It looks like you're using an Ad Blocker.

Please white-list or disable AboveTopSecret.com in your ad-blocking tool.

Thank you.

 

Some features of ATS will be disabled while you continue to use an ad-blocker.

 

Maths and pi test.

page: 1
14

log in

join
share:

posted on Jul, 17 2009 @ 05:57 PM
link   
Okay this will have some relevance and is open for everyone, some rough calculations based on the width of a tower (64 metres) and rounding off the ball of smoke to a spherical shape (a diameter of 150 metres, win some lose some here but a pretty good estimate) What is the volume?.......



If you think this will be a far from accurate estimate please state before hand, like I say it will bear some significance when enough reply
.




posted on Jul, 17 2009 @ 06:30 PM
link   
I sure am wondering about the accuracy and significance of such a calculation. I'm not sure what the conditions are inside the cloud of dust - is there a lot of air in there? Is the air (and contents) compressed? Without knowing these details, an indication of volume seems somewhat useless. But I'm curious as to what your plans are, and I do loveth the maths.. so here's a formula:V = 4/3 × pi ×r^3 (see this mathy page).

So for a sphere with a diameter of 150m / radius of 75m, there is a volume of 1767145.9 m^3. A sphere with a diameter of 222m has a volume of 5728719.3 m^3. For fun, the average volume between both spheres is 3747932.6 m^3.

What happens next



posted on Jul, 17 2009 @ 06:41 PM
link   
Well if we keep it perfectly spherical with a diameter of 150 metres = 75 metres radius, now iirc it`s radius x3 x 4 3rds then again times by pi 3.14159265 = 1767145.865625 cubic metres.

Hope that`s right lol.

EDIT: P.S.

P.S. I will see if a few more guys reply, then i`ll reveal why I done this
.

[edit on 16/07/2009 by Seventh]



posted on Jul, 17 2009 @ 07:11 PM
link   
reply to post by scraze
 


4/3(Pi * r^3) Scraze has the formula and the math right.



posted on Jul, 17 2009 @ 07:13 PM
link   
I have to go now so i`ll reveal why I done this, for now i`ll keep it short and sweet as there is a lot more to this than meets the eye. We have here just under 1,800,000 cubic metres of smoke very shortly after the initial impact, so at this time there wasn`t that much burning to add to the volume.

Grade A Jet fuel burns at the rate of 1 cubic metre ca. 10,000 cubic metres of smoke emitted gases, a fully loaded 767 - 200ER maximum fuel capacity is 90 cubic metres for a 12,000 km flight. This day it`s pay load was 11000 gallons which = around 40 cubic metres, 34 cbm for flight to L.A. + 6 cbm for 1 hour reserve.

It had used 6cbm in 1 hour flying = 34 cbm left, which in turn = 340,000 cbm of smoke gases, roughly 20% of the amount of the respective smoke ball we calculated the volume of, and at the very least does not fit in with the aviation fuel burning for long enough to weaken the towers theory.


Where the smoke came from that increased the what should have projected amount of smoke gases, is a completely other story
.



posted on Jul, 18 2009 @ 02:42 AM
link   

Originally posted by Seventh
It had used 6cbm in 1 hour flying = 34 cbm left, which in turn = 340,000 cbm of smoke gases, roughly 20% of the amount of the respective smoke ball we calculated the volume of, and at the very least does not fit in with the aviation fuel burning for long enough to weaken the towers theory.


Why are you repeating this complete nonsense? I corrected you in a previous post. Nobody has a theory that aviation fuel burned for the entire time the towers were on fire. Perhaps FEMA might have originally suspected it, but the NIST report has been available for a number of years now.

Why don't you read it and find out what the actual "official story" is instead of repeating something you heard off a truther site as if it was actually accurate (it's not).



posted on Jul, 18 2009 @ 02:58 AM
link   

Originally posted by Seventh
It had used 6cbm in 1 hour flying = 34 cbm left, which in turn = 340,000 cbm of smoke gases, roughly 20% of the amount of the respective smoke ball we calculated the volume of, and at the very least does not fit in with the aviation fuel burning for long enough to weaken the towers theory.


Your number, (340,000), is only the correct volume when you don't take into account the smoke expanding, mixing with the surounding air, moving with heat currents and the winds in the area.

In a real world application, the "volume" of the smoke would be much greater.



posted on Jul, 18 2009 @ 07:33 AM
link   

Originally posted by exponent

Originally posted by Seventh
It had used 6cbm in 1 hour flying = 34 cbm left, which in turn = 340,000 cbm of smoke gases, roughly 20% of the amount of the respective smoke ball we calculated the volume of, and at the very least does not fit in with the aviation fuel burning for long enough to weaken the towers theory.


Why are you repeating this complete nonsense? I corrected you in a previous post. Nobody has a theory that aviation fuel burned for the entire time the towers were on fire. Perhaps FEMA might have originally suspected it, but the NIST report has been available for a number of years now.

Why don't you read it and find out what the actual "official story" is instead of repeating something you heard off a truther site as if it was actually accurate (it's not).


Let me re-phrase what I originally wrote.... `Not all the fuel ignited on impact, thus the smoke cloud would not be at full potential due to the fuel being dispersed where ever`.

There, hope this clears this up, thanks for correcting me though.



posted on Jul, 18 2009 @ 07:40 AM
link   

Originally posted by XTexan

Originally posted by Seventh
It had used 6cbm in 1 hour flying = 34 cbm left, which in turn = 340,000 cbm of smoke gases, roughly 20% of the amount of the respective smoke ball we calculated the volume of, and at the very least does not fit in with the aviation fuel burning for long enough to weaken the towers theory.


Your number, (340,000), is only the correct volume when you don't take into account the smoke expanding, mixing with the surounding air, moving with heat currents and the winds in the area.

In a real world application, the "volume" of the smoke would be much greater.


Like I say it`s impossible to get an accurate assessment, but when a tried and trusted formula is at hand it`s worth a shot, the picture I used for this is a few seconds after impact, and the point i`m trying to get over here is.. As we all know there were 3 explosions simultaneously as the 2nd plane hit, and as the smoke gases formula reveals something added to the smoke substantially, reworked the original to show how big the smoke cloud would be with a 36 cbm pay load...........




posted on Jul, 18 2009 @ 10:18 AM
link   
reply to post by exponent
 


I think that pretty much debunks the "damage in the lobby was caused by jet fuel", as if anyone who seriously looked into things still thinks that. I'm not sure if thats where he was going with it...



posted on Jul, 18 2009 @ 10:23 AM
link   
interesting post for sure, thanks for sharing it.



new topics

top topics



 
14

log in

join