Hardfire: Physics of 9/11

[Furcifer]

I don't need to think about it in the slightest. I have presented the absolute facts in the post you are responding to:

ALL the energy is used in change of velocity of the 2 bodies of mass. Simple physics fact. Zero used in the deformation of materials. End of story.

Used? Please, think about this. Take your time. How is it used?

 

[rwguinn]

I don't need to think about it in the slightest. I have presented the absolute facts in the post you are responding to:

38.67e6 kg drops 3.6576m impacting a floor of mass 2.47e6 kg at 8.47m/s
Energy 'usage' due to conservation of momentum = 83.33e6 J
Resultant velocity of whole mass = 7.96m/s

What we have is 2 bodies of mass changing velocity, both actions consuming energy.

1) 38.67e6 kg decelerating from 8.47m/s to 7.96m/s (consumes 5.01e6 J)
2) 2.47e6kg accelerating from rest to 7.96m/s (consumes 78.32e6 J)

Add them up, we get 5.01e6 + 78.32e6 = 83.33e6 J

ALL the energy is used in change of velocity of the 2 bodies of mass. Simple physics fact. Zero used in the deformation of materials. End of story.

Wow. Just ... wow
BTW--you now have the whole thing moving at 7.96 m/sec. What stops it?

 

[Toke]

What we have is 2 bodies of mass changing velocity, both actions consuming energy.

Shouldnt one of them release energy?

 

[alienentity]

Don't worry. The Mackey post was about energy loss at collision upper part C/lower part A and damages associated with it, and I just repeated/clarified again that damages are often split 50/50 between objects C/A involved. As Mackey in his Hardfire show suggests damages C/A are split 0/100, it seems we have different opinions.
If Mackey can repost his ideas in sub-posts, why can't I? This is a friendly and lively discussion!

Right. Explain how this relates to Femr2's model, as Mackey was discussing.

You've missed the point. It was not about your C vs A theory.

 

[Furcifer]

I'm going to clarify something for you that may be eluding you. In an elastic collision, both kinetic energy and momentum are conserved. In inelastic collisions, conservation of momentum holds, but kinetic energy is not conserved.

Hope this helps.

 

[femr2]

You guys can argue about choice of words for as long as you like, but it does not change the fact that 100% of the energy exchange is accounted-for by change in velocity of the two bodies of mass.

None of the energy exchange within a conservation of momentum calculation applied to an inelastic collision between two bodies of mass is involved in deformation of materials.

I'll add, that as Ryan has incorrectly interpreted conservation of momentum in his model, he is going to have to go back and modify it.

 

[Furcifer]

Wow. Just ... wow

I know, when i read it I heard the collective sound of a million face palms smacking at once.

I think he's getting it though. At any moment, his will be the loudest sound joining the echo of a million face palms.

 

[twinstead]

I'll add, that as Ryan has incorrectly interpreted conservation of momentum in his model, he is going to have to go back and modify it.

Hmm. Whom to believe, whom to believe...You alone have it all figured out do you? And all the other folks on this forum, many who deal with this stuff for a living, don't know what they are talking about, do they?

Pardon me if I get a second opinion.

 

[Furcifer]

None of the energy exchange within a conservation of momentum calculation applied to an inelastic collision between two bodies of mass is involved in deformation of materials.

Egads man, balance your kinetic energy equation again:

1/2(38.67)(8.47)² + 0 = 1/2(38.67)(7.96)² + 1/2(2.47)(7.96)²
1387.1 = 1225.09 + 78.25

1387.1 = 1303.43???? + (83.75)

That is the energy you added to balance the equation. It is the energy of deformation. It can't be anything else. There's no magic involved here.

I suggest you read up on this and get back to us. You either lack the fundamental skills involved or you're being obstinant. This is very basic stuff. I had assumed you made the program and you were pretty well versed in science and physics. As of right now, I am beginning to wonder. There are certainly many people capable of using computer programs that don't understand the physics engines that drive them. You might be such a case.

