Breaking Balance (Part A)
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If we assume the classic balance puzzle (Twelve balls and a scale) is phrased like this:
You have 12 coins that are identical except for one counterfeit that is heavier or lighter. Can you use a balance scale three times to determine the counterfeit coin and say if it was heavier or lighter?
Can you design an object that could properly be called a "balance scale" that could solve the above problem with a single weighing instead of three?
Obviously any such object is a far cry from the intention of the original problem, and may be hard to draw or describe. I just think us Engineers deserve a little fun at the expense of of you math-riddle folk sometimes.
Part B and Part C will contain something like spoilers as to the two ways I thought of answering this, so if you want to take an unbiased stab at this, I would avoid those for now.
And before anybody mentions it, coins are always better than "balls" because they don't roll off the table and get lost under your desk. It is befuddling to me that so many people do these puzzles with balls.
open-ended weighing physics
asked Jan 26 at 14:35
Dark ThunderDark Thunder
3549
$endgroup$
add a comment |
$begingroup$
If we assume the classic balance puzzle (Twelve balls and a scale) is phrased like this:
You have 12 coins that are identical except for one counterfeit that is heavier or lighter. Can you use a balance scale three times to determine the counterfeit coin and say if it was heavier or lighter?
Can you design an object that could properly be called a "balance scale" that could solve the above problem with a single weighing instead of three?
Obviously any such object is a far cry from the intention of the original problem, and may be hard to draw or describe. I just think us Engineers deserve a little fun at the expense of of you math-riddle folk sometimes.
Part B and Part C will contain something like spoilers as to the two ways I thought of answering this, so if you want to take an unbiased stab at this, I would avoid those for now.
And before anybody mentions it, coins are always better than "balls" because they don't roll off the table and get lost under your desk. It is befuddling to me that so many people do these puzzles with balls.
open-ended weighing physics
asked Jan 26 at 14:35
Dark ThunderDark Thunder
3549
$endgroup$
1
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Some of us math-riddle folks are engineers ... but I think the engineering answer is to challenge the requirements! You'll need a better story than that! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:03
add a comment |
$begingroup$
If we assume the classic balance puzzle (Twelve balls and a scale) is phrased like this:
You have 12 coins that are identical except for one counterfeit that is heavier or lighter. Can you use a balance scale three times to determine the counterfeit coin and say if it was heavier or lighter?
Can you design an object that could properly be called a "balance scale" that could solve the above problem with a single weighing instead of three?
Obviously any such object is a far cry from the intention of the original problem, and may be hard to draw or describe. I just think us Engineers deserve a little fun at the expense of of you math-riddle folk sometimes.
Part B and Part C will contain something like spoilers as to the two ways I thought of answering this, so if you want to take an unbiased stab at this, I would avoid those for now.
And before anybody mentions it, coins are always better than "balls" because they don't roll off the table and get lost under your desk. It is befuddling to me that so many people do these puzzles with balls.
open-ended weighing physics
asked Jan 26 at 14:35
Dark ThunderDark Thunder
3549
$endgroup$
If we assume the classic balance puzzle (Twelve balls and a scale) is phrased like this:
You have 12 coins that are identical except for one counterfeit that is heavier or lighter. Can you use a balance scale three times to determine the counterfeit coin and say if it was heavier or lighter?
Can you design an object that could properly be called a "balance scale" that could solve the above problem with a single weighing instead of three?
Obviously any such object is a far cry from the intention of the original problem, and may be hard to draw or describe. I just think us Engineers deserve a little fun at the expense of of you math-riddle folk sometimes.
Part B and Part C will contain something like spoilers as to the two ways I thought of answering this, so if you want to take an unbiased stab at this, I would avoid those for now.
And before anybody mentions it, coins are always better than "balls" because they don't roll off the table and get lost under your desk. It is befuddling to me that so many people do these puzzles with balls.
open-ended weighing physics
open-ended weighing physics
asked Jan 26 at 14:35
Dark ThunderDark Thunder
3549
asked Jan 26 at 14:35
Dark ThunderDark Thunder
3549
asked Jan 26 at 14:35
Dark ThunderDark Thunder
3549
asked Jan 26 at 14:35
Dark ThunderDark Thunder
3549
asked Jan 26 at 14:35
Dark ThunderDark Thunder
3549
3549
1
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Some of us math-riddle folks are engineers ... but I think the engineering answer is to challenge the requirements! You'll need a better story than that! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:03
add a comment |
1
$begingroup$
Some of us math-riddle folks are engineers ... but I think the engineering answer is to challenge the requirements! You'll need a better story than that! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:03
1
1
$begingroup$
Some of us math-riddle folks are engineers ... but I think the engineering answer is to challenge the requirements! You'll need a better story than that! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:03
$begingroup$
Some of us math-riddle folks are engineers ... but I think the engineering answer is to challenge the requirements! You'll need a better story than that! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:03
add a comment |
5 Answers
5
active
oldest
votes
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I have a fun solution:
We can use the idea of water pressure to solve this question. Imagine a mechanism with 12 pistons each with water inside. The 12 pistons are connected to each other. Now, place a coin on each piston. Due to a difference in weight, one of the pistons will be at a different height from the others. This allows you to identify the counterfeit coin, and also whether it is heavier or lighter. Technically it's a 12 pan hydraulic balance. :)
answered Jan 26 at 16:00
Ong Yu HannOng Yu Hann
66410
$endgroup$
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I love it! You might have to get clever to make the small weight difference noticeable for an operator but as a basic approach that's perfectly sound.
