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David Chandler jumps the shark

jaydeehess said:
IMHO, mass is a red herring anyway. KE increases much greater due to increased velocity than by increase in mass.
NT completely ignored that.
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I am not a physicist.... but isn't isn't he gravity PE of a mass related to the height it can drop? Mass isn't increasing but as it drops V does. F=ma
 
jaydeehess said:
Hah! LOL. True.
Well, while mass increases due to more destroyed structure, its also losing some mass due to shedding off to the sides. Velocity however, of any the blodly H thing that stays within the tower footprint is increasing proportional to 'g', and increasing KE proportional to the square of v. There is a possible argument for steady state mass if shedding equals gain, but there is NO argument that velocity doesn't increase.
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I got a B in "intro to physics" in high school. I never attended a college (or cared to). This is so obvious, one has to wonder what demographic they expect to accept their half assed excuse for an argument.
 
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Notconvinced said:
..... dumb......as conservation of momentum insists.....
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Conservation of momentum was satisfied when the top of the towers hit the ground. You’ll have to find another excuse to seem less ignorant. Have you tried no planes?
 
Grizzly Bear said:
Nope. Wrong. Conservation of momentum kicks in. You dont lose much velocity at all with each of the floor impacts even in the most optimistic case.. The deltaV just with the first impact is -0.54 meters per second (meaning still around 8.09 meters per second as a starting velocity rather than 0) after only the first hit (using the WTC 1 case scenario). So no, theres barely any loss in kinetic energy, and in fact... there is a net gain in potential energy by the time you reach the successive floors since youre calculating for an addition 3.8 meter drop for each additional floor. The math is already part of that
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I said you lose KE. The lower KE then has to drive it's way through additional resistance while at the same time depleting itself of the equivalent KE. If one story below is crushed, then one story above is crushed, and in that crushing, their net KE is zero. This means your next impact has the KE of the initial falling mass, minus the momentum of one floor. The natural collapse loses KE with each iteration.
 
Notconvinced said:
I said you lose KE. The lower KE then has to drive it's way through additional resistance while at the same time depleting itself of the equivalent KE. If one story below is crushed, then one story above is crushed, and in that crushing, their net KE is zero. This means your next impact has the KE of the initial falling mass, minus the momentum of one floor. The natural collapse loses KE with each iteration.
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What happened to the mass? Did gravity give it the day off?

I really hope you've just pulling our chains because, wow............
 
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Notconvinced said:
I said you lose KE. The lower KE then has to drive it's way through additional resistance while at the same time depleting itself of the equivalent KE. If one story below is crushed, then one story above is crushed, and in that crushing, their net KE is zero. This means your next impact has the KE of the initial falling mass, minus the momentum of one floor. The natural collapse loses KE with each iteration.
Click to expand...

Going out on a limb; you never took physics. Or, proof there is no practical knowledge of physics used to compose this silly BS.


I am afraid to tell you, gravity is used to destroy buildings in CD; small amounts of explosives are used to start the collapse in CD, but gravity (E=mgh released) is used to destroy the building. Thus on 911, fire started the collapse, and gravity destroyed the building, when E=mgh was released.

Please explain this special 911 truth physics, use some math, add some numbers, joules, or TNT equivalent.


Did you make this up, or get it from 911 truth sources?

What do you have on Flights 77 and 93; any KE stuff?
 
Notconvinced said:
If one story below is crushed, then one story above is crushed, and in that crushing, their net KE is zero.
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The math shows otherwise. The first impact only dissipates a small portion of the accumulated potential energy into kinetic energy. See below, you have a choice of seeing my work or skipping the to the results:

Started with 15 stories, and the impact/failure of the floor below adds to the falling mass:
p = momentum = m x v
m1 = mass of the top 15 stories
m2 = mass of the top 16 stories = aprox. (16/15) x m1
v1 = velocity before the additional mass is added = 8.63 m/s
v2 = velocity after the mass is added (unknown)

Momentum is conserved, so:
p = m1 x v1 = m2 x v2 = (31/30) x m1 x v2
Solving for v2:
V2=V1*(15/16)
V2=8.63*(15/16)
V2~ 8.09 m/s

We want to find DeltaT for an accurate calculation of the instantaneous acceleration caused by impact with the floor so:

d = 1/2 (Vi+Vf) x t
d=.46 meters Vi=8.63 m/s Vf=8.09m/s t=?

