Re: Who needs a quenching gap ?

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

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

Tesla list wrote:
 
> Original poster: "Finn Hammer" <f-hammer-at-post5.tele.dk>

> I believe it is possible to determine the coupling from this trace, how
> is that done?

To have an idea, I list below where is the first notch of the primary
voltage for the first optimum coupling coefficients. Look at the
comments
at the end, facts that I have just observed:

First series:
These are the most usual modes, with total energy transfer at the 1st
envelope notch.

mode	k		cycles (primary)
1,2	3/5	= 0.600	1.0
2,3	5/13	= 0.385	1.5
3,4	7/25	= 0.280	2.0
4,5	9/41	= 0.220	2.5
5,6	11/61	= 0.180	3.0
6,7	13/85	= 0.153	3.5
7,8	15/113	= 0.133	4.0
8,9	17/145	= 0.117	4.5
9,10	19/181	= 0.105	5.0
10,11	21/221	= 0.095	5.5
11,12	23/265	= 0.087	6.0
12,13	25/313	= 0.080	6.5
13,14	27/365	= 0.074	7.0
14,15	29/421	= 0.069	7.5
15,16	31/481	= 0.064	8.0
16,17	33/545	= 0.061	8.5
17,18	35/613	= 0.057	9.0
18,19	37/685	= 0.054	9.5
19,20	39/761	= 0.051	10.0
20,21	41/841	= 0.049	10.5

Second series:
There modes result in total transfer at the -second- envelope notch. 
I don't list the modes equivalent to the 1st series.

mode	k		cycles (primary)
1,4	15/17	= 0.882	2.0
2,5	21/29	= 0.724	2.5
4,7	33/65	= 0.508	3.5
5,8	39/89	= 0.438	4.0
7,10	51/149	= 0.342	5.0
8,11	57/185	= 0.308	5.5
10,13	69/269	= 0.257	6.5
11,14	75/317	= 0.237	7.0
13,16	87/425	= 0.205	8.0
14,17	93/485	= 0.192	8.5
16,19	105/617	= 0.170	9.5
17,20	111/689	= 0.161	10.0
19,22	123/845	= 0.146	11.0
20,23	129/929	= 0.139	11.5
22,25	141/1109= 0.127	12.5
23,26	147/1205= 0.122	13.0
25,28	159/1409= 0.113	14.0
26,29	165/1517= 0.109	14.5
28,31	177/1745= 0.101	15.5
29,32	183/1865= 0.098	16.0
31,34	195/2117= 0.092	17.0
32,35	201/2249= 0.089	17.5
34,37	213/2525= 0.084	18.5
35,38	219/2669= 0.082	19.0
37,40	231/2969= 0.078	20.0
38,41	237/3125= 0.076	20.5
40,43	249/3449= 0.072	21.5

In general: a=integer, b=a+odd integer:

mode=a,b;  k=(b^2-a^2)/(b^2+a^2); full primary cycles=b/2
Or k~=1/(2*cycles), as mentioned in other posts.

Note the curious fact that it's possible to have total energy transfer
at the 1st notch (modes a,a+1), at the second notch (modes a,a+3), or
at the nth notch (modes a,a+2*n-1).
Close to each optimum k for total transfer at the 1st notch there are
two values of k the result in total transfer at the second notch. 
Close to these ks there are other values that result in total transfer
at the 3rd noth, and so on. There is also a set of high optimum
couplings
corresponding to modes 1,1+odd integer, and the families that appear
around them.
I don't believe, however that a real spark gap is sensitive enough
to the primary energy to quench precisely where the primary energy
disappears, because the differences among the primary energies at all
the
envelope notches is small. But there is a tendency for this.

Antonio Carlos M. de Queiroz