Randi - You've Been Had! TS1234 Email to Randi

[R.Mackey]

Huh? Walk me through this Mackey.

Glad to. Thank you for entering an actual discussion. This way, we can both learn something.

First, free-fall time in a vacuum is 9.2 sec, but air resistance adds to that, in relation to the ratio of surface area to mass.

Air resistance adds very little to that. As I have noted here dozens of times, the tower is not falling in a free-stream of still air. The only air drag on falling objects is on the sides.

Even if it was, air drag creates a drag force proportional to the square of the falling object's velocity. Because this is a non-linear relationship, the drag will be a minor correction until the velocity becomes reasonably close to the terminal velocity of the object falling. A steel girder will have a terminal velocity in excess of 110 meters per second. Free-fall in vacuum for 9.2 seconds only gets you to 90 meters per second. So even at worst, air drag is minor.

But remember that we are concerned with an object that falls on itself and accelerates, not one that falls at "free-fall" speeds. If we assume the top of the tower hits ground in 15 seconds, and accelerates at a constant rate, then its speed at impact is only 56 meters per second. Again, even if it fell in a free stream of air which it didn't, this means the drag force is at most only a quarter that of gravity. And remember, this is for the very top which moves the fastest. The overwhelming majority of the tower never gets to those speeds. When we also take into account the fact that it didn't fall in a free-stream, we may safely neglect drag entirely.

You have this backwards. If free-fall is 66% of actual fall time, then 33% is left for deformation, not 66%. If you adjust your free fall time to be 12 seconds (more reasonable considering air), then you have 3/15 or 1/5 or 20% of PE available.

No, I don't. You see, the fact that it slows down 33% means that 1 - (0.66)² = 54% is left for deformation.

Likewise, slowing to 12 seconds (a 25% slow down), not supported by the video but let's do it anyway, leaves 1 - (0.75)² = 43% available for deformation. Plenty.

Think of it this way. How long does it take you to stop your car when you're driving 50 kph? Now try it from 100 kph. Is the stopping distance the same, or longer? Why is this? (It's because energy scales with the square of velocity.)

This is imagining that the entire PE of the building is going to be directed toward the work. Thus you are imagining an entire tower on top of the actual tower, the "trash compactor" model of LARED, debunked by Wood.

http://www.democraticunderground.co...w_mesg&forum=125&topic_id=49321&mesg_id=50155

Wood is severely out of her depth.

The "lost" energy is all going towards deformation. Where else can it go?

 

[TruthSeeker1234]

Are you daft? It SUBTRACTS.

Falling rocks aren't slowed terribly by drag. By your logic, avalanches aren't dangerous.

Air resistance SUBTRACTS from fall time???????

 

[rwguinn]

Glad to. Thank you for entering an actual discussion. This way, we can both learn something.


Air resistance adds very little to that. As I have noted here dozens of times, the tower is not falling in a free-stream of still air. The only air drag on falling objects is on the sides.

Even if it was, air drag creates a drag force proportional to the square of the falling object's velocity. Because this is a non-linear relationship, the drag will be a minor correction until the velocity becomes reasonably close to the terminal velocity of the object falling. A steel girder will have a terminal velocity in excess of 110 meters per second. Free-fall in vacuum for 9.2 seconds only gets you to 90 meters per second. So even at worst, air drag is minor.

forgiddaboudit, RM
These guys are convinced that velocity in air of a steel bar= velocity in air of pulverized concrete=velocity in air of a feather.
After all, didn't Gallelio prove....
sigh.

But remember that we are concerned with an object that falls on itself and accelerates, not one that falls at "free-fall" speeds. If we assume the top of the tower hits ground in 15 seconds, and accelerates at a constant rate, then its speed at impact is only 56 meters per second. Again, even if it fell in a free stream of air which it didn't, this means the drag force is at most only a quarter that of gravity. And remember, this is for the very top which moves the fastest. The overwhelming majority of the tower never gets to those speeds. When we also take into account the fact that it didn't fall in a free-stream, we may safely neglect drag entirely.


