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[blissy] you forgot to press the croissant button! |
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I can not imagine how it would be done mechanically. |
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1) Detect whenever there are 3 in a row or 3 in two rows or whatever
2) Eject just those three |
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Mechanical turk counts as "mechanical," right? |
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And the proposed mechanism is...? |
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I have no doubt this is mechanically feasible, by
way of each jewel having a key mechanism in
place, with the lock having x solutions where (x) is
the number of types of jewels involved, and a
solution is
defined as three identical keys lining up. |
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Once the mechanism unlocked, the board would
be tilted and the jewels weighted such that they
fall out by gravity. |
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The swapping mechanism would have to be
approximated, most likely by making it so that any
jewel can be pulled out, but once one is gone,
only adjacent jewels can be removed, and the
drop mechanism would only trigger when all jewels
were in place (triggering the drop mechanism may
involve pulling a lever after each move). |
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Positioning and loading the extra jewels would be
difficult, so you would most likely use a long throw
lever that also picked and oriented them as stages
before feeding them to a drop in place mechanism
at the top. |
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What might prove difficult would be over-length
plays where four or five jewels in a row were
triggered, as that adds a layer of mechanism
complexity. Also a problem is that the jewel
reserve would have to be significantly larger than
the board size in order for the newly loaded
jewels to approximate randomness. |
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Sorry, I sick xandie. I push twice this time round. |
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