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David Chandler jumps the shark

DGM said:
Good job Tony.

Now explain what this has to do with the mass that actually was in motion.

You guys remind me of the dogs in the move "up". Except you expect us to follow the squirrels.
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WTF are you even talking about ?? Is this a US thing that I am missing or are you callin me a dog?
 
gerrycan said:
WTF are you even talking about ?? Is this a US thing that I am missing or are you callin me a dog?
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No. In the movie "up" the dogs are constantly easily distracted by the squirrels.

The comparison is you expect people to get distracted by the mass of the individual levels of the towers when only the ones in motion mater. If anything your data only shows the further the collapse progresses the less likely it will be to stop.
 
gerrycan said:
WTF are you even talking about ?? ...?
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Don't worry, it is physics, and fantasy. AE911T CD is fantasy, the gravity collapse of the WTC due to fire is physics. Remember, physics is banned for members, true believers, and followers of 911 truth delusional inside job and BS CD club of woo.

Why is no steel damaged by thermite on 911? Is David Chandler a thermite CD pusher, or silent explosives conspiracy theorist?
 
DGM said:
No. In the movie "up" the dogs are constantly easily distracted by the squirrels.

The comparison is you expect people to get distracted by the mass of the individual levels of the towers when only the ones in motion mater. If anything your data only shows the further the collapse progresses the less likely it will be to stop.
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Surely we need to know the mass of the individual levels to work out what mass is in motion though. That's all I was asking for.
There was a scottish film from 30 years ago called "restless natives". It was all about a wolfman and a guy just like you robbing US tourists in the highlands. But I am not calling you a thief.
ETA https://www.youtube.com/watch?v=tM9rUC18JFs
 
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gerrycan said:
Surely we need to know the mass of the individual levels to work out what mass is in motion though. That's all I was asking for.
There was a scottish film from 30 years ago called "restless natives". It was all about a wolfman and a guy just like you robbing US tourists in the highlands. But I am not calling you a thief.
ETA https://www.youtube.com/watch?v=tM9rUC18JFs
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No problem with knowing the mass. The important point is, does the upper section come to a complete stop at first and then subsequent levels. If the answer is no then there is no real chance the collapse will arrest.

Nationally, if your numbers show this not to be true I will retract this statement.
 
gerrycan said:
Surely we need to know the mass of the individual levels to work out what mass is in motion though. That's all I was asking for.
]
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I believe the point is that if the initial upper falling mass was enough to fail the first floor it encounters then that mass is enough to fail the next one especially considering that if it does not come to a full stop at first impact it will also impact further down at greater velocity, AND, the mass of debris that comes from that first destroyed floor has accelerated one level and contributes to the next impact as well,,,,, repeat 90+ times.

The exact mass of each subsequent contributing floor is rather immaterial since the original falling mass at the velocity it was going, was enough to fail the first floor lower.
 
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jaydeehess said:
I believe the point is that if the initial upper falling mass was enough to fail the first floor it encounters then that mass is enough to fail the next one especially considering that if it does not come to a full stop at first impact it will also impact further down at greater velocity, AND, the mass of debris that comes from that first destroyed floor has accelerated one level and contributes to the next impact as well,,,,, repeat 90+ times.

The exact mass of each subsequent contributing floor is rather immaterial since the original falling mass at the velocity it was going, was enough to fail the first floor lower.
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The velocity is the driver really though. So, given that I previously presumed the top block falling at freefall for 13.5ft before impacting the floors below, what would a more realistic figure be for a real world calculation?
Obviously, in a fire/damage scenario, there would be no possibility of a whole floor disappearing and allowing a freefall of 13.5 ft for the top block.
Does 10mph sound fair?
 
gerrycan said:
The velocity is the driver really though.
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I keep making that point, yes.

