RE: H/D ratio

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: Jim Lux <jimlux-at-earthlink-dot-net> 

At 11:59 AM 1/16/2004 -0700, you wrote:
>Original poster: "Luke" <Bluu-at-cox-dot-net>
>4.5:1 is good I get it.
>So I can build my secondary 2" wide and 9" tall with an input of 900
>watts.
>Seems we have been here before.
>
>If 2" coil is bad and 4 is better and 6 is better and 8 is even better
>then a 50" dia coil would be incredible.
>
>So maybe I should just go with a 50" diameter coil so my height would
>work out to be 225".  I could set this up and fire off some long arcs in
>my back yard powering it with 900 watts.
>
>These are both extremes.
>Where is the middle ground?
>Ok here is where you give examples of coils that are in the 900watt
>range and tell me the diameters and heights.
>NO NO NO
>Don't go there please.
>
>I am looking for WHY the middle ground is good and HOW you or they have
>figured out where that middle ground.
>
>I re word this question in every way I can think of only to be greeted
>with a well meaning response that tells me this works well.  I WANT TO
>KNOW WHY!!


Why the middle is good:

Too small for the power -> arcs over because it's too short for the 
voltage/power
Too big for the power -> Can't put enough charge on the top load to get it 
to breakdown.

You have a certain amount of energy to put into the system on each half 
cycle. With 900W in a 60Hz system, that's about 7.5 Joules per "bang".

Let's consider JUST the topload capacitance, since toploads and secondaries 
are usually built to be commensurate in size.

We want the topload to be big enough so that the voltage can get to a 
reasonable value before it breaks down from the surface field.

We want the topload small enough that the limited amount of energy we've 
got can just get it to that voltage.

So, we have 7.5 Joules.  Let's try a toroid 8 feet in diameter with 24 inch 
minor diameter, about 100 pF capacitance.  Energy = 1/2 * C * V^2, so,
V = sqrt(15/100) MV or V = 387 kV..  The breakdown voltage of something 24 
inches in diameter is about 1 Megavolt, so your spark won't start.

Now, let's try a toroid of a more reasonable size, say, 4 feet in diameter 
with 12inch minor diameter. That's about 50 pF, so the voltage will be 1.4 
times greater, or about 541 kV.  AND, the breakdown voltage of the toroid 
will be halved, to 500 kV, so now, the voltage is higher than the 
breakdown, so the spark will start.

Now lets try even smaller: 2 foot toroid, with 6" dryer duct, for 26 
pF.  This would get to a voltage of 760kV

In reality, the breakdown voltages are a bit lower, and you've got to add 
in the capacitance of the secondary (which reduces the voltage)

Taking your example of a 50" diameter secondary, 225" high, the capacitance 
of the secondary alone would be about 100 pF, and once you hung a 120"x48" 
topload on it (another 128 pF), you'd only be getting about 260 kV, well 
below the megavolt kinds of breakdown.

Now.. if you take your 900W and charge up a BIG capacitor so you can get 
900J bangs every second or so, you might have something (2.8MV).. but that 
wouldn't really be a tesla coil, but more of an impulse transformer.

In the too big case what you've done is changed the system from a tesla 
coil designed to make sparks into a radio transmitter with an antenna.