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Is the Internatiomnal space station actually within an ordinairy airplane..

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posted on Feb, 17 2016 @ 12:38 PM
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Just because a video might be shortened or several pieces put together isn't proof of what it contains is false.

If that were the rule then almost all videoed events are false.




posted on Feb, 17 2016 @ 01:02 PM
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a reply to: roadgravel

Correct, but this vid doesn't prove it wasn't shot in a plane during parabolic flight.



posted on Feb, 17 2016 @ 01:29 PM
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originally posted by: DutchMasterChief
a reply to: roadgravel

Correct, but this vid doesn't prove it wasn't shot in a plane during parabolic flight.



And it doesn't prove it was.

The likelihood of an empty ISS orbiting is so close to zero that for reality it is zero.



posted on Feb, 17 2016 @ 01:58 PM
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a reply to: roadgravel




And it doesn't prove it was.


I never said it did. Others did say that it proves that it wasn't shot in a plane.



posted on Feb, 17 2016 @ 09:05 PM
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a reply to: DutchMasterChief

it kind of does..

between the cuts the footage gets upto almost 1 minute long of zero gravity footage..

it is physically impossible for the vomit comet to maintain zero gravity for that long..

even if you consider that the ace pilot was able to put the aircraft into free fall after introducing zero g to the occupants to maintain zero g for the initial regular 25 seconds. by the time the remaining 30+ seconds of free fall zero g is over, the aircraft will be well well over mach 1.. and you expect the ace pilot to safely pull 2g-2.5g out of that without hitting the ground??



posted on Feb, 18 2016 @ 10:18 AM
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a reply to: choos




between the cuts the footage gets upto almost 1 minute long of zero gravity footage..


True, but the floating water, which would be the hardest to fake, is never seen longer than 32 sec.




even if you consider that the ace pilot was able to put the aircraft into free fall after introducing zero g to the occupants to maintain zero g for the initial regular 25 seconds. by the time the remaining 30+ seconds of free fall zero g is over, the aircraft will be well well over mach 1.. and you expect the ace pilot to safely pull 2g-2.5g out of that without hitting the ground??


How did you get to a speed of mach 1?



posted on Feb, 18 2016 @ 10:52 AM
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a reply to: DutchMasterChief


How did you get to a speed of mach 1?


because in a parabolic dive [ theres a hint in the name ] - the airfraft has to descend FASTER than the " weightless " crew

i use " weightless " as they are infact falling and still accelerating at 9.8m/s2

its just that if the aircraft falls faster they SEEM to be weightless

thats why there are limits on how long parabolic dives can last



posted on Feb, 18 2016 @ 11:03 AM
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a reply to: ignorant_ape




because in a parabolic dive [ theres a hint in the name ] - the airfraft has to descend FASTER than the " weightless " crew


I am not in need of your hint. But just for fun, what hint is in the name that has to do anything with what you said right now?

If the plane was descending faster they would be stuck to the ceiling......

Again I ask, what is that mach 1 based on? Who says they weren't flying a bit above stall speed before the dive?


edit on 18-2-2016 by DutchMasterChief because: (no reason given)



posted on Feb, 18 2016 @ 04:52 PM
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Where is this plane's hanger and what air field is it flown out of. Can't be a commercial airport or it wouldn't be secret after all these years.



posted on Feb, 18 2016 @ 08:14 PM
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a reply to: DutchMasterChief


Again I ask, what is that mach 1 based on? Who says they weren't flying a bit above stall speed before the dive?


it wont be abit above stall speed as that would make their zero g time less..

using the parabolic flight path enables them to begin zero g while they are still going upwards it also enables them greater control to maintain zero g..

staying a bit above stall speed means they wash off nearly all of their speed as they go up (which has stilll yet to begin the zero g experience) and just before stall try to attempt to maintain zero g without stalling because as soon as they stall they dont have control of the craft. and not to mention near stall speeds the aircraft becomes quite unresponsive.. so i dont know how you are going to maintain zero g at that speed.

and also stall speed for a large aircraft is not exactly low especially at very high altitude.

and the aircraft will need to fall faster than free fall because inside there is no air resistance, so they would need to maintain zero g within the aircraft by flying faster than free fall downwards. if the aircraft was to fall at free fall speeds the occupants would be flat against one of the walls most likely the cockpit walls.

p.s. the mach 1 is based on falling at 9.8m/s^2 for 30 seconds.
edit on 18-2-2016 by choos because: (no reason given)



posted on Feb, 20 2016 @ 08:57 AM
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a reply to: choos

Ok, this is a pretty complicated subject and I've been doing some research to get a better understanding.




it wont be abit above stall speed as that would make their zero g time less..