0.46=1/2(8.63-8.09)t
0.46=1/2(0.54)t
0.46=0.27t
t= 0.46/0.27
t= 1.7 seconds

A=dV/dT
dV=0.54 m/s dT=1.7 seconds

A=0.54/1.7
A= 0.31 m/s2This means the floor fails by an overwhelming margin since the upward resistance only reduces the acceleration by 0.31 m/s/s when the force hitting the floor is 7.99 times that when the mass is at rest.

=========================
Since the floor failed we have to restart the equation to account for the initial velocity (8.09 m/s) falling another floor height (3.8 meters) and find out what the maximum possible force exerted on the next floor will be

16 stories worth of mass is the new starting point for the North tower falling on the floor below
d=distance, g=acceleration of gravity, t=time, v=velocity
d = 0.5 g x t²

Solve for time when g=9.81 m/s2, d= 3.8 meters:
3.8=0.5(9.81)*t23.8=4.905*t23.8/4.905=t20.775=t2(SQRT)0.775=t
0.88 Seconds = t (again)

Solve for v:
VF=VI+gt
Since initial velocity is now 8.09 m/s due to the dampening effect of the first impact:
VF=8.09+(9.81*0.88)
VF= 18.78 m/s

  • So the VF value was 8.63 m/s by the moment of impact with the first floor.
  • You lost 0.54 m/s with the resistance of the floor slab before it gave out, giving a new VI for the next impact of 8.09 m/s
  • Then for the next floor impact your VF increases to 18.78 m/s

The math shows a net gain in velocity. So let's recalculate the instantaneous acceleration once more. Assuming a floor has the capacity to stop the mass completely. (DeltaV=-18.78 m/s). Depth of the floor is assumed to be 18 inches or ~.46 meters. Find time:

d = 1/2 (Vi+Vf) x t
D= .46 meters Vi=18.78 m/s Vf=0 m/s t=?

.46=1/2(18.78+0)*t
.46= 9.39t
t= .46/9.39
t= 0.049 seconds

Acceleration:
A=deltaV/DeltaT
Since we count the direction of gravity as positive and the force required to stop the mass is in the negative direction dV=-18.78 m/s

dT=.049 seconds
A=-18.78/0.049
A= -383.265 m/s2In other words the dynamic load at the 2nd floor impact is ~39.06G

But again, the floor's capacity as estimated by NIST is about 6x the static load. And our calculated dynamic load is 6.5x that. So it's apparent that the floor again doesn't have nearly the capacity to bring the mass to a stop. Meaning it fails before an ~39g amplification is met. Let's find something closer:

p = momentum = m x v
m1 = mass of the top 16 stories
m2 = mass of the top 17 stories = aprox. (17/16) x m1
v1 = velocity before the additional mass is added = 18.78 m/s
v2 = velocity after the mass is added (unknown)

Momentum is conserved, so:
p = m1 x v1 = m2 x v2 = (31/30) x m1 x v2
Solving for v2:
V2=V1*(16/17)
V2=18.78*(16/17)
V2~ 17.675 m/s

We want to find DeltaT for an accurate calculation of the instantaneous acceleration caused by impact with the floor so again:

d = 1/2 (Vi+Vf) x t
d=.46 meters Vi=18.78 m/s Vf=17.675 m/s t=?

0.46=1/2(18.78-17.675)t
0.46=1/2(1.105)t
0.46=0.55t
t= 0.46/0.55
t= 0.836 seconds

A=dV/dT
dV=1.105 m/s dT=0.836 seconds

A=1.105/0.836
A= 1.32 m/s2
And as before, since the direction of gravity is considered positive, the upwards direction will be considered negative here so the value for acceleration relative to the direction of gravity in this final value should read:

A= -1.32 m/s2


First impact Values:
  • Starting mass: 15 stories
  • Ending mass: 16 stories
  • Initial velocity: 0 m/s
  • Final Velocity after one floor height drop: 8.63 m/s
  • Final Velocity after one floor resists: 8.09 m/s
  • Velocity loss from impact: -0.54 m/s
  • Load capacity estimate of one floor: 6G
  • Dynamic Load exerted: 7.99G

Second impact values:
  • Starting mass: 16 stories
  • Ending mass: 17 stories
  • Initial velocity: 8.09 m/s
  • Final Velocity after one floor height drop: 18.78 m/s
  • Final Velocity after one floor resists: 17.675 m/s
  • Velocity loss from impact: -1.105 m/s
  • Load capacity estimate of one floor: 6G
  • Dynamic Load exerted: 39.09G

You'll notice that the velocity is still increasing quite rapidly, and the force of the impact first increases 8 fold, then increases to 39 fold because only a portion of the energy is dissipated before the floor gives out. In other words your claim that the impact completely dissipates the KE release is incorrect under the consideration of conservation of momentum and newtons' laws.


