Home > Archive > Cobol > May 2004 > more junk deleted
You are viewing an archived Text-only version of the thread.
To view this thread in it's original format and/or if you want to reply to
this thread please [click here]
|
|
| Richard 2004-05-24, 1:30 am |
| robert.deletethis@wagner.net (Robert Wagner) wrote
>
> Here's more junk. The Oklahoma City Federal building (partially) collapsed
> because something sheared a structural beam in its center. The pressure required
> to do that was more than 5,000 psi. A large ammonium nitrate bomb 40 feet away
> can deliver less than 100 psi.
You are correct: is is more junk.
You write two figures with units of psi and relate the two as if they
had anything to do with each other: they don't.
The first is, as you say, the shear force required to shear a beam and
'square inches' refer to the section of the beam. eg an 'H' beam 10"
x 10" of 1/10" steel would have 3 square inches of steel and thus
would require 3 x 5000 lbs force to shear it. ie 15,000 lbs.
The bomb pressure works on a completely different set of 'square
inches', the side of the beam. If the beam was 8 foot long then the
side area would be 960 square inches (8' x 12 x 10"). At 100 psi this
would result in 96,000 lbs of force.
I don't know what actual sizes were used in the building, but anyone
with two clues can work out what is required.
| |
| Robert Wagner 2004-05-25, 6:30 am |
| riplin@Azonic.co.nz (Richard) wrote:
>robert.deletethis@wagner.net (Robert Wagner) wrote
>
required[color=darkred]
away[color=darkred]
>
>You are correct: is is more junk.
>
>You write two figures with units of psi and relate the two as if they
>had anything to do with each other: they don't.
>
>The first is, as you say, the shear force required to shear a beam and
>'square inches' refer to the section of the beam. eg an 'H' beam 10"
>x 10" of 1/10" steel would have 3 square inches of steel and thus
>would require 3 x 5000 lbs force to shear it. ie 15,000 lbs.
>
>The bomb pressure works on a completely different set of 'square
>inches', the side of the beam. If the beam was 8 foot long then the
>side area would be 960 square inches (8' x 12 x 10"). At 100 psi this
>would result in 96,000 lbs of force.
The force of an explosion decreases as the cube of the distance. If a bomb
asserts 100,000 psi at a distance of one foot, at 40 feet it will asset
100K/40^3 = 100K/64K = 2psi.
The only square inches of interest are the ones at the sheer point. Pressure on
the other eight feet are wasted. That's why they make shape charges.
Most civilians have no first-hand experience with explosives larger than a
firecracker. All their 'knowledge' was gleaned from Hollywood special effects.
As a result, we have Urban Legends such as the one about an exploding toilet
propelling a man across the room, which is impossible.
Both OKC and WTC involved a large machine delivering fire or explosion that
destroyed a building. People believe it because they've seen the same in movies
hundreds of times. The theatrical term for this is 'deus ex machina'.
| |
| Richard 2004-05-25, 5:30 pm |
| robert.deletethis@wagner.net (Robert Wagner) wrote
> away
[color=darkred]
[color=darkred]
> The force of an explosion decreases as the cube of the distance. If a bomb
> asserts 100,000 psi at a distance of one foot, at 40 feet it will asset
> 100K/40^3 = 100K/64K = 2psi.
First you claim that '40 feet away .. 100psi'. Now you claim that was
only 2psi. This seems entire consistent with "If the facts don't fit
the theory then change the facts".
In fact your assertion about the cube root of the distance is wrong,
the pressure wave decreases at approximately the square root.
> The only square inches of interest are the ones at the sheer point. Pressure
> on the other eight feet are wasted.
No. Quite wrong. The pressure wave creates a force by operating over
an area. This force creates a shear force by operating against the
points where the beam is held rigidly. If the beam were unrestrained
then the force (96,000lb) would accelerate the beam and move it
somewhere. The rigid attachment to the ground and roof prevents that.
The beam, however, is not infinitely strong. If the force trying to
move the beam away exceeds the strength of the attachments then the
beam will move away (it will break).
Your assertion about 'being wasted' is simply wrong. That would be
like saying: "if you want to push a car then you need to push on the
wheels, pushing on anywhere else (such as the boot) is wasted".