During the upward part of the parabole that creates "zero gravity" there is actually deceleration, so it is flying relatively slow once it reaches the top of the parabole.


(Contrary to popular misconception, the 0 g freefall phase of flight begins as the aircraft climbs, and does not occur solely as the aircraft descends. Although the aircraft has upward velocity during the initial 0 g phase, its acceleration is downward: the upward velocity is decreasing.)



and the aircraft will need to fall faster than free fall because inside there is no air resistance, so they would need to maintain zero g within the aircraft by flying faster than free fall downwards. if the aircraft was to fall at free fall speeds the occupants would be flat against one of the walls most likely the cockpit walls.


Again how could the plane be falling faster than its occupants if they are floating freely inside of it. They are obviously falling at the exact same speed, that's the whole point.


Parabolic flight generates freefall by following a trajectory wherein the acceleration of the aircraft cancels the acceleration due to gravity (Figure 1), along the aircraft vertical (z) axis. Essentially, if the aircraft and its occupants "fall" together at 9.81 m/s2, "0 g" is achieved, where there is no reaction force





p.s. the mach 1 is based on falling at 9.8m/s^2 for 30 seconds.


What initial velocity do you base this on?


During such parabolic flight an aircraft flies a trajectory that provides freefall for up to 40 seconds


Like I said before, it can be pushed to at least 40 seconds, they base this on a F-94.


Between 1955 and 1958, a refined approach in the F-94 fighter allowed a variety of medical experiments to be performed during 30 to 40 seconds of freefall [3]. Between 1957 and 1959, the much larger C-131B cargo transport allowed simultaneous experiments on multiple subjects [4] and sufficient room for Mercury program astronauts to train (Figure 2), although this slower, propeller-driven aircraft could only produce parabolas with 10 to 15 seconds of freefall.


Note that an F-94 has a lower top speed than an A300.

www.ncbi.nlm.nih.gov...

edit on 20-2-2016 by DutchMasterChief because: (no reason given)



posted on Feb, 20 2016 @ 09:20 AM
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originally posted by: DutchMasterChief

During the upward part of the parabole that creates "zero gravity" there is actually deceleration, so it is flying relatively slow once it reaches the top of the parabole.


not really

the quote you have taken has broken apart the airplanes velocity into two parts, the horizontal velocity and the vertical.
and is referring to the vertical part only.

so while the aircraft is trying to maintain its airspeed, its vertical component of the airspeed is decreasing as it begins to nose down reaching the peak then the vertical component will rise again as it is facing down.



Again how could the plane be falling faster than its occupants if they are floating freely inside of it. They are obviously falling at the exact same speed, that's the whole point.


this is after the parabola, since the parabola flightpath will only give about 25 seconds of apparent weightlessness.

and free fall speeds is fixed for different frontal surface areas because of wind resistance, since inside the aircraft there is no wind resistance the occupants would fall at 9.8m/s^2. meaning the aircraft needs to drop at the rate which will sooner or later pass free fall speeds.

to explain your quote,

what the vomit comet does is that it tries to maintain airspeed while bringing the nose down while climbing.
this has the effect of making the floor of the aircraft move away from the occupants.
the aim of the pilot is to move the floor away at the same rate that the occupants fall, so that both the aircraft is falling at 9.8m/s^2 and so are the occupants.

p.s. free fall is not falling at 9.8m/s^2 it initially is but eventually wind resistance stops acceleration, leaving them at constant speed.



p.s. the mach 1 is based on falling at 9.8m/s^2 for 30 seconds.


What initial velocity do you base this on?

i base it on about 250 knots, its a wild guess but im not trying to be accurate.


Like I said before, it can be pushed to at least 40 seconds, they base this on a F-94.


the problem with that is that 40 seconds can be reached by smaller more powerful aircraft (better power to weight).

generally the bigger the aircraft the less zero g time achievable..

unless they filmed all ISS scenes in a F-94 fighter jet??


Note that an F-94 has a lower top speed than an A300.



note its also much smaller and lighter.
edit on 20-2-2016 by choos because: (no reason given)



posted on Feb, 20 2016 @ 09:22 AM
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a reply to: roadgravel




Where is this plane's hanger and what air field is it flown out of. Can't be a commercial airport or it wouldn't be secret after all these years.