$endgroup$
– Dark Thunder
Jan 26 at 16:07
add a comment |
$begingroup$
This might be not what an engineer likes to hear, but
one could go to outer space (no air friction, no graviational potential) and form a equally spaced ring out of the balls/coins.
Then,
the counterfeit one will break the symmetry of the center of mass of the circle.
If the counterfeit is lighter than the rest
the circle will contract away from it
else
it will contract towards it.
answered Jan 26 at 14:46
A. P.A. P.
3,95411148
$endgroup$
$begingroup$
Amazing! Looking past the nit-pick that you don't have a "balance scale" object, can you actually differentiate between those two outcomes? I'm really asking, I can't tell.
$endgroup$
– Dark Thunder
Jan 26 at 15:59
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@DarkThunder I replaced the drawings by simulations, so that it's easier to see the difference.
$endgroup$
– A. P.
Jan 26 at 17:52
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Would low earth orbit be enough, or would tidal forces mess that up?
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– deep thought
Jan 26 at 18:32
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@deepthought Of course the exact answer would depend on how small the mass differences are, but I would say that tidal forces don't have much influence on the result. As you can see in the figures the outcome of the test is rather local, while tidal forces would have a more long-range effect, which would probably make the circle more elliptical. And judging from the fact that with a similar setup it is possible to detect gravitational waves without being disturbed by tidal forces, it must be possible to measure very small mass differences of the coins.
$endgroup$
– A. P.
Jan 26 at 19:14
add a comment |
$begingroup$
Another answer, which is probably closer to what the OP intended:
One could think of just extending a 3-pan balance to a 12-pan balance by just adding 3 equally spaced arms into each gap, so that we end up with arms spaced by 30°.
The problem with this is that
you can't distinguish whether one coin is lighter or whether the coin on the other side is heavier. This results from the symmetry of the balance. Hence we have to create a balance without any pair of coins lying on opposing sides. This sounds like the balance could never be balanced, but the 3-pan balance is an example which fulfills this criterion and is still balanced (if all coins are equal).
To make a
12-pan balance fulfilling the criterion out of it, we can just add the additional arms in other angle combinations than (30°, 30°, 30°). To make sure that at the same time the balance is intrinsically balanced we can group triples of arms to regular 3-pan balances. As long as they share their center of mass the whole construction is balanced. In the following drawing the triangles representing the 3-pan balances are rotated by 15° relative to each other, having the coins sit on the corners of an imaginary 24-gon.
answered Jan 26 at 22:25
A. P.A. P.
3,95411148
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2
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That is the exact thing I came up with, so you win a cookie (sorry, I don't know how this forum works, I'm assuming it's a cookie). The YouTube channel Numberphile had a video on balancing a centrifuge not that long ago that dealt with surprisingly similar issues, which I thought was neat.
$endgroup$
– Dark Thunder
Jan 27 at 0:54
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@DarkThunder If it's the right answer, you can click the tick near the top of the answer to accept it as correct.
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– ZanyG
Jan 27 at 3:52
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@DarkThunder I just saw that video, and then I come here and see your problem. Neat!
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– LinuxBlanket
Jan 27 at 13:37
add a comment |
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You could have some kind of
Metal tube with 12 springs attached to it and a plate at the end of each spring. The heaviest/lightest coin would be the one that's lowest/highest
answered Jan 26 at 17:35
J. DionisioJ. Dionisio
1007
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Haha, yeah I guess you could. Each coin is balanced against its own spring and doesn't interact with each other. Not what I had in mind but that is literally the point of asking this sort of question.
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– Dark Thunder
Jan 26 at 21:25
add a comment |
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What about
putting smaller balances on larger balances
for example you can obtain
one large 3-pan balance, three medium 2-pan balances, six small 2-pan balances, and stack them in the obvious way.
Or even
four different sizes of 2-pan balances, with four extra known-genuine coins.
answered Jan 26 at 18:58
deep thoughtdeep thought
3,6861940
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1
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One could weld 4 3-pan balances together and say this is one device. :-)
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– A. P.
