Kinetic energy calculation
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From: Next door
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
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
Last edited by Zoyberg; 30th May 2013 at 11:11.
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From: Hamburg
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)
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)
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From: UK
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 ...
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
... 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
Last edited by Bellerophon; 30th May 2013 at 12:39.
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From: canada
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?
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From: huntsvegas
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?
Oh yeah what is the formula for that kinetical energy, could somebody post that again?
Last edited by kenneth house; 30th May 2013 at 21:00.
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From: Hampshire
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!
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!