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# Ejecting steel: horizontal displacement

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posted on Sep, 14 2006 @ 09:20 PM
Can anyone tell me how I would calculate how far the steel can reach without any force other than gravity when it falls from a height of lets say 400 meters?

I'm being told that the formula to calculate that is (in dutch) h + v²/2 g

h=height
v=acceleration at start
g=gravitational acceleration

so for 400 meters, that would be..

400 + 0²/2 9.8

or about 400 meters of horizontal displacement without an outside force, what am I doing wrong? or is that correct?

posted on Sep, 14 2006 @ 09:23 PM
No, you won't get any horizontal distance if something is dropped straight down and acted on by only gravity. The horizontal displacement would be zero.

The equation for distance traveled is:

d = (1/2)at^2+vt

where a=acceleration, t=time and v=initial velocity

[edit on 9-14-2006 by Valhall]

posted on Sep, 14 2006 @ 09:29 PM
Thank you for clearing that up

posted on Sep, 14 2006 @ 09:51 PM
No no no.

dy=1/2at^2 will yield one dimensional distance travelled under constant acceleration in time, Valhall's suggestion only adds initial velocity. Horizontally, you're looking at a [rapid initial acceleration which we'll best not consider here and subsequent] constant velocity, yielding a horizontal displacement of dx=vt

First equation for dy=400m yields t=9s which you then enter into the second yielding dx=v times 9s. You see we're missing a variable here. Using for example dx=200m (how distant was the Wintergarden?), v would have been 22m/s.

[Disclaimer: These rules apply only in a vacuum and under standard earth acceleration]

In case anyone's confused:

dx = horizontal distance
dy = vertical distance
t = time in seconds
a = acceleration
v = horizontal constant velocity

[edit on 14-9-2006 by nQuire]

posted on Sep, 14 2006 @ 10:01 PM
For horizontal launch, the formula to work out distance achieved is:

d = v*SQRT(2h/g)

Where,
v is lateral velocity at launch
h is height from the ground
g is acceleration due to gravity

As Val said, gravity won't affect the horizontal displacement. I assume you hope to prove that the collapse of the twin towers wouldn't be sufficient to hurl the debris as far as was observed if gravity was the only force acting in the system. In order to do that, you'd have to perform an individual case study on a piece of debris. You'd need to know the exact mechanism by which the piece was ejected horizontally, and then show how that mechanism could not provide sufficient impulse to accelerate the debris to the required launch velocity. As you can see, nowhere near as simple as you perhaps first assumed.

posted on Sep, 14 2006 @ 11:53 PM
You'd also need to know from which height it emanated, or you'd be forced to always assume maximum possible height to make any valid point.

posted on Sep, 15 2006 @ 06:00 AM

Originally posted by nQuire
No no no.

dy=1/2at^2 will yield one dimensional distance travelled under constant acceleration in time, Valhall's suggestion only adds initial velocity. Horizontally, you're looking at a [rapid initial acceleration which we'll best not consider here and subsequent] constant velocity, yielding a horizontal displacement of dx=vt

First equation for dy=400m yields t=9s which you then enter into the second yielding dx=v times 9s. You see we're missing a variable here. Using for example dx=200m (how distant was the Wintergarden?), v would have been 22m/s.

[Disclaimer: These rules apply only in a vacuum and under standard earth acceleration]

In case anyone's confused:

[edit on 14-9-2006 by nQuire]

No one but you is confused. This is what he asked:

Can anyone tell me how I would calculate how far the steel can reach without any force other than gravity when it falls from a height of lets say 400 meters?

And that is what I answered. Now - if you want to answer a different question, that's all well and good, but blame the disconnect on some one else!