 

[femr2]

Shouldnt one of them release energy?

38.67e6 kg drops 3.6576m impacting a floor of mass 2.47e6 kg at 8.47m/s
KE 'usage' due to conservation of momentum = 83.33e6 J
Resultant velocity of whole mass = 7.96m/s

What we have is 2 bodies of mass changing velocity, both actions require energy. (It still requires energy to decelerate a moving object)

1) 38.67e6 kg decelerating from 8.47m/s to 7.96m/s (requires 5.01e6 J)
2) 2.47e6kg accelerating from rest to 7.96m/s (requires 78.32e6 J)

Add them up, we get 5.01e6 + 78.32e6 = 83.33e6 J

No deformation of materials. 100% of the energy exchange accounted for by velocity change of the 2 bodies of mass.

78.32e6 J is applied to the impacted mass. That is your release.

 

[psikeyhackr]

Yes and stacking pizza boxes or making a oscillating toy in your living room IS supposed to prove something. And you guys wonder why people laugh at you...

.
Who is stacking pizza boxes?

Well I get to laugh at the people who don't notice that the oscillation of the TOY changes with the mass and distribution of mass and then those people don't ask about the distribution of mass in the WTC.

And then they talk about PHYSICS!

Of course if the pizza and the box weight 4 pounds then the one on the bottom would have to support 436 pounds. It would most likely have to be reinforced which would make it heavier. So stacking pizza boxes would present the same design problem that all skyscraper designers must confront. How to distribute the steel to make it support its own weight. And oscillate in the wind without collapsing.

Your ignorance is slightly amusing. Very slightly.

psik

 

[femr2]

Egads man, balance your kinetic energy equation again:

1/2(38.67)(8.47)² + 0 = 1/2(38.67)(7.96)² + 1/2(2.47)(7.96)²
1387.1 = 1225.09 + 78.25

1387.1 = 1303.43???? + (83.75)

That is the energy you added to balance the equation. It is the energy of deformation. It can't be anything else. There's no magic involved here.

I suggest you read up on this and get back to us. You either lack the fundamental skills involved or you're being obstinant. This is very basic stuff. I had assumed you made the program and you were pretty well versed in science and physics. As of right now, I am beginning to wonder. There are certainly many people capable of using computer programs that don't understand the physics engines that drive them. You might be such a case.

You are ignoring the energy required to decelerate the initial mass from 8.47 to 7.96m/s.

A simple mistake for you to make, but there it is, as I clearly state on my calcs.

 

[alienentity]

1387.1 = 1225.09 + 78.25

1387.1 = 1303.43???? + (83.75)

1387.1 = 1303.43 [1303.34] + (83.75)[83.76]

slight correction.

 

[Furcifer]

I'll add, that as Ryan has incorrectly interpreted conservation of momentum in his model, he is going to have to go back and modify it.

Do you seriously think it's him that's wrong here? I mean come on, you have to give the guy a little credit just because of what he does. I'm not saying take his word for it, but you need to at least say to yourself "He probably knows more than me, we are in a dispute, perhaps I should take a closer look"

Everything is right there for you, you just need to take the step. Forget what you think you know and look at what we are teling you. Wipe your mind clean. Go Zen. Start from scratch. You had a potential energy. You converted this into kinetic energy and worked out a velocity. Now you have a collision. There is no change in potential (ideally). You know you have to conserve momentum and kinetic energy for it to be elastic. If they both aren't conserved then you have an inelastic collision. Inelastic, what defines and inelastic collision? A loss of kinetic energy. Where does it go? Let's see, potential? No. Chemical? Nuh uh. Hmmm....Is it released into space?....No it isn't released into space. Another dimension? No, that violates more laws than we are already trying to violate. Where then? Where can I put that energy? Heat? Yah, heat is friction, we've got that. But that's not enough. Sound? Yah, sound is good. But that is tiny, even smaller than friction. Let's see..........Oh wait, it takes energy to bend something. Yah, it takes a lot of energy to bend steel. Work. Oh yah, work! There's work! invloved here in bending all that steel and smashing concrete. That must be where it goes. Eureka!!!!

