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On the use of tvsa

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  • dnorric
    Hi all To pass the interview on ou that we all recently listened to through the lie detector simply do the following. In the windows mixer under recording
    Message 1 of 42 , Feb 6, 2003
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      Hi all
      To pass the interview on ou that we all recently listened to through the
      lie detector simply do the following. In the windows mixer under recording
      properties set the source to wave which means that what ever you hear is
      recorded when you press the record button on your recording software. I
      personally used cool edit to record it to file, and then passed it through
      the lie detector.
      I know this is not strictly tesla related but in the area we are in it
      helps to have an upper hand.

      Damian
    • Bert Hickman
      Jim and Jet, It certainly sounds like a reasonable approach - I don t see why it shouldn t work as long as the resistance of the wattmeter s current input
      Message 42 of 42 , Mar 4, 2003
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        Jim and Jet,

        It certainly sounds like a reasonable approach - I don't see why it
        shouldn't work as long as the resistance of the wattmeter's current
        input resistance can be kept small compared to the resistance of the
        individual wires.

        Best,

        -- Bert --
        --
        -------------------------------------------------
        Where Small Change is created "Electromagically"
        Coin crushing by powerful electromagnetic fields
        Stoneridge Engineering: http://www.teslamania.com
        -------------------------------------------------

        Jim Farrer wrote:
        > Jet and all,
        >
        > I read your emails on multiplying the wattage of a watt meter, and felt
        > unqualified to reply. But I did have an idea. This is not a rigorous idea,
        > but I feel it should work. It seems very logical (and simple) to me, but I'm
        > not an engineer - quite. (Quite a BIG quite!)
        >
        > I believe the wattmeter has two wires in, and two wires out.
        >
        > You will have to vary the lengths of the wires I'm, going to mention, I'm going
        > to use 1 foot long (or 1 meter long, or 10 meters; whatever) as an example.
        >
        > Make one heavy wire into and out of the meter.
        > For the other wire, make 10 wires in parallel, each one foot long, smalller
        > diameter wires. (How big? Just big enough to safely carry the wattage your
        > meter will handle). Take one of these wires, split it in the middle, and
        > splice in the other two terminals of the meter.
        >
        > Since the resistances of these ten wires are identical, and they are in
        > parallel, each should carry one tenth of the current! Viola, as this pseudo
        > frenchman is wont to exclaim! I myss-speled it on poorpose.
        >
        > Even if you have a non-resistive load, and thus current and voltage out of
        > phase (bad power factor), I'd think this would work. My simple mind *demands*
        > that if one has 9 conductors in palallel, and a 10th one also in that same
        > parallel group (with the wattmeter in series with this 10th wire), the power
        > should split evenly between these 10 wires, thus 50 watts shown on the meter
        > should indicate 500 watts total dieivered to the load.
        >
        > Not quite perfect. Idea as presented assumes ZERO impedance for the watthour
        > meter; and nothing can be zero impedance. But if the impedance of the meter is
        > only 1% of the impedance of the 1 foot wire it splits, then it will be about
        > 1% off. Thus, use 20 foot long wires to get nearer perfection.
        >
        >
        > Jim Farrer
        >
        >
        > Jet Black wrote:
        >
        >>I have a problem (actually I have _many_ problems but thats too big an ask
        >>for the list ;> ) that has been _slightly_ bothering me for almost a week.
        >>Can I "bridge" my 100 Watt DC analog Marion Electric meter to read 500 or
        >>1000 watts fsd ?
        >>
        >>I asked Google how to attempt this feat and it came up with no sensible
        >>answers , a few interesting texts on basic electricity stuff that I
        >>bookmarked for review , so I hit my olde book collection , Audels Radiomans
        >>Guide , (printed in India 1966) usually never lets me down for sensible
        >>hands on answers , its tells me how to "shunt" a DC volt-ammeter and gives
        >>a formula to apply , seems "bridges" were built by a chap called
        >>"wheatstone" (silly me).I also checked out an earlyier edition of Audels
        >>"New Electric Library" vol 1 , nice red leather cover printed NY in 1931 ,
        >>WOW!!! the list "elders" had some funky text books to learn off , I'll save
        >>this book for bare bones construction methods or in case of accidental time
        >>travel.The Newnes Electrical pocket book from the UK in 1960 gives me
        >>ammeter shunts , voltmeter shunts , voltage divider & capacitor voltage
        >>divider shunts.Nothing on DC Watt meters.... but my mind has begun to start
        >>ticking as I re read these texts & write this email.....
        >>
        >>Another look at my wattmeter , "FS=1MA" is printed on the face 3/4 hidden
        >>by the bezel.Time to grab my multi meter briefcases , my digital
        >>capacitance & inductance meters send the meter FSD , I'll try my digital
        >>ohm meter on its lowest scale ,the meter reads 94 ohms the wattmeter reads
        >>60 watts , that makes the current flowing thru it ~798 Milliamps.So if P=IV
        >>, I get 75V as my meters output voltage ??? Err theres only a 9V alky or
        >>lithium battery in my meter , well the earlier numbers make some
        >>sense.....if I take "FS=1MA" to be FSD = 1000 Milliamps.So the internal
        >>resistance of my wattmeter is 100 Ohms ????
        >>
        >>Back to the Indian version of Audels.. , shunting all seems to be about
        >>ratios , I want 1000W out of a 100W meter , 10:1
        >>ratio.So I put 100 Ohms /(10-1) to get a shunt resistor of 11.1111 Ohms ,
        >>I'll see what my box of bulky ceramic resistors has in it , I feel a 1/4
        >>watt resistor will not be up to the task :) The stack of 10 x 120 ohm 10
        >>watt resistors is not looking like an easy job to parallel up befor elunch
        >>, I'll try these 2 x 5 Watt 5 Ohm resistors in series instead , just
        >>quickly , NO , that is not the answer.Fortunately nothing got toasted on
        >>the test "tap" but my meter went FSD really fast & the DC fans that I was
        >>testing slowed down for a split second.
        >>
        >>Am I attempting something that cannot be done or am I overlooking a vital
        >>piece of knowledge about the shunting of analog meters , like properly
        >>matching the wattage of the resistors to the load power ? Any guidance
        >>is most welcome my brain is tapped & my stomach is empty , time for food &
        >>more thought.
        >>
        >>JB
        >>
        >>PS How do I get rid of the condensation buildup resulting from the 4 x
        >>50Watt Peltier coolers I have attached to a piece of box aluminium 100mm x
        >>100mm x 300mm x 4mm thick ? I'm using it as a part time intercooler for the
        >>air intake to my PC , it drops the case temperature great for a while , but
        >>then the condensation starts , unlike my high performance car (where this
        >>200+Watt cooler will be fitted to), I feel my PC does not need water
        >>injection to make it run well.
        >>
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