> Most civilians have no first-hand experience with explosives larger than a
> firecracker. All their 'knowledge' was gleaned from Hollywood special effects.
And yours is gained from ... ??
> As a result, we have Urban Legends such as the one about an exploding toilet
> propelling a man across the room, which is impossible.
It is not impossible to propel a man across a room with an explosion,
even one in a toilet. What is unlikely is that he will be in one
piece afterwards.
> Both OKC and WTC involved a large machine delivering fire or explosion that
> destroyed a building. People believe it because they've seen the same in movies
> hundreds of times. The theatrical term for this is 'deus ex machina'.
I was working in the East End of London when the Bishopsgate Bomb was
set off. Six tons of fertiliser in a truck. It extensively damaged
the Hong Kong and Shanghai Bank and NatWest building and everything
within 400 metres.
It wasn't a movie, I went to see it.
| |
| Robert Wagner 2004-05-25, 7:31 pm |
| riplin@Azonic.co.nz (Richard) wrote:
>robert.deletethis@wagner.net (Robert Wagner) wrote
>
>
>First you claim that '40 feet away .. 100psi'. Now you claim that was
>only 2psi. This seems entire consistent with "If the facts don't fit
>the theory then change the facts".
Let me restate .. If a bomb asserts 1M psi at its core, at 40 feet it will
assert 1M/40^3 = 1M/64K = 20psi.
>In fact your assertion about the cube root of the distance is wrong,
>the pressure wave decreases at approximately the square root.
A cursory Web search and common sense contradicts that. The concussion is
expanding in a sphere, not a circle. Must I provide cites? Just enter 'explosion
cube distance'.
effects.[color=darkred]
>
>And yours is gained from ... ??
US Marine Corps, Reconnaissance and Special Operations, Southeast Asia, Cuba and
others in the early '60s. (Second and Third Force Recon out of Camp Lejeune NC
and Okinawa.) We were watching Khymer Rouge long before the Killing Fields in
the '70s. I wrote the 'bad intelligence' blamed for the Bay of Pigs failure.
>I was working in the East End of London when the Bishopsgate Bomb was
>set off. Six tons of fertiliser in a truck. It extensively damaged
>the Hong Kong and Shanghai Bank and NatWest building and everything
>within 400 metres.
The OKC bomb 'damaged' buildings within a one mile radius. The Big Tom explosion
in Jersey City during WW I "broke every window in lower Manhattan [half a mile],
and broke windows as far away as Times Square [two miles]". Look it up.
Windows are easy to break. Poorly supported masonry is easy to topple or loosen.
But air bursts cannot shear structural beams beyond 5 meters.
| |
| Richard 2004-05-25, 11:30 pm |
| robert.deletethis@wagner.net (Robert Wagner) wrote
>
> Let me restate .. If a bomb asserts 1M psi at its core, at 40 feet it will
> assert 1M/40^3 = 1M/64K = 20psi.
And now you have a different figure entirely. You seem to just be
inventing 'facts' to fit your theory.
>
> A cursory Web search and common sense contradicts that. The concussion is
> expanding in a sphere, not a circle. Must I provide cites? Just enter 'explosion
> cube distance'.
First of all these explosions were not 'air blasts'. They didn't
expand in a shere, at most it was a semi-sphere. In some cases there
were buildings on both sides which further constrained the explosion
making it less than a semi-sphere.
Secondly it only reverts to 'cube-root' _after_ all the explosive
material has been consumed. This does happen in uncontained air
blasts at some distance. In the case of fertilizer bombs the
explosion is relatively slow compared with the distances involved and
in these cases the pressure wave is closer to following the square
root. What happens is that the 'pressure wave' expands with the area
of the sphere, not its volume, until the material is all consumed. If
you assume speed of sound then you can calculate the time to cover 40
feet and compare that with the time taken for all the material
(including that now within the whole sphere) to fully be consumed.
> But air bursts cannot shear structural beams beyond 5 meters.
It wasn't an 'air burst'. Your assertion about structural beams and
distance is a vast generalisation that may, or may not, be applicable
in any particular case, and it certainly isn't true. For example at
Hiroshima the air burst did shear structural beams well beyond 5
meters distance.
