Re: Ball lightning caught on tape!
- From: John Fields <jfields@xxxxxxxxxxxxxxxxxxxxx>
- Date: Tue, 04 Oct 2005 09:45:46 -0500
On Tue, 4 Oct 2005 23:11:59 +1300, Jasen Betts
<jasen-b@xxxxxxxxxxxxxxxxxx> wrote:
>On 2005-10-03, ehsjr <ehsjr@xxxxxxxxxxxxxxxx> wrote:
>
>>>>While i'm on the topic, another clip from the show had a lightning bolt
>>>>strike a wet soccer field knocking something like 5 players down and a
>>>>few sent the hospital in ambulances. Is this just a clip of under payed
>>>>soccer stars sucking up to the camera? It seems like the field,
>>>>especially being wet, would be incredibly conductive, but not
>>>>latteraly. Besides, Why would the electricity bother to run through a
>>>>human bady only to go back to it's startting point, the ground? Seems
>>>>pretty ify to me.
>>>
>>>
>>> contrary to popular belief electric current does not exclusively take the
>>> path of least electrical resistance (or the shortest path) if it always
>>> took the path of least resistance turning on the toaster would make the
>>> lights go out.
>>
>> Say what?
>
>read it again... if electricity always took the path with the lowest
>resistance and since a toaster has a lower resistance than most lightbulbs.
>turning on the toaster would offer a lower path or resistance and the
>electricity would flow that way.
>
>This is a form of argument called reductio ad absurdum, showing that soome
^^^^^^^^^^^^^^^^^^^^
reducto ad absurdium
>claim is false by showing that it requires some absurdity.
---
Not _some_ absurdity, but by reducing it to the absurd.
---
>>> a soccer player with his legs apart will experience that voltage.
>>>
>>>
>>> What electricity does do is take the path that provides the lowest voltage
>>> drop
>>
>
>> Right. And it also takes the path the provides the highest
>> voltage drop, and all paths of all voltage drops in between.
>> So the statement is meaningless.
>
>> Source---+---[Z1]---+---Return
>> | |
>> +---[Z2]---+
>> | |
>> +---[Z3]---+
>> | |
>> ~ ~
>> | |
>> +---[Zn]---+
>
>
>I guess I could have expressed that better. (but i'm not sure what to add)
---
All you have to do is show that the charge which flows out of a
voltage source is shared by loads connected to it in parallel:
<--11A
+---[+10V-]---+
| |
| 1A---> |
+----[10R]----+
| |
| 10A--> |
+----[1R]-----+
--
John Fields
Professional Circuit Designer
.
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