9/11 Physics from Non-Experts

[R.Mackey]

One basic misconception. I am calculating actual load not design/code/permitted load. There is 75% reserve. What the building can carry is never what the building does carry. Think of your home. Think of your office. Nothing is loaded to the max permitted by code. If it is it is usually modified so that there will still be some reserve strength.

Ah, there's that mythical safety factor of 4 again. Not correct, not supported, and irrelevant for this calculation.

Sorry, you're just plain wrong.

Let me try again with this hypothetical example. You've been brought in to help finish the WTC towers. The structure is finished, but nothing else -- you're riding a crane or climbing a rope ladder to get to any given floor. Deciding the concrete slab and exposed (still not fire-proofed) trusses above are ugly, your first suggestion is to put a layer of 1" plywood above and below. Plywood on the floor to allow attachment of temporary walls and such without damaging the concrete, and plywood above to insulate the noise from the next floor and give you a working surface to suspend lights and fire sprinklers.

Your brilliant idea has just consumed 28% of the building's permitted load. And you don't even have windows, stairs, or paint yet.

Yes, gypsum is more common than plywood, and gypsum was used extensively in the WTC, but it's even heavier.

Your numbers are insane. If you want help finding out why, I suggest you look at my previous posts. It may not be easy, but science is almost never about taking the path of least resistance. Get to work.

 

[GregoryUrich]

GregoryUrich,

I'd be curious to hear how you came to your conclusions that WTC3, WTC4, WTC5 and WTC6 combined would weigh the same amount as one tower.

I dispute this value, and I'm going to provide calculations for why.

First, let's clear up the size of the buildings.

You assert:



However, a quick glance at Wikipedia reveals otherwise.

The footprint for each tower was 3,969m², with a total floor area of 436,590m².

For now we will ignore the fact that, by a floor average, the WTC towers would weigh much more as they were much taller structures, thus needed much heavier construction at the bottom. We'll assume an equal weight across total floor area for the much smaller buildings.

Glancing at number of floors, it appears you've already made a mistake:

WTC3 - 22 stories
WTC4 - 9 stories
WTC5 - 9 stories
WTC6 - 7 stories (oops!)

Now let's look at total floor area:

I couldn't get a figure for WTC3, but I dude some crude measurements on a scaled site plan, and got a building width of 20m with a long dimension of 120m. This gives us a total footprint of 2,400m² - about 60% of the footprint of one tower. That's obviously quite different to what you claim, so I'd like to know how you came by your calculation.

I couldn't find a value for WTC4 either, but it's slightly smaller than WTC5, so I decided to use the same floor area - which gives you a bit more than is actually the case.

For WTC5, Wikipedia tells us the average floor area per floor was 11,000m² which is about what you claimed in floor area (nearly 3x one tower).

WTC6 we get a total floor area from Wikipedia which works out at 7,136m² per floor - again only a little less than you proposed (although you had an extra floor!).

Now, if we add all of those floor areas together... we get:

WTC3: 52,800m²WTC4: 99,000m² (over estimate)
WTC5: 99,000m²WTC6: 49,953m²
Add all those up...

300,753m²
Now, according to you, this 300,000m² of floor space should weight about the same as the 436,000m² of floor space on one WTC tower, despite the fact that you're talking 45% more floor area!

And this is not even taking into account the much greater average mass per floor that would be found in the towers due to their heights.

Care to explain your figures?

By my considerations, each WTC tower should probably weigh AT LEAST half again as much as WTC3, WTC4, WTC5 and WTC6 combined.

Then there's your basement figure, which is GREATER than the mass of one tower. Now, I don't know how much of the WTC site was covered by the underground plaza, but I know it wasn't the entire thing. So let's for now, assume it was the entire site (perhaps Gravy can give us a better idea of how much of the WTC site was covered by the underground plaza). Assuming the entire 16 acre WTC site had six levels of basement, that's only 388,000m², still LESS than the total floor area of one tower. Yet you assert it weighs 20% more!