Jan 26 at 19:17
1
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@A.P. But I'm cheap and I was planning on borrowing the equipment, so I need to return it in good condition! :-) :-)
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– deep thought
Jan 26 at 19:23
2
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Definitely cheaper than flying to outer space.
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– A. P.
Jan 26 at 19:25
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Nesting balances was one of my ideas too. Can you do it with only 2-pan balances but without additional genuine coins?
$endgroup$
– Dark Thunder
Jan 26 at 20:28
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I have a fun solution:
We can use the idea of water pressure to solve this question. Imagine a mechanism with 12 pistons each with water inside. The 12 pistons are connected to each other. Now, place a coin on each piston. Due to a difference in weight, one of the pistons will be at a different height from the others. This allows you to identify the counterfeit coin, and also whether it is heavier or lighter. Technically it's a 12 pan hydraulic balance. :)
answered Jan 26 at 16:00
Ong Yu HannOng Yu Hann
66410
$endgroup$
$begingroup$
I love it! You might have to get clever to make the small weight difference noticeable for an operator but as a basic approach that's perfectly sound.
$endgroup$
– Dark Thunder
Jan 26 at 16:07
add a comment |
$begingroup$
I have a fun solution:
We can use the idea of water pressure to solve this question. Imagine a mechanism with 12 pistons each with water inside. The 12 pistons are connected to each other. Now, place a coin on each piston. Due to a difference in weight, one of the pistons will be at a different height from the others. This allows you to identify the counterfeit coin, and also whether it is heavier or lighter. Technically it's a 12 pan hydraulic balance. :)
answered Jan 26 at 16:00
Ong Yu HannOng Yu Hann
66410
$endgroup$
$begingroup$
I love it! You might have to get clever to make the small weight difference noticeable for an operator but as a basic approach that's perfectly sound.
$endgroup$
– Dark Thunder
Jan 26 at 16:07
add a comment |
$begingroup$
I have a fun solution:
We can use the idea of water pressure to solve this question. Imagine a mechanism with 12 pistons each with water inside. The 12 pistons are connected to each other. Now, place a coin on each piston. Due to a difference in weight, one of the pistons will be at a different height from the others. This allows you to identify the counterfeit coin, and also whether it is heavier or lighter. Technically it's a 12 pan hydraulic balance. :)
answered Jan 26 at 16:00
Ong Yu HannOng Yu Hann
66410
$endgroup$
I have a fun solution:
We can use the idea of water pressure to solve this question. Imagine a mechanism with 12 pistons each with water inside. The 12 pistons are connected to each other. Now, place a coin on each piston. Due to a difference in weight, one of the pistons will be at a different height from the others. This allows you to identify the counterfeit coin, and also whether it is heavier or lighter. Technically it's a 12 pan hydraulic balance. :)
answered Jan 26 at 16:00
Ong Yu HannOng Yu Hann
66410
answered Jan 26 at 16:00
Ong Yu HannOng Yu Hann
66410
answered Jan 26 at 16:00
Ong Yu HannOng Yu Hann
66410
answered Jan 26 at 16:00
Ong Yu HannOng Yu Hann
66410
66410
$begingroup$
I love it! You might have to get clever to make the small weight difference noticeable for an operator but as a basic approach that's perfectly sound.
$endgroup$
– Dark Thunder
Jan 26 at 16:07
add a comment |
$begingroup$
I love it! You might have to get clever to make the small weight difference noticeable for an operator but as a basic approach that's perfectly sound.
$endgroup$
– Dark Thunder
Jan 26 at 16:07
$begingroup$
I love it! You might have to get clever to make the small weight difference noticeable for an operator but as a basic approach that's perfectly sound.
$endgroup$
– Dark Thunder
Jan 26 at 16:07
$begingroup$
I love it! You might have to get clever to make the small weight difference noticeable for an operator but as a basic approach that's perfectly sound.
$endgroup$
– Dark Thunder
Jan 26 at 16:07
add a comment |
$begingroup$
This might be not what an engineer likes to hear, but
one could go to outer space (no air friction, no graviational potential) and form a equally spaced ring out of the balls/coins.
Then,
the counterfeit one will break the symmetry of the center of mass of the circle.
If the counterfeit is lighter than the rest
the circle will contract away from it
else
it will contract towards it.
answered Jan 26 at 14:46
A. P.A. P.
3,95411148
$endgroup$
$begingroup$
Amazing! Looking past the nit-pick that you don't have a "balance scale" object, can you actually differentiate between those two outcomes? I'm really asking, I can't tell.
$endgroup$
– Dark Thunder
Jan 26 at 15:59
$begingroup$
@DarkThunder I replaced the drawings by simulations, so that it's easier to see the difference.
$endgroup$
– A. P.
Jan 26 at 17:52
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Would low earth orbit be enough, or would tidal forces mess that up?
