Jump to content
The Education Forum

Did NASA Use 1/6 Gravity Simulators to Fake Apolllo ?


Recommended Posts

I'm looking forward to your next set of bogus equations very much Dave , as I have always found physics and math equations so interesting ... :lol:

I have a feeling by the time this particular debate is over with , you are gonna need several pints of extra cold Guinness ! :mellow:

That's the thing about equations, they have a happy knack of being verifiable, regardless of how dull you find them! I can assure you the equation is completely and 100% valid, no matter how bogus you think it is.

It's the figures you plug into that equation are measurable and hence subject to error. If I've made an error in measuring them I'll happily admit to it (though like I said, I used a wide margin for error to take that into account).

So while you're waiting for your think-tank to do the maths, what figures have you measured for the height of the drop, and the time of the drop?

Link to post
Share on other sites
  • Replies 93
  • Created
  • Last Reply

Top Posters In This Topic

That's the thing about equations, they have a happy knack of being verifiable, regardless of how dull you find them! I can assure you the equation is completely and 100% valid, no matter how bogus you think it is.

It's the figures you plug into that equation are measurable and hence subject to error. If I've made an error in measuring them I'll happily admit to it (though like I said, I used a wide margin for error to take that into account).

So while you're waiting for your think-tank to do the maths, what figures have you measured for the height of the drop, and the time of the drop?

Dave ... It's seems as if I mis-spoke in my last post to you ... It was not the equations that my "think tank" had a problem with , it was indeed your figures which seem to be in error .... So I stand corrected .

Speaking of my "think tank " it looks as if they are extremely busy and haven't had the time to do the maths yet .... So it might be me that needs those extra cold pints of Guiness , instead of you . :huh:

As for your question , I do believe the height estimate of the drop is the same as your figure ... As for the time , I will have to get back to you after I light a fire under a certain "think tank" :idea

Link to post
Share on other sites
Dave ... It's seems as if I mis-spoke in my last post to you ... It was not the equations that my "think tank" had a problem with , it was indeed your figures which seem to be in error .... So I stand corrected .

Speaking of my "think tank " it looks as if they are extremely busy and haven't had the time to do the maths yet .... So it might be me that needs those extra cold pints of Guiness , instead of you . :huh:

As for your question , I do believe the height estimate of the drop is the same as your figure ... As for the time , I will have to get back to you after I light a fire under a certain "think tank" :idea

No probs. Thanks for the update.

There's plenty of Guinness in the world to go round, so help yourself. It being a bank holiday Friday evening here in England, I'll bedoing the same. :o

Link to post
Share on other sites
  • 3 weeks later...
That's the thing about equations, they have a happy knack of being verifiable, regardless of how dull you find them! I can assure you the equation is completely and 100% valid, no matter how bogus you think it is.

It's the figures you plug into that equation are measurable and hence subject to error. If I've made an error in measuring them I'll happily admit to it (though like I said, I used a wide margin for error to take that into account).

So while you're waiting for your think-tank to do the maths, what figures have you measured for the height of the drop, and the time of the drop?

Dave ... It's seems as if I mis-spoke in my last post to you ... It was not the equations that my "think tank" had a problem with , it was indeed your figures which seem to be in error .... So I stand corrected .

Speaking of my "think tank " it looks as if they are extremely busy and haven't had the time to do the maths yet .... So it might be me that needs those extra cold pints of Guiness , instead of you . :cheers

As for your question , I do believe the height estimate of the drop is the same as your figure ... As for the time , I will have to get back to you after I light a fire under a certain "think tank" :flame

Duane

How is your "think tank" getting on? Three weeks and counting since I posted my conclusions.

Maybe they liked the Guinness more than their own conclusions?

;)

Link to post
Share on other sites
That's the thing about equations, they have a happy knack of being verifiable, regardless of how dull you find them! I can assure you the equation is completely and 100% valid, no matter how bogus you think it is.

It's the figures you plug into that equation are measurable and hence subject to error. If I've made an error in measuring them I'll happily admit to it (though like I said, I used a wide margin for error to take that into account).

So while you're waiting for your think-tank to do the maths, what figures have you measured for the height of the drop, and the time of the drop?

