Re[2]: WIND_TALK digest 63

From: Geoffrey.Boehm@wj.com-DeleteThis
Date: Fri Aug 25 1995 - 11:40:51 PDT


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From: Geoffrey.Boehm@wj.com-DeleteThis
Subject: Re[2]: WIND_TALK digest 63
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     When I studied electricity in school, the favorite analogy was to
     think of it as water flowing through a pipe - voltage was water
     pressure, current was water flow etc. This was a reasonable approach
     because most students had a lot more experience with water than
     electricity, and water is both less abstract and more visible.
     
     How ironic that there are actually windsurfers who find electrical
     objects more intuitive than water. Only in silicon valley.

______________________________ Reply Separator _________________________________
Subject: Re: WIND_TALK digest 63
Author: James.Paugh@Eng.Sun.COM-DeleteThis (Jim Paugh) at INTERNET
Date: 8/25/95 11:17 AM

Jim Paugh asks:
     
> "What I would like to hear is an explanation of why slack current occurs
> after say, a low tide. For instance, today, low tide at Crissy is at
> 5:03pm, but low slack is at 7:29pm! How is it that slack current occurs
> 2.5 hours after low tide? "
     
Will Estes writes:
     
> The best explanation I have heard on this was Ken Poulton's a few
> months ago on the wind_talk list. He compared tidal dynamics to
> electron flow within a wire. At the outside, resistance to the flow
> is greater than in the middle. So at the point where the flow stops
> on the outside, the flow is still moving at the center.
>
> Consider the analogy to the Bay: at the edges we have relatively
> shallow mud flats that offer a lot of resistance to the current. In
> the center you have a deep shipping channel where the water flows more
> freely. So at the low tide, the resistance of the mud flats on the
> edges just matches the current flow, and you have a zero current at the
> edge. But in the middle of the channel the ebb is still continuing,
> and it doesn't stop until a later time, because there is less
> resistance to stop the outward flow.
     
Kirk Lindstrum writes:
     
> It has to do with the shape of the bay. If you are an EE, then you can
> model the bay as a bunch of capacitors in each big, deep section and
> each "straight" as an inductor with the largest inductor at the Golden
> Gate. The "system" is driven by tide height (Voltage input) at the
> Golden Gate. Voltage is the same as tide height and current is the same for
> both with one being water and the other electrons.
     
Oh, NO! The dreaded EE analogy! Picture a blank stare on my face, eyes
glazed over, mouth agape :^o
     
Thanks guys, for trying, but I think I'll stick to the tidal wave rolling
through the Golden Gate analogy!
     
~Jim



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