$endgroup$
– deep thought
Jan 26 at 18:32
$begingroup$
@deepthought Of course the exact answer would depend on how small the mass differences are, but I would say that tidal forces don't have much influence on the result. As you can see in the figures the outcome of the test is rather local, while tidal forces would have a more long-range effect, which would probably make the circle more elliptical. And judging from the fact that with a similar setup it is possible to detect gravitational waves without being disturbed by tidal forces, it must be possible to measure very small mass differences of the coins.
$endgroup$
– A. P.
Jan 26 at 19:14
add a comment |
$begingroup$
This might be not what an engineer likes to hear, but
one could go to outer space (no air friction, no graviational potential) and form a equally spaced ring out of the balls/coins.
Then,
the counterfeit one will break the symmetry of the center of mass of the circle.
If the counterfeit is lighter than the rest
the circle will contract away from it
else
it will contract towards it.
answered Jan 26 at 14:46
A. P.A. P.
3,95411148
$endgroup$
$begingroup$
Amazing! Looking past the nit-pick that you don't have a "balance scale" object, can you actually differentiate between those two outcomes? I'm really asking, I can't tell.
$endgroup$
– Dark Thunder
Jan 26 at 15:59
$begingroup$
@DarkThunder I replaced the drawings by simulations, so that it's easier to see the difference.
$endgroup$
– A. P.
Jan 26 at 17:52
$begingroup$
Would low earth orbit be enough, or would tidal forces mess that up?
$endgroup$
– deep thought
Jan 26 at 18:32
$begingroup$
@deepthought Of course the exact answer would depend on how small the mass differences are, but I would say that tidal forces don't have much influence on the result. As you can see in the figures the outcome of the test is rather local, while tidal forces would have a more long-range effect, which would probably make the circle more elliptical. And judging from the fact that with a similar setup it is possible to detect gravitational waves without being disturbed by tidal forces, it must be possible to measure very small mass differences of the coins.
$endgroup$
– A. P.
Jan 26 at 19:14
add a comment |
$begingroup$
This might be not what an engineer likes to hear, but
one could go to outer space (no air friction, no graviational potential) and form a equally spaced ring out of the balls/coins.
Then,
the counterfeit one will break the symmetry of the center of mass of the circle.
If the counterfeit is lighter than the rest
the circle will contract away from it
else
it will contract towards it.
answered Jan 26 at 14:46
A. P.A. P.
3,95411148
$endgroup$
This might be not what an engineer likes to hear, but
one could go to outer space (no air friction, no graviational potential) and form a equally spaced ring out of the balls/coins.
Then,
the counterfeit one will break the symmetry of the center of mass of the circle.
If the counterfeit is lighter than the rest
the circle will contract away from it
else
it will contract towards it.
answered Jan 26 at 14:46
A. P.A. P.
3,95411148
edited Jan 26 at 17:51
answered Jan 26 at 14:46
A. P.A. P.
3,95411148
answered Jan 26 at 14:46
A. P.A. P.
3,95411148
answered Jan 26 at 14:46
A. P.A. P.
3,95411148
3,95411148
$begingroup$
Amazing! Looking past the nit-pick that you don't have a "balance scale" object, can you actually differentiate between those two outcomes? I'm really asking, I can't tell.
$endgroup$
– Dark Thunder
Jan 26 at 15:59
$begingroup$
@DarkThunder I replaced the drawings by simulations, so that it's easier to see the difference.
$endgroup$
– A. P.
Jan 26 at 17:52
$begingroup$
Would low earth orbit be enough, or would tidal forces mess that up?
$endgroup$
– deep thought
Jan 26 at 18:32
$begingroup$
@deepthought Of course the exact answer would depend on how small the mass differences are, but I would say that tidal forces don't have much influence on the result. As you can see in the figures the outcome of the test is rather local, while tidal forces would have a more long-range effect, which would probably make the circle more elliptical. And judging from the fact that with a similar setup it is possible to detect gravitational waves without being disturbed by tidal forces, it must be possible to measure very small mass differences of the coins.
$endgroup$
– A. P.
Jan 26 at 19:14
add a comment |
$begingroup$
Amazing! Looking past the nit-pick that you don't have a "balance scale" object, can you actually differentiate between those two outcomes? I'm really asking, I can't tell.
$endgroup$
– Dark Thunder
Jan 26 at 15:59
$begingroup$
@DarkThunder I replaced the drawings by simulations, so that it's easier to see the difference.
$endgroup$
– A. P.
Jan 26 at 17:52
$begingroup$
Would low earth orbit be enough, or would tidal forces mess that up?
$endgroup$
– deep thought
Jan 26 at 18:32
$begingroup$
@deepthought Of course the exact answer would depend on how small the mass differences are, but I would say that tidal forces don't have much influence on the result. As you can see in the figures the outcome of the test is rather local, while tidal forces would have a more long-range effect, which would probably make the circle more elliptical. And judging from the fact that with a similar setup it is possible to detect gravitational waves without being disturbed by tidal forces, it must be possible to measure very small mass differences of the coins.