Then there's WTC7, which you assert to be 50% of the weight of one tower, yet at 174,000m², it has only 40% of the floor area.

So, it would appear to me that you have OVER estimated the weight for EVERYTHING EXCEPT the towers, and over estimated quite considerably.

Please explain where you got your figures?

-Gumboot

1. NIST gives 8 floors for wtc6. Argue with them not me.
2. The wtc towers has 99 elevators all in the core. The core was mostly elevator shafts == air. So floor area in the towers is missleading. In the lower buildings the usable floor space to floor area is much higher.
3. wtc3-6 were conventional designs which weigh more than a tube/core design. It's hard to guess how much but you can look at the ESB relative to the wtc tower to get an idea.
4. The basement floors had 8" reinforced concrete floors. Much heavier than the lightweight 4" floors in the towers. There was room for 2000 cars down there. 6' x 12' x 2000. That's 144,000 sq ft just for parking.
5. The floor space for WTC7 was 185,800 sq ft (again NIST--don't argue with me) which is more than half the usable floor space of wtc tower.
6. What did the Con Edison (no pun intended) substation and it's contents weigh?
7. Here's one supporting your argument. I'm not sure wtc7 had a basement similar to wtc1. Since the basement levels are proportionally heavier, that's why I guessed 50%.
8. Regarding the subterranean complex, I just doubled my value because that's what you guys are saying (i.e. everything weighs more than my calculations). If I was more careful I would have used 1.8 x my value.

Why not calculate with your values and we can compare?

 

[GregoryUrich]

There you go again

Ah, there's that mythical safety factor of 4 again. Not correct, not supported, and irrelevant for this calculation.

Sorry, you're just plain wrong.

Let me try again with this hypothetical example. You've been brought in to help finish the WTC towers. The structure is finished, but nothing else -- you're riding a crane or climbing a rope ladder to get to any given floor. Deciding the concrete slab and exposed (still not fire-proofed) trusses above are ugly, your first suggestion is to put a layer of 1" plywood above and below. Plywood on the floor to allow attachment of temporary walls and such without damaging the concrete, and plywood above to insulate the noise from the next floor and give you a working surface to suspend lights and fire sprinklers.

Your brilliant idea has just consumed 28% of the building's permitted load. And you don't even have windows, stairs, or paint yet.

Yes, gypsum is more common than plywood, and gypsum was used extensively in the WTC, but it's even heavier.

Your numbers are insane. If you want help finding out why, I suggest you look at my previous posts. It may not be easy, but science is almost never about taking the path of least resistance. Get to work.

1. No myth, NIST gives original design documents with 100 psf. I use 25 psf so I have 75 psf left before I'm over code. Why are you having so much trouble understanding this?
2. I've used 5" normal concrete floors over the entire core on every floor. The core was mostly air as 99 elevator shafts were placed there. That certianly out weighs your stairs, paint, gypsum and windows by at least a factor of 10.

 

[R.Mackey]

What makes you think I'm having comprehension problems?

I've already showed you your sources of error, and I've already explained why your conclusions -- including that every blasted thing apart from structural steel and concrete fits into a mere 27 lb / ft² -- are preposterous.

I understand you've cherry-picked the NIST report, avoided a proper accounting of many materials that aren't calculated as part of the floor load, and invented a safety factor of 4, which I suspect you've cribbed from Gordon Ross. This is all wrong, and it reflects in your final answers.

Really, there's not much more to say.

 

[rwguinn]

If I may intervene--kick my tail out if I'm out of line, but:
The most important mass in the towers was that which was AT and ABOVE the damaged area.
Like runway behind you and altitude above you, mass below the damage does nothing for collapse.
It is the weight on the damaged and remaining columns and beams that caused collapse, and that energy that kept it running.
Let's concentrate on the point

 

[gumboot]

1. NIST gives 8 floors for wtc6. Argue with them not me.

1. 
   
   
   Picture here.
   
   (foreground building on the left).
   
