Finally !!! ASK QUESTIONS, DEMAND ANSWERS.

Bitstream · Apr 5, 2006

Quote from kjkent1:

Yep. In the words of Newton: "F=ma"; force = mass times acceleration. A commercial airliner that hits the ground head on at 550 miles per hour results in all of that velocity being instantaneously transferred into acceleration. The max weight of a 757 on takeoff is about 240,000 lbs., So, allowing for some jet fuel burn, if the plane hit the ground and weight only 220,000 lbs., then the resulting explosive force released on impact is about 60,000 tons, or 6 kt., which is about three times larger than the smallest nuclear device in the U.S. military arsenal, i.e., a 2kt tactical anti-tank round, originally designed to stop a large scale tank assault into Western Europe by the USSR.

The fact that there was nothing left of three of the 9/11 flights is simply the result of their head on collisions with a really big object -- the Earth in one case, the WTC towers in the other two.

As for the 4th 9/11 aircraft, well, the Pentagon Building obviously has quite a bit more reinforcement than anyone would have expected. So despite its mundane appearance, it too is a "really big object," and it stopped the attacking aircraft cold, resulting in the plane's almost complete and instantaneous annihilation.

ja, and almost all da bodies were recovered and identified in zero time.

JAGUARBONE · Apr 5, 2006

So it is possible for two 6 TON engines to VAPORIZE in the Pentagon fire?

Quote from esuperbo:

The layman has very little idea of the the incredible energy involved in a typical airline crash... at many incidents you're lucky to find something more than foot or two square.

Bitstream · Apr 5, 2006

Quote from JAGUARBONE:

So it is possible for two 6 TON engines to VAPORIZE in the Pentagon fire?

Quote from esuperbo:

The layman has very little idea of the the incredible energy involved in a typical airline crash... at many incidents you're lucky to find something more than foot or two square.

More...

and for da plane not only to survive da impact but be able to make it trough da other side of da penta wall and leave a hole there also..[missiles do that, as far as i know, not planes].

but of course da engine, bang...wooooosh...pile of sand.

EqtTrdr · Apr 5, 2006

Quote from Bitstream:

and for da plane not only to survive da impact but be able to make it trough da other side of da penta wall and leave a hole there also..[missiles do that, as far as i know, not planes].

but of course da engine, bang...wooooosh...pile of sand.

yes...lol

the engines would vanish.. poof.. pixie dust

but the nose of the aircraft would continue on to make another huge hole in the next layer of the building..

0h man... are we getting reamed or what?

kjkent1 · Apr 5, 2006

Quote from esuperbo:

Pabst, I'd have to agree with you. My father ran a (nameless) international airport for a number of years, when I was a kid I got to hang out at a ton of airshows at the VIP booth. Needless to say this led to a huge interest in aviation - but I still think you need to get out more if you're read every NTSB report, lol!. The layman has very little idea of the the incredible energy involved in a typical airline crash... at many incidents you're lucky to find something more than foot or two square.

kjkent, you're doing a good job of sounding intelligent but (forgive me) you don't know what the hell you're talking about.

1. Mass multiplied by velocity is momentum, not force. You stated force is mass multiplied by acceleration, then blew the calcs.
2. Then somehow this force reverts units back into weight, and then this weight is somehow now a measure of energy! (kt is measuring TNT energy equivalents, not any real unit)

In short, you're on the right track, but your math sucks.

First, one Kiloton of TNT is 10^9 thermal calories (not diet). So it is a real unit of measurement.

Second, at the instant of impact with the Earth, all of the aircraft's velocity is decelerated to zero (in relativistic terms, whether something is accelerated or decelerated releases or uses the same amount of energy), thus, this satisfies the "a" in Newton's formula.

As for the 'm" in the formula, that is a constant mass value at moment of impact, but you are correct, that I blew the calculations here. I won't even bother to start the conversion from lbs. of weight to kg of mass, or the conversion of 550 mph to meters per second squared (about 14.5). But, at the end of the conversation, the result is still a really big explosion -- one that is difficult to reproduce absent a small nuclear chain reaction.

Anyway, if I could do the math correctly, I wouldn't be able to overbill anymore, and that would put me at a distinct disadvantage, so I'm glad that I can't.

