- How many watts are provided through a motorcycle "hot" wire? I know that W=VxI but I don't know the amperage either!
Current is a function of resistance and voltage so, for a fixed V, I varies depending on what is plugged in. When you are building something that supplies power the wattage specified is the maximum rating not a fixed value that it always supplies.
- Are all "hot" wires equal to each other? do they all carry the same voltage/amperage? Just so I know if I can tap into any supply on the wiring loom.
As a general rule all power wires will be at 12V (there will be some special ones that you don't want to mess with, anything related to the ECU, sensor, ignition, etc. but being a bike it is more likely to be carby than EFI and you probably don't have too many of those sorts of things to worry about). Not all wires have the same current rating, pull 10amps from your instrument cluster lighting wire and things will melt. Thick wires can carry more current than thin, if in doubt run a heavy wire from your alternator/battery and use a relay.
- Since vehicle wiring (on most motorcycles) is designed in a way that assumes electrons flow from positive to negative, is my schematic possible? Or do I need to run my components the other way round? Surely if a fuse holder was the first thing after the positive feed it would negate the effect of having one in the first place because it would be the last thing to be reached by a surge?
Electrons always flow from negative to positive. Current can be thought of as either a flow of negative or positive charge (and in fact in some cases it is a flow of positive charge). Current flowing from +ve to -ve is called conventional current and it works exactly the same as electron flow, the only time it matters which one you use is when you are doing maths and even then you only have to be consistent. Both are equally correct and as long as you don't try to work with both at the same time you are fine.
The current flowing in a circuit is the same at every point in that circuit at any given time, it doesn't flow through one lightbulb, then the next, then the fuse, it flows through everything at once. This might be easier if you stop thinking of electricity as a flow of electrons, think of it as a field which forms between two things of different potential allowing them to equalise. The field forms instantaneously when the conductors are connected along the entire length of the circuit and in that moment charged particles throughout the circuit begin moving to equalise the charge. It isn't electrons leaving one end of the battery and flying through the circuit to the other end, it is electrons (or plasma or ions or any mobile charge carrier) from everywhere moving due to the influence of the electric field. In fact individual electrons typically move at a speed of several cm/minute so the electrons leaving the -ve end of the battery could easily take many minutes to reach the +ve side.
The fuse should be located close to the +ve of the battery because the rest of the bike is connected to the -ve. Imagine the insulation on a wire is damaged and it touches part of the bike, if the wire is connected to -ve then nothing happens because the two things are at the same potential, now imagine the wire is connected to +ve, you have just created a short and you want a fuse in the circuit. You can't put a fuse between the bike and the -ve so you have to put a fuse between the +ve and every wire that will carry a voltage.
I have included a picture of my schematic (apologies for amateur drawings) and a schematic of my bikes wiring loom. I know that I will probably need higher than a 1/2 watt resistor for the LED but as I don't know how to calculate the amps going through the live wires I'm not sure which one to use.
I can't see the images, but the important value is the current rating of the LED. Find what current the LED wants (I) and the voltage of your bike (V, if your bike is 12V then use V=15 or so, the charging voltage for a 12V lead acid battery is 13.8V and you will get the occasional surge), calculate the resistance
using R=V/I (EDIT: this is wrong, see below) and the power using P=VI. I doubt you will need more than a 1/2W resistor if you are just trying to light up a status LED.