Dave ... It's seems as if I mis-spoke in my last post to you ... It was not the equations that my "think tank" had a problem with , it was indeed your figures which seem to be in error .... So I stand corrected .

Speaking of my "think tank " it looks as if they are extremely busy and haven't had the time to do the maths yet .... So it might be me that needs those extra cold pints of Guiness , instead of you . :cheers

As for your question , I do believe the height estimate of the drop is the same as your figure ... As for the time , I will have to get back to you after I light a fire under a certain "think tank" :flame

Duane

How is your "think tank" getting on? Three weeks and counting since I posted my conclusions.

Maybe they liked the Guinness more than their own conclusions?

;)

Just maybe your figures were not in error as Duane seems to think. Much harder to rebut something when the something you are trying to rebut is correct.

Link to post
Share on other sites
Just maybe your figures were not in error as Duane seems to think. Much harder to rebut something when the something you are trying to rebut is correct.

I suspect they have actually looked at the numbers and come up with similar estimates to myself, then realised they're in error about their original conclusion. If so, my suspicion is that the premise will change to "Ah, well they must have faked it by changing the playback speed of the video". Which is in direct contradiction to the original argument being put forth on Youtube, which was along the lines of "it was obviously filmed on Earth - you can tell just by looking at the footage and how seeing how quickly the objects drop".

However, I may be doing Duane's "think tank" an injustice, so I'll reserve judgement until they're forthcoming with their figures. If the figures are never forthcoming: well, that in itself will speak volumes.

Link to post
Share on other sites
  • 3 weeks later...

My "think tank" finally found the time to crunch the numbers to rebutt your figures ... I am posting what was sent to me verbatum .

Bag drop

Basic equations

V2 = V02 + 2a∆S

V = V0 + at

y = y0 + ス (V0 + V)t

y = y0 + V0 t + ス at2

So he claims it can’t be on Earth since the “-“ implies that the bag should need to go upward and he claims that’s not what we see in the video since we see it falling... How the hell does he think the bag got out of his backpack? His method requires too much precision and you sure don’t get that with the video…

Here is how we do it Greer!

∆S = 1,4 m (1,25 m is way too short if we consider the boots and the actornaut’s height but we’ll include it at the end)

tvideo = 0,83 s

We need to propose an initial velocity which will be used for both scenarios starting from -3 m/s to 1m/s. this will allow us to get the final velocity and therefore the theoretical time. And that time can be compared to the video.

so with 1,4 m in Earth’s scenario

V2 = V02 + 2a∆S

V2 = (-3)2 + 2(9,8)(1,4)

V2 = 36,44

V = √(36,44) = 6,04 m/s

Then let’s use Eq#2 to check out the theoretical time of descent:

V = V0 + at

t theoretical = (V- V0)/a = (6,04 – (-3))/ 9,8 = 0,92 second

Comparing for Earth scenario theoretical and video (experimental) time : 0,92 ≠ 0,83.

Now let’s compare with the other initial velocities

V0 = -2m/s, V0 = -1m/s, V0 = 0 m/s, V0 = 1m/s

t t = 0,78 s, t t = 0,65 s, t t = 0,53 s, t t = 0,44 s

So since we know NASA slowed down the video by nearly 50% to create the illusion of a lesser gravity, if we multiply our results by 2, it gives us:

1,84 s ; 1,56 s ; 1,30 s ; 1,06 s ; 0,88 s

So the closer one to the video time, it’s with initial velocity : 1 m/s which give us a theoretical time of 0,88 second when the time measured in the experimental video is 0,83 second. So basically, depending on the real initial height, for example, if it’s higher, the theoretical time for the Earth scenario will be higher, but if it’s really 1,25 m as suggested by Greer, than we have as theoretical time : 0,413 s but if we consider the slow-mo, the real deal is rather 0,83 s.

Now for the Moon scenario same thing but with G = 1,6 m/s2, same thing so let’s cut short and only look at the results. Whoever doubts them, do it yourself.