   
   
   

   The wtc towers has 99 elevators all in the core. The core was mostly elevator shafts == air. So floor area in the towers is missleading. In the lower buildings the usable floor space to floor area is much higher.

   
   
   My values are based on TOTAL floor area, not just rentable space. Non-rentable floor area weighs stuff too, dontcha know.
   
   
   
   

   wtc3-6 were conventional designs which weigh more than a tube/core design. It's hard to guess how much but you can look at the ESB relative to the wtc tower to get an idea.

   
   
   Yes but the WTC towers were over 10 times as high. As you know, structural components at the base increase as structure height increases.
   
   
   
   

   [*]The basement floors had 8" reinforced concrete floors. Much heavier than the lightweight 4" floors in the towers. There was room for 2000 cars down there. 6' x 12' x 2000. That's 144,000 sq ft just for parking.

   
   
   Well I allowed an enormous area for the basement (4.1 million square feet). I imagine the actual area was a fraction of this. I also imagine parking took up a significant chunk of the basement.
   
   
   
   
   

   [*]The floor space for WTC was 185,800 sq ft (again NIST--don't argue with me) which is more than half the usable floor space of wtc tower.

   
   ?
   
   
   

   [*]What did the Con Edison (no pun intended) substation and it's contents weigh?

   
   The con ed station is not part of WTC7's structure. WTC7 was built on top of it.
   
   
   
   

   [*]Regarding the subterranean complex, I just doubled my value because that's what you guys are saying (i.e. everything weighs more than my calculations). If I was more careful I would have used 1.8 x my value.
   
   Why not calculate with your values and we can compare?

   
   
   
   Calculate my values? I did. Did you not read the post you replied to?
   
   I suspect you didn't, because despite your response you failed to provide the answers I was asking for.
   
   Well okay, that's not true. You said for the basement area you "guessed".
   
   Now please explain how you came to the conclusion that the other WTC site buildings (excluding WTC7) weighed as much as one WTC tower?
   
   -Gumboot

 

[R.Mackey]

Begging your pardon rwguinn, but the topic of discussion is this letter by Mr. Urich, and it concerns primarily the total mass of the Towers. Mr. Urich has stated that he is not using this argument in support of any alternative conspiracy theory, and so far we should give him the benefit of the doubt.

I did explain in this post why any change in mass will not support any alternative collapse or crumbling theory, pre-emptively.

 

[Minadin]

In the meantime, let’s take a look at the loads. What loads are there outside the core? This is office space: moveable lightweight partitions, desks, chairs, computers, maybe a file cabinet. Think of a cubicle 5’x5’ = 25 sqft. Using my value 25 psf, we are up to 625 lbs for a cubicle. This could include 100 lb desk, 100lb partition, 100lb file cabinet/books etc, 25 lb computer, 200 lb person, and 100 lbs left over. Then we have corridor space probably a minimum of 10 sqft per cubicle (another 250 lbs). Are you getting my picture? Now I have 350 lbs extra per cubicle with a possible 800 cubicles per floor. This leaves 140 tons per floor for other stuff (i.e. safes, mainframes, etc.).

Sorry I'm late in replying to this - it's from yesterday - but these values are grossly underestimated. Just last week, I had to figure the load imposed by a filing system for a doctor's tenant space in one of the office buildings we're currently working on. The system itself was a relatively moderate load of 1,200 lbs over an area of roughly 150 sq. ft. - 8 lb per - but - the paper files that would be placed in these units, assuming they would only ever be filled to about 70% capacity, was an additional 26,000 lbs, or ~ 173 lbs per square foot. Now, you might think that this is an inappropriate example because what we're talking about is a more dense than normal storage system designed to save space. But, the point is, that we had to add structural support to the floor system in this area, because we typically use 100 - 120 lb/sq. ft. as a "rule of thumb" for those loads.