So, given that I previously presumed the top block falling at freefall for 13.5ft before impacting the floors below, what would a more realistic figure be for a real world calculation?
Obviously, in a fire/damage scenario, there would be no possibility of a whole floor disappearing and allowing a freefall of 13.5 ft for the top block.Does 10mph sound fair?
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Free fall of the upper block becomes possible when columns bend then fail due to tilt. They no longer line up so there is nothing to slow that upper block.
Some energy may be transferred as columns bend but the off-axis loading guarantees they don't offer much resistance. Now with this the fall isn't quite 4 meters either
Perhaps 0.8g (average acceleration) for 0.75 height of a floor.

At any rate, unless that mass comes to a stop at first impact, before continuing through, its going to hit subsequent floors even harder.
 
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Notconvinced said:
You get to attribute p (high kinetic load) to the first floor(s) of falling building, but the moment you have impact, the v of that lower floor is roughly zero, as it is simply crushed against itself. The added PE is nowhere near the PE you got when x number of floors got to descend through y vertical space. So you get less KE after impact, less to no PE released, and a system which comes to a halt well above the ground.
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Let me break down this paragraph to make it easier to reply:


Notconvinced said:
You get to attribute p (high kinetic load) to the first floor(s) of falling building, but the moment you have impact, the v of that lower floor is roughly zero, as it is simply crushed against itself.
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The highlighted part is misguiding. At the very instant of contact, it's at zero velocity, but as the impacting floor pushes on it due to its inertia, it gains velocity *very* quickly, basically instantly. Maybe it will help you if you think about it in terms of a bullet hitting a block of wood: the block of wood gains velocity very quickly too when impacted, as the bullet's inertia pushes it. The resulting theoretical velocity can be calculated by the conservation of momentum formulas.

That velocity is unlikely to be reached(*) because there are several energy sinks, one of which is the energy inverted in breaking the connections of the floors to the columns. But still, given that these connections are designed to only support the static load of one floor (times a factor of safety) and not such a dynamic load, the energy loss due to that alone will be quite small, so I don't expect that to be an important factor.

(*) It can be reached for other reasons. A partially elastic collision may make the elastic part compensate the loss of energy and make the bottom floor reach the same theoretical velocity as in a perfectly inelastic collision. But that's nitpicking. Almost surely that didn't happen.

To summarize, the final velocity of the impacted floor an instant after impact will not be zero, so it will have an initial KE (which depends on the initial energy of the impacting floor) that gets added to its initial PE, both available to keep crushing, AND the impacting floor will still have some of its KE available and the rest of its PE.


Notconvinced said:
The added PE is nowhere near the PE you got when x number of floors got to descend through y vertical space.
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Actually, my initial assumption was that one floor alone is able to break the connections by impacting on the next floor, making your x = 1. Under that assumption, the impacted floor alone has enough energy to continue the process, even more if we consider that the impacting floor still has PE.

But let's take the case of x = 5 floors as the minimum number required to fall for the height of 1 floor and break the connections. Under the assumption that the loss of mass due to lateral ejection is about 20% of the mass of one floor on each impact, we have the following situation: the total mass in movement after the first impact is now 6 floors minus 20% of one floor = 5.8 floors, falling for the height of 1 floor, regardless of any loss of energy (we can assume they start again with 0 velocity, in Judy Wood's Billiard Ball model's fashion). Since this floor didn't resist 5 floors falling on it for the height of one floor, there's no way the next floor will resist 5.8 floors falling on it for the same height. Much less if they have an initial velocity, as will be the case.

In practice, x is probably 1, and almost surely not bigger than 2. Again, these connections are not designed for such a dynamic load as that of one floor falling on it.


Notconvinced said:
So you get less KE after impact, less to no PE released, and a system which comes to a halt well above the ground.
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The highlighted is false. The impacted floor will be detached, and since it's no longer held, its PE will start converting into KE. Meanwhile, the impacting floor has about the same PE as the impacted one. Actually more if it's a block rather than a single floor.

The conclusion that the system comes to a halt, is wrong for all the reasons above.
​


Notconvinced said:
You've misunderstood. I think of the columns as the axis, as that is the longest dimension. I see the floor (my definition) as being crushed upon the columns, so net KE decreases.
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I'm not sure what your definition is, but if it departs from the reality of the towers' collapse, it's not useful for discussion purposes.


