Okay let's try again, first divide the 12 balls into 3 groups.....let's name them AAAA, BBBB & CCCC
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* r0 s9 x, O, |2 ?, y( A* m2 cFirst weight....AAAA ^ BBBB, if balance, then problem in C balls, see my post above. If not balance, continue below.TVBNOW 含有熱門話題,最新最快電視,軟體,遊戲,電影,動漫及日常生活及興趣交流等資訊。. Q# Y8 Y0 k9 l* E6 q
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From the first weight, let's say AAAA is heavier then BBBB, then either the problem ball is heavier among AAAA, or lighter among BBBB, but all C balls are normal.TVBNOW 含有熱門話題,最新最快電視,軟體,遊戲,電影,動漫及日常生活及興趣交流等資訊。( Q+ s3 q$ j) L# l$ F' _: ?# A7 X& `
9 y/ D$ _, L8 G ~0 i3 p公仔箱論壇2nd weight....AAAB ^ ACCC, there will be 3 scenarios:TVBNOW 含有熱門話題,最新最快電視,軟體,遊戲,電影,動漫及日常生活及興趣交流等資訊。" y( e" k: J8 I. M8 i3 [; ?4 c
(1) if AAAB is heavier then ACCC, for sure the problem ball is a heavier ball and it's among AAA. Take 2 of these and weight against each other, the heavier ball is the problem, if balance then it's the remaining ball.TVBNOW 含有熱門話題,最新最快電視,軟體,遊戲,電影,動漫及日常生活及興趣交流等資訊。 F0 l' d! m/ o$ g0 ?
(2) if ACCC is heavier then AAAB then weight the A ball in ACCC against a normal C ball, if balance then the problem ball is the B ball in AAAB and it's a lighter ball. If heavier then this is the problem ball. (note it cannot be lighter)
- Y- m/ P: Y! ?' Y( X(3) if balance then the problem ball is one of the 3 B balls not touched in the 2nd weight and it's a lighter ball. Take 2 of these B balls and weight against each other. If balance then the other B ball is the problem. If not balance then the lighter ball is the problem. |