TGregg · Apr 5, 2006

Quote from kjkent1:

Second, at the instant of impact with the Earth, all of the aircraft's velocity is decelerated to zero (in relativistic terms, whether something is accelerated or decelerated releases or uses the same amount of energy), thus, this satisfies the "a" in Newton's formula.

One wonders if Esuperbo forgot that acceleration is a change in the velocity vector. Even going in a circle at a constant velocity is acceleration according to Newton physics, thus we have centripetal acceleration and no such thing as centrifugal force.

esuperbo · Apr 6, 2006

Quote from TGregg:

One wonders if Esuperbo forgot that acceleration is a change in the velocity vector. Even going in a circle at a constant velocity is acceleration according to Newton physics, thus we have centripetal acceleration and no such thing as centrifugal force.

Acceleration is _the_ rate of change of the velocity vector, not _a_ change. And going in a circle produces acceleration because of the fact that the velocity vector is constantly changing - it's magnitude is the only thing that remains the same. The whole concept of "no such this as centrifugal force" is superficial understanding - from the reference frame of the rotating object there is certainly something we could call centrifugal force. However, the same cannot be said for the reference frame observing the rotation itself. Come on, this is high school physics.

Sorry kjkent, but if there was instantaneous acceleration to zero miles per hour, that would produce infinite acceleration and hence infinite force. Easiest way to get an estimate here is to approximate the plane as a point load and the distance it takes to decelerate as the length of the plane. You've got a starting velocity of 550 mph, ending of 0, and a distance of a plane length - solve for acceleration and you're there.

Your original point is certainly valid though - a plane crash produces an obscene amount of energy!

kjkent1 · Apr 6, 2006

Quote from esuperbo:

Sorry kjkent, but if there was instantaneous acceleration to zero miles per hour, that would produce infinite acceleration and hence infinite force. Easiest way to get an estimate here is to approximate the plane as a point load and the distance it takes to decelerate as the length of the plane. You've got a starting velocity of 550 mph, ending of 0, and a distance of a plane length - solve for acceleration and you're there.

Your original point is certainly valid though - a plane crash produces an obscene amount of energy!

Thanks for acknowedging my original point.

Can you help all of us mere mortals and calculate the actual energy released, from the crash in terms of kilotons of TNT (1 Kiloton of TNT = 4.184 Ã 10^12 Joules)?

esuperbo · Apr 6, 2006

kj, I don't mean to come off as a prick, I'm just an engineering dork. Apologies if I rubbed you the wrong way in my attempt to figure out the correct way to calculate this.

I'd say the easiest way to get an estimate of the energy released would be to simply assume that the kinetic energy before impact is completely disappated.

Kinetic Energy = 0.5mv^2
m= 100,000 kg (~225,000 lbs)
v= 250 m/s (~560 mph)

So the total kinetic energy of the plane is 3,125,000,000 J. Assuming this is completely disappated in the crash (and I fail to see how a part of the plane would continue merrily along) this means a plane crash is equivalent to approximately 0.75 kg of TNT.

So not quite a nuclear bomb, but certainly enough to blow a plane into tiny pieces!

kjkent1 · Apr 6, 2006

Quote from esuperbo:

kj, I don't mean to come off as a prick, I'm just an engineering dork. Apologies if I rubbed you the wrong way in my attempt to figure out the correct way to calculate this.

I'd say the easiest way to get an estimate of the energy released would be to simply assume that the kinetic energy before impact is completely disappated.

Kinetic Energy = 0.5mv^2
m= 100,000 kg (~225,000 lbs)
v= 250 m/s (~560 mph)

So the total kinetic energy of the plane is 3,125,000,000 J. Assuming this is completely disappated in the crash (and I fail to see how a part of the plane would continue merrily along) this means a plane crash is equivalent to approximately 0.75 kg of TNT.

So not quite a nuclear bomb, but certainly enough to blow a plane into tiny pieces!

You haven't rubbed me the wrong way. I just wanted to know the real answer. I was thinking that you'd have to do a bunch of other calculations so as to convert lbs. into mass, and also mph into m/s^2, before you could come up with an accurate answer.

But, if you're correct, then as you say, it's still enough to incinerate the aircraft, which destroys many of the proposed conspiracy theories.

Thanks!