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.

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!

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.

Boffin exercise.
 

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).

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.

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.

F-4J
 

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 ?;)

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.:)

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 ' ??

reckoning the wires strength
 

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

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.

: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:)

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.

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?

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.

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.

47 or 33? (Your posts #35 #33) That is the question.
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

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.

Energy and frames
 

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.

hospital law ? ambulance theorem ? :ouch:

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

Home

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

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