So for ∆S = 1,4 m the results are

T : 4,17 s ; 3,07 s ; 2,08 s ; 1,32 s ; 0,83 s

Now let’s suppose again that the initial height is 1,25 m :

T = 0,77 s

Conclusion: Bad case to solve the problem if they’re either on the Moon or on Earth. It could be one or the other since there’s not enough precision due to the quality of the video. A better case would be to analyze the hammer-feather video.

See below.

Feather-hammer video

∆S = 1,6 m (but more likely to be way more than that considering the boots and the actornaut’s height)

t video = 1,2 seconds (according to what Bert did with video Mach, and I agree with him)

Considering that Scott is 1,83 m (6’) (at least that’s what they claim at NASA, can’t trust them though)

Earth scenario:

Let’s find out the V (velocity when the objects touch the ground)

V2 = V02 + 2a∆S

V2 = ?, V02 = 0 , a = 9,8 m/s2, ∆S = 1,6 m

V2 = 0 + 2(9,8)(1,6)

V2 = 31,36

V = 5,6 m/s

Now let’s use Eq2 to get the time it would give us on Earth

t theoretical = (V- V0)/a = 5,6/9,8 = 0,57 second (or 0,571428571 s)

t video = 1,2 seconds ≠ 0,57 second BUT! We know they slowed down the video by at least 2 times so : 0,55 x 2 = 1,14 seconds. Still not 1,20 you might say? But let’s try on the Moon to see what we got.

Moon scenario:

Same thing as above but using G = 1,6 instead of 1,62 since we don’t use more than 1 decimal for Earth G.

V = 2,26 m/s (or 2,2627417 s)

Used in Eq#2

t theoretical = (V- V0)/a = 2,26/1,6 = 1,41 seconds (or 1,4125 s). Not quite 1,20 s and we can’t blame it on air-resistance can we?

Conclusion: definitely not on the Moon. What choice we got left? Ah yeah! Earth!

Post-conclusion: Anyone noticed that both objects in the hammer-feather video are not even at the same initial height even though they reach the floor at the same exact precise time as claimed by NASA , which is almost their biggest gun that they sent people to the Moon? Might just be a little detail of no importance…(sarcasm)

Link to post
Share on other sites
My "think tank" finally found the time to crunch the numbers to rebutt your figures ... I am posting what was sent to me verbatum .

Thanks for finally getting back to me on that one Duane. I'll post a detailed rebuttal in due course: but initial inspection shows whoever did the maths made a fundamental error that shows their lack of objectivity when analysing the footage.

So since we know NASA slowed down the video by nearly 50% to create the illusion of a lesser gravity, if we multiply our results by 2, it gives us:

Do they really expect to be taken seriously using statements such as this? Have they ever heard of the scientific method? They have just invented a "fudge factor" to get the result they want. Extremely misleading. They should be analysing the footage as is and seeing where the results lead them, not starting from a position that the footage was faked and fixing the numbers to try and prove their case.

Link to post
Share on other sites
Do they really expect to be taken seriously using statements such as this? Have they ever heard of the scientific method? They have just invented a "fudge factor" to get the result they want. Extremely misleading. They should be analysing the footage as is and seeing where the results lead them, not starting from a position that the footage was faked and fixing the numbers to try and prove their case.

That's quite a slick reply there Dave ... You have already pretended to debunk the math before you even supply any figures ... Nice trick ! :)

Link to post
Share on other sites
Do they really expect to be taken seriously using statements such as this? Have they ever heard of the scientific method? They have just invented a "fudge factor" to get the result they want. Extremely misleading. They should be analysing the footage as is and seeing where the results lead them, not starting from a position that the footage was faked and fixing the numbers to try and prove their case.

That's quite a slick reply there Dave ... You have already pretended to debunk the math before you even supply any figures ... Nice trick ! :)

Sorry Duane, but Angeazreal (or whoever formed your "think-tank") has tricked YOU. If you understood anything at all about science, you'd know why. Do you really think it is an objective approach to say "Ah, we know that NASA changed the film speed, so I'll just multiply all the figures I've measured by 2. Oh look, multiplying the figures by 2 proves it wasn't filmed on the moon! So, they MUST have changed the film speed! Ergo, I was right!"