$endgroup$
– A. P.
Jan 26 at 19:14
$begingroup$
Amazing! Looking past the nit-pick that you don't have a "balance scale" object, can you actually differentiate between those two outcomes? I'm really asking, I can't tell.
$endgroup$
– Dark Thunder
Jan 26 at 15:59
$begingroup$
Amazing! Looking past the nit-pick that you don't have a "balance scale" object, can you actually differentiate between those two outcomes? I'm really asking, I can't tell.
$endgroup$
– Dark Thunder
Jan 26 at 15:59
$begingroup$
@DarkThunder I replaced the drawings by simulations, so that it's easier to see the difference.
$endgroup$
– A. P.
Jan 26 at 17:52
$begingroup$
@DarkThunder I replaced the drawings by simulations, so that it's easier to see the difference.
$endgroup$
– A. P.
Jan 26 at 17:52
$begingroup$
Would low earth orbit be enough, or would tidal forces mess that up?
$endgroup$
– deep thought
Jan 26 at 18:32
$begingroup$
Would low earth orbit be enough, or would tidal forces mess that up?
$endgroup$
– deep thought
Jan 26 at 18:32
$begingroup$
@deepthought Of course the exact answer would depend on how small the mass differences are, but I would say that tidal forces don't have much influence on the result. As you can see in the figures the outcome of the test is rather local, while tidal forces would have a more long-range effect, which would probably make the circle more elliptical. And judging from the fact that with a similar setup it is possible to detect gravitational waves without being disturbed by tidal forces, it must be possible to measure very small mass differences of the coins.
$endgroup$
– A. P.
Jan 26 at 19:14
$begingroup$
@deepthought Of course the exact answer would depend on how small the mass differences are, but I would say that tidal forces don't have much influence on the result. As you can see in the figures the outcome of the test is rather local, while tidal forces would have a more long-range effect, which would probably make the circle more elliptical. And judging from the fact that with a similar setup it is possible to detect gravitational waves without being disturbed by tidal forces, it must be possible to measure very small mass differences of the coins.
$endgroup$
– A. P.
Jan 26 at 19:14
add a comment |
$begingroup$
Another answer, which is probably closer to what the OP intended:
One could think of just extending a 3-pan balance to a 12-pan balance by just adding 3 equally spaced arms into each gap, so that we end up with arms spaced by 30°.
The problem with this is that
you can't distinguish whether one coin is lighter or whether the coin on the other side is heavier. This results from the symmetry of the balance. Hence we have to create a balance without any pair of coins lying on opposing sides. This sounds like the balance could never be balanced, but the 3-pan balance is an example which fulfills this criterion and is still balanced (if all coins are equal).
To make a
12-pan balance fulfilling the criterion out of it, we can just add the additional arms in other angle combinations than (30°, 30°, 30°). To make sure that at the same time the balance is intrinsically balanced we can group triples of arms to regular 3-pan balances. As long as they share their center of mass the whole construction is balanced. In the following drawing the triangles representing the 3-pan balances are rotated by 15° relative to each other, having the coins sit on the corners of an imaginary 24-gon.
answered Jan 26 at 22:25
A. P.A. P.
3,95411148
$endgroup$
2
$begingroup$
That is the exact thing I came up with, so you win a cookie (sorry, I don't know how this forum works, I'm assuming it's a cookie). The YouTube channel Numberphile had a video on balancing a centrifuge not that long ago that dealt with surprisingly similar issues, which I thought was neat.
$endgroup$
– Dark Thunder
Jan 27 at 0:54
$begingroup$
@DarkThunder If it's the right answer, you can click the tick near the top of the answer to accept it as correct.
$endgroup$
– ZanyG
Jan 27 at 3:52
$begingroup$
@DarkThunder I just saw that video, and then I come here and see your problem. Neat!
$endgroup$
– LinuxBlanket
Jan 27 at 13:37
add a comment |
$begingroup$
Another answer, which is probably closer to what the OP intended:
One could think of just extending a 3-pan balance to a 12-pan balance by just adding 3 equally spaced arms into each gap, so that we end up with arms spaced by 30°.
The problem with this is that
you can't distinguish whether one coin is lighter or whether the coin on the other side is heavier. This results from the symmetry of the balance. Hence we have to create a balance without any pair of coins lying on opposing sides. This sounds like the balance could never be balanced, but the 3-pan balance is an example which fulfills this criterion and is still balanced (if all coins are equal).
