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Please tell me how this scenario would work in either [2] was heavy or [5] was light.

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Hey DKing, I'll try my best. I'll use as much of the above example as possible so as not to confuse things.

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This is for if [2] is Heavy.

Now we go back and look at the First Comparison which was set up as follows back in Part 1:

[1] [2] [3] [4] ^ {5} {6} {7} {8}

But, this time the Scale is "NOT" Balanced. That will mean the "different" Bullet is in:

[1] [2] [3] [4] ^ {5} {6} {7} {8}

Since the Right side of the Scale goes Up then [1] [2] [3] [4] is Heavy or {5} {6} {7} {8} is Light.< !--color--> And, we know for sure that <9> <10> <11> <12> are all alike.

For the Second Comparison set it up as follows:

[1] [2] {5} <9> ^ [3] [4] <10> <11>

Now the Right side of the Scale goes Up, and we know that either [1] or [2] must be Heavy because [3] & [4] are on the Right Side and they were on the side that went Down in the First Comparison. We also know {5} can't be heavy because of the First Comparison. And <9> <10> <11> are already known to be alike from the First Comparison.

Now the Third Comparison:

[1] ^ [2] would tell me which one is Heavy.

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This is for if {5}is Light.

Now we go back and look at the First Comparison which was set up as follows back in Part 1:

[1] [2] [3] [4] ^ {5} {6} {7} {8}

But, this time the Scale is "NOT" Balanced. That will mean the "different" Bullet is in:

[1] [2] [3] [4] ^ {5} {6} {7} {8}

The Right side of the Scale goes Up so [1] [2] [3] [4] is Heavy or {5} {6} {7} {8} is Light.< !--color--> And, we know for sure that <9> <10> <11> <12> are all alike.

For the Second Comparison set it up as follows:

[1] [2] {5} <9> ^ [3] [4] <10> <11>

Now the Right side of the Scale goes Down, and we know that either [3] or [4] "might be" Heavy because [1] & [2] are on the other side and they were on the side that went Down in the First Comparison. We also know {5} "might be" Light because it was on the Side that went Up in both the First Comparison and Second Comparison. And <9> <10> <11> are already known to be alike from the First Comparison.

Now the Third Comparison:

[3] ^ [4] would tell me if one of them is Heavy.

If they Balance then {5} is Light.

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Hey DKing, Did that help?

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You guys all blew it. you dont need to weigh any of them on the scale. load all bullets into equal brass and powders. shoot all through a chrony. the one that goes the fastest is the lightest, or the slowest is the heaviest. And I did not touch your darn scale once.

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moving, say, the right set to the right to achieve balance.

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This is a very old Logic Problem from one of my Professors when I was in school. I've modified it slightly to be of interest to those of us who Reload.

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You have a Balance Beam Scale which has a 6" diameter Pan on each side.

And you have 12 - 0.308" 165gr Bullets which look as if they are all exactly the same. Eleven of them have the EXACT same weight and one is either slightly heavier or lighter.

The difference in weight can not be felt. You do not have access to any other objects of a known weight.

What is the minimum number of weighings it takes to know for absolute certainty which Bullet is different and if it is heavier or lighter?

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It took me 30 minutes just to remember the Problem. If you do figure it out, I'll encourage you to hold onto the way to do it until everyone else that expresses an interest in "thinking through it" has responded, but do respond with the number of weighings it took you.

Best of luck to you folks.

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Sorry, you said "balance beam scale", and the teeter totter type qualifies Also, when one weighs, say, a powder charge for the first time you are seeking a state of balance, and when you achieve that by moving the weights that is "one weighing".

How about a poll asking for a vote on whether I got the right answer given the problem as stated?

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... Hey AC, ...It seems that I accidentally mis-lead you though. ... We really aren't "weighing" the Bullets. It is a "Comparison Scale" - a pan on each side - like the old scales in the Assayor's Office...

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