do you all agree we are talking groundspeed here? I stand to be corrected...

It depends what you are interacting with.

An aircaft in flight only interacts with the air around it- so the only meaningful velocity and thus KE is relative to the air.

For an Aircraft landing on an Aircraft Carrier, the KE needed to be absorbed will be that RELATIVE to the Aircraft carrier.

Speed relative to the ground has no special significance unless one is talking about contact WITH the ground, such as landing on a runway.

KE varies as the square of velocity, but velocity is RELATIVE- the same object has different measures of KE depending on what you are measuring Velocity against.

Misunderstanding this leads to such things as the "Great Downwind Turn" myth.

[awblain]

The energy is frame dependent, and the difference in energy is frame dependent. In changing from speed V1 to V2, the energy change in two frames, one moving at deltaV to the other, with all velocities in a straight line are

V1-squared - V2-squared

and

(V1-deltaV)-squared - (V2-deltaV)-squared.

These differ by 2.(V2-V1).deltaV

As Wizofoz says, the only frame that matters for stopping is the one in which the ambulance is parked. The only one that matters for flying along is the one in which the wind is stationary.

[roulishollandais]

hospital law ? ambulance theorem ? :ouch:

[Wizofoz]

As Wizofoz says, the only frame that matters for stopping is the one in which the ambulance is parked. The only one that matters for flying along is the one in which the wind is stationary.

Here's a Guy that clearly doesn't get that-

Home (http://www.dynamic-soaring-for-birds.co.uk/)

The more informed among you might care to email him and explain his misapprehensions.

Be warned, I've tried and failed...

[Natstrackalpha]

If its a US registered aircraft use American units like pounds, and feet not this pussy stuff like meters and knots


Knots . . ?

Anyway,

The F4 landing on the carrier is F=MA as the F4 is in fact Aing its F-ing M all the way down the . . deck, forcing the wire to a force greater than the original hook up force.

So there is a secondary accellerating force - whilst the Massive F4 is being ground to a halt by the poor old wire up to and until the Accel Force is dissipated. see that? All due, as you all know, to the fact that you have to apply full chat, in case the `lil ol wire gives or in case you have not hooked up - on touchdown

Sorry to sound so technical - do I get a degree now?:}

[awblain]

Right, Natstrackalpha,

The wire not only has to absorb the kinetic energy, but also has to do work against the running engines.

The aircraft moves ~100m to stop (?), and has ~10 tons (?) of thrust, so the work done is ~10MJ: a modest, but not insignificant, correction.

You might not get a degree, but that's definitely worth some coursework credit.

[roulishollandais]

Since Machinbird's reckoning exercize proposition (KE reckoning when the F-4 was catching the wire) which seemed easy (I used the easiest frame : the Ship), the question of absorbing the Bird's energy came on the table.

I discovered that we have to use a frame independant from the Ship to "stop" the Bird at the same speed that the Ship. But since you are using that frame , new factors (oscillation, first shock, thrust, geometry of wires, position of the Bird, Ship's speed modification, natural wind, crosswind, etc.) appeared of significant possible Megajoules.

I searched too if the frames were galilean, it seems they are.

Which additional terms would help to coherence of different frames and methods?

@ Machinbird and your Navy's apilots Friends
Which shocks did you experience with Ship doing its own wind ? Is it the same - or not - than natural wind? Reckoning and theories must correspond to experience.

[awblain]

The frames are Galilean: at 200kt, relativistic corrections are at the one-in-a-trillion level.

The only frames that matter are i) that of the deck/concrete on which you stop, and ii) that of the moving air in which you are flying. The ship should be steaming relative to the wind so that its angled deck has a direct headwind of at least 30kt.

You can pick any other frame you'd like too; but you don't need to know what the energies are in any of those.

[Natstrackalpha]

You might not get a degree, but that's definitely worth some coursework credit.
http://images.ibsrv.net/ibsrv/res/src:www.pprune.org/get/images/statusicon/user_online.gif

Excellent!

~10 tons (?) of thrust

The Panavia Tornado (as flown by us Teabags, yet not renowned for carrier deck landings) has 15,000LBS static thrust and 18,000LBS with afterburner.

Stored energy, how to control it - ans= fly the aircraft. click







see? easy!

[awblain]

Do I still get some credit for a reasonable guess?

Looking it up, it seems that a Phantom has around 2x12,000lb dry thrust, which I make a total of about 11 tonnes. It also seems that the 100m distance to stop is within about 10%, perhaps a little generous, so 10MJ to run the wire out against the engines is I think within 10% of the total. If it's in afterburner, about 50% more.

That makes sense, given the time it would take to reach 150kts (~75m/s) from a standing start should be about 8s, and it stops in ~2s, so the kinetic energy is about four times the work done by the engines during the stop.

[VolLibre]

Any hang glider pilot who has landed a modern glider in a tailwind, as I have, will have had the physics of it rubbed in quite effectively, usually via the face...

[awblain]

I don't see the energy involved in landing a large aircraft being substantially different from landing a fast aircraft, other than having relatively less specific thrust, and this needing longer to add enough energy to climb away instead.

"Stabilized" to my mind means having steadily declining potential energy and constant kinetic energy: airspeed is steady in gust-free air; rate of descent is steady. The kinetic energy you carry to the ground is then the energy that goes into the air behind the spoilers and alongside the reverse-thrusting engines, and into the brakes. If this "stable" condition isn't met by 1000-2000 feet above the ground, leaving a minimum amount of things to do and avoiding excessive workload near the ground, then you go around and try again.

I'm not sure why your heavy aircraft leaves its steady progress to the ground? Is it a change in wind, or to the degree of a microburst? A change in thrust or drag, or setting up a dutch roll? If none of these things are a big change, and can be quickly countered by the controls and throttles, then the energy is about the same, you're still "stable" in terms of energy, and you should still be safe to land. If you choose to add energy and go around, you need add only a little - just enough per second to overcome the sink rate.

Even as the wheels touch, you're still flying. A fraction of 1% of the energy of the aircraft is sapped away to spin the wheels up, leading to a change by a half of that amount in groundspeed, but you're still almost flying, and still have the whole length of the runway to add power and go around. At that point, if you're happy to decide to brake instead, all the kinetic energy is safely dumped into the wheels and the airflow behind, and you can drive off safely.