Elmo said:
Soldering anything with a melting point close to that of solder to something that's as big a heatsink as an 18650 cell is going to be damn near impossible with an iron. It could probably be done with a micro torch but you'll have to heat the cell up a lot before you ever touch the zinc wire to it or else the zinc will melt while the cell stays well below the melting point of the solder.
I'd do a lot of practicing on dud cells before I tried that on a good one.
What I have found is that pure lead wire works pretty well as a fuse and can be soldered. Its finicky, but can be done. It melts barely above lead/tin solder and can be had in nice sizes. My first tests show that a .015" lead wire will trip at 6-7A.
My thoughts on the conductivity of fuse material is that it doesn't make any difference. My reasoning is that we care about the energy that is being dissipated by that fuse (piece of wire) in relation to that metals melting point. In order to blow a fuse, you obviously need to heat that metal to its melting point and the energy to do so is directly linked to the temperature at which at it melts. For example, copper melts at a relatively high point, and lead at a low one. Heat dissipation is proportional to the difference in temperatures between the two items. In our case, the fuse gets hot, and the air and batteries themselves absorb that heat.
Now for the sake of example and reasoning, I will be making some general statements such just as I did with saying lead melts at a lower temperature than copper. At this point I don't care exactly how much lower, just that it is lower.
For our copper fuse lets say that it takes 2 watts of energy to heat that wire up to its melting point at that it happened at exactly 10 amps. since W/A=I that gives me a voltage drop of 0.2V across that fuse for a resistance of 0.02 Ohms.
Now since lead melts at a lower temperature, it may only take 1 watt to heat a lead fuse of the same length to its melting point. In our imaginary test this also happened at exactly 10 amps giving us a voltage drop of 0.1V and resistance of .01 Ohms.
But wait, based on the above, the lead fuse would have to have a lower resistance than the copper one. And copper is more conductive than lead which is the opposite of what we need. Lets make this simple and say that copper is twice as conductive. The math for resistance is easy. Its just cross sectional area X Length X a conductivity constant for that metal. Our example fuses are of equal length so that can be removed. So to get half the resistance, with a metal that's half as conductive, I just need a wire with 4X the cross sectional area. So to conclude that, if my wire is not as conductive, just use a bigger one.
Besides not glowing red hot, low melting point fuses also help reduce power loss. Since we don't need as high of resistance to generate the heat required to melt the fuse. the energy lost due to fuse resistance lowers.
For example, lets say we have a 14S 100P pack, and we want to draw 2800w from it. lets assume we are at 4.0v per cell which means we need 50A from the pack, spread over 100P that's 0.5A per cell
Power(W) = Resistance (R) x Current (I)^2
Example 1: copper fuse 0.02 Ohms
0.02 X 0.5^2 = 0.005W per cell
Example 2: Lead fuse 0.01 Ohms
0.01 X 0.5^2 = 0.0025W per cell
so 0.0025W, who cares right? well take that per cell we have (1400) and now we get a loss of 7W in the Copper fused pack and 3.5W in the Lead pack. You may not care, but this is the kind of stuff I care about.
As a brand new member(but lurker) feel free to ignore my rambles until I show my tests, setup, and exact math
Remember, this is dead reckoning, not exact figures