Notconvinced said:
I said you lose KE. The lower KE then has to drive it's way through additional resistance while at the same time depleting itself of the equivalent KE.
Click to expand...
Again, refer to the quantified values above. They directly contradict your claim without the need to quantify the KE released by the impact.

Notconvinced said:
If one story below is crushed, then one story above is crushed, and in that crushing, their net KE is zero.
Click to expand...

Potential energy and Kinetic energy have absolutely nothing to do with the condition of an object(s). It's proportional to the mass of the object(s) and the square of its/their velocity:

K.E. = 1/2 mv2
And you're ignoring gravitational acceleration which is a constant.

The only way the KE is completely dissipated is if the floor/structure can absorb that "energy" release without failing. In the real event the kinetic energy was released only on parts of the structure at a time. The calculations - while using a simple scenario - take this into account. It's also worth noting that even if one discounts a net "gain" in the falling mass keeping the initial mass as a constant, the minute loss of velocity from the first impact effectively means that the next impact will still be larger than the first and the end result is a failure of the next floor since the PE value is still experiencing a net gain that's higher than the rate of loss. I could go on with subsequent floor impacts in this scenario but it's unnecessary considering your argument is already shot down.

Notconvinced said:
This means your next impact has the KE of the initial falling mass, minus the momentum of one floor. The natural collapse loses KE with each iteration.
Click to expand...
And the "minus" value is demonstrated to be a value less than the net "positive". Rendering your claim false
 
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beachnut said:
I am afraid to tell you, gravity is used to destroy buildings in CD; small amounts of explosives are used to start the collapse in CD, but gravity (E=mgh released) is used to destroy the building. Thus on 911, fire started the collapse, and gravity destroyed the building, when E=mgh was released.
Click to expand...

Explosives are used to start the collapse in CD, and the collapse proceeds because the structure has already been systemically pre-weakened by other means, cuts, charges, incendiaries, etc...

911 was like this except the pre weakening occurred at and after the planes hit, and then during the collapse itself, thereby walking the structure down at an increasing velocity.
 
Notconvinced said:
I said you lose KE.
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KE is a derived value equal to 1/2 mv2In order to lessen KE either the mass or its velocity must go down.
We are assuming the mass remains the same so its velocity that must go down. If it does not reduce to zero then that velocity will add to the increase in velocity as the mass continues to be acellerated. That means the KE of that mass will be greater when it hits the next lower level than it was when it hit the first floor.

The lower KE then has to drive it's way through additional resistance while at the same time depleting itself of the equivalent KE.
Click to expand...
I dont know what you are talking about but its not physics. Looks more like you've made it up out of thin air.

If one story below is crushed, then one story above is crushed, and in that crushing, their net KE is zero.
Click to expand...

????
Again: KE is a function of mass and the velocity of that mass. If KE is zero then one of those two factors must be zero. Which do you propose is zero, mass or velocity?

This means your next impact has the KE of the initial falling mass, minus the momentum of one floor. The natural collapse loses KE with each iteration.
Click to expand...

So
1/2 m1v12 - m2v2=KE?

You can no more subtract momentum from energy than you can subtract time from length.

Its become very apparent you either have no education in physics, or you did not understand it when you took it in high school.
 
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Notconvinced said:
Explosives are used to start the collapse in CD, and the collapse proceeds because the structure has already been systemically pre-weakened by other means, cuts, charges, incendiaries, etc...

911 was like this except the pre weakening occurred at and after the planes hit, and then during the collapse itself, thereby walking the structure down at an increasing velocity.
Click to expand...

the WTC towers were on fire. Fire started the collapse; steel fails in fire. Steel fails faster than wood. Why do we use steel?

You came unarmed to a physics fight.

When things fall down, they accelerate at g...

