# Never Bury Electrical Cable

Page 2 of 2
• posted on November 29, 2006, 12:29 am

Make that 120.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on November 29, 2006, 4:33 am
Bert Byfield wrote:

If you guys are getting all technical, wouldn't it only *change* direction 119 times?
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on November 29, 2006, 3:54 pm

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.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on November 29, 2006, 4:45 pm
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"
2 4 \\ /\\ /\\ / \\ / \\ / \\ / \\/ \\/ \\/ 1 3 5
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:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on November 29, 2006, 5:21 pm
snipped-for-privacy@artisticphotography.us says...

You've only show 2-1/2 cycles. Try:
|<-- 3 cycles -->| 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.
--
Keith

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on November 30, 2006, 3:54 am

The direction of current flow changes when the graphed line crosses the zero line, not at the peaks, but otherwise I agree. the peak.
--
No dumb questions, just dumb answers.

Larry Wasserman - Baltimore, Maryland - snipped-for-privacy@charm.net
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on November 30, 2006, 7:17 pm
On Wed, 29 Nov 2006 21:54:56 -0600, snipped-for-privacy@fellspt.charm.net () wrote:

For 120VAC RMS, the peak voltage is 169V. That's the highest hot-to-neutral voltage.
--
25 days until the winter solstice celebration

Mark Lloyd
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on November 30, 2006, 7:42 pm
|<-- 3 cycles -->|
/\\ /\\ /\\ --- 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) - 1 because your terminating point isn't a change of direction.
Therefore for 2 seconds, it would be 119 changes of direction (but 120 cycles)..
snipped-for-privacy@fellspt.charm.net wrote:

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on November 30, 2006, 8:20 pm
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
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
• posted on November 30, 2006, 7:27 pm
wrote:

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

Mark Lloyd