Re: Measurements using field probe (fwd)

---------- Forwarded message ----------
Date: Wed, 8 Jul 1998 06:44:49 +0000
From: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: Measurements using field probe (fwd)


  The AC peak should equal the DC peak? This would be true when the AC is
rectified and stored. However, that would depend on the probe and the
instrumentation. I have not been able to find in my literature where a field
mill measures the potential. 

  What are you using for a probe? What is the equation used?

  John Couture


At 04:30 PM 7/7/98 -0600, you wrote:
>---------- Forwarded message ----------
>Date: Tue, 7 Jul 1998 08:24:08 -0600
>From: "D.C. Cox" <DR.RESONANCE-at-next-wave-dot-net>
>To: Tesla List <tesla-at-pupman-dot-com>
>Subject: Re: Measurements using field probe (fwd)
>Hi John:
>The field mill meter actually measures the potential difference and is
>usually used with a calibration voltage.  We use 150 kV DC to calibrate our
>units, and then the potential is directly proportional to the distance from
>the measured terminal.  Wouldn't the AC peak be equal to 1.4 x Erms?  This
>would mean the ACpeak should be equal to the DCpeak.  Example, if you use a
>common 1N4007 to rectify the AC line which is 120 VAC Erms, then the DC
>reading is around 170 volts.  This is the same as measuring the AC line
>with a meter than is not a true RMS meter and gives only the peak AC
>reading --- again around 170 volts.