To make a
12-pan balance fulfilling the criterion out of it, we can just add the additional arms in other angle combinations than (30°, 30°, 30°). To make sure that at the same time the balance is intrinsically balanced we can group triples of arms to regular 3-pan balances. As long as they share their center of mass the whole construction is balanced. In the following drawing the triangles representing the 3-pan balances are rotated by 15° relative to each other, having the coins sit on the corners of an imaginary 24-gon.
answered Jan 26 at 22:25
A. P.A. P.
3,95411148
$endgroup$
2
$begingroup$
That is the exact thing I came up with, so you win a cookie (sorry, I don't know how this forum works, I'm assuming it's a cookie). The YouTube channel Numberphile had a video on balancing a centrifuge not that long ago that dealt with surprisingly similar issues, which I thought was neat.
$endgroup$
– Dark Thunder
Jan 27 at 0:54
$begingroup$
@DarkThunder If it's the right answer, you can click the tick near the top of the answer to accept it as correct.
$endgroup$
– ZanyG
Jan 27 at 3:52
$begingroup$
@DarkThunder I just saw that video, and then I come here and see your problem. Neat!
$endgroup$
– LinuxBlanket
Jan 27 at 13:37
add a comment |
$begingroup$
Another answer, which is probably closer to what the OP intended:
One could think of just extending a 3-pan balance to a 12-pan balance by just adding 3 equally spaced arms into each gap, so that we end up with arms spaced by 30°.
The problem with this is that
you can't distinguish whether one coin is lighter or whether the coin on the other side is heavier. This results from the symmetry of the balance. Hence we have to create a balance without any pair of coins lying on opposing sides. This sounds like the balance could never be balanced, but the 3-pan balance is an example which fulfills this criterion and is still balanced (if all coins are equal).
To make a
12-pan balance fulfilling the criterion out of it, we can just add the additional arms in other angle combinations than (30°, 30°, 30°). To make sure that at the same time the balance is intrinsically balanced we can group triples of arms to regular 3-pan balances. As long as they share their center of mass the whole construction is balanced. In the following drawing the triangles representing the 3-pan balances are rotated by 15° relative to each other, having the coins sit on the corners of an imaginary 24-gon.
answered Jan 26 at 22:25
A. P.A. P.
3,95411148
$endgroup$
Another answer, which is probably closer to what the OP intended:
One could think of just extending a 3-pan balance to a 12-pan balance by just adding 3 equally spaced arms into each gap, so that we end up with arms spaced by 30°.
The problem with this is that
you can't distinguish whether one coin is lighter or whether the coin on the other side is heavier. This results from the symmetry of the balance. Hence we have to create a balance without any pair of coins lying on opposing sides. This sounds like the balance could never be balanced, but the 3-pan balance is an example which fulfills this criterion and is still balanced (if all coins are equal).
To make a
12-pan balance fulfilling the criterion out of it, we can just add the additional arms in other angle combinations than (30°, 30°, 30°). To make sure that at the same time the balance is intrinsically balanced we can group triples of arms to regular 3-pan balances. As long as they share their center of mass the whole construction is balanced. In the following drawing the triangles representing the 3-pan balances are rotated by 15° relative to each other, having the coins sit on the corners of an imaginary 24-gon.
answered Jan 26 at 22:25
A. P.A. P.
3,95411148
answered Jan 26 at 22:25
A. P.A. P.
3,95411148
answered Jan 26 at 22:25
A. P.A. P.
3,95411148
answered Jan 26 at 22:25
A. P.A. P.
3,95411148
3,95411148
2
$begingroup$
That is the exact thing I came up with, so you win a cookie (sorry, I don't know how this forum works, I'm assuming it's a cookie). The YouTube channel Numberphile had a video on balancing a centrifuge not that long ago that dealt with surprisingly similar issues, which I thought was neat.
$endgroup$
– Dark Thunder
Jan 27 at 0:54
$begingroup$
@DarkThunder If it's the right answer, you can click the tick near the top of the answer to accept it as correct.
$endgroup$
– ZanyG
Jan 27 at 3:52
$begingroup$
@DarkThunder I just saw that video, and then I come here and see your problem. Neat!
$endgroup$
– LinuxBlanket
Jan 27 at 13:37
add a comment |
2
$begingroup$
That is the exact thing I came up with, so you win a cookie (sorry, I don't know how this forum works, I'm assuming it's a cookie). The YouTube channel Numberphile had a video on balancing a centrifuge not that long ago that dealt with surprisingly similar issues, which I thought was neat.
$endgroup$
– Dark Thunder
Jan 27 at 0:54
$begingroup$
@DarkThunder If it's the right answer, you can click the tick near the top of the answer to accept it as correct.
$endgroup$
– ZanyG
Jan 27 at 3:52
$begingroup$
@DarkThunder I just saw that video, and then I come here and see your problem. Neat!
