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Mark Tarrabain wrote:
> When A and C are both high, A+C will therefore be low, so the left
> piston will be down. So the bottommost switch on the left piston will
> already be down to receive A&B and produce a carry out. If only one of
> A and C are high, the left piston will be up, so the left bottom switch
> and the switch above it will be up and therefore be ready to receive
> whatever B happens to be, which again must be the carry.
Oops... after rereading what I wrote, I see made a typo there, and I can
see how it could be very, very confusing.
What I meant to say was that the bottommost switch on the left piston
will already be down to receive A & _C_, which is high and will produce
a carry out.
>> Mark
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Message is in Reply To:
 | | Re: A better full adder!
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| (...) Thanks. Here's how it works: Consider only the right piston for a moment. The output of this subsystem (which will be driving the left piston) is simply (A&~C)|(~A&C), which is an XOR operator, or a half-adder. So what drives the left piston (...) (21 years ago, 28-Jun-03, to lugnet.technic, lugnet.robotics)
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