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A simple question that many don't know the answer to

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posted on Jun, 28 2010 @ 08:53 PM
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Originally posted by spy66
The mass of the coin is not equal all around. That means your equation would only work in a vacuum? The force up is not equal to the force down with normal G.

Since there are only two sides to a coin. Its logic that a even number would be the right answer. There are only two possibilities.

But how would you know beforehand which side the coin would land on before you tossed it?

Your equation does not tell you that.

[edit on 27.06.08 by spy66]


It doesn't tell you unless you always know how much force is applied.

Since weight is only the amount of pull gravity has on an object which mass weight can substitute for mass.

You forget it landing on its edge. That's a third possibility.




posted on Jun, 28 2010 @ 08:55 PM
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Nice equation, but how are you going to know what the inputs are between the time the person begins the flip and when it's caught?

The answer is, you won't. So there goes your dream of 100% predictable coin tosses.

It's better to just have a double head coin.



posted on Jun, 28 2010 @ 08:56 PM
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Originally posted by Gentill Abdulla

Originally posted by spy66
The mass of the coin is not equal all around. That means your equation would only work in a vacuum? The force up is not equal to the force down with normal G.

Since there are only two sides to a coin. Its logic that a even number would be the right answer. There are only two possibilities.

But how would you know beforehand which side the coin would land on before you tossed it?

Your equation does not tell you that.

[edit on 27.06.08 by spy66]


It doesn't tell you unless you always know how much force is applied.

Since weight is only the amount of pull gravity has on an object which mass weight can substitute for mass.

You forget it landing on its edge. That's a third possibility.


If it lands on the edge, it still need to flip to one side. That has to be added to your equation. Because your equation is not complete until the coin shows heads or tail.

If you really want to know how many sides a coin has got. It has more than three sides. The coin has two more sides as well. That it can land on.

[edit on 27.06.08 by spy66]



posted on Jun, 28 2010 @ 08:57 PM
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Originally posted by harrytuttle
Nice equation, but how are you going to know what the inputs are between the time the person begins the flip and when it's caught?

The answer is, you won't. So there goes your dream of 100% predictable coin tosses.

It's better to just have a double head coin.


Probably just have all the other variables in the equation set out.

Then input force last when you get a reading. And find the answer.
(Like how they do it in sports science they get the force)



posted on Jun, 28 2010 @ 09:00 PM
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Originally posted by spy66


If it lands on the edge, it still need to flip to one side. That has to be added to your equation. Because your equation is not complete until the coin shows heads or tail.


By landing on its edge I mean like this...

Heads Tails
_____ _____ Edge l

Sorry about the size.

[edit on 28-6-2010 by Gentill Abdulla]

[edit on 29-6-2010 by Gentill Abdulla]



posted on Jun, 28 2010 @ 09:05 PM
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Originally posted by valhala
reply to post by Phlynx
 


Give me all input variables you think we need and I'll write a proggy


Coin mass, gravity, air pressure, angle of flip, what side it starts on, pressure of thumb with flip, and which area of the coin is hit with the thumb. Those would all be some variables. Also wind speed and direction.



posted on Jun, 28 2010 @ 09:12 PM
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reply to post by Gentill Abdulla
 


Let's have a look at (part) of variables here:

Definition: Coin is a 3D object in 3D space devised of 3 edges (sorry don't have proper name for acctual 3D body in english - tube with closed edges
)

Object: Coin with 3 edges is tossed to the air. What are the probabilities to hit either of edge?

Scenario 1: (environment we live in; with gravity): To hit 3rd edge (and for coin to stay there) possibility is as low as 1%. So for the other two possibilities we're left with 99% (49.5~% per head/tail)

Scenario 2: (Vacum or space; no gravity) The coin would spin for a long time but would eventualy stop on one of 3 possible sides and stay there. Possibility 1/3 of any edge to hit.


Yeah ... I should define what's up or down in gravity free environment.

[edit on 28.6.2010 by valhala]



posted on Jun, 28 2010 @ 09:14 PM
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reply to post by Phlynx
 


I'll have a look a these but I'm sure that's not all



posted on Jun, 28 2010 @ 09:16 PM
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Originally posted by Gentill Abdulla

Originally posted by spy66

Originally posted by Gentill Abdulla

Originally posted by spy66
The mass of the coin is not equal all around. That means your equation would only work in a vacuum? The force up is not equal to the force down with normal G.

Since there are only two sides to a coin. Its logic that a even number would be the right answer. There are only two possibilities.