$endgroup$
– LinuxBlanket
Jan 27 at 13:37
2
2
$begingroup$
That is the exact thing I came up with, so you win a cookie (sorry, I don't know how this forum works, I'm assuming it's a cookie). The YouTube channel Numberphile had a video on balancing a centrifuge not that long ago that dealt with surprisingly similar issues, which I thought was neat.
$endgroup$
– Dark Thunder
Jan 27 at 0:54
$begingroup$
That is the exact thing I came up with, so you win a cookie (sorry, I don't know how this forum works, I'm assuming it's a cookie). The YouTube channel Numberphile had a video on balancing a centrifuge not that long ago that dealt with surprisingly similar issues, which I thought was neat.
$endgroup$
– Dark Thunder
Jan 27 at 0:54
$begingroup$
@DarkThunder If it's the right answer, you can click the tick near the top of the answer to accept it as correct.
$endgroup$
– ZanyG
Jan 27 at 3:52
$begingroup$
@DarkThunder If it's the right answer, you can click the tick near the top of the answer to accept it as correct.
$endgroup$
– ZanyG
Jan 27 at 3:52
$begingroup$
@DarkThunder I just saw that video, and then I come here and see your problem. Neat!
$endgroup$
– LinuxBlanket
Jan 27 at 13:37
$begingroup$
@DarkThunder I just saw that video, and then I come here and see your problem. Neat!
$endgroup$
– LinuxBlanket
Jan 27 at 13:37
add a comment |
$begingroup$
You could have some kind of
Metal tube with 12 springs attached to it and a plate at the end of each spring. The heaviest/lightest coin would be the one that's lowest/highest
answered Jan 26 at 17:35
J. DionisioJ. Dionisio
1007
$endgroup$
$begingroup$
Haha, yeah I guess you could. Each coin is balanced against its own spring and doesn't interact with each other. Not what I had in mind but that is literally the point of asking this sort of question.
$endgroup$
– Dark Thunder
Jan 26 at 21:25
add a comment |
$begingroup$
You could have some kind of
Metal tube with 12 springs attached to it and a plate at the end of each spring. The heaviest/lightest coin would be the one that's lowest/highest
answered Jan 26 at 17:35
J. DionisioJ. Dionisio
1007
$endgroup$
$begingroup$
Haha, yeah I guess you could. Each coin is balanced against its own spring and doesn't interact with each other. Not what I had in mind but that is literally the point of asking this sort of question.
$endgroup$
– Dark Thunder
Jan 26 at 21:25
add a comment |
$begingroup$
You could have some kind of
Metal tube with 12 springs attached to it and a plate at the end of each spring. The heaviest/lightest coin would be the one that's lowest/highest
answered Jan 26 at 17:35
J. DionisioJ. Dionisio
1007
$endgroup$
You could have some kind of
Metal tube with 12 springs attached to it and a plate at the end of each spring. The heaviest/lightest coin would be the one that's lowest/highest
answered Jan 26 at 17:35
J. DionisioJ. Dionisio
1007
answered Jan 26 at 17:35
J. DionisioJ. Dionisio
1007
answered Jan 26 at 17:35
J. DionisioJ. Dionisio
1007
answered Jan 26 at 17:35
J. DionisioJ. Dionisio
1007
1007
$begingroup$
Haha, yeah I guess you could. Each coin is balanced against its own spring and doesn't interact with each other. Not what I had in mind but that is literally the point of asking this sort of question.
$endgroup$
– Dark Thunder
Jan 26 at 21:25
add a comment |
$begingroup$
Haha, yeah I guess you could. Each coin is balanced against its own spring and doesn't interact with each other. Not what I had in mind but that is literally the point of asking this sort of question.
$endgroup$
– Dark Thunder
Jan 26 at 21:25
$begingroup$
Haha, yeah I guess you could. Each coin is balanced against its own spring and doesn't interact with each other. Not what I had in mind but that is literally the point of asking this sort of question.
$endgroup$
– Dark Thunder
Jan 26 at 21:25
$begingroup$
Haha, yeah I guess you could. Each coin is balanced against its own spring and doesn't interact with each other. Not what I had in mind but that is literally the point of asking this sort of question.
$endgroup$
– Dark Thunder
Jan 26 at 21:25
add a comment |
$begingroup$
What about
putting smaller balances on larger balances
for example you can obtain
one large 3-pan balance, three medium 2-pan balances, six small 2-pan balances, and stack them in the obvious way.
Or even
four different sizes of 2-pan balances, with four extra known-genuine coins.
answered Jan 26 at 18:58
deep thoughtdeep thought
3,6861940
$endgroup$
1
$begingroup$
One could weld 4 3-pan balances together and say this is one device. :-)
$endgroup$
– A. P.
Jan 26 at 19:17
1
$begingroup$
@A.P. But I'm cheap and I was planning on borrowing the equipment, so I need to return it in good condition! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:23
2
$begingroup$
Definitely cheaper than flying to outer space.
