Never Bury Electrical Cable

Draw a graph of 60cps sine wave for 1 second. It will consist of 60 cycles in the form of full sine waves. Each peak, top and bottom, is a change of direction. So: 120.

I'll simplify this because of the formatting to 3 cycles using straight lines. Assume starting at the "top" of a cycle. I have numbered the "changes"

At the first and last positions, there is not "change in direction". Therefore the number of changes of direction = (cycle X 2) - 1

Only be starting mid-cycle do you get cycleX2

But I think we are going a little overboard on this.

Bert Byfield wrote:

You've only show 2-1/2 cycles. Try:

|| 2 4 6 \\ /\\ /\\ /\\ / \\ / \\ / \\ / \\ / \\/ \\/ \\/ \\/ 1 3 5

Perhaps, but you're still not counting right. Using your arithmetic we'd be missing a cycle every two seconds.

Well, 12-16 million illegal aliens think they've got it figured out. Most of the rest of us know we've got something good when everybody else wants a piece of it.

The direction of current flow changes when the graphed line crosses the zero line, not at the peaks, but otherwise I agree. the peak.

For 120VAC RMS, the peak voltage is 169V. That's the highest hot-to-neutral voltage.

Consider that the current will VERY SELDOM be turned on at the exact moment of zero-crossing, and left on for exactly 1 second.

||

/\\ /\\ /\\

--- 1 / 2 / 3 / 4 ------------ center line \\/ \\/ \\/

Starting at point 1, a complete cycle is achieved at point 2. The next cycle is completed at point 2, the third cycle is point 4

You do not have a change of direction at point 1 Change of direction at point 2 Change of direction at point 3 You do not have a change of direction at point 4 (does not cross the line, mere reaches it).

Therefore you only have 2 changes of direction. This = (cycles X 2) -

Therefore for 2 seconds, it would be 119 changes of direction (but 120 cycles)..

snipped-for-privacy@fellspt.charm.net wrote:

Try that again, 1/4 cycle farther along the phase :-)

If the segment between 1 and 1.01 does not contain a zero-crossing, than there must be a zero-crossing to the left of 1. If that is true, and we have three complete cycles, then there must also be a zero-crossing to the left of 4.

To demonstrate this, chop up your three-cycle diagram into three separate 1-cycle diagrams.

If you have 1-2, 2-3, and 3-4, do the zero-crossings at 2 and 3 just dissapear?

Do your diagram again for 6 cycles. How many direction changes are there? Where did the extra one come from?

--Goedjn