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
\\ /\\ /\\ /
\\ / \\ / \\ /
\\/ \\/ \\/
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:
|<-- 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
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?
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