No, I don't. You see, the fact that it slows down 33% means that 1 - (0.66)² = 54% is left for deformation.

Likewise, slowing to 12 seconds (a 25% slow down), not supported by the video but let's do it anyway, leaves 1 - (0.75)² = 43% available for deformation. Plenty.

Think of it this way. How long does it take you to stop your car when you're driving 50 kph? Now try it from 100 kph. Is the stopping distance the same, or longer? Why is this? (It's because energy scales with the square of velocity.)

but...but...but
you can't use that liitle ^2 there, can you? it is obvious via common sense that a semi and a motorcycle will stop in the same distance from the same speed, idn't it?

Wood is severely out of her depth.

The "lost" energy is all going towards deformation. Where else can it go?

no kiddin'
sigh...
To paraphrase Robert A. Heinlein --an individual who cannot deal with math is a best a sub-human beast who might possibly be trusted to learn not to soil himself...

 

[TruthSeeker1234]

Mackey -

the fact that it slows down 33% means that 1 - (0.66)2 = 54% is left for deformation.

OK, so then with my numbers, 1 - (0.80)2 = .36 = 36% left for deformation.

Next we have to subtract the percentage of mass which landed outside the footprint. As mentioned, the "dust" which "puffed" out of the towers in all directions was very dense, falling nearly as rapidly as steel. Could we not estimate the density of this "dust" from its fall time? Could we not estimate its volume in comparison to the intact tower? Considering density and volume, could we not estimate the mass which landed outside the footprint?

 

[R.Mackey]

OK, so then with my numbers, 1 - (0.80)2 = .36 = 36% left for deformation.

Fair enough, except your numbers imply a collapse time from the top of 11.5 seconds. I'm not sure that number is supported by the video.

Another thing to consider is that we're consider comparison vs. "free fall" from the top of the structure, not the middle where it was hit. And given the huge clouds of dust and smoke, it's quite difficult to estimate when the timer should be stopped.

My only point with this argument is to show that "near free fall" is a specious argument. Slowing down the collapse even by two or three seconds means that an enormous release of energy took place as it fell -- 35% to over 50% of the total collapse energy. Thus, any argument based on it happening "too fast" is bunk.

Next we have to subtract the percentage of mass which landed outside the footprint. As mentioned, the "dust" which "puffed" out of the towers in all directions was very dense, falling nearly as rapidly as steel. Could we not estimate the density of this "dust" from its fall time? Could we not estimate its volume in comparison to the intact tower? Considering density and volume, could we not estimate the mass which landed outside the footprint?

Much of that dust is going to be fine drywall, relatively light. Also the size of the cloud does not indicate how dense the cloud is. Given that the dust is opaque, I don't think you can make a valid estimate -- there are just too many variables.

Furthermore, we similarly have a hard time estimating just how much damage was caused during the collapse versus after the pieces hit the ground. We just don't have enough visibility into the collapse. Regardless, as my simple math shows, the amount of energy expended during collapse is huge -- greater than 3 x 10¹¹ Joules per tower -- and thus the phenomenon seen is entirely credible.

Likewise, if this isn't enough energy, then you need to find an absolutely enormous source of energy other than gravity, and one that leaves no telltale signatures. Explosives would have been heard, shattered windows, killed people standing nearby and been so large as to have been utterly impractical in the first place. I can't think of any other candidates.

 

[Roger_Harris]

The "lost" energy is all going towards deformation. Where else can it go?

Correct, I believe, and I think analyzing the collapse time vs. free-fall time in terms of "lost" energy is not the proper analysis. The thing that would slow the collapse is the transfer of momentum from the falling section to the next impacted floor (and somewhat to floors below that, through the columns): Since the original mass is increased by the mass of the impacted floor, the velocity must decrease in accordance with the momentum conservation law. But from the videos, that appears to be exactly what's happening in the early stages of the collapse, since it's clear that the debris falling outside the building is falling faster than the collapse is proceeding. What happens, though, is that more and more mass gets to moving faster and faster as the collapse proceeds, until the momentum that's being transferred is too small a percentage of the total momentum to slow the falling mass by any significant percentage -- "near" free-fall.