The silly part about the BS posted by you; you can't provide how much KE is lost in the collisions. Not much KE was lost in the colliions of the first 10 or 20 floors because the collapse timing, the collapse velocity and acceleration matches a simple momenutum model based on each floor as a unit. We take the top section and drop it on the lower section and base the timing only on momentum of the floors added to the collapse front.
Since the real world collapse time of the WTC matches close a momentum model, not much KE is lost. Each floor fails instantly when the top section reaches it, this adds, what over 2,000,000 or is it 4,000,000 pounds to the mass which is already over 29,000,000 pounds and moving down.
A floor fails at over 29,000,000 pounds, and a floor is not going to stop a 29,000,000 pound mass moving, thus the only thing that happens at each floor is a slight reduction in velocity, and converstion of momentum.
The best part about debating or dicussing with you, no matter how much BS you post, you can't present the joulles lost in heat, and damage to the lower floor; plus it appears you have no respect for the fact the collisions are not like two cars hitting head on, but these collisions are in a gravity field... going down.

Anyway, mass ejected on lower floors might not slow down the collapse, because stuff hit and ejected would not have to be accelerated to the new velocity based on conseravtion of momentum. So how much energy is lost eject stuff; The tracking of energy, momentum and forces quickly becomes a big mess; this is why we break things down into "block", which confusses MT, but it the stuff models can be made of.

There is my BS for the night, have to go save the world from ignorance...
 
Notconvinced said:
Thank you for your service, and I'm certain you're not dumb. What is intriguingly not intuitive to many (including myself for many years) is that the impact of the floors actually decreases the kinetic energy as conservation of momentum insists. The mass of the towers isnt changing, perhaps getting smaller as bits fall or are blown apart, but never getting larger.
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Hold on right there.

The mass of the towers isn't initially moving, therefore it has no kinetic energy.

When one floor falls, it acquires kinetic energy. If the impact energy is enough as to make another floor fail, then there are two moving floors, minus the mass that may escape which isn't much. The total mass of the moving parts, and thus its kinetic energy, therefore increases. It's all about potential energy being transformed into kinetic. Basic physics.


Notconvinced said:
Except you got it wrong, because while the first floor may not have been able to "stop" the upper section, it's kinetic energy would be significantly depleted. And the next impact would do even more "stopping" while reducing the kinetic load again.
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I told you before and I'll tell you again.

Your physics claims are valid for this case:

momentum-sideways.png


but not for this other one:

momentum-vertical.png


Because you're failing to take into account gravity, whose presence implies that PE gets released and transforms into KE, and that's not factored in in your reasoning.
 
pgimeno said:
I told you before and I'll tell you again.

Your physics claims are valid for this case:

[qimg]http://www.formauri.es/personal/pgimeno/xfiles/11-s/momentum-sideways.png[/qimg]

but not for this other one:

[qimg]http://www.formauri.es/personal/pgimeno/xfiles/11-s/momentum-vertical.png[/qimg]

Because you're failing to take into account gravity, whose presence implies that PE gets released and transforms into KE, and that's not factored in in your reasoning.
Click to expand...

Thank you. Our differences in understanding this problem is because I see the horizontal lines as representing entire sections of tower (including inner and outer columns) and many of you (I'm fairly sure) see them as only representing floors (portions of mass) which might be freed into instantaneous PE.

I don't see the floors as being able to be liberated into PE, because I see all of the work that must be exerted on the entirety of the structure below it to do so, and consequently all of the work that must be used (subtracted) from the initial KE.
 
Units!

Originally Posted by jaydeehess
So
1/2 m1v12 - m2v2=KE?
Click to expand...
Click to expand...

Notconvinced said:
No

1/2 ((14/15)m1(14/15)v1))2=KE

Originally Posted by jaydeehess
You can no more subtract momentum from energy than you can subtract time from length.
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Really?? :eye-poppi
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Yes, really. Kinetic energy has the units of mass times velocity squared; in the MKS system, that would be kilograms times (meters2/seconds2).

Momentum, on the other hand, has units of mass times velocity (kilogram-meter/second in MKS).

When you use the expression
1/2 ((14/15)m1(14/15)v1))2you are talking about a quantity that is mass2*velocity2, or kilogram2*meter2/seconds2
Pray tell, what is this new quantity you've stumbled on? It's not energy, and it's not momentum.

I haven't seen abject floundering like this in a long time. Carry on, this is hilarious.
 
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