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9/11 Physics from Non-Experts

oh yah, saw that earlier today. :) the ramifications of "net force=zero" are far reaching, and its implications have yet to be fully understood.
but i now have this image of a giant hand holding a meteor, and another giant hand with a red throbbing thumb holding the WTC. the blurb over head is just "$@*&!!!"
 
3bodyproblem said:
but i now have this image of a giant hand holding a meteor, and another giant hand with a red throbbing thumb holding the WTC. the blurb over head is just "$@*&!!!"
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Don't forget the other WTC tower, bent in half from the first bad attempt at hammering it in :)
 
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R.Mackey said:
:D Not "funded," we had to supply the toothpicks.

A good engineer is often saddled with lame requirements... it's part of the training!
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When I was at Carleton U. in Ottawa the 1st yr engineers had to build a bridge using popsicle sticks and wood glue. No nails, no thread or rope. It had to span, IIRC, 20 inches and had to support a set load.

The 3rd yr students entered a team in the concrete canoe contest that had teams from many universities compete. The canoe had to be built of concrete, carry a two man crew and be paddled over a short course. First accross the finish line wins or it was a beat-the-clock thing, I don't recall.

Most canoes sank, some very soon after launch, others because they were not stable enough and tipped, but a few made it. It is a lesson in constraints to construction. But of course engineers such as RMackey realise that. A doofus such as the subject of this thread, would not.
 
Monza said:
Concerning the video in the original post and the man's lack of understanding the difference between statics and dynamics, he is not alone. I remember watching a video where a group of about five or six people attempted a mass bungee jump from a bridge. It wasn't a bungee "jump", rather they swung from the bridge like a giant pendulum.

The lead daredevil had summed the weights of the people and selected a bungee cord designed to withstand this total. Because Fnet = 0, all should be fine, right? Well, he didn't acount for the dynamic (centripetal) force and the cord broke, resulting in severe injury to some of the people.

I think the pendulum problem is about the first example one learns in studying dynamics. Nevertheless, some people never do.
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I saw that too. It was a few years back but very, tragically, instructive. I too could not believe that the guy was so very stupid about this. All he needed to do was contact the local university engineering school and they could have calculated the strength of the wire he needed. That wire broke as they passed the lowest point of the arc and slammed them all into the river. IIRC one girl died, the others all badly injured.

But NetForceZero demonstrates that he would have calculated the strength required exactly as the above daredevil did.
 
jaydeehess said:
All he needed to do was contact the local university engineering school and they could have calculated the strength of the wire he needed. That wire broke as they passed the lowest point of the arc and slammed them all into the river. IIRC one girl died, the others all badly injured.
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University nothing, I did calculations like that in highschool. A Coyote-and-Roadrunner problem. My teacher always liked to make very "involved" questions for his exams. There was one, the Coyote is on a cliff, with a rope, ready to swing down, and jump onto the Roadrunner. So you had to figure out if the rope breaks as he swings (and before he planned to let go), and if so/if not, when he goes into projectile motion, does he land where the Roadrunner is eating the plate of seeds. For extra credit, calculate the whole time it takes the Coyote to get there, and based on the rate the Roadrunner is eating, is the Roadrunner still there?

Of course, if you knew the teacher, you could just assume the Coyote got the Roadrunner and work backwards. He hated that little *BEEP!*

:)
 
quick-draw-mcshill.png
 
this is a copy of post 282 but with diagrams!!!!!!!

For Yandross, this is a situation in which each floor is blown sequentially.
If d= distance from the level of initial collapse

it will take 0.78 seconds for the upper section to reach the next floor, 3 meters below initial collapse
1295946493150c8fb0.bmp


and if the next floor is blown at t=0.39 seconds after the initial collapse begins(one half the time for the upper section to reach the next floor)
1295946493150d6681.bmp



d=0.5(9.8)t2 for upper section

d= 3 + 0.5(9.8)(t-0.39)2 for the next floor

The upper section mass will meet the floor and columns of the next floor when both of these are equal.

0.5(9.8)t2=3 + 0.5(9.8)(t-0.39)2
4.9t2=3 + 4.9(t-0.39)23=4.9t2 - 4.9(t-0.39)23/4.9 = t2 - (t-0.39)20.612 = t2 - (t-0.39)20.612 = t2 - t2 + 0.78t - 0.152

0.612 = 0.78t - 0.152
t=0.979

Self check:

At 0.979 s the upper section has fallen
d= 0.5(9.8)(0.979)2d= 4.70meters

At t=0.979 the second level has been falling for (0.979-0.39) = 0.589 s
d = 3 + 0.5(9.8)(0.589)2d = 4.70 meters

The upper section has collided with the next level's floor and column mass
The upper section is moving at v=at
v=9.8(0.979) =9.69 m/s

The mass of the second floor is moving at
v=9.8(0.589) = 5.77 m/s

1295946493150dacce.bmp
_____________________________________

Now conservation of momentum

Assume that the upper section is 10 times the mass of one floor.

The upper section's momentum at the time it meets the falling mass of the next floor is;
pu=10m(9.69)
whereas the momentum of the next floor's falling mass is
pn=m(5.77)

If they collide inelastically then by conservation of momentum we have
pt=pu + pn
pt=10m(9.69) + m(5.77)

Now we know that pt=11mv where v is the velocity of the now combined masses so;
11mv=10m(9.69) + m(5.77)
11mv=m(96.9 + 5.77)
v=9.33 m/s

The total mass is now slower, by 3.7%, than the original upper section would have been in an unimpeded free fall.

