Kinetic energy calculation
 

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

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

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!

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

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

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)

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)

The government health warning made my scribbles illegible!

And yes, Zoyberg; your figures seem about right.

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

Kinetic energy calculation
 

Nerds.......

Nerds and Proud of It!!!
 

No Nerds, No Birds

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 :)

lomapaseoDo 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?

Ft-pounds is all I understand.

yep, y'all are
 

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?

i'm goin metric inch by inch

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

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

Frames of Reference
 

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

@ Bye and Hydromet

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