On Jun 25, 2007, at 1:13 PM, Michel Jullian wrote:
----- Original Message -----
From: "Horace Heffner" <[EMAIL PROTECTED]>
To: <vortex-l@eskimo.com>
Sent: Monday, June 25, 2007 3:47 AM
Subject: Re: [Vo]:Air threads
On Jun 24, 2007, at 3:50 PM, Michel Jullian wrote:
I had some trouble coming to terms with this observation myself
when I first experimented with corona discharges, but emitter
current is equal to plate current at all times, not just on
average, even though the ions take ages (milliseconds) to cross the
gap. Try it if you don't believe me.
But we may not have corona discharges here.
Agreed, but corona generated ions are, just like droplets, slow air
flying charge carriers, i.e. the conduction mode in which you seem
to expect emitter and plate currents not to be equal, except in
steady state. I maintain that they are equal at all times,
including when the very first carrier starts crossing the gap.
Of course if you add a grid or a ring, or any interfering grounded
object in addition to the plate, as in your new circuit diagram
below, things are different. Maybe I should have made it clear that
what I said only applies when, apart from the plate, there are no
other grounded objects around (I thought that was implicit since
none was shown on the original circuit diagram).
The ground is always around. It always gets some piece of the action
unless it is screened.
A more experienced friend gave me the answer to this apparent
paradox: the plate charge is equal and opposite to the sum of the
charges on the emitter and in the air.
False. This is true only on average for drops.
No, it is true at all times. The discharge device as a whole
remains neutral at all times, like a capacitor or any component, so
current in = current out.
Like angels on the point of a pin. Sigh. If you can't accept the
ground is around I can see why you think this.
OK, so it appears you want to utterly dismiss the possibility of
drops or filaments, and the effect of the coupling of an AC signal to
other paths through ground, or through a conducting filament. You
seem to want to consider only DC steady state.
No, what I said applies to instantaneous currents, whatever the
waveform, whether steady state or not.
OK, in that case you need to consider the fact the field of the drop
couples to the ground, inducing charge there, as well as the plate.
The coupling to the plate increases as the drop approaches the plate
and correspondingly decreases with the ground. As the drop falls the
current increases in the plate.
Similarly, the current into the forming drop varies with the rate of
expansion of the surface area of the drop. When a drop breaks off
that changes dramatically. An AC signal is generated from the tip.
That signal capacitively couples to ground as well as to the plate.
It will be seen as stronger in R2 than in R1+R3 below (gosh if R1 and
R3 are used we need yet another resistor to measure the summed
current. Budget creep begins. 8^)
Within those confines
a ring-pass-through set-up should still be interesting. Bill did say
he had successfully done that.
---------------
| |
Emitter V |
. |
|
. |
|
Ring O . O |
| o-----C1----o AC Signal (Optional)
. | |
| |
Plates ___ ___ | T1=== AC
| | | |
| | R4 |
| V3 o | P
| | | |
| R3 | |
| | | |
-o-R1-o--G---R2-o-
| | |
o o o
V1 G V2
Fig. 2 - Circuit diagram for ring pass-through drop/thread
detection
Fig. 2 is again a diagram of the ring pass through concept. It shows
that there is at least one alternate current path through the ring
and thus R4 to ground. This takes fixed font Courier to view.
In the case of ions, at some ratio of distances, but still
maintaining the stream through the ring to the plates as Bill did, I
would expect a significant amount of the ions go to the ring and not
to the plates (but not so for a true conductive filament). From a
corona R4 would take a significant proportion of the current R1+R3+R4
= R2.
Not if the low current corona emitted a linear string of ions,
which is my favorite theory currently.
What prevents the ions from being attracted radially to the ground
ring and thus spread? Also, what prevents the ions from flying apart
due to mutual repulsion?
Further, of those charges that make it to the plates, I would
not expect a sharp change in current from R3 to R1 as the emitter and
ring are passed from left to right over the plates, but would for a
filament.
Ah, this I agree with, the landing area of the flying charges would
necessarily be a few mm wide so when crossing the border we could
have current on both plates at the same time, which could not
happen with a molecular sized filament.
OK, so we agree this is a valid test for a filament. This is good.
It seems to me an AC signal could be placed on the filament, if such
exists, to make measuring the proportion of signal on each plate
easier. Measuring DC currents is comparatively tough. The
conductivity of a filament is better than for air. I'll quote some
old stuff now for reference.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
- - - - -
"I'm using an old negative ion generator as the power supply. It puts
out 10uA maximum, and maybe 10KV to 15KV."
at http://amasci.com/weird/unusual/airhard.html.
"- I connected a microamp meter in series with the plate. It
indicated zero. When I let the other HV wire create one furrow in the
mist, the meter indicated zero UA. When I brought the cable close, so
there were maybe 50 to 70 furrows being drawn along the mist, the
meter started flickering, indicating approx. 0.5uA. These ion-
streams, if that's what they are, are each delivering an electric
current in the range of 10 nanoamperes or less. Jeeze. No wonder
nobody ever notices them."
at http://amasci.com/weird/unusual/airexp.html.
Assuming a gap of about an inch for "up close" it sounds like the
resistance of the filaments is about R = V/I = (10^4 V)/(10^-8 A) =
10^12 ohm. If there were 50 of them though, as above, there would be
a .5 uA signal, which would be readily detected. Might take a pre-
amp to get it cleanly though. I don't think you would get a signal
like that though an air gap from a fine point.
Having looked at signals through a 10 m tygon tube of flowing
electrolyte, and seeing their dependence on flow rate, I would expect
some surprises regarding the signals that would be transmitted
through such filaments, assuming they exist. Anyway, moving on ...
"In pure water, sensitive equipment can detect a very slight
electrical conductivity of 0.055 µS/cm at 25°C."
See: http://en.wikipedia.org/wiki/Water#Electrical_conductivity
This gives us an estimate of the filament cross sectional area:
A = 1/((1E12 ohm/inch)*(0.055E-6 S/cm)) = 4.6E-9 m^2
and radius:
r = (A/(2 Pi))^(1/2) = 2.7E-5 m
and diameter:
d = 2*r = 5.4E-5 m = 5.4 E-3 cm
about the thickness of a human hair. So, the filament is over 10,000
water molecules wide if this is correct.
If flow is moving at the top end speed Bill estimated, about 10 MPH,
we get a flow rate per filament of:
F = (4.6E-9 m^2)*(10 MPH) = 2E-8 m^3/s = 0.02 cm^3/sec
or about 1.2 cm^3 per minute, which sounds a bit high from the
description, but maybe very roughly in the ballpark.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
End old material
OK, with a radius of 2.7E-5 m, or less if conductivity of the
material is higher than for pure water, a really high frequency
signal would be good because it will travel mostly by skin effect.
Capacitively coupling the signal into the circuit above the power
supply sounds like the cheapest and easiest way to go. A small cap
made from some thick glass and some foil would probably work. Could
use a signal generator for input, maybe one which can amplitude
modulate. Run the signal into a step up transformer. It would be
neat to be able to get an audio output from the detection side for
experimenting. Feed it into a stereo? Maybe this is overkill.
*Something* has to be done to improve the current measuring ability
though.
Regards,
Horace Heffner
