Safety match activation energy

I saw a YouTube video where someone took nichrome wire and took a wall outlet attached to wire and managed to light a whole series of matches. I am assuming his mains was 220V, 50Hz since he didn't have any kind of plug I had ever seen before (so not US plug).

I am not particularly interested in replicating this experiment, but rather though I would have fun doing some calculations on how it works.

Now the resistance he measured was around 300 ohms. I don't know much of how to calculate stuff in AC so it could be everything I did below is dead wrong, but this is what I got so far:

Area under a sine wave from 0 to pi radians is equal to 2, so multiplying that by the voltage of 220VAC would mean there is a total of 440V available to the mihcrome wire during one half of the sine wave. Maybe? Or is this a dead wrong assumption.

And then using (V*V)/R I calculated that there was 645 watts of power (mostly dissipated as heat I would imagine), and given that each half sine wave (counting positive and negative) happens 100 times per second (twice the full wave frequency of 50Hz) then taking 645/100 would leave about 6.45watts or 6.45 joules per second.

I looked online and couldn't find anyway where the activation energy of common safety matches was listed, but I would imagine each match must have an activation energy less then or equal to that created by his nichrome wire and mains setup because those matches lit up very quickly. Does anyone know what the minimum activation energy is aside from like experimenting and playing with a bench power supply until a match finally lights?

would leave about 6.45watts

I don't think so. 240VAC is the RMS voltage which means it is the voltage equivalent of a DC supply. So, the power dissipated by the wire will be V*V/R = 240*240/300 = 192W.


Its the temperature that starts the match burning, which is related to the activation energy of the reaction since enough chemical has to react to sustain a thermal chain-reaction.

With sinusoidal AC into a pure resistance the rms power is 1/2 the peak power dissipation - the instantaneous power dissipation is

(cos(2wt)+1) . Vrms^2/R

Where the voltage is sqrt(2).cos(wt) Vrms

Nice exercise in trigonometry to prove it I feel.

Ok, so let me see if this math looks a little more correct to you guys.

Vrms is .707 * Mains voltage so Vrms is going to be 155.54 since his video said mains was 220V.

w (angular frequency) = 314.16 = 50 * 2 * pi

And looking at the video I estimated it took maybe 1 second (or sightly under) from when he flipped the power strip on to when matches lit up so lest set t = 1.

And as he showed in video R = 276 ohms.

Using those formulas I calculated the instantaneous power dissipation at t = 1 to be 1.01 watts, and the voltage to be about 153.24 Volts.

And if RMS power is half peak power dissipation, then I would need to run he instentaneous formula over a range of times I suppose and find max then divide that by two.

I ran a script and determined the max instantaneous power is 2.08 watts, so then I would assume the RMS power is 1/2 of that or about 1.04 watts which is evidently enough to ignite a match.

You don't need any of the angular frequency stuff. The specified voltage of an AC supply is its RMS voltage which will dissipate the same heat in a resistor as a DC supply of the same voltage. If the values are now 220VAC and 276 ohms, the power dissipated in the wire is 220*220/276 = 175 watts.


Ok good to know and thanks!

Mains voltages and currents are always quoted as rms, so 220V mains is 220V rms, or 311V peak for a sinusoid. For a heating wire with a large thermal mass the calculation for ac mains or dc power is the same.

With pure resistances rms is just a way of describing the average power, since power is proportional to voltage or current squared.

If the wire is thin and has low thermal mass its temperature will cycle at twice the mains frequency as there's enough time between half-cycles for it to cool to some extent. Thus for very thin wires (like light bulb filaments) the max temperature with ac will be higher than for dc of the same rms voltage due to this temperature oscillating about a mean level. The effect can still be small even for blub filaments as 10ms is a very short time.

wes000000: I saw a YouTube video where someone took nichrome wire and took a wall outlet attached to wire and managed to light a whole series of matches. I am assuming his mains was 220V, 50Hz since he didn't have any kind of plug I had ever seen before (so not US plug).

The 2 pin plug he used is known as a Europlug and I'd guess his mains supply was 240/250V AC, which is common throughout most of Europe. The socket he was using is not a UK socket, but looks like the type used in most of mainland Europe.

Don't you think it would make more sense of you got yourself some fine nichrome wire, a box of whatever type of match you are interested in and a variable voltage DC power supply. With those you could conduct your own empirical tests. Using a sledge hammer to crack a walnut (a metaphor for the video shown) doesn't confirm the energy required, it merely demonstrated that enough was used.

As far as maths is concerned Watts is the product of voltage and current --- time does not come into the equation

If you look on pyro forums , there will be information on commercisl e matches.

They have range of firing currents around 20 1 ohm.

However they often have a sensitive prime on the wire

For experimentation , bare e matches can be had with 42 swg nichrome wire which are much easier to use as that guage wire is very fiddly to work with.