Home |
Search |
Today's Posts |
#81
|
|||
|
|||
On Tue, 16 Nov 2004 07:39:50 +0000, Kevin Aylward wrote:
Rich The Philosophizer wrote: On Mon, 15 Nov 2004 12:30:13 -0600, John Fields wrote: On Mon, 15 Nov 2004 04:43:22 GMT, Rich The Philosophizer wrote: I came up with "neocon" because of election hangover, and probably confused names because of the frenzy. --- "Neocon" as a ****raction for "neoconservative", or what? No, a contraction. I think he meant the ****raction, just as in "Good evening, ****stable, I was not speeding at all." I know that's what he meant, and I was making a point. Thanks again, Kevin. ?:^| Rich |
#82
|
|||
|
|||
On Tue, 16 Nov 2004 07:39:50 GMT, "Kevin Aylward"
wrote: Rich The Philosophizer wrote: On Mon, 15 Nov 2004 12:30:13 -0600, John Fields wrote: On Mon, 15 Nov 2004 04:43:22 GMT, Rich The Philosophizer wrote: I came up with "neocon" because of election hangover, and probably confused names because of the frenzy. --- "Neocon" as a ****raction for "neoconservative", or what? No, a contraction. I think he meant the ****raction, just as in "Good evening, ****stable, I was not speeding at all." --- BINGO! Give that man a see-gar! :-) -- John Fields |
#83
|
|||
|
|||
On Tue, 16 Nov 2004 08:04:27 GMT, Rich The Philosophizer
wrote: On Tue, 16 Nov 2004 07:39:50 +0000, Kevin Aylward wrote: Rich The Philosophizer wrote: On Mon, 15 Nov 2004 12:30:13 -0600, John Fields wrote: On Mon, 15 Nov 2004 04:43:22 GMT, Rich The Philosophizer wrote: I came up with "neocon" because of election hangover, and probably confused names because of the frenzy. --- "Neocon" as a ****raction for "neoconservative", or what? No, a contraction. I think he meant the ****raction, just as in "Good evening, ****stable, I was not speeding at all." I know that's what he meant, and I was making a point. --- Hmm... and what was your point then? I thought you were just being your ****rary little self. -- John Fields |
#84
|
|||
|
|||
Hi- If you are not using any "electricity" then turn off your main circuit
breaker. You should not notice any difference. Jim Paul Burridge wrote in message ... The power company run a line to my house. They supply me with electricity. This amounts to a 230V, 65A facility at the distribution board in a cupboard under the stairs. I run all my stuff from that board. The board contains several RCBOs that trip-out in the event of any leakage current being sensed. If current in = current out; they're happy and won't trip. Because they don't trip out, I conclude I don't use any current. The voltage supplied is 230VAC RMS. Since this is alternating between equal positive and negative half-cycles, the average level of this voltage supply is zero. I use no current and they effectively supply no voltage. Why do I get billed for electricity usage when I clearly can't have used any? -- "What is now proved was once only imagin'd." - William Blake, 1793. |
#85
|
|||
|
|||
There is no cancellation because the postive and negative peaks do not
occur at the same time. Paul Burridge wrote in message . .. On Sat, 13 Nov 2004 11:42:11 +0000, Scott wrote: Huh? It's only 5:30AM here and I just got up but, the ONLY time you aren't consuming power is at the zero crossing of the voltage and current sine waves (assuming a purely resistive load where I and E are in phase). Since you are paying for power, which is P=I X E, during the negative half cycle, you have, for example, -168 Volts X -1 Amp = +168 Watts...try it on a calculator...negative times a negative is positive. Thanks, Scott. So you're basically agreeing with me. I owe the power co. for the positive cycles they send me; they owe *me* for the negative ones. Since they are equal and opposite, they cancel each other out. Overall, then, zero billing justified. We are being conned!!! |
#86
|
|||
|
|||
Huh? It's only 5:30AM here and I just got up but, the ONLY time you aren't consuming power is at the zero crossing of the voltage and current sine waves (assuming a purely resistive load where I and E are in phase). Since you are paying for power, which is P=I X E, during the negative half cycle, you have, for example, -168 Volts X -1 Amp = +168 Watts...try it on a calculator...negative times a negative is positive. P=I^2R, so which direction the current is flowing is irrelevant, as the squaring removes any negatives, and R is always positive. Would be interesting if you could get a true -R though! (not like a tunnel diode, which just has a small region where increasing V decreases I, but it's still positive) |