Also, the spaces aren't going to uniformly be the lightest type of open office space imaginable. You're going to have multiple tenants on many floors, each with different loading conditions. Demising walls between tenants will go from the floor to the underside of the decking - those add significant weight, in addition to real (not cubicle) partitions they might have in between various offices, conference rooms, etc. Some of these financial offices likely had filing rooms very similar to the one I describe above.

You'll have to support the ceilings of the storey below as well, and in some areas those are likely to be gypsum drywall instead of ACT. Then, there's all the other things which are attached to the underside of the floor - things like sprinklers, HVAC ductwork, lighting, localized water heaters, etc. That stuff isn't weightless, either.

 

[R.Mackey]

Bingo, and thanks for your insight Minadin.

I'm working at a disadvantage because I don't work with buildings.

However, in my line of work (NASA), we do perform weight calculations. Very, very, very sensitive ones. Millions of dollars are often spent to eliminate mere hundreds of grams!

If I've learned anything from working with spacecraft architects and proposal teams, it's that everyone underestimates weight. No matter how careful, thorough, and experienced. Everyone.

ETA: Well, everyone except bodybuilders...

 

[rwguinn]

Begging your pardon rwguinn, but the topic of discussion is this letter by Mr. Urich, and it concerns primarily the total mass of the Towers. Mr. Urich has stated that he is not using this argument in support of any alternative conspiracy theory, and so far we should give him the benefit of the doubt.

I did explain in this post why any change in mass will not support any alternative collapse or crumbling theory, pre-emptively.

I done Tole you to tell me to butt out if i'm outta line!
And the post on change in mass not supporting, etc was where I started going astray, since I agree there, too.
I agree with you that with his suspected cooked numbers everything else he does is suspect--I just lost track of where the argument started--
My apologies, sir. I'll sit and watch some more now...

 

[beachnut]

Did you expect more from a paper/letter in the special journal of 9/11 from Dr Jones?

 

[Furcifer]

In going back to your summation notation, which you have defended rather vehemently, i would like to point out that if indeed there were 12 grades of steel ranging 4-1/4 inch thickness, the effect of smoothing this over the entire length of the building greatly reduces the overall mass. Correct?

 

[beachnut]

In going back to your summation notation, which you have defended rather vehemently, i would like to point out that if indeed there were 12 grades of steel ranging 4-1/4 inch thickness, the effect of smoothing this over the entire length of the building greatly reduces the overall mass. Correct?

I found on the upper floor there were mixes of steel from 1/4 to 13/16 in thickness. You can not model it in a linear fashion. You have to have the real figures. The building pieces are all numbered. There is a list somewhere. His thickness could be off over 200 percent in some sections. His errors will get bigger as we go. He should of studied more for the final.

 

[Corsair 115]

This wouldn't be a big deal if it were not for the fact that it has caused confusion in calculating fuel loads on airplanes (with nearly catastrophic results at least once - a crash landing by a very experienced pilot is all that saved the lives of many) and that it has caused all manner of confusion for many people trying to comprehend the difference in gasoline prices on the opposite side of the border.

In fairness, the aircraft incident you refer to took place in 1985 when the transition was still relatively new. I think we've got the hang of it now.

We now return you to your regularly scheduled thread topic...

 

[Furcifer]

I found on the upper floor there were mixes of steel from 1/4 to 13/16 in thickness. You can not model it in a linear fashion. You have to have the real figures. The building pieces are all numbered. There is a list somewhere. His thickness could be off over 200 percent in some sections. His errors will get bigger as we go. He should of studied more for the final.

Exactly. I would have used at least the three sections of the WTC independently. Us math guys love triangles, non better than the memory triangle 3,4,5. Use 3 grades of steel in the first (lower) section, 4 grades of steel in the second section, and 5 grades of steel in the third section. The mass ratio for the sections should be easy to find, I'm assuming one of our architects would have a pretty good idea.

 

[jaydeehess]

Is this all not a rather inconsequential problem, that is the calculation of the mass of a tower?