Keeping what secret exactly?

Hypothetically, if this plane actually used by ESA, is used for faking ISS footage, what would the problem on a commercial airport. No outsiders can look inside of the plane.


Operating out of Bordeaux-Mérignac Airport, the aircraft usually flies up to a specially assigned air corridor above the Gulf of Gascogne. Flying level at about 6000 metres the weightlessness manoeuvre – known as a parabolic arc – begins when the aircraft is sent into a steep 45 degree climb on full engine thrust.


www.esa.int...


edit on 20-2-2016 by DutchMasterChief because: (no reason given)



posted on Feb, 20 2016 @ 09:26 AM
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a reply to: choos




the problem with that is that 40 seconds can be reached by smaller more powerful aircraft.


Just a quick reply,the F-94 is not more powerful, since it is way smaller than an A300, yet it is slower....




this is after the parabola, since the parabola flightpath will only give about 25 seconds of apparent weightlessness.


But you used it as an argument refering to the actual "zero gravity" period. So that's some BS.




the quote you have taken has broken apart the airplanes velocity into two parts, the horizontal velocity and the vertical. and is referring to the vertical part only.


It is flying in an upwards angle. So any vertical deceleration means horizontal deceleration too.
edit on 20-2-2016 by DutchMasterChief because: (no reason given)



posted on Feb, 20 2016 @ 09:33 AM
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originally posted by: DutchMasterChief
a reply to: choos




the problem with that is that 40 seconds can be reached by smaller more powerful aircraft.


Just a quick reply,the F-94 is not more powerful, since it is way smaller than an A300, yet it is slower....




this is after the parabola, since the parabola flightpath will only give about 25 seconds of apparent weightlessness.


But you used it as an argument refering to the actual "zero gravity" period. So that's some BS.



yep i meant power to weight ratio is higher.

also.. im mistaken regarding the free fall..

technically while inside the aircraft they would be free falling as there is barely any air resistance.

i was thinking free fall as in sky diving free falling.

so yes, the aircraft would be falling at the same free fall acceleration of the occupants inside which would be 9.8m/s^2



But you used it as an argument refering to the actual "zero gravity" period. So that's some BS.


you might be misunderstanding..

the top part o the parabola flight path is the "normal" zero g flight experience.

i was considering free fall AFTER the parabola flight path.
thats where i begin the 30 seconds of falling at 9.8m/s^2..

during the top parabola flight path the aircraft is still flying and i am assuming that this is the initial velocity of about 250 knots.
edit on 20-2-2016 by choos because: (no reason given)



posted on Feb, 20 2016 @ 09:39 AM
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a reply to: choos




yep i meant power to weight ratio is higher.


No, otherwise it would be faster, ignoring aerodynamics. All that matters here is that the A300 is faster, period.



posted on Feb, 20 2016 @ 09:48 AM
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a reply to: DutchMasterChief

Comparing the F-94 to the A300 is like comparing a Corvette to the family station wagon. The F-94, in aerodynamic terms was much more powerful than the A300, and could stay in free fall longer because it was designed that way.

Fighters are designed to withstand more positive and negative Gs than transports. That means it could stay in a zero G situation and pull more Gs to recover. The A310 used pulls 1.8Gs at the bottom of the parabola. They're limited to 2.5Gs maximum. An F-94 pulled 6+ routinely with no problems.



posted on Feb, 20 2016 @ 09:49 AM
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originally posted by: DutchMasterChief
a reply to: choos




yep i meant power to weight ratio is higher.


No, otherwise it would be faster, ignoring aerodynamics. All that matters here is that the A300 is faster, period.


it also weighs a hell of alot more..

it also isnt purely about speed.. its about maintaining flight at zero g.

a large heavy aircraft has enough trouble maintaining high altitudes and it needs to climb at speed as well?



posted on Feb, 20 2016 @ 09:49 AM
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a reply to: DutchMasterChief

No, there are a few dozen other factors that matter. Top speed is great on paper, but doesn't mean much.



posted on Feb, 20 2016 @ 09:54 AM
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a reply to: Zaphod58

The A300 I was talking about is reinforced specially for withstanding G forces.




Comparing the F-94 to the A300 is like comparing a Corvette to the family station wagon.


Bad example since in this case the family station wagon is actually faster than the Corvette....




The F-94, in aerodynamic terms was much more powerful than the A300, and could stay in free fall longer because it was designed that way.


Aerodynamically more powerful? What is that?



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