If he's wanting horizontal distance travel for a projectile (trajectory) motion, then he needs the following:

either initial vertical velocity (vy) and initial horizontal velocity (vx) OR the initial velocity vector (v) and the angle from the horizontal it is launching (b).

vx(0) = vcos(b)
vy(0) = vsin(b)

and then he'll need the vertical acceleration (ay) and the horizontal acceleration (ax), or again, the accleration vector along with the angle from the horizontal it is acting.

ax = acos(b) = 0 (should be zero)
ay = asinb(b) = gravity = g

y(t) = - (1/2)g*t^2 + vy(0)*t + h

where h is the original height from where the object is launched. Solve this equation for t at y(t) = 0 (in other words get the time for the object to hit the ground). Plug that time into the equation for horizontal distance traveled which is

x(t) = vx(0)*t

and solve for x...that will give the horizontal distance traveled at the time the object hits the ground (of course, these are all for a vacuum, but close enough).

Please note as the debris pile at the bottom gets bigger, you'll have to solve your timetag for y no longer equal to 0, but to the height of the debris pile (it won't make that big of a difference, but if you're going down this avenue, you might as well do it right).

[edit on 9-15-2006 by Valhall]

posted on Sep, 15 2006 @ 06:08 AM
Jeez.

I guess you haven't noticed, but wecomeinpeace came to the same conclusions as I did, only he combined the two formulas I provided into one. And as it was kind of implicit what Shroomery wanted to calculate - horizontal distance covered when falling 400m - I explained to him/her in simple terms. No need to complicate matters by looking at dynamic vectors, or is there?

So, who's confused now?

[edit on 15-9-2006 by nQuire]

posted on Sep, 15 2006 @ 06:34 AM
nQuire,

If we're going to start reading his mind, then let's read all of it. His initial question seems to imply there should be no outward force when a floor collapses on an object (floor, structure, shishkabob - whatever) below when falling under nothing but the force of gravity.

That's not true. It would take an pure elastic collision in order to impart no outward force on the impacted object. But - as you can see in the videos of the collapsing buildings - these were not pure elastic collisions.

For every action there is an equal and opposite reaction. And as soon as the impact became less than pure elastic via the failures of the individual components of the impacted floor (including but not limited to failed structural elements, pulverized concrete and sheetrock, snapped bolts and welds, and squished shishkabobs) there is now the ability for the equal and opposite reaction to cause upward and outward motion of freed components.

Much like smacking the crap out of say a 2"x2"x2" concrete cube with a hammer. Better have your safety glasses on - because some of it's coming back at you. The individual chunks freed in the impact will launch upward and outward in some - probably very predictable - trajectory angle. Also, just like the ejecti radius of an impact crater - while in that instance the energy imparted to the freed components is great enough to liquidize them and cause a fluid reaction, it is the very same phenomenon working on the concrete cube and hammer instance, as well as the floor smacking a floor instance - just of varying severity.

[edit on 9-15-2006 by Valhall]

posted on Sep, 15 2006 @ 05:51 PM

Originally posted by wecomeinpeace
For horizontal launch, the formula to work out distance achieved is:

d = v*SQRT(2h/g)

Where,
v is lateral velocity at launch
h is height from the ground
g is acceleration due to gravity

As Val said, gravity won't affect the horizontal displacement. I assume you hope to prove that the collapse of the twin towers wouldn't be sufficient to hurl the debris as far as was observed if gravity was the only force acting in the system. In order to do that, you'd have to perform an individual case study on a piece of debris. You'd need to know the exact mechanism by which the piece was ejected horizontally, and then show how that mechanism could not provide sufficient impulse to accelerate the debris to the required launch velocity. As you can see, nowhere near as simple as you perhaps first assumed.

Yes, I assumed it would not be as simple as the guy who presented the formula would have me belief.

For instance, to make such a calculation I would assume the weight of the object needs to be know. And if you want to calculate the 9/11 case in particular, you would need to know the force created by the collapsing floors.

But then again, as this is just air displacement created by the floors and hypothetically not by explosions, wouldn't it be negligible anyway compared to the weight of the steel that is being pushed away?

Thanks for all the formulas, they already gave me a headache, but I will try to get my head around them because I want some basic numbers so I can atleast show the absurdity of it instead of just claiming it.

As for the observed horizontal movement of the steel beams, I've heard numbers like 500 feet (David Ray Griffin) as well as 300 feet.

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