 

[TruthSeeker1234]

Fair enough, except your numbers imply a collapse time from the top of 11.5 seconds. I'm not sure that number is supported by the video.

I was using 15 second collapse time, and 12 seconds as free fall time with air resistance.

Another thing to consider is that we're consider comparison vs. "free fall" from the top of the structure, not the middle where it was hit. And given the huge clouds of dust and smoke, it's quite difficult to estimate when the timer should be stopped.

As I said, some of the building didn't hit the ground for days.

My only point with this argument is to show that "near free fall" is a specious argument. Slowing down the collapse even by two or three seconds means that an enormous release of energy took place as it fell -- 35% to over 50% of the total collapse energy. Thus, any argument based on it happening "too fast" is bunk.

No, that's not your only point.

Much of that dust is going to be fine drywall, relatively light. Also the size of the cloud does not indicate how dense the cloud is. Given that the dust is opaque, I don't think you can make a valid estimate -- there are just too many variables.

I was suggesting an approach which you did not acknowledge. I suggested that we estimate the density of the clouds based upon how fast they fell. Then we could estimate the volume based on the size of the buildings. From that, we could get a handle on how much mass went outside the tower.

Likewise, if this isn't enough energy, then you need to find an absolutely enormous source of energy other than gravity, and one that leaves no telltale signatures. Explosives would have been heard, shattered windows, killed people standing nearby and been so large as to have been utterly impractical in the first place. I can't think of any other candidates.

There were lots of broken windows on buildings which were not hit by rubble. There were windows blown out of firetrucks which were not hit by rubble.

Explosives heard? Plenty of them were heard at various times. The lack of a definitive "crack" at the beginning of the "collapse" may well be explained by high energy weapons, as suggested by Wood and Reynolds.

In fact, as I think about it, the star wars beam solution answers a number of questions.

 

[beachnut]

I was using 15 second collapse time, and 12 seconds as free fall time with air resistance.

As I said, some of the building didn't hit the ground for days.

No, that's not your only point.


I was suggesting an approach which you did not acknowledge. I suggested that we estimate the density of the clouds based upon how fast they fell. Then we could estimate the volume based on the size of the buildings. From that, we could get a handle on how much mass went outside the tower.




There were lots of broken windows on buildings which was not hit by rubble. There were windows blown out of firetrucks which were not hit by rubble.

Explosives heard? Plenty of them were heard at various times. The lack of a definitive "crack" at the beginning of the "collapse" may well be explained by high energy weapons, as suggested by Wood and Reynolds.

In fact, as I think about it, the star wars beam solution answers a number of questions.

the beam weapon is pure nuts, with no proof these two people are just plain nuts.

I did not say beam weapons do not exist, in I think there has been a test of an air borne laser weapon

But the gravity collapse of the WTC has plenty of energy, you only have to ask an engineer (me) or a CD expert. You will have to find the CD guy since you do not trust me as a EE.

They debunk Dr Jones, Morgan believes no planes hit anything.

So you hitch your wagon to two idiots, one with no planes, the other who messes up basic momentum using balls.

 

[R.Mackey]

I was using 15 second collapse time, and 12 seconds as free fall time with air resistance.

Nope. Your figure was 80%, meaning the collapse was slowed by 20% from its theoretical maximum of 9.2 seconds, which works back to exactly 11.5 seconds.

You can't compute free-fall time with air resistance without making more specific and very important assumptions.

As I said, some of the building didn't hit the ground for days.

Yes, there was quite a lot of dust created. Changes nothing... anything turned to dust has already participated in the crumbling, and has expended its energy. There is no significant loss.

No, that's not your only point.

I didn't realize you were a mindreader as well! Please, elucidate.

I was suggesting an approach which you did not acknowledge. I suggested that we estimate the density of the clouds based upon how fast they fell. Then we could estimate the volume based on the size of the buildings. From that, we could get a handle on how much mass went outside the tower.