Circular reasoning, and poor science, designed to prop up a faulty initial assumption. Don't fret sunshine, the numbers will be crunched very shortly!

Link to post
Share on other sites

It wasn't ange who supplied those numbers , it was a physics engineer who obviously didn't fall for nasa's tricks .

So don't you fret either sunshine , cuz whatever numbers you crunch here , they will be refuted again ... No matter how long it takes me to get them.

Link to post
Share on other sites
My "think tank" finally found the time to crunch the numbers to rebutt your figures ... I am posting what was sent to me verbatum .

Bag drop

Basic equations

V2 = V02 + 2a∆S

V = V0 + at

y = y0 + ス (V0 + V)t

y = y0 + V0 t + ス at2

So he claims it can’t be on Earth since the “-“ implies that the bag should need to go upward and he claims that’s not what we see in the video since we see it falling... How the hell does he think the bag got out of his backpack? His method requires too much precision and you sure don’t get that with the video…

Here is how we do it Greer!

∆S = 1,4 m (1,25 m is way too short if we consider the boots and the actornaut’s height but we’ll include it at the end)

tvideo = 0,83 s

We need to propose an initial velocity which will be used for both scenarios starting from -3 m/s to 1m/s. this will allow us to get the final velocity and therefore the theoretical time. And that time can be compared to the video.

so with 1,4 m in Earth’s scenario

V2 = V02 + 2a∆S

V2 = (-3)2 + 2(9,8)(1,4)

V2 = 36,44

V = √(36,44) = 6,04 m/s

Then let’s use Eq#2 to check out the theoretical time of descent:

V = V0 + at

t theoretical = (V- V0)/a = (6,04 – (-3))/ 9,8 = 0,92 second

Comparing for Earth scenario theoretical and video (experimental) time : 0,92 ≠ 0,83.

Now let’s compare with the other initial velocities

V0 = -2m/s, V0 = -1m/s, V0 = 0 m/s, V0 = 1m/s

t t = 0,78 s, t t = 0,65 s, t t = 0,53 s, t t = 0,44 s

So since we know NASA slowed down the video by nearly 50% to create the illusion of a lesser gravity, if we multiply our results by 2, it gives us:

1,84 s ; 1,56 s ; 1,30 s ; 1,06 s ; 0,88 s

So the closer one to the video time, it’s with initial velocity : 1 m/s which give us a theoretical time of 0,88 second when the time measured in the experimental video is 0,83 second. So basically, depending on the real initial height, for example, if it’s higher, the theoretical time for the Earth scenario will be higher, but if it’s really 1,25 m as suggested by Greer, than we have as theoretical time : 0,413 s but if we consider the slow-mo, the real deal is rather 0,83 s.

Now for the Moon scenario same thing but with G = 1,6 m/s2, same thing so let’s cut short and only look at the results. Whoever doubts them, do it yourself.

So for ∆S = 1,4 m the results are

T : 4,17 s ; 3,07 s ; 2,08 s ; 1,32 s ; 0,83 s

Now let’s suppose again that the initial height is 1,25 m :

T = 0,77 s

Conclusion: Bad case to solve the problem if they’re either on the Moon or on Earth. It could be one or the other since there’s not enough precision due to the quality of the video. A better case would be to analyze the hammer-feather video.

I'll stick to the video in question for the time being, since that is what we initially agreed to analyse. We can look at the hammer drop later.

∆S = 1,4 m (1,25 m is way too short if we consider the boots and the actornaut’s height but we’ll include it at the end)

tvideo = 0,83 s

My initial guesstimate at the height was actually 1.25m +- 0.2m, so his guess at the height falls within my range. Incidentally you did agree with me on the height.

(Note: when measuring things like this, it is always wise to provide an estimate of how accurate your measurements are. Does he mean 1.4 metres, 1.40 metres, 1.41 metres, 1.45 metres?)

However, to keep things simple, let's go with his estimate of the height without error bars. I notice he has also agreed with my estimate of the time at 0.83 seconds. So, we have:-

Height s = 1.4 m

Time t = 0.83 s

We need to propose an initial velocity which will be used for both scenarios starting from -3 m/s to 1m/s. this will allow us to get the final velocity and therefore the theoretical time. And that time can be compared to the video.