The criteria is not whether you think it should happen or whether you
have a better idea, it is whether there is enough force or not. You
specified some figures and I showed that those conditions were enough
to shear some beams.
| |
| Robert Wagner 2004-05-26, 10:30 pm |
| riplin@Azonic.co.nz (Richard) wrote:
>robert.deletethis@wagner.net (Robert Wagner) wrote
>First of all these explosions were not 'air blasts'. They didn't
>expand in a sphere, at most it was a semi-sphere. In some cases there
>were buildings on both sides which further constrained the explosion
>making it less than a semi-sphere.
>
>Secondly it only reverts to 'cube-root' _after_ all the explosive
>material has been consumed. This does happen in uncontained air
>blasts at some distance. In the case of fertilizer bombs the
>explosion is relatively slow compared with the distances involved and
>in these cases the pressure wave is closer to following the square
>root. What happens is that the 'pressure wave' expands with the area
>of the sphere, not its volume, until the material is all consumed. If
>you assume speed of sound then you can calculate the time to cover 40
>feet and compare that with the time taken for all the material
>(including that now within the whole sphere) to fully be consumed.
Your claims are credible, especially those about the low velocity of ANFO. The
only way to settle this is to conduct an experiment. Wait .. someone already
did. He found a similar building, albeit without similar neighboring buildings,
and set off a similar bomb. It didn't come close to shearing the structural
support.
> For example at
>Hiroshima the air burst did shear structural beams well beyond 5
>meters distance.
The explosive force was several orders of magnitude greater than chemical
explosives.
>The criteria is not whether you think it should happen or whether you
>have a better idea, it is whether there is enough force or not. You
>specified some figures and I showed that those conditions were enough
>to shear some beams.
Others have shown by experiment that the force was insufficient.
Error correction: 'Big Tom' should have been 'Black Tom'. The distance from
Liberty Park in Jersey City to Times Square should have been 4-5 miles rather
than 2 miles.
| |
| Richard 2004-05-27, 3:30 am |
| robert.deletethis@wagner.net (Robert Wagner) wrote
> and set off a similar bomb. It didn't come close to shearing the structural
> support.
How was it determined that 'it didn't come close' ? How would you
tell the difference between 20% and 99% ?
> The explosive force was several orders of magnitude greater than chemical
> explosives.
So, you agree that you were wrong that an air burst couldn't do that ?
As I said it is not whether you like it or not, it is entirely whether
the forces are sufficient or not.
> Others have shown by experiment that the force was insufficient.
Well, no, they could not actually show that. They don't know _exactly_
what force was required, or what _exact_ force was available at that
point. By doing many experiments they may come up with 'unlikely' or
'possible' or similar.
You had quoted figures which were sufficient.
> Error correction: 'Big Tom' should have been 'Black Tom'. The distance from
> Liberty Park in Jersey City to Times Square should have been 4-5 miles rather
> than 2 miles.
And dividing by the cube of 5 miles gives what ????
| |
| Robert Wagner 2004-05-27, 1:30 pm |
| riplin@Azonic.co.nz (Richard) wrote:
>robert.deletethis@wagner.net (Robert Wagner) wrote
>
>
>How was it determined that 'it didn't come close' ? How would you
>tell the difference between 20% and 99% ?
By inspecting damage to intermediate objects. Both the test and real buildings
had weaker beams on their periphery, midway between the explosion and the core
column. In both cases peripheral beams were not damaged by the explosion. If it
couldn't damage them, 'it didn't come close' to shearing the stronger and more
distant core beam.
>
>Well, no, they could not actually show that. They don't know _exactly_
>what force was required, or what _exact_ force was available at that
>point. By doing many experiments they may come up with 'unlikely' or
>'possible' or similar.
Steel manufacturers publish tables giving the force required. Theory and
experiment give the force available. The difference is not subtle, it's on the
order of 100 (required) to 1 (available).
>And dividing by the cube of 5 miles gives what ????
Barely enough to break windows. The Black Tom incident involved more than two
million pounds exploding over a period of several hours. The biggest single
explosion came from 100,000 pounds of TNT. Although it was in a residential
area, only two adults and a baby were killed (from falling out of a crib).
|
|
|
|
|