I did acknowledge it. I also said it was foolish. I do not believe you can make a reasonable mass estimate of the cloud. We have dust samples and debris studies, but they were taken much later. We do not have a real-time estimate of the cloud composition, size, or density. Feel free to prove otherwise if you know of someone who thinks differently.

There were lots of broken windows on buildings which was not hit by rubble. There were windows blown out of firetrucks which were not hit by rubble.

You'll have to prove that. But that's not the point. All windows within a klick or so would have been blown out, if an explosive releasing over 3 x 10¹¹ Joules was employed. Believe me, we'd know.

Explosives heard? Plenty of them were heard at various times. The lack of a definitive "crack" at the beginning of the "collapse" may well be explained by high energy weapons, as suggested by Wood and Reynolds.

In fact, as I think about it, the star wars beam solution answers a number of questions.

Beam weapons are fantasy. The only thing even close to deployment is the Boeing ABL (AirBorne Laser), which is a measly megawatt class weapon. It would require over 80 hours of continuous firing to generate the kinds of energies seen here.

Really, if you're going to retreat to your comic books, there's little point in this discussion.

 

[Gravy]

In fact, as I think about it, the star wars beam solution answers a number of questions.

Yes. Mainly it answers the question, "How can we tell a 9/11 crackpot from a garden-variety 9/11 denier."

TS, you never explained how you reconcile this statement

I am a skeptic in the tradition of the Amazing Randi. Randi became famous as a magician who explained how his tricks did not violate physics.

9/11 was a trick.

With this statement

You are wrong. It is you who have been “taken,” not I…

I do not discuss conspiracy theories.

James Randi.

Have you revised your opinion of yourself since Mr. Randi took you behind the woodshed?

 

[Roger_Harris]

In fact, as I think about it, the star wars beam solution answers a number of questions.

Hint: If you need imaginary technology to answer the questions, your answers are raising more questions than they are answering.

Of course, the reason that conspiracy hucksters like Wood and Reynolds are pushing imaginary technology like beam weapons and mini fusion bombs is that the traditional explosives and/or thermite hypothesis just won't withstand scrutiny. Those "answers" are absurd, but nobody can prove they don't exist.

 

[TruthSeeker1234]

Let's see. Multiplying 200 ft x 200 ft x 1300 ft i get 52,000,000 cu ft for the total volume of a tower. If the tower is 90% air, this is 5,200,000 cu ft of solid stuff.

During the "collapse", the dust outside the tower looks to be at least as big as the tower. So the volume of the dust is 52,000,000 cu ft. If the dust is 10 percent as dense as steel, this means all of the mass was converted to dust. We see some steel left over at the end, so this can't be right. Especially considering that the "dust" fell almost as fast as the steel, indicating that it is very dense.

Perhaps the dust was only half the size of the whole tower, although it sure looks bigger. If the dust is half the size of the tower, and is has a density 16% that of steel, that would mean that 80% of the mass of the tower was converted to dust.

I think that matches our observations fairly well, that is, it looks like around 20% of the mass of the towers survived, mostly steel and aluminum.

 

[defaultdotxbe]

If the dust is 10 percent as dense as steel, this means all of the mass was converted to dust.

id be surprised if it was as high at 1%

Especially considering that the "dust" fell almost as fast as the steel, indicating that it is very dense.

what collapse were you watching? the dust took a heck of a lot longer to settle than the rest of the debris

 

[R.Mackey]

Let's see. Multiplying 200 ft x 200 ft x 1300 ft i get 52,000,000 cu ft for the total volume of a tower. If the tower is 90% air, this is 5,200,000 cu ft of solid stuff.

OK. 1.67 million cubic meters is the more accurate figure. Using your 90% hypothesis, that also gives you an average density of 3000 kg / m³ or 182 pounds per cubic foot of the solid stuff. Feels a tad high but pretty reasonable.