Notconvinced said:
Trying to see things your way here.... Crushed columns along the axial path would likely release their floor (YOUR definition) upon the next one. What is the KE of JUST the floorspan falling one story? Taking into account the KE it would lose crushing desks, people, cabinets, etc....
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No column crushing happens at this stage. The KE loss you refer to is inherent to inelastic impacts. Once again, your assessment applies to the horizontal case, but gravity adds the ingredient that allows the collapse to continue all the way to the bottom.

I suggest you to google "progressive collapse" or "disproportionate collapse" to learn on how the real world structures behave. Same for gerrycan.
 
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gerrycan said:
The velocity is the driver really though. So, given that I previously presumed the top block falling at freefall for 13.5ft before impacting the floors below, what would a more realistic figure be for a real world calculation?
Obviously, in a fire/damage scenario, there would be no possibility of a whole floor disappearing and allowing a freefall of 13.5 ft for the top block.
Does 10mph sound fair?
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Why? Using Newtonian physics, no calculations are necessary to understand why the mass and momentum are increasing, and I don't think your input numbers really matter if you intend to use the new Chandler/Szamboti physics instead. Just declare your conclusion.
 
jaydeehess said:
I keep making that point, yes.



Free fall of the upper block becomes possible when columns bend then fail due to tilt. They no longer line up so there is nothing to slow that upper block.
Some energy may be transferred as columns bend but the off-axis loading guarantees they don't offer much resistance. Now with this the fall isn't quite 4 meters either
Perhaps 0.8g (average acceleration) for 0.75 height of a floor.

At any rate, unless that mass comes to a stop at first impact, before continuing through, its going to hit subsequent floors even harder.
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I think you are erring in favour of your own viewpoint a bit there, but fair enough, let's suppose 0.8g at 2.25m. Wouldn't this reduce the velocity at impact from 19mph to somewhere south of 13.5mph.
You still think there's enough energy there for the collapse not to arrest? Velocity is the squared value in the equation after all.
 
gerrycan said:
The velocity is the driver really though. So, given that I previously presumed the top block falling at freefall for 13.5ft before impacting the floors below, what would a more realistic figure be for a real world calculation?
Obviously, in a fire/damage scenario, there would be no possibility of a whole floor disappearing and allowing a freefall of 13.5 ft for the top block.
Does 10mph sound fair?
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2 notes:
  1. One, G has to be in constant consideration because gravitational acceleration is a constant for this case.
  2. Two, 10 mph is an arbitrary number. If you want something accurate you'll have to do something to get closer to the real number but below is my [simple] calc to see the result difference:

(Simple calc = ideal limit case)

10 mph = ~4.4704 m/s

Assume a floor has the capacity to stop the mass completely. (DeltaV=-4.4704 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=4.4704 m/s Vf=0 m/s t=?

.46=1/2(4.4704+0)*t
.46= 2.235t
t= .46/2.235
t= 0.206 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=-4.4704 m/s

dT=.206 seconds
A=-4.4704/0.206
A= -21.7 m/s2


or ~2.21 g's (21.7/9.81) where 9.81 is the "g" constant.
Which is akin to modeling a bit over double the static load. The floors were estimated to be able to hold 6 floors worth as a static load so this number places it above that threshold and well

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. (15/16) x m1
v1 = velocity before the additional mass is added = 4.4704 m/s
v2 = velocity after the mass is added (unknown)

Momentum is conserved, so:
p = m1 x v1 = m2 x v2 = (15/16) x m1 x v2
Solving for v2:
V2=V1*(15/16)
V2=4.4704*(15/16)
V2~ 4.2 m/s

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=4.4705 m/s Vf=4.2m/s t=?