But how would you know beforehand which side the coin would land on before you tossed it?

Your equation does not tell you that.

[edit on 27.06.08 by spy66]


It doesn't tell you unless you always know how much force is applied.

Since weight is only the amount of pull gravity has on an object which mass weight can substitute for mass.

You forget it landing on its edge. That's a third possibility.


If it lands on the edge, it still need to flip to one side. That has to be added to your equation. Because your equation is not complete until the coin shows heads or tail.




By landing on its edge I mean like this...

Heads Tails
_____ _____ Edge l



That edge also has two sides. That are a bit smaller then the edge you refer to. The odds for the coin to land on the edge of the edge is larger, because of the size of that side of the coin, and because of rotation(Momentum).



posted on Jun, 28 2010 @ 09:18 PM
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Originally posted by spy66

That edge also has two sides. That are a bit smaller then the edge you refer to. The odds for the coin to land on the edge of the edge is larger, because of the size of that side of the coin, and because of rotation(Momentum).



But it is still a possibility is it not?

(I think we should stop with the long quotes. You and me)



posted on Jun, 28 2010 @ 09:25 PM
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off-topic post removed to prevent thread-drift


 



posted on Jun, 28 2010 @ 09:27 PM
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Originally posted by TheRealJesus

So your saying no one can pass the speed of light? I already did.



Never said that.



posted on Jun, 28 2010 @ 09:27 PM
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Originally posted by Gentill Abdulla

Originally posted by spy66

That edge also has two sides. That are a bit smaller then the edge you refer to. The odds for the coin to land on the edge of the edge is larger, because of the size of that side of the coin, and because of rotation(Momentum).



But it is still a possibility is it not?

(I think we should stop with the long quotes. You and me)


Your equation is right if you either want heads or tails. If the coin shows heads before you toss it. And you want heads to show when it lands. The coin must flip a even number.
The third side doesn't count since it wasn't a part of the experiment.



posted on Jun, 28 2010 @ 09:31 PM
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Originally posted by spy66


Your equation is right if you either want heads or tails. If the coin shows heads before you toss it. And you want heads to show when it lands. The coin must flip a even number.
The third side doesn't count since it wasn't a part of the experiment.



When I meant rotations I meant rotating back to heads when it is at heads. So one rotation would be back at heads.

But if it did not complete the rotation it would be at half a rotation.
Meaning tails.


Then then if it completed more than half but not one full rotation it would be at it's edge(but unless it is exactly stable at the edge then it will fall).



posted on Jun, 28 2010 @ 09:33 PM
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off-topic post removed to prevent thread-drift


 



posted on Jun, 28 2010 @ 09:39 PM
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reply to post by Gentill Abdulla
 


You have to count the fall as a rotation.

If a coin rotates half way, Its between two sides. If it lands on the edge and falls down it will complete the rotation, It wont matter what side it falls down on. It will complete a rotation either way?



posted on Jun, 28 2010 @ 09:41 PM
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Originally posted by spy66
reply to post by Gentill Abdulla
 


You have to count the fall as a rotation.

If a coin rotates half way, Its between two sides. If it lands on the edge and falls down it will complete the rotation, It wont matter what side it falls down on. It will complete a rotation either way?





The fall is the final rotation.

Unless it lands PERFECTLY it will fall and complete a rotation.

Which is why it is so hard to create this effect.

Are we at an agreement?



posted on Jun, 28 2010 @ 09:49 PM
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Who really likes to know anyway ?

When I flip a coin it is to let faith decide.


IMO the only ones that want to know, are the ones that plan to cheat.



posted on Jun, 28 2010 @ 09:52 PM
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Originally posted by Sinter Klaas
Who really likes to know anyway ?

When I flip a coin it is to let faith decide.


IMO the only ones that want to know, are the ones that plan to cheat.


I don't cheat!!

a lot

But seriously I was just wondering.

[edit on 28-6-2010 by Gentill Abdulla]



posted on Jun, 28 2010 @ 09:55 PM
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Here are my results:

10 flips
Begin on heads:
H
T
H
H
T
H
H
H
T
T
------
Results 6 Heads: 4 Tails

10 flips
Begin on Tails:
T
H
H
T
H
T
H
H
T
H
-----
Results 6 Heads: 4 Tails

Make of it what you will.

Micellaneous data: Coin used was a 2000 Maryland Quarter Chosen at random from my change bucket

[edit on 28-6-2010 by In nothing we trust]



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