Originally posted by C46driver
Originally posted by backinblack
reply to post by Human_Alien
To some, like Weed. I must be on one side or the other, black & white is all they see..
Like most pilots weed is paying attention to ops manuals, SOP's and the written word. If we go down that conspiracy road we probably won't live to fly
another day.
Let's forget op specs and SOPs and such and get down to the job of designing an aircraft to fit the mission profile that fits witness statements.
WEIGHT CARRYING CAPACITY. Photos show trails that are about the same size as one half of one horizontal stabilizer on a Boeing 757. Full span of the
horizontal stab on a 757 is 49'. So we have a trail 12.5' in diameter. So the cross section area is 6.25^2 * pi, or 123 square feet, or 11.4 square
meters. Witnesses describe trails that go from horizon to horizon. Line of sight from 40,000' is 265 miles each way, or 530 miles, or 853,000 meters.
So the volume of the trail is 9,724,000 cubic meters. To get enough opacity to be clearly visable from 40,000' away, you would need on the order of
20% by volume of the sprayed substance. Since most claim it's aluminum (Al), let's use that. So you need 1,944,800 cubic meters of aerosolized Al.
For ease of doing this stuff in my head, let's round to 2 million m^3. Since density is given in grams/cubic centimeter, we multiply 2.8 g/cm^3 times
the volume of the Al and get 5,600,000 grams or 56,000 kg. Now pictures of "chemtrails" often show four trails being left, so to leave 4 visible
trails of Al from horizon to horizon, you would need to haul 224,000 kg. Since cost is always an object in aircraft design, we would like to use
something already flying as a starting point. Oh-Oh - big problem. Nothing flying today can haul that payload. A 757 maxes out at 114,000 kg. The
C-5, although much bigger, is limited to 118,000 kg. The king of the heavy lifters, the Russian Antanov An-124, can go a hefty 136,000 kg. Houston,
we have a problem.
SIZE. We need to carry 8 million cubic meters of the stuff. That is, if the particles are compressed to a solid, a block measuring 200 meters on
each side. Drat! Another problem. the 757 interior is only 36 meters long and 3.5 meters wide. And we need an aircraft cabin 200 meters long and
200 meters wide. Maybe we should talk to the Russians. Nope, no help there. The Antanov cabin interior is also only 36 meters long.
WING LOADING. To be able to fly and not have the wings fall off, we need to keep wing loading, that is the total weight of the beast divided by the
wing area, at about the 700 kg/m^2 level. The 757 is 661 kg/m^2. If the empty aircraft is really light, let's say it can carry its own weight, and
the only thing I've ever flown that could do that was the Douglas A-1 Skyraider, gross weight would be about 450,000 kg, so we need about 650 square
meters of wing area. In order to operate at altitude we will need an aspect ratio (span/chord) length of 8 or so. So we use the formula 8x*x=650,
and see that we need a wing span of 200 meters and a chord of 25 meters. Holy embiggenate Batman! That's 600 feet plus change. That's four 757s
wingtip to wingtip. Or maybe I should say winglet to winglet if we're talking the 757-200WL or 757-300 aircraft.
Has anybody noticed a 600 foot long, 1200 foot wide(including cabin), and 600 foot tall behemoth cruising the friendly skies lately?
Living an empirically based reality is such a buzz kill.
edit on 30-1-2011 by 4nsicphd because: not enough sleep to spell. On layover in
Dubai