WILL IT FLY?

Eh. Is why I posted what I did above. Not posting any of this to be a prick, or piss you off, or anything of the sorts, was a discussion piece. Yes, people 'hate' this debate, because there's a few fiercely debated viewpoints... and that's kinda what I put it here for, to see what was said / defended / thrown out the window, etc, for the types of people that like thinking about stuff like this. Some people like that kinda brain-wracking debate, that's the kinda stuff that gets me goin, the philosophical mindless debates. If its not something you enjoy, that's aight, its a personal preference.
 
I was enjoying watching this debate, but since it's almost over I must prolong my enjoyment by posing the following question:

You have 10 bottles of pills, where each bottle contains 10 pills. All the bottles and all the pills are identical in size, shape, and color. However, the pills come in two different weights. 9 of the bottles contain only pills of one weight, while the remaining bottle contains only pills of the second weight. In addition, you are told the specific weight of individual pills from the 9 bottles, and the specific weight of individual pills from the 1 bottle.

You have at your disposal a scale (not a counterbalance, just a flat scale). What is the fewest number of weighings necessary to be able to conclusively determine which bottle contains the 10 pills of the "odd" weight?
 
/head asplode

Between the UFO Haiti thread on bluegartr and this, its all too much!
 
I was enjoying watching this debate, but since it's almost over I must prolong my enjoyment by posing the following question:

You have 10 bottles of pills, where each bottle contains 10 pills. All the bottles and all the pills are identical in size, shape, and color. However, the pills come in two different weights. 9 of the bottles contain only pills of one weight, while the remaining bottle contains only pills of the second weight.

You have at your disposal a scale (not a counterbalance, just a flat scale). What is the fewest number of weighings necessary to be able to conclusively determine which bottle contains the 10 pills of the "odd" weight?
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Let me see if I understand the question. I don't really care if I'm wrong on these, lol.

10 bottles filled with 10 pills each?

9 of those bottles have 10 pills with weight "A"?
1 bottle has 10 pills with weight "B"?
 
furax said:
then stick a rocket in my ass and put me on a fucking treadmill with my arms extended out like wings, and I'll eventually take off too.
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Furax, you just got me a whole lot of strange looks from my coworkers laughing my ass off at that. The mental imagery is just too much!!
:grin:
 
Jotaru said:
You aren't trying to 'launch' anything, just accelerate it to the point where the air pressure over the wings sucks it off the runway. If it will accelerate, it will be at 10mph one second... 20mph the next.. so on and so forth. Eventually it will just 'fly', there is no real 'launching' involved. If it doesn't accelerate, it just sits there at 0mph and looks retarded.
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That is where your failing. You can accelerate something to 10,000,000 mph, how far is it actually moving in relation to the ground? You're not telling us if the ground is accelerating, too. Being vague is no substitute for a counterpoint. The pressure you're referring to is created by drag, which is not related to acceleration, but to velocity, air pressure at a given altitude and surface area (among other things). So we have a plane accelerating, we don't know its velocity, we don't know the runway distance, we don't know how much drag is being exerted on the plane. There's no way to tell if it will fly, but it will definitely move.
 
Archain said:
Let me see if I understand the question. I don't really care if I'm wrong on these, lol.

10 bottles filled with 10 pills each?

9 of those bottles have 10 pills with weight "A"?
1 bottle has 10 pills with weight "B"?
Click to expand...

Yes, and you know what A and B are. How many weighings to determine which bottle contains the "B" pills?
 
Jotaru said:
Eh. Is why I posted what I did above. Not posting any of this to be a prick, or piss you off, or anything of the sorts, was a discussion piece. Yes, people 'hate' this debate, because there's a few fiercely debated viewpoints... and that's kinda what I put it here for, to see what was said / defended / thrown out the window, etc, for the types of people that like thinking about stuff like this. Some people like that kinda brain-wracking debate, that's the kinda stuff that gets me goin, the philosophical mindless debates. If its not something you enjoy, that's aight, its a personal preference.
Click to expand...

Yeah man, this is the only thing keeping me awake at work, lol. Literally, I had my head on my desk.
 
Kalia said:
I was enjoying watching this debate, but since it's almost over I must prolong my enjoyment by posing the following question:

You have 10 bottles of pills, where each bottle contains 10 pills. All the bottles and all the pills are identical in size, shape, and color. However, the pills come in two different weights. 9 of the bottles contain only pills of one weight, while the remaining bottle contains only pills of the second weight. In addition, you are told the specific weight of individual pills from the 9 bottles, and the specific weight of individual pills from the 1 bottle.

