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Not if the voltage and current are always 90 degrees out of
phase which is a fact of physics for pure standing waves. There is no power, instantaneous or otherwise, in pure standing waves. The cosine of 90 degrees is *always* zero. -- ************************************************** *******************************8 Cecil, your technical and mathematical skills vastly exceed mine... My poor little monkey brain has to work in the concrete, not the abstract... I sort of understand the concept of the standing wave being the instantaneous vector product of two colliding EM wave fronts... I accept the fact that MATHEMATICALLY; P = V*I* cos (theta).... And being that theta is DEFINED as 90 degrees the cos = 0, so the instantaneous power is zero... But here is where the math breaks down... On my herd of 100KW RF generators... Every time the load failed for even a split second, those fire breathing dragons would blow a hole through the quarter inch thick slabs of copper that made up the line, instantaneously... To prove that mathematically there is no power contained in a standing wave collides with my, farm boy, real world, physics experiments... The voltage peak on that line when that standing wave forms is real, it has mucho power (esp when driven by the dragon breath of a 100KW triode with 15KV on the plate), and it blows the hole in the line at the same distance from the load every time (proof of standing wave theory)... This is why i have a problem accepting that a standing wave contains no power - regardless of the prominence of the men who wrote the textbooks.. cheers ... denny |
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