It looks like you're using an Ad Blocker.

Thank you.

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

Help ATS via PayPal:

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

page: 18
14
share:

posted on Feb, 23 2016 @ 06:02 AM

I'm no good at math but I think you are forgetting terminal velocity here, like I pointed out several times before. But hey, just keep ignoring that while posting your flawed calculations.

You also seem to agree that the plane would not be near topspeed at the top of the parabole.

Did you take this into account?

Use the terminal velocity formula, v = the square root of ((2*m*g)/(ρ*A*C)). Plug the following values into that formula to solve for v, terminal velocity.[1] m = mass of the falling object g = the acceleration due to gravity. On Earth this is approximately 9.8 meters per second per second. ρ = the density of the fluid the object is falling through. A = the projected area of the object. This means the area of the object if you projected it onto a plane that was perpendicular to the direction the object is moving. C = the drag coefficient. This number depends on the shape of the object. The more streamlined the shape, the lower the coefficient. You can look up some approximate drag coefficients here.

www.wikihow.com...

The fatal flaw, with the 9.8 m/s2 you keep using in your "calculations"

The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by ɡ0 or ɡn, is the nominal gravitational acceleration of an object IN A VACUUM near the surface of the Earth

Furthermore, what if the parabola is made larger so that the top section is bigger?
edit on 23-2-2016 by DutchMasterChief because: (no reason given)

posted on Feb, 23 2016 @ 08:32 AM

originally posted by: DutchMasterChief

I'm no good at math but I think you are forgetting terminal velocity here, like I pointed out several times before. But hey, just keep ignoring that while posting your flawed calculations.

why do you keep bringing up terminal velocity?? i have told you i am ignoring terminal velocity because:
1, i cant confirm terminal velocity of an a300
2, if i was to guess terminal velocity it would be lower than the velocity required to maintain 9.8m/s^2 downwards for 42 seconds.

You also seem to agree that the plane would not be near topspeed at the top of the parabole.

Did you take this into account?

im assuming the vomit comet would be around 250 knots at the top of the parabola.. do you realise if it was doing mach 0.82 /0.86 at the top of the parabola it would mean its final velocity (after 1 minute of zero g) would be even greater.

The fatal flaw, with the 9.8 m/s2 you keep using in your "calculations"

you need to understand this..
terminal velocity works AGAINST your argument.. thus why i have ignored it.

if the aircraft will most likely hit terminal velocity before the 1 minute zero g experience will be over..
have you wondered why i am saying that this 1 minute zero g experience is impossible in the vomit comet and why im calculating these ridiculously high velocities??

Furthermore, what if the parabola is made larger so that the top section is bigger?

the only way to complete a larger parabola AND maintain zero g through it will be to maintain a HIGHER airspeed during the maneuver..

are you sure you want to increase the airspeed even more??

p.s. my calculation of final velocity is a little wrong as maintaining 45 degrees nose down might make the occupants touch the floor they might have to use varying angle nose down. final net velocity should be about 1563km/hr if they were to maintain a 500km/hr horizontal velocity component throughout the dive.

1563 is still far too excessive for an aircraft designed for subsonic flight.

posted on Feb, 23 2016 @ 11:14 AM

So even after it was pointed out to you that the number you base your calculations is based on falling in a vacuum, and that you are ignoring several other factors, you still insist that your speed calculations are correct?

posted on Feb, 23 2016 @ 11:26 AM

ISS is a international thing (hint: it's name? International Space Station.

What and the "Coalition of the Willing" didn't invade the Middle East on spurious WMDs "evidence"

Just because other countries are involved doesn't mean they aren't in collusion.

posted on Feb, 23 2016 @ 04:19 PM
No.

Tell me, how is it the ISS always manages to appear exactly where these telescopes expect it to be? That wouldn't be the case for random observers scattered around if it were just a "plane" flying in the atmosphere; topocentric would reveal the hoax.

posted on Feb, 23 2016 @ 08:12 PM

originally posted by: ngchunter
No.

Tell me, how is it the ISS always manages to appear exactly where these telescopes expect it to be? That wouldn't be the case for random observers scattered around if it were just a "plane" flying in the atmosphere; topocentric would reveal the hoax.

As someone who has watched the ISS hundreds of times (i even stop people in the street to point it out when I can. Most people are interested)..i am blown away by those videos. Absolutely incredible to capture so much detail from the ground.

The other images I can't stop staring at are the ones where it and/or the shuttle are silhouetted by the against the sun. Stunning.

posted on Feb, 23 2016 @ 08:29 PM

originally posted by: DutchMasterChief

So even after it was pointed out to you that the number you base your calculations is based on falling in a vacuum, and that you are ignoring several other factors, you still insist that your speed calculations are correct?

what you are not understanding is that i am PURPOSEFULLY IGNORING alot of factors..

why am i ignoring these factors?? because it would make your scenario of the ISS filmed inside the ISS absolutely IMPOSSIBLE.

i realise that falling at 9.8m/s^2 for 42 seconds can only be accomplished in a vacuum.. this only leads to your scenario being MORE unlikely..

ive tried to work out an ideal situation to SATISFY YOUR belief.. and you are the one now finding all the holes with YOUR belief..

all those factors that you have mentioned would only slow the aircraft down making it harder to reach those velocities.. what you dont get is that the aircraft must reach those velocities in order to maintain zero g for the occupants for 42 seconds.

new topics

top topics

14