$endgroup$
– A. P.
Jan 26 at 19:25
$begingroup$
Nesting balances was one of my ideas too. Can you do it with only 2-pan balances but without additional genuine coins?
$endgroup$
– Dark Thunder
Jan 26 at 20:28
add a comment |
$begingroup$
What about
putting smaller balances on larger balances
for example you can obtain
one large 3-pan balance, three medium 2-pan balances, six small 2-pan balances, and stack them in the obvious way.
Or even
four different sizes of 2-pan balances, with four extra known-genuine coins.
answered Jan 26 at 18:58
deep thoughtdeep thought
3,6861940
$endgroup$
1
$begingroup$
One could weld 4 3-pan balances together and say this is one device. :-)
$endgroup$
– A. P.
Jan 26 at 19:17
1
$begingroup$
@A.P. But I'm cheap and I was planning on borrowing the equipment, so I need to return it in good condition! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:23
2
$begingroup$
Definitely cheaper than flying to outer space.
$endgroup$
– A. P.
Jan 26 at 19:25
$begingroup$
Nesting balances was one of my ideas too. Can you do it with only 2-pan balances but without additional genuine coins?
$endgroup$
– Dark Thunder
Jan 26 at 20:28
add a comment |
$begingroup$
What about
putting smaller balances on larger balances
for example you can obtain
one large 3-pan balance, three medium 2-pan balances, six small 2-pan balances, and stack them in the obvious way.
Or even
four different sizes of 2-pan balances, with four extra known-genuine coins.
answered Jan 26 at 18:58
deep thoughtdeep thought
3,6861940
$endgroup$
What about
putting smaller balances on larger balances
for example you can obtain
one large 3-pan balance, three medium 2-pan balances, six small 2-pan balances, and stack them in the obvious way.
Or even
four different sizes of 2-pan balances, with four extra known-genuine coins.
answered Jan 26 at 18:58
deep thoughtdeep thought
3,6861940
answered Jan 26 at 18:58
deep thoughtdeep thought
3,6861940
answered Jan 26 at 18:58
deep thoughtdeep thought
3,6861940
answered Jan 26 at 18:58
deep thoughtdeep thought
3,6861940
3,6861940
1
$begingroup$
One could weld 4 3-pan balances together and say this is one device. :-)
$endgroup$
– A. P.
Jan 26 at 19:17
1
$begingroup$
@A.P. But I'm cheap and I was planning on borrowing the equipment, so I need to return it in good condition! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:23
2
$begingroup$
Definitely cheaper than flying to outer space.
$endgroup$
– A. P.
Jan 26 at 19:25
$begingroup$
Nesting balances was one of my ideas too. Can you do it with only 2-pan balances but without additional genuine coins?
$endgroup$
– Dark Thunder
Jan 26 at 20:28
add a comment |
1
$begingroup$
One could weld 4 3-pan balances together and say this is one device. :-)
$endgroup$
– A. P.
Jan 26 at 19:17
1
$begingroup$
@A.P. But I'm cheap and I was planning on borrowing the equipment, so I need to return it in good condition! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:23
2
$begingroup$
Definitely cheaper than flying to outer space.
$endgroup$
– A. P.
Jan 26 at 19:25
$begingroup$
Nesting balances was one of my ideas too. Can you do it with only 2-pan balances but without additional genuine coins?
$endgroup$
– Dark Thunder
Jan 26 at 20:28
1
1
$begingroup$
One could weld 4 3-pan balances together and say this is one device. :-)
$endgroup$
– A. P.
Jan 26 at 19:17
$begingroup$
One could weld 4 3-pan balances together and say this is one device. :-)
$endgroup$
– A. P.
Jan 26 at 19:17
1
1
$begingroup$
@A.P. But I'm cheap and I was planning on borrowing the equipment, so I need to return it in good condition! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:23
$begingroup$
@A.P. But I'm cheap and I was planning on borrowing the equipment, so I need to return it in good condition! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:23
2
2
$begingroup$
Definitely cheaper than flying to outer space.
$endgroup$
– A. P.
Jan 26 at 19:25
$begingroup$
Definitely cheaper than flying to outer space.
$endgroup$
– A. P.
Jan 26 at 19:25
$begingroup$
Nesting balances was one of my ideas too. Can you do it with only 2-pan balances but without additional genuine coins?
$endgroup$
– Dark Thunder
Jan 26 at 20:28
$begingroup$
Nesting balances was one of my ideas too. Can you do it with only 2-pan balances but without additional genuine coins?
$endgroup$
– Dark Thunder
Jan 26 at 20:28
add a comment |
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Some of us math-riddle folks are engineers ... but I think the engineering answer is to challenge the requirements! You'll need a better story than that! :-) :-)
$endgroup$
– deep thought
Jan 26 at 19:03