Very poor. There's no need at all to "guess" what the initial velocity, since we can calculate it. Look at the following equation of motion.

s=ut+1/2at2

What we know:

s = 1.4m

t = 0.82s

aearth=9.8ms-2

amoon=1.6ms-2

What we don't know:

u (initial velocity)

Just plug the figures in and calculate the initial velocity - simple!

So since we know NASA slowed down the video by nearly 50% to create the illusion of a lesser gravity, if we multiply our results by 2, it gives us:

Logical fallacy! Your "Physics Engineer" is simply begging the question. I know you think this is a Clavius term, but take a minute or two to educate yourself and find out what the term actually means.

Basically, "the argument fails to prove anything because it takes for granted what it is supposed to prove.", which is exactly what your think tank has done here. He's made the assumption that the video speed was halved to create the illusion of lesser gravity, and used that assumption to prove the same. Does not compute, Duane old boy. Your think tank have been laughed out of the debating hall.

Let's get to the crux of the matter. You don't understand the maths behind all of this, and have no way of knowing objectively who is right and who is wrong. I don't have the slightest problem with that, at least you made the effort to get "independent" analysis, even if it was fundamentally flawed. You don't need to be skilled at mathematics to realise the fundamental error he has made here, anyone non-mathemtcailly minded person possessing critical thinking skills should be able to see the error in his logic.

On top of that, you yourself stated earlier in this thread that the astronauts were on wires, and that the bag falls were clearly in 1g. Nothing at all to do with video being slowed down! That is what the whole purpose of this discussion as about: you stated that clearly the bag drops were in 1g, and not on the moon.

Source

They got the 1/6 g effects down very well for the astro-NOTS on the wires , but what do you think happened with those plastic bags that fell out of their pockets at 1 g speed ? ... I guess that part wasn't in the script ?

Source

greenmagoos did analyse the videos and put them on YouTube just like he got them from nasa ... and what they show ( in spite of all your fancy tap dancing ) are objects which fall at 1 g speed in an allegedly 1/6 g environment ... The Apollo videos speak for themselves .... and the word they all speak , very loudly and very clearly is ... F A K E !

So what do we have? Your think tank trying to use circular reasoning to prove the film speed was halved, which directly contradicts what you yourself say the vidoes shows (bag drops clearly in 1g). What happened when your think tank looked at the "filmed on the moon" scenario? They calculated the time of the drop at 0.77 seconds! This compares with a measured time of 0.83 seconds! (The measured time is actually 0.80 seconds, since there are 25 frames per second, not 24. No biggy.) Given that we guessed the height, and the errors in measuring the time, that is a perfectly acceptable result, well within experimental error! So what did he do? Tried to bluster his way out of it, saying it wasn't conclusive! IN OTHER WORDS, HE COULD NOT PROVE THE FILM WASN'T SHOT ON THE MOON! The only thing he managed was proving you wrong about the film being obviously shot in 1g - he agrees it wasn't. His conclusion is that it was either shot on the moon, or they filmed it on earth and slowed the speed down.

Sorry Duane old bean, but you've been debunked by your own think tank.

Link to post
Share on other sites

I sent your reply to my think tank ... They never saw my post about me thinking the bag dropped in 1 g , so that was not even part of the equation for them ... but it's ok if they debunked me , as it's debunking you that is more important .

I said that it looked like the bag dropped in 1 g because it fell too fast for it to have been in 1/6 g ... If you watch objects that are thrown on purpose in the Apollo videos , they move in slow motion ... Where as the bag dropping out of his pocket was not done on purpose , and it didn't fall in slow motion ... So if the astronots are moving in the slow motion of 1/6 g and the stuff they throw or drop ( like the hammer and feather ) move in the slow motion of 1/6g , then why didn't the bag drop out of his pocket in the slow motion of 1/6 g ?

That was the point I was making before all the math figures came into play .... but not knowing anything about this kind of math , it really is pointless for me to continue this ... and of course you knew that I knew nothing about this kind of math and that's why you chose to play this particular game .... Anything to make me look "ignorant" , right Dave ? .. Or maybe anything to get "one up" in the Apollo game would be more like it .