During the "collapse", the dust outside the tower looks to be at least as big as the tower. So the volume of the dust is 52,000,000 cu ft. If the dust is 10 percent as dense as steel, this means all of the mass was converted to dust. We see some steel left over at the end, so this can't be right. Especially considering that the "dust" fell almost as fast as the steel, indicating that it is very dense.

Problems here. I don't know how to eyeball the dust cloud. Saying it's comparable to the size of the intact tower is fair, but there's potential for at least a full order of magnitude of error here.

The dust on the ground is probably the same mass as the tower, or perhaps lighter, since it's made of the same stuff. The dust in air is going to be very hard to estimate.

I don't buy that the dust suspended in air is 10% as dense as steel, though. That would mean a dust density of 780 kg per cubic meter (assuming 10% of the density of A36 structural steel), which means each kilogram of air (density of air is about 1.27 kg / m³) would have to suspend 603 times its own weight in dust.

Not likely. I suggest you're definitely off by at least two full orders of magnitude, and probably three.

Perhaps the dust was only half the size of the whole tower, although it sure looks bigger. If the dust is half the size of the tower, and is has a density 16% that of steel, that would mean that 80% of the mass of the tower was converted to dust.

If I assume the dust cloud is the size of the tower, and use a more credible (but still high) estimate of 600 grams of dust for every kilogram of air in the cloud, that gives me almost exactly 1 million kilograms of dust. That's a mere 0.2% of the tower's mass.

I think that matches our observations fairly well, that is, it looks like around 20% of the mass of the towers survived, mostly steel and aluminum.

No, it doesn't. I also question, as I have from the first day you arrived here, the source of your "observation." I believe much, much more than 20% of the mass of the towers fell to the ground. By the estimate above, I would say more than 99% survived.

 

[defaultdotxbe]

Have you revised your opinion of yourself since Mr. Randi took you behind the woodshed?

unfortunately i think its more likely that he has revised his opinion of randi

 

[Gravy]

I think that matches our observations fairly well, that is, it looks like around 20% of the mass of the towers survived, mostly steel and aluminum.

You and the six-foot-tall bunny? Good grief, TS, is none of this sinking in?

 

[Horatius]

If the tower is 90% air, this is 5,200,000 cu ft of solid stuff.
...

If the dust is 10 percent as dense as steel, this means all of the mass was converted to dust. We see some steel left over at the end, so this can't be right.
...

If the dust is half the size of the tower, and is has a density 16% that of steel, that would mean that 80% of the mass of the tower was converted to dust.

I think that matches our observations fairly well, that is, it looks like around 20% of the mass of the towers survived, mostly steel and aluminum.

And I think you've made your estimates with the answer you wanted to find in mind.

From this paper at Implosion World, they state (see page 5, discussion of Assertion #3) that typical human-inhabited buildings are about 30% structural elements and contents, so your 10% value is off by about a factor of 3.

Other people have discussed the density of the dust in the air. There are so many possible variables on this that there's no way, short of some serious reasearch in a lab perhaps, of getting a realistic estimate of this value. So once again, you're free to pull out any number you want to get the answer you're looking for.

So your calculation is basically worthless.

 

[Moochie]

Have you revised your opinion of yourself since Mr. Randi took you behind the woodshed?

When, exactly, did this occur, and how old was TS?

M.

 

[ImaginalDisc]

Air resistance SUBTRACTS from fall time???????

I can't believe what a ninny you are. It subtracts from the acceleration.

 

[CurtC]

...so this can't be right.

Hey - BS1234 said something that's correct!

Especially considering that the "dust" fell almost as fast as the steel, indicating that it is very dense.

It's clear now that you have no idea what "dense" means in relation to physics.

I think that matches our observations fairly well, that is, it looks like around 20% of the mass of the towers survived, mostly steel and aluminum.

I'm all for knowledgeable estimation when it makes sense. But in this case, you have cascaded multiple estimates, each with an error range of an order of magnitude or more. The final result you get from that process is meaningless. If you want to tighten up each of those estimates that build to the conclusion, that would be a good idea. You may want to start with figuring out exactly what you mean by "dense" when you're referring to a cloud of dust.