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*To*: tesla-at-pupman-dot-com*Subject*: RE: Odd numbers*From*: "John H. Couture" <couturejh-at-worldnet.att-dot-net> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)*Date*: Wed, 13 Sep 2000 21:06:14 -0600*Delivered-To*: fixup-tesla-at-pupman-dot-com-at-fixme

Sundog - Changing the voltage and not the watts can also be checked with the JHCTES Ver 3.1 computer program. For example changing from 15KV/30ma to 4KV/113ma will give you the same 450 watts with the JHCTES program but the spark length will be reduced from 28.7 to 25.2 inches. This is based on empirical data extracted from actual tests. However, the spark lengths shown will vary depending on the construction of the coil. If you use the joules equation you need the efficiency. The efficiency for the 15KV/30ma (21000 peak) 450 watts J = .5 * .01*10^-6 * 21000^2 = 2.21 joules for 28.7 inch spark The input watts = 450/120 bks = 3.75 watts /break This is 2.21/3.75 = .59 * 100 = 59 % efficiency Assuming the same number of joules and efficiency The 4000 volts (5600 peak) and 113 ma 450 watts C = 2.21/5600^2 = .07 uf primary cap This gives a 25.2 inch spark There is not much difference in spark length at low wattages. Note that the primary cap must be selected to agree with the primary voltage available. As you say the larger currents will cause more losses because of the heating. For this reason the 59 % I show for the 4000 volt coil may be lower. This would give a lower spark length. Tesla coil design can get complicated and a TC program can be a great help. John Couture ------------------------------ -----Original Message----- From: Tesla list [mailto:tesla-at-pupman-dot-com] Sent: Wednesday, September 13, 2000 10:31 AM To: tesla-at-pupman-dot-com Subject: Odd numbers Original poster: "sundog" <sundog-at-timeship-dot-net> Hi All!, I had myself an idea (Quick! Run and hide!), and plugged same #'s into Wintesla. I was looking at the J per bang figure, and since the more J you push, the bigger the bang. J=.5*C*V^2 Right? Good. That formula itself tells me that if you maintain the same current (30mA in the case of my example NST's), the higher voltage units have a better watts/joule 'rating'. IE, the 15/30 charges it's reso cap to ~4.9J (did I do that right? sounds reasonable, but it it *iss* wrong, it's consistantly wrong). The same size bang, for a 4kv tranny takes ~115mA (aroung 450 watts) to do the same thing, as it's charging a 76nf cap. There's the basis. Now, let's say I want a bigger bang, say, 1J bigger. I'll just increase the current. The 15kv'er will see less of an increase, and the 4kv will see a fairly chunky increase just to meet the same J per bang. But each is still ~ the same wattage. For TC's, it's been said that a higher primary voltage is better. i'm guessing it's because a 5J bang at 4KV will sink a lot of current, but a 5J bang at 15kv will have less current (I see the correlation of voltage&capacitance of the tank cap there), for the same size bang. Less current=less heat=less losses in the gap. Okay, it's beginning to click. Now, when the gap fires, you get a healthy dose of voltage/current into the primary. What makes a stronger magnetic field? A higher voltage or current, or is it just a combination of both (IE, is a 12kv30ma energizing a selenoid of wire just as strong as a 100v/3.6A?) Both are 360 watts. Okay, if it's the voltage, I can see a 28kv1.5kva PT being worthwhile. If it's the current, I can see MOT's or a 7.2kv20-50kva PT being a solution. On either end of the examples, quenching is a problem. The 28kv system can ignite a long arc just by it's potential. The 7.2kv'er will try to powerarc just by sheer current (several hundred MA at least) A side-question: The low voltage/high current approach would need a cap with a good DT/VT, as it's got a *lot* of juice to move very quickly, whereas the HV caps (smaller in capacitance by far), could be a bit "slower" i guess. ? Soo, it all boils down to : more intense magnetic field by current or voltage (if the total wattage is the same)? Hmm..re-read it all, and It all looks to boil down to the total input wattage. Only real variable I can see is the V & I. I'll be researching this on my own, but freely welcome comments :) now my head hurts... Sundog

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