0.46=1/2(4.4704-4.2)t
0.46=1/2(0.28)t
0.46=0.14t
t= 0.46/0.14
t= 3.29 seconds

A=dV/dT
dV=0.28 m/s dT=3.29 seconds

A=0.28/3.29
A= 0.085 m/s2 (up direction resisting the falling mass)

In other words, if we use 4.47 m/s as the baseline (in a perfect impact scenario) you're loss in volocity from the first impact is about -0.28 m/s which means the falling mass is still traveling at a new initial velocity of 4.2 m/s (9.395 mph)

As an aside... if we use the average acceleration (2/3G) as the base line that's 6.54 m/s/s
So...
Solve for time when g=6.54 m/s2, d= 3.8 meters:
3.8=0.5(6.54)*t23.8=3.27*t23.8/3.27=t20.775=t2(SQRT)1.16=t
1.077 Seconds = t

Solve for v:
VF=VI+gt
Since initial velocity is 0:
VF=gt

G=6.54 m/s2 t=1.077 seconds
VF=6.54*1.077

VF= 7.04 m/s
or 15.75 mph

So.... just to summarize:

Your suggestion: (10 MPH) = 4.47 m/s
If we use 2/3G (6.54 m/s/s) the velocity at impact = 7.04 m/s (~36.5% greater)

My other results using the max baseline case:
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
  • 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
  • Dynamic Load exerted: 39.09G

or ~45% greater first impact velocity assuming freefall (the dynamic load multiplier is higher though)

As an aside.... it's time consuming so I have no plans currently to quantify the second one for the extra stuff. But gravitational acceleration has to be included in subsequent impacts meaning again, the net change in velocity by each subsequent floor hit is higher than the previous. Use my earlier result as a case in point.

Sorry for the lengthy post people.... feel free to nitpick the maths
 
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gerrycan said:
Surely we need to know the mass of the individual levels to work out what mass is in motion though. That's all I was asking for. ...
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Before computing the mass on each level, you need to understand what part of the mass got into motion upon impact.

As a matter of observation, a very significant proportion of the perimeter columns were merely pushed outward and toppled over as the floor slabs inside were stripped loose and rushed down - their mass plays little role in the CoE/CoM considerations. Also, a very significant proportion of the core columns remained standing for a few seconds and toppled over later as the floor slabs inside were stripped loose and rushed down - their mass plays little role in the CoE/CoM considerations.
 
gerrycan said:
The velocity is the driver really though.
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Well, no: energy is the driver and velocity is the result. Estimating the velocity just unnecessarily complicates the problem, in fact, because then you have to try to figure out how much force there is in the impact. I admit I thought you would need to do that, too, but reading Bazant's 2002 energy argument was like "doh!" -- if you can't dissipate the energy due to gravity, then the collapse proceeds, and the velocity is an irrelevant detail.
 
gerrycan said:
So am I. Absolute guff. (do you use that word over there?)
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Never saw the word "guff" here. But I am in Germany. Looked it up ... aha.

So in the world of Chandler/Szamboti/gerrycan physics, math and Conservation of Momentum/Conservation of Energy are guff?
That says a lot.

I think you fail to understand something extremely basic:

Every part of the tower's top that is moving down has ALREADY lost practically all structural contact with the lower, still standing part.
In other words, by the time you see the top part falling, ALL columns have ALREADY failed and are no longer supporting the top.
All these failed columns are passing each other - the lower ends of the top part have ALREADY moved below the top ends of the bottom part.
All the descending mass, so far as it isn't falling outside of the bottom part's perimeter, keeps dropping mostly on floor slabs. So the collapse dynamics is totally dominated by the response of the floor slabs to this dynamic loading - not by the structural strength of the standing columns!

95% of each story is air, and 98% of the horizontal area is air, except for the 4 inches of light-weight concrete
What's going to stop most of the falling rubble and steel from accelerating at practically g during the 13 ft drop between slabs?
I am asking this for the 3rd time now.
(Usually, when truthers stubbornly pretend they don't see my repeated questions, I interprete this as them realizing that a true and honest answer to them will defeat them, and that is why the run run run away from such question.)
 
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