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ArcticMyst Security by Avery

Taser core ?

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the voltage isn't directly related to the amount of energy in the spark.

6yi41uz.png


Energy = Voltage × Current × Time

That is undeniably a direct relation.
 
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Benm

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Oh on the stove thing: I have a gas stove with an electric convection oven in one unit, so it has mains power. The sparks are somehow generated from mains, it's just a switching mechanism that activates the sparks when you rotate and push down on the burner adjustment knob. It also fires all 4 burner igniters simultanously no matter which burner you're turning on.

I do remember the older mechanical ones on gas ranges that had no electrical connections, which had an igniter (push)button that required quite a bit of force to operate. Perhaps there were battery powered ones too, i've never seen one but would not be surpirsed.


On voltage and sparks: the voltage is required to jump a certain distance through an air gap, several kV per mm of distance typically. After the spark is established the voltage -usually- drops quite rapidly but the arc is sustained: Arcs are pretty good conductors and can draw enormous amounts of power/current if it is available. You can sometimes see this when switches are operated on electrical substations and such.

Something like that ebay device might generate 400 kV (although i serously doubt it) before the spark strikes, but once struck it is limited to whatever power i can provide (30 watts or so according to the input requirements assuming 100% efficiency).
 
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That claim on Ebay of 400/1000 kV is obviously garbage. Estimating the voltage based on the distance the arc can jump it's actualy around 15-20 kV.

30 Watts in a 1mS Spark is a very high energy spark. As proved by the very loud crack it makes.
 

Benm

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I'm not overly sure about the duration of the sparks this thing produces - perhaps i should get one to figure it out. Downside is i have no further application for it.

As for the voltage: that would be diffcult to measure really. You'd need a number of very high valued resistors with high voltage ratings to start doing anything there. These things are available (like 10 kV, 100 MOhm or 1 GOhm models) but cost around $10 each, and you'd need 40 to 100 of them if the claimed voltage is real.

Realistically i don't think the voltage is that high, 20 kV or so seems to be more realistic (comparable to a CRT TV anode driver). Those things are quite dangerous to experiment with though, one used in a big tv can output dozens of watts continously under normal operation.
 
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6yi41uz.png



Energy = Voltage × Current × Time

That is undeniably a direct relation.

Absolutly. You need all 3 components. Voltage on it's "own" is not directly related, which is what I posted.
 
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Voltage on it's "own" is not directly related, which is what I posted.

6yi41uz.png


That's NOT what you said. Here's what you said, which is demonstrably false:

the voltage isn't directly related to the amount of energy in the spark.

Even if you HAD said that, you're still wrong. It would be akin to saying "increasing your horsepower doesn't mean more speed", because maybe you put a fucking grizzly bear in the back seat and you haven't told us. Colloquially, statements are prefaced with an unsaid "all else equal," because if something else changed, we expect you to tell us about it.

Voltage IS directly related to energy. If it's in the equation, it's directly related.

Maybe what you MEANT to say, is "voltage is not the only contributor to the energy of a spark"
 
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You are quoting what I said correctly but saying it's wrong. You need to think about it a bit.

The voltage isn't directly related to the amount of energy in the spark.

That is correct. Only the Voltage "and" the Current "and" the Time are directly related.

Stop making assumptions and read what I actually said.

What if voltage is doubled but time is halved... Energy is constant. Proof that voltage is not directly related.
 
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Benm

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Well, lets not get it to semantic details there.

If you have a single spark, the energy in it is the product of voltage, current and time. Multiplication of those units gives you a value in joules. Output power of single-shot devices like photo flashes or defilibrators are usually quotes in joules per shot/pulse/etc.

If you have a continous arc (like in a jacobs ladder, or inside a CRT tube etc) the figure is quoted in watts, since it lasts very long time no longer is a factor and it is just the product of voltage and current, hence watts.

This is important from a safety standpoint: For a single shot device the harm is in the amount of joules, regardless if those are delivered in a microsecond or millisecond. When touching your hands across a capacitor at high voltage the energy stored dictates the outcome. If it is 1 kV at 100 uF or 10 kV at 1 uF probably does not matter that much, both are 100 joules and good for a severe shock.

When struck by an electrical arc time becomes a factor. If you somehow get your fingers into an arc that is running 10 watts (say 1kV at 10 mA or 10 kV at 1 mA) the damage accumulates over time. This sort of exposure will probably not kill you immediately, but the amount of tissue damage increases over time.

Things like getting struck by lighning are good examples of the first scenario, getting your fingers across mains voltage might be more towards the second, especially in low mains voltage countries like the US.
 

Benm

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I think Cyparagon is the one you are referring to when talking about the assumption.

I don't disagree that much though: consider how the weight of a block of concrete depends on its width. If you make it half as wide, but twice as long or twice as tall, it would be the same.

Does the weight of a concrete block depend on it's width? Not if you compensate for it by adjusting length or heigth, but otherwise it does.

The amount of energy released from a spark is the product of voltage, current and time. So by only adjusting one of those parameters, leaving the others constant, there is a direct relation.

With sparks it's more complex though: if you lower the voltage across a spark gap it will simply not fire at all, and you cannot just assume you can raise the voltage across it to any value you desire keeping the current constant: It's not a perfect resistor let alone perfect current sink.
 




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