|
Post by petereldergill on May 30, 2007 12:23:26 GMT -4
Thats a fascinating piece. Orbital mechanics are "funny"...the most direct path requires some very counterintuitive thinking. Common sense definitely does not apply up there. I tell my students to reject common sense and trust the mathematics when I am teaching them probability and statistics. Two good examples of this are "The Birthday Problem" *Given a group of n people, what is the probability that at least two of them share the same birthday and "The Monty Hall Problem" * You have a choice to pick one of three doors, and there is a prize behind one of them. "Monty Hall" opens one of the doors you haven't chosen and shows that the prize is not behind that door. You are given the choice to stay with your choice or switch to the unkown door. Do you switch or not? Both of these problems are very counterinuitive. The first one probably because we very rarely know a large group of people and their birthdays (at 60 people, the probability is about 90% or so, I'd have to look up the exact figure) The second, you are supposed to change you decision and the odds are 2:1...common sense indeed Pete
|
|
|
Post by JayUtah on May 30, 2007 12:27:20 GMT -4
File under Self-Contradictory and Hypocritical and leave it at that.
|
|
|
Post by JayUtah on May 30, 2007 12:41:12 GMT -4
I tell my students to reject common sense and trust the mathematics when I am teaching them probability and statistics.
The whole reason science exists is because common sense (i.e., intuition) is so often wrong.
|
|
Jason
Pluto
May all your hits be crits
Posts: 5,579
|
Post by Jason on May 30, 2007 13:07:11 GMT -4
There is a card in a hat. The card is either the Ace of Spades or the King of Spades (50/50 probability of either). You take an identical Ace of Spades and throw it into the hat. You shake the hat around so both cards are mixed. Then you draw a card from the hat without looking. The card you drew is an Ace of Spades. What is the probability that the original card was an Ace of Spades?
The answer is a counter-intuitive 2/3.
|
|
|
Post by petereldergill on May 30, 2007 15:37:48 GMT -4
Conditional probability....ugghh..the bane of my existence...(a bit dramatic, I know..)
I wonder if your example is similar to the Monty Hall problem?
Pete
|
|
|
Post by gillianren on May 30, 2007 15:57:26 GMT -4
Yet "proof" of the vast conspiracy is routinely offered in the form of YouTube, etc. videos showing poor-quality and heavily restricted snippets of whatever the CTs want to wave their arms about. Is there such a thing as three-faced? Some aspects of the Goddess are, but not in a negative way. Just three literal faces. No, "conniving and devious" will do it, I think.
|
|
rocky
Earth
BANNED
Posts: 212
|
Post by rocky on May 30, 2007 17:17:39 GMT -4
I don't feel like you've explained this satisfactorily.
(reply #351)
I see a difference in the motion of the astronauts in the two clips.
In the Apollo 11 clips I'd bet they just halved the speed with no wire supports. In the later footage I'd bet that wire supports were used and the speed was probably sixty or sixty five percent of normal. This is just a rough estimation of course.
You people are the experts. Just explain it. Remember that there are probably lots of people with science backgrounds reading this thread.
|
|
|
Post by BertL on May 30, 2007 17:25:10 GMT -4
Explain what? Explain your observation "hey if you speed it up it kind of looks like it's done on earth"? That's, roughly, because of the gravity. If you speed up footage, things will fall down faster (because of a higher framerate rate - Frames Per Second - , but an equal amount of frames). If you speed it up the right way, there is a point where the film has sped up to a framerate where gravity behaves as 'fast'/'heavy' as it behaves on earth.
You are saying that if an object takes one second to drop on earth and two seconds to drop at another body, then you can film the object dropping on earth and then slow it down at a rate (not twice as slow; it doesn't slow down proportionally, your FPS has to be √X times as large, X being the difference between one gravity and another). But the opposite is also possible: you could film an object dropping on the other body, then speed it up so that it looks like it's falling on earth.
EDIT: The explanation is actually very simple. Changing the framerate also changes the perceptual gravity in a particular clip. It works with some gravitational consequences (like dropping objects), but other byproducts (like center of mass and other stuff) do not match up simply because you changed the framerate. That's why sometimes it does look like it is on earth when sped up, and sometimes it doesn't.
