Wow... that's the best one yet... Santos Mahogany!!
Amps are amps... If it were V*A, the'd have a wattage rating.
Ok... so to interject another silly question into this...
Any reason I shouldn't just stick this kick on top of my dibi and skip all the stuff with fuses? It's exactly the size of the top of any battery, and will fit right in the wood tube (because it's exactly the size of the battery). I'd just need to maybe solder a wire for the negative to run to the cap or something, since the wood wall won't provide the same ground as the metal tube?
Will it fit? It's about 1/2" tall. (someone asked about kicks earlier, and so while I had pulled apart my itaste 134, I decided to pull apart one of my mech mods too for pix).

If I ever get to run my own company, I'm hiring BuzzKilla to head up the Customer Service department.![]()
What confuses me about it is that obviously when a fuse blows it heats up (consumes power). It must take a certain amount of power to do this and power = volts x amps. What I didn't think about is that the voltage drop across the fuse is not the same as the supply voltage (i.e. neither 4 V or 12 V but rather something more like 0.1 V or 0.01 V). Apparently the voltage drop doesn't vary (much) no matter what the supply voltage is so the only thing that can change the power is the amps.
I'll have to think about this more. I'm not entirely convinced. I know, I know... it works so what's there to argue with? Well I just don't quite understand it to my satisfaction yet.
Hahahaha,
You should see me at the office. My boss tries his best to stay composed during the client meetings.
It's jaw dropping fun.
But CS has never been for me. My girlfriend had to take over that part when we started an Etsy account together.
She won't let me respond to the emails
I'm no longer going to add fuel to this fire. I've done my share in derailing this thread.
pdib, my b-day is on St. Patrick's Day
I would like OliverDibi to at my doorstep before then
Sent from my Nexus 4
I 2nd that... Even if you can, it doesn't mean you should.You also dont want a kick on this device. For one it would no longer be a mechanical mod, and for two they only provide 15watts. I love my 35watt vape and you can't get that with a kick or DNA.
It'll fit. Just need to use and 18500 battery instead of an 18650.Not to mention that the kick is pretty tall and wouldn't actually fit in pdibs mod..
Hahahaha,
You should see me at the office. My boss tries his best to stay composed during the client meetings.
It's jaw dropping fun.
But CS has never been for me. My girlfriend had to take over that part when we started an Etsy account together.
She won't let me respond to the emails
I'm no longer going to add fuel to this fire. I've done my share in derailing this thread.
pdib, my b-day is on St. Patrick's Day
I would like OliverDibi to at my doorstep before then
Sent from my Nexus 4
You also dont want a kick on this device. For one it would no longer be a mechanical mod, and for two they only provide 15watts. I love my 35watt vape and you can't get that with a kick or DNA. Not to mention that the kick is pretty tall and wouldn't actually fit in pdibs mod.

What confuses me about it is that obviously when a fuse blows it heats up (consumes power). It must take a certain amount of power to do this and power = volts x amps. What I didn't think about is that the voltage drop across the fuse is not the same as the supply voltage (i.e. neither 4 V or 12 V but rather something more like 0.1 V or 0.01 V). Apparently the voltage drop doesn't vary (much) no matter what the supply voltage is so the only thing that can change the power is the amps.

I 2nd that... Even if you can, it doesn't mean you should.
It'll fit. Just need to use and 18500 battery instead of an 18650.
OK, let me try to explain it:
Short version: The power dissipation in a resistor is equal to the square of the current times the resistance. This is all the fuse (being a low-ohm resistor) cares about: It's resistance and the current flowing through it. Of course if you know the current flowing through it and and its resistance, you can also calculate the resulting voltage drop across it, but that isn't necessary.
Long version: The fuse is very low-ohm resistor in series with a substantially higher-ohm load. Because it's a series circuit, the current through both resistors is intrinsically the same, and how much current is flowing is is primarily determined by the (relatively high) resistance of the load, where you will then find almost all of the voltage drop. If you do the Volts x Amps (or I^2 * R) calculation for both the load and the fuse, the load is consuming almost all the power and the fuse is consuming almost none (which is why it doesn't blow). Now what happens if we short the load? The current flowing in the circuit will rise until something limits it somehow. Assuming a perfect power supply and wiring, the only non-negligible resistance left in the circuit is the fuse. So that's where all the voltage drop is forced to go. You'll have a whopping current through (and all your voltage drop at) the fuse, which will then be "consuming" some crazy amount of power, which cause it to blow.
Incidentally, in the absence of a fuse (or other current limiting device) the battery becomes the fuse, but batteries typcally "blow" less gracefully than fuses do.![]()
I like the last line.
T