Given an initail failure that has the upper section dropping down onto the next floor, it is the floor, meaning the concrete/trusses/truss seats, where that upper section mass would be landing.

Quite obviously that mass would far overwhelm the floor's strength. So that floor buckles, breaks and tears loose from the truss seats.

The only way that futher collapse will halt would be if the falling mass is to be reduced enough to allow the strength of another floor to not fail under this load.

In a more typical building, not a tube-in-tube such as the towers, the initial collapse would not have occured in the fashion we saw that had the upper block moving at once. A post and beam construction would have given the upper section more distributed resistance to initial failure and resulted in a more widely assymetric initial collapse. A shearing along a plane of posts would be likely which would see most of the mass falling preferentially away from the building.

Even with the south tower in which the upper section pivoted and tilted, the pivot point was very near the center of the building at the core columns. It could simply not pivot about anything between the core and the perimeter since there are no vertical columns in that space. This meant that when the pivot failed(core columns bend to the point of fracture) the center of mass of the upper section is still above the lower section and thus, again, most of the mass of the upper section(which is in the first place greater than that of the upper section of the north tower) will fall upon the flooring of the next intact level.

One may then point out that this would have the columns standing while the floors crash down through the tower. Well now we have two things occuring in regards to the core columns. First they now have no lateral support from the connection to the perimeter and will buckle. secondly they are being buffeted by the collapse of thousands of tons of dense material and will be losing horizontal connections between core columns , the elevator motors, air conditioning units, power transformers and many other very dense objects will be falling through the tower's core.

In fact the destruction of floor systems occuring first would certainly fit the observed collapse which sees dust expellation from floors ahead of the tearing away of the perimeter columns.

This is not a 'pancake' collapse'. It is much more complex and chaotic than that. One other thing this is very much not, is a collapse that has the upper section falling upon the load bearing vertical structural members.

 

[Furcifer]

Is this all not a rather inconsequential problem, that is the calculation of the mass of a tower?

Yes, absolutely. But Loose Ends leads to Loose Change.

 

[GregoryUrich]

I found on the upper floor there were mixes of steel from 1/4 to 13/16 in thickness. You can not model it in a linear fashion. You have to have the real figures. The building pieces are all numbered. There is a list somewhere. His thickness could be off over 200 percent in some sections. His errors will get bigger as we go. He should of studied more for the final.

You guys aren't very careful. My ratio is from the external columns. The 13/16 is from core columns. The ratio for core columns is 7" plate thickness at the bottom up to 13/16" (if that's the value you propose). Then we have the cross section for the heaviest core columns 3 X 7 " X 52" which you can compare to your dimensions for the core column at the top. The ratio will be much higher than 16:1 as I have scaled it. By the way the different grades of steel have very small differences in density. Someone can check that out if they want.

 

[rwguinn]

You guys aren't very careful. My ratio is from the external columns. The 13/16 is from core columns. The ratio for core columns is 7" plate thickness at the bottom up to 13/16" (if that's the value you propose). Then we have the cross section for the heaviest core columns 3 X 7 " X 52" which you can compare to your dimensions for the core column at the top. The ratio will be much higher than 16:1 as I have scaled it. By the way the different grades of steel have very small differences in density. Someone can check that out if they want.

I'll even let you use .283 lb/in^3.
Just get it right. So far, the professionals are showing you up for the uninformed amateur that you apparently are.
Try answering the questions. This one-track, circular reasoning is making me dizzy.

 

[beachnut]

You guys aren't very careful. My ratio is from the external columns. The 13/16 is from core columns. The ratio for core columns is 7" plate thickness at the bottom up to 13/16" (if that's the value you propose). Then we have the cross section for the heaviest core columns 3 X 7 " X 52" which you can compare to your dimensions for the core column at the top. The ratio will be much higher than 16:1 as I have scaled it. By the way the different grades of steel have very small differences in density. Someone can check that out if they want.

Wrong! sorry you have shown your weakness. Research. Come back when you get it right. Sorry, but you should have read NIST, not cherry picked it.