I could go to the moon, film myself walking, then go back to earth, speed it up twice and say "Look I found the fake NASA studio where they faked it!" Basically doing the same you think NASA did, only the other way around.
|
|
|
Post by Data Cable on May 30, 2007 17:27:09 GMT -4
In the Apollo 11 clips I'd bet they just halved the speed with no wire supports. Care to provide any evidence to support your speculation? Care to provide any evidence to support your speculation? Ok. You're looking at actual footage of actual astronauts actually walking on the moon. Explained.
|
|
Bob B.
Bob the Excel Guru?
Posts: 3,072
|
Post by Bob B. on May 30, 2007 17:35:37 GMT -4
and "The Monty Hall Problem" * You have a choice to pick one of three doors, and there is a prize behind one of them. "Monty Hall" opens one of the doors you haven't chosen and shows that the prize is not behind that door. You are given the choice to stay with your choice or switch to the unkown door. Do you switch or not? … you are supposed to change you decision and the odds are 2:1...common sense indeed I’ve thought about this question before and I think the odds depends on whether Monty acted randomly or if he knows where the prize is and purposely opens a door behind which there is no prize. The problem does not make it clear which case is applicable, though I suppose the latter is implied. In the latter case the odds are 1/3 that you picked the door with the prize and 2/3 that you did not. When Monty intentionally goes to a door without the prize and opens it, the probability you were right with your initial pick is still 1/3. The odds that the prize is behind the remaining door you did not pick are therefore 2/3. You double your chances of winning by switching your pick. (This of course assumes Monty will always open a door without the prize regardless of whether you were right the first time or not.) If the former case, the odds are 1/3 that you picked the door with the prize, 1/3 that Monty will open the door with the prize (since he doesn’t know were it is either), and 1/3 that it is behind the door neither of you selected. In the 2/3 of cases in which Monty doesn’t open the door with the prize, you have a 1/2 chance the prize is behind the door you picked and a 1/2 chance it is not. The odds are the same so it doesn’t matter whether you switch your selection or not. In the case where Monty acts randomly without prior knowledge of where the prize is, the situation described in the problem will happen only 2/3 of the time. If we are to assume the situation described in the problem occurs 100% of the time, then that can only be true if Monty knows were the prize is his actions are premeditated. edit spelling
|
|
Jason
Pluto
May all your hits be crits
Posts: 5,579
|
Post by Jason on May 30, 2007 17:52:22 GMT -4
The card-in-the-hat problem has a fairly simple answer. Given the question there are only three possible outcomes, all equally probable: 1. The original card was a King and you drew the same Ace that you threw in. 2. The original card was an Ace and you drew the same Ace that you threw in. 3. The original card was an Ace and you drew the original card.
Two of the three possible conditions have an Ace in the hat as the original card, therefore the odds are 2 in 3 that the original card was an Ace.
|
|
Jason
Pluto
May all your hits be crits
Posts: 5,579
|
Post by Jason on May 30, 2007 18:06:24 GMT -4
With the Monty Haul problem, you increase your chances of winning by switching only if Monty knows where the prize is and always opens a non-prize door, whether you've picked the right door to begin with or not.
Say you win if you pick the Ace of Spades out of a deck of playing cards. You pick a card but don't look at it. The dealer then looks at all the other cards, eliminates fifty of them that are not the Ace of Spades, leaving one card face down in front of him. He then offers to switch cards with you. It becomes pretty obvious you should switch in this case. The odds that you picked the right card were 1 in 52. The odds that the Ace of Spades is the single remaining card in the dealer's hand is 51 in 52. Same principle.
|
|
|
Post by captain swoop on May 30, 2007 18:23:52 GMT -4
I'd volunteer for that. My repertoire covers Teesside (and various sub-dialects), Hey, I just noticed U are in redcar, I am up the hill in Guisborough!
|
|
|
Post by JayUtah on May 30, 2007 20:33:22 GMT -4
This is just a rough estimation of course.
No, it's pure speculation.
You people are the experts. Just explain it.
The explanation is that you're speculating and have provided no actual evidence. That situation imposes no obligation on us or anyone else.
Remember that there are probably lots of people with science backgrounds reading this thread.
I'm one of them. That doesn't change the fact that you're speculating.
|
|
|
Post by gillianren on May 30, 2007 23:00:48 GMT -4
I don't feel like you've explained this satisfactorily. Yes, well, I don't think you've ever explained anything satisfactorily, but that's not going to alter your behaviour, now, is it?
|
|