I got to hand to you though ... You are very good at this game and very quick with your 'rebuttals' ... Kind of makes me wonder who your think tank might be .

Link to post
Share on other sites
but it's ok if they debunked me , as it's debunking you that is more important .

Ah yea...screw the truth. What is important is thinking the Hoaxer worldview wins.

How open minded of you Duane.

Another chance to actually learn something blown.

Ignorance is bliss eh?

Link to post
Share on other sites
I sent your reply to my think tank ... They never saw my post about me thinking the bag dropped in 1 g , so that was not even part of the equation for them ... but it's ok if they debunked me , as it's debunking you that is more important .

Fair enough. The point I was making was, you were using this video as proof they obviously filmed in 1g. I think we can both agree that it's either filmed on the moon, or filmed on earth and slowed down. (There are other indicators to show the latter isn't the case, we can look at them later if you wish).

I said that it looked like the bag dropped in 1 g because it fell too fast for it to have been in 1/6 g ... If you watch objects that are thrown on purpose in the Apollo videos , they move in slow motion ... Where as the bag dropping out of his pocket was not done on purpose , and it didn't fall in slow motion ... So if the astronots are moving in the slow motion of 1/6 g and the stuff they throw or drop ( like the hammer and feather ) move in the slow motion of 1/6g , then why didn't the bag drop out of his pocket in the slow motion of 1/6 g ?

That's where analysis of the data using maths comes in. For it to have been filmed on the moon, it must have had a slight initial downward velocity from the point when it dropped. That appears to be backed up on the video - look at it very carefully, and the bag appears to roll forward along the astronauts sidepouch before it drops.

That was the point I was making before all the math figures came into play .... but not knowing anything about this kind of math , it really is pointless for me to continue this ... and of course you knew that I knew nothing about this kind of math and that's why you chose to play this particular game .... Anything to make me look "ignorant" , right Dave ? .. Or maybe anything to get "one up" in the Apollo game would be more like it .

Duane, I'm not interested in making you look ignorant about maths. I was interested in analysing the footage. I did that and provided my conclusions: you automatically rejected that before you'd even had the results back form your own think tank. Their analysis says it was either filmed on earth and slowed down, or filmed on the moon. Hence, you can't use this footage as proof that it wasn't filmed on the moon. That was my motivation for going along with this. You can call it "one up" if you like, but that's because the maths bore out what I knew all along. Yes, the dice were loaded in my favour, but that's because I'm convinced the footage was filmed on the moon.

I got to hand to you though ... You are very good at this game and very quick with your 'rebuttals' ... Kind of makes me wonder who your think tank might be .

I'll take that as a compliment since I consist of a think-tank of one - yours truly! I have an analytical mind when it comes to science, I'm good at maths, I enjoy solving puzzles. I've taught myself a lot about photography, and how light and shadows interact, in the last few years.

A few years ago I didn't know why or if Percy was wrong about things like missing fiducials, shadows pointing in wrong direction, mountains appearing to move from one photo to the next. That made me uncomfortable, for one of two reasons: either Apollo was a sham, or I was having the wool pulled over my eyes by Percy. It was only by increasing my own knowledge of things I previously knew little or nothing about that I was able to decide between the two, from a position of knowledge. I really do urge you to do the same. That way you can come to a conclusion on Apollo photos from an objective stance, rather than a subjective one.

I've no interest in making you look stupid, or ignorant about anything. However, I'm convinced your batting for the wrong side, so if you put yourself up as some kind of defender or proposer of either Percy or White, then that makes the message you're putting across fair game as far as I'm concerned. I realise you can sometimes take this as some kind of personal attack on you, probably because you have a lot of emotional investment in the Apollo hoax. It isn't. It's the message you're putting across that I'll keep attacking.

"Once more unto the breach, dear friends, once more; Or close the wall up with our English dead."

Happy debating! :)

Link to post
Share on other sites

Please sign in to comment

You will be able to leave a comment after signing in



Sign In Now
×
×
  • Create New...