You have at your disposal a scale (not a counterbalance, just a flat scale). What is the fewest number of weighings necessary to be able to conclusively determine which bottle contains the 10 pills of the "odd" weight?
Click to expand...

I got it in 7 weighs. Assuming the B pills were not a significant enough weight difference to be felt by hand. You could get lucky and get it in 4, though.
 
Twice. you weigh the one bottle of the pills whose weight you do know to get the weight of said bottles. Once this is done put all 10 on you can solve for the weight of the 10th bottles pills
 
Tarnak said:
Once. You know the weight of the other 9. So when you put all 10 on the scale you subtract the weight of the 9 to get the value of the 10th.
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That doesn't tell you which bottle contains the odd weight though. It only tells you the weight of the odd pilled bottle. Which you already knew anyway, because as mentioned previously you know the values of A and B.

Is it the 2nd bottle in the line if you line them up? The 8th bottle? The 1st bottle?
 
Tarnak said:
Twice. you weigh the one bottle of the pills whose weight you do know to get the weight of said bottles. Once this is done put all 10 on you can solve for the weight of the 10th bottles pills
Click to expand...

Nice, we now have a value we already knew. The question is which of the bottles contains the pills with different weight.

Interesting though, would placing all 10 on the scale and removing them one at a time, in the hope you get lucky, constitute a single weighing? Or would removing a bottle from the scale mean a new weighing was initiated?
 
Glad you feel the same way Furax, didn't want things getting out of hand :)

I'll try not to be vague, the question is initially meant to be worded as such to.. get a rise out of people, generally. Assuming the runway is plenty long enough, the question really just boils down to whether the aircraft will accelerate or not. In conventional passenger aircraft, the only thing that really matters is how fast the air travels past the wings (over/under). If the treadmill causes the plane to remain stationary, no air would pass over/under, so it wouldnt generate lift. If the plane could accelerate while on the treadmill (in relation to the air , eventually it would accelerate to the point where the wind passing over/under the wing would generate sufficient lift to pull the airplane off the ground. Generally I just believe that it would accelerate regardless of the treadmill, but were I to be proved wrong I would certainly acknowledge. I have yet to pull the physics calculations on it, as its more a thought experiment, and a discussion piece, its just what I believe (and generally what a few friends of mine believe from where I take flying lessons from. Then again, we have never taken off from a moving runway ;)
 
re pill bottles:

With luck it could be as few as 3. However that's just by weighing them individually and getting lucky and having the heavier be one of the first 3. Scientifically it would take between 4-6. Though I probably missed some super psycho Kalia method.

1st 2) 5 bottles and 5 bottles. Discard the lighter set.

2nd 2) 2 bottles and 2 bottles. If they are even then its the last. If they are not then discard all but the heaviest 2

3rd 2) Weigh the last 2. Heavier is the winnar.

4-6 by that method.

I suppose it could be as few as 3 weighings to start if you went 3 by 3 by 3 and they were all even. If not it could still take up to 6.
 
Like Tarnak said, screw the scale. Since the pills are the exact same size and shape (volume and surface area), the only difference to account for weight is density. Take one from each bottle, and drop them in a tub of water (keeping track of which came from where of course). Whichever one sinks more slowly is the odd man out. And even if they are water soluable pills, because of the lower density, it'll dissolve faster relative to its beginning weight and still sink more slowly than the others (of what's left of them).

Only concern is if they're horse pills will be hard to tell. Might have to use something more viscous than water. Hmm...

Edit: Damn you tarnak taking my water dropping idea while I was typing it!!!!!
 
You're in the middle of the desert, so you don't have any water.


Eticket is thinking good, but you can optimize it further. The answer is quite surprising.
 
I havent done the workings on it yet, but is it possible to :

Empty them all out, but assign them in groups, so you know which came from which bottles

Put 1 of Bottle1 into BottleX, 2 of Bottle2 into BottleX, 3 of Bottle3 into BottleX, etc etc... up to all 10 of Bottle10 into BottleX

... and then go from there. Fawk, I ain't working the rest out if its not a somewhat decent solution to the problem. Drinking a Starbucks now, I'll come back to it. Maybe. Seems you could do it in 1-2 weighings this way.
 
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