#87
|
|||
|
|||
On Fri, 19 Nov 2004 08:26:54 -0500, "Dave VanHorn"
wrote: Huh? It's only 5:30AM here and I just got up but, the ONLY time you aren't consuming power is at the zero crossing of the voltage and current sine waves (assuming a purely resistive load where I and E are in phase). Since you are paying for power, which is P=I X E, during the negative half cycle, you have, for example, -168 Volts X -1 Amp = +168 Watts...try it on a calculator...negative times a negative is positive. P=I^2R, so which direction the current is flowing is irrelevant, as the squaring removes any negatives, and R is always positive. Would be interesting if you could get a true -R though! (not like a tunnel diode, which just has a small region where increasing V decreases I, but it's still positive) --- Overunity?! -- John Fields |
#88
|
|||
|
|||
Would be interesting if you could get a true -R though! I spent years perfecting a bar magnet with only a North pole. (Monopolar) I set it down on a table for a moment and it promptly took off for Antarctica. I haven't seen it since. I've been making them with only South poles for the last year, but have them all tied down. I'm going to release them on December 24 and give Santa a big surprise. |
#89
|
|||
|
|||
On Fri, 19 Nov 2004 08:26:54 -0500, Dave VanHorn wrote:
Huh? It's only 5:30AM here and I just got up but, the ONLY time you aren't consuming power is at the zero crossing of the voltage and current sine waves (assuming a purely resistive load where I and E are in phase). Since you are paying for power, which is P=I X E, during the negative half cycle, you have, for example, -168 Volts X -1 Amp = +168 Watts...try it on a calculator...negative times a negative is positive. P=I^2R, so which direction the current is flowing is irrelevant, as the squaring removes any negatives, and R is always positive. Would be interesting if you could get a true -R though! (not like a tunnel diode, which just has a small region where increasing V decreases I, but it's still positive) Instead of wires, use glass tubes full of plasma. ;-) Cheers! Rich |
#90
|
|||
|
|||
Was anybody keeping count of the number of postings to this thread BEFORE it
became personal and acrimonious? Seems like some posters lack the necessary sense of humour? They seem to hate anythingthey say be challenged? Then, as a reaction they descend in childish name calling and attempts at derision. Shame! Sticking to the subject. "Why electricity (for our antique radios of course) is/is not free". Well, hmm! The tube heaters use full wave, but what about those pulses of one way rectified half wave AC for the B+? (Primarily in non transformer radios!). Intended pun; non power transformer radios don't have a primary! :-) Personally I'd like to 'rectify'? my high electricity cost! Our consumption is recorded by a 60 cycle analog AC meter on the outside of my house, which is owned by the power company and read and billed monthly. Maybe I could get those positive half cycles and then not 'return' the negative ones, as someone has already suggested, and reduce electricity consumption that way? Joking of course :-) What good would half cycles be to respectable AC operated equipment? So anybody got any other 'practical' ideas, in addition to burning my non electric wood stove during the winter, to reducing my electrical heating cost? Our domestic electricity presently costs about 9 cents Canadian per kilowatt hour. That's roughly 7 cents US and roughly 4 UK New Pence, per unit/kilowatt hour. This part of Canada is a pretty small and somewhat widespread market. I believe that in Ontario in central Canada, a much larger population and population density, it is, or has been, due to political pressure following a botched attempt to privatize the electrical system? substantially less than that at around 5 cents; even though much of the energy is generated, by hydro power, in Labrador in this particular part of Canada! Thinking about it. |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|