You can plug in any numbers you want about when the next floor is blown but the velocity of the upper block will always be greater than anything that starts moving at a later time and therefore will always catch the slower mass. In fact the worst case senario for catching that mass that starts moving after the upper block does, is the case where the next floor gets blown just as the upper block gets to it. In that case, after a drop of 3 meters the upper block would be doing 7.64 m/s and the next floor's mass is not yet moving(its has been blown from its connection to the lower structure just as the upper block reachs it) so its velocity and momentum is zero.

Now the combined mass of 11m will have a new momentum (as per conservation of momentum)of
11mv=10m(7.64) + m(0.0)
giving us a velocity of the combined mass of
v=(10/11)7.64
v= 6.95 m/s
a 9% drop in velocity from what the upper section would be doing if it had been in unimpeded free fall.
 
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Has anyone passed these video physics courses on to real physics classes?
http://www.youtube.com/watch?v=sBfkAzWbnvM
http://www.youtube.com/watch?v=sbyu0zuxo5A
http://www.youtube.com/watch?v=jYMDsBN3t80 at 3:52 he does his best Sam Kinison for Physics.
http://www.youtube.com/watch?v=yZ-byshtnq4

Speaking of Sam, http://www.youtube.com/watch?v=Xfi4s8cjLFI Yandros needs to do another video. http://www.youtube.com/watch?v=tMjM7292eHk Why is Yandros Physics video like Sam Kinison's Wild Thing? F Net = 0 Is a constant. k
 
Blackadder_no said:
Ok, this one here apparently comes from a Norwegian truther, and although it may be a case of irony (but honestly, nothing from that camp surprises me anymore), I feel that this one ranks up there with the chicken wire experiments...

[qimg]http://i8.tinypic.com/4p9g7cg.jpg[/qimg]
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A) The empty carton is not carrying a gravity load until the mass is put on it. It did at one time though, contain milk. Since the designed load, the milk, had been removed, perhaps they should try it with the designed load inside. I dare say that the extra mass it could then carry would be significantly less

2) Now try picking up that 30Kg mass a few cm above the carton and let it drop. .................
 
Sylvester1592 doesn't seem to have understood my paper

For those who are interested, my article is at:

...journalof911studies.com/letters/wtc_mass_and_energy.pdf

Sylvester wrote:

Ulrich uses a total mass of 235.000.000kg instead of 510.000.000kg to reduce the total potential energy by 60%. He claims not to know where the numbers come from to calculate it, yet he references it in his paper including the Port Authority (which claims a weight of 500.000 tons). Afterwards he claims an 80% overestimation, when he shows a little over 60%. I think he got confused here with the requirements Ross is looking for to keep the tower in existence. He does not differentiate between the different types of steel used or any other elements, even though he has references to them. He picks the lowest number and then fudges his conclusions. Smoke and mirrors by supposedly being accurate showing a floor by floor analysis, when after all the assumptions and estimations an analysis based on the average center of gravity would suffice. All of this to increase the ratio of the number of floors one floor could hold up. Even this scenario would not save the towers... I would think, nice try but no prize
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After reading my paper, I think most people will understand that:

  1. The point of the paper is that the values provided by Port Authority, Wikipedia, etc. are not backed up by any data.
  2. I calculate the weight of the towers based on design data provided by FEMA and NIST.
  3. All data is clearly referenced.
  4. My paper has nothing to do with Ross.
  5. My value for the total mass of one tower is 254,000 metric tons not 235,000 as stated by Sylvester.
  6. Sylvester is insinuating that my purpose is to reduce the potential energy. The lower potential energy (which happens to be very close to FEMAs) was not my goal, rather it was my result.
  7. The mass given by Ashley, Bazant and
    Zhou, and Wikipedia (500,000 short tons) is nearly 80% more than the actual mass of one tower. Do the math Sylvester.

Thus if anyone is confused it is surely Sylvester.

Nonetheless Sylvester does have one point. I use the density of A35 steel while many grades of steel were used in the towers. I'm not really sure if higher grade steels are more dense, but I doubt it. Anyone?

Others have pointed out a few error sources in my calculation and I have found a few myself which I will share here:

  1. I have omitted the hat truss and antenna.
  2. I included all steel in the scaling by floor when actually none of the floor assembly steel should be scaled

All of these errors affect except the antenna effect only the PE and not by a huge amount. I am interested in any comments and will be updating my paper soon.
 
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When you calculated the mass of the towers, did you factor in live load?

-Gumboot
 
That which we do not understand is suspect...

3Body, commenting my article, wrote:

i must admit i stumbled on this by accident, and i was looking for clarification here. i find his summation notation suspect
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Derive it yourself. Just relate the ratio for steel mass = 16:1 to the height. If your can't derive it yourself then I find your education suspect.
 
GregoryUrich said:
What should the floor area be?
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The floors had a dimension of 208ft by 208ft, with a core area 88ft x 138ft.

This gives a total floor area of 43,264ft, an outside core area of 31,120ft, and an inside core area of 12,144ft.

-Gumboot
 
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