That's as simple as I can make it for you.

 

[Furcifer]

1387.1 = 1303.43 [1303.34] + (83.75)[83.76]

slight correction.

lol i can't do maths in my head no more.

 

[femr2]

Conservation of momentum between two bodies of mass:

38.67e6 kg drops 3.6576m impacting a floor of mass 2.47e6 kg at 8.47m/s
KE 'usage' due to conservation of momentum = 83.33e6 J
Resultant velocity of whole mass = 7.96m/s

What we have is 2 bodies of mass changing velocity, both actions require energy. (It still requires energy to decelerate a moving object)

1) 38.67e6 kg decelerating from 8.47m/s to 7.96m/s (requires 5.01e6 J)
2) 2.47e6kg accelerating from rest to 7.96m/s (requires 78.32e6 J)

Add them up, we get 5.01e6 + 78.32e6 = 83.33e6 J

No deformation of materials. 100% of the energy exchange accounted for by velocity change of the 2 bodies of mass.

Ryan is wrong.
End of story.

 

[Grizzly Bear]

Conservation of momentum between two bodies of mass:

38.67e6 kg drops 3.6576m impacting a floor of mass 2.47e6 kg at 8.47m/s
KE 'usage' due to conservation of momentum = 83.33e6 J
Resultant velocity of whole mass = 7.96m/s

What we have is 2 bodies of mass changing velocity, both actions require energy. (It still requires energy to decelerate a moving object)

1) 38.67e6 kg decelerating from 8.47m/s to 7.96m/s (requires 5.01e6 J)
2) 2.47e6kg accelerating from rest to 7.96m/s (requires 78.32e6 J)

Add them up, we get 5.01e6 + 78.32e6 = 83.33e6 J

No deformation of materials. 100% of the energy exchange accounted for by velocity change of the 2 bodies of mass.

Ryan is wrong.
End of story.

Just a question... are you treating the building as a system or as a single unit of structure? If the former, what happens when individual parts are unable to offer sufficient resistance to stop the collapse?

 

[Myriad]

You are ignoring the energy required to decelerate the initial mass from 8.47 to 7.96m/s.

A simple mistake for you to make, but there it is, as I clearly state on my calcs.

Try this one to do a reality check on the principles you're applying: a car massing 1000 kg going 30 m/sec crashes into the front of a stationary 7-11 and subsequently comes to a stop.

The car has (.5 * 1000 * 30^2) = 450,000 Joules of kinetic energy.

It takes 450,000 Joules of energy to slow the car to a stop.

So, there's no energy left to deform the materials of the car or the store. So, the car and the store suffer no damage.

Is that analysis correct? If not, what's wrong with it?

Respectfully,
Myriad

 

[Furcifer]

You are ignoring the energy required to decelerate the initial mass from 8.47 to 7.96m/s.

A simple mistake for you to make, but there it is, as I clearly state on my calcs.

1/2(38.67)(8.47)2 + 0 = 1/2(38.67)(7.96)2 + 1/2(2.47)(7.96)2

No, I can clearly see it on one side going 8.47 m/s and going 7.96 m/s (in a very short period of time). But, oh my, look at that second mass! It was going 0 now it's going 7.96 m/s.

I guess the big mass gave some of it's energy to the little mass. You're ignoring that! There's no hidden fees in the big mass lending the little mass velocity. It does so because the laws of physics say if they want to get together and not violate the laws of thermodynamics, they can't destroy energy. Kinetic or otherwise.

 

[Furcifer]

KE 'usage' due to conservation of momentum = 83.33e6 J

The words "usage" and "conservation" should kinda irk you. I mean, don't you get the slightest feeling they don't go together in the same sentence? I think you do. I think you know or you wouldn't put quotation marks around it.