The wire gauge you need is a function of several things.
- The acceptable voltage drop or power loss (that appears to be the
only thing considered in the website you linked). The voltage drop (and power loss) is proportional to wire length and inversely
proportional to the cross-sectional area of the wire- in other words
inversely proportional to the square of the wire diameter (assuming constant current).
- The acceptable temperature rise. This is a function of the number
of current-carrying wires bundled together, the environment (maximum ambient temperature
and air pressure or altitude, for example), the insulation type, the
wire type (some types of wire are plated to withstand higher
temperatures than bare copper without corroding).
- Regulatory requirements and other considerations- for example, the
wire may be rated for 200°C insulation, but you might not want the
wire to run that hot.
- Fusing- the fuse or circuit breaker should protect the wire in the
case of faults such as overload or short circuit.
Very short lengths of wire can depend on heat sinking through the ends (indeed, in a vacuum, that may be the main heat loss mechanism), but usually that's not taken into account.
Normally you'd run through a checklist such as the above to make sure ALL the requirements are satisfied simultaneously, so you might find that using PTFE insulated wire allows you to use AWG 18 wire, but because of the voltage drop limitation you'll have to use AWG 12 wire.
First, compute the power you need.
This will consist of radiated and conducted heat loss.
The radiated power can be computed from Stefan's Law. Assume an emissivity of 1.0 to be on the conservative side; you will also need the wire diameter in order to compute the radiative area.
This is likely to be relatively low. If you can keep the wire in a high vacuum, this will be the total power you need from the battery.
Conducted heat loss requires knowledge of the material surrounding the wire, such as air, water, insulating material, etc. Assuming still air, use Newton's Law. You need to plug in a "heat transfer coefficient" ... assuming the wire is horizontal, here's how to calculate it. For the value of k the thermal conductivity of air, use k = 0.024 W/m/K. You will need the wire diameter D, you can assume a Prandtl Number of 0.8 and a Rayleigh number of about 1E6 as a starting point, and you must make reasonable assumptions about the air temperature.
If the air is moving, cooling will increase and things get much more complex and less accurately calculable as calculating the heat transfer coefficient gets more difficult.
If this computation is too much, bypass it by experimenting with a metre of wire and a variable power supply.
This is likely to give a much higher power requirement than radiative cooling at the temperature given.
Add these figures and you now have a power requirement. Given 12V, you can calculate the required current, and thus the life of a car battery.
You can also calculate the resistance you need as a load. Compare that with the resistance for 1km of your chosen diameter of Nichrome wire and you will see that you cannot power a continuous 1km length from only 12V. Which means you need multiple short lengths of wire connected in parallel, or a much higher voltage than 12V.
EDIT : The new information changes things A LOT. You'll achieve this temperature with much less power because the wires are in close proximity so they are heating each other. Model the heated surface as a plane when calculating radiation losses and heat transfer coefficients.
Parallel 2m wires are the way to go, and in a 3m^2 (2m*1.5m) blanket they will only be 3mm apart. Why parallel? Because with one single wire, a single break will stop the entire blanket, and be difficult to find and repair, while with parallel wires, a few breaks won't even be noticed. As well as allowing much safer voltages around your children!
You'll also have to revise that temperature downwards - if the wire's at 110C the surface of the blanket could be around 100C, causing burns.
Best Answer
There are two effects going on. The heat sinking effect of the connections and the temperature coefficient on the wire.
Initially the wire is all at the same temperature.
You turn the power on and it starts to heat up.
The heating is determined by the electrical power dissipation in the wire, for any given section of the wire Power = Current * Voltage. All parts of the wire will have the same current. For a given length the Voltage = Current * Resistance giving Power = Current squared * Resistance.
Initially all the wire has the same resistance and so the heating is even along the length of the wire.
The heat flows from hotter to objects to cooler ones (this is the first law of thermodynamics). In this case the connection points are cooler and so heat flows from the ends of the wire to the connectors cooling the ends slightly. Since the very ends are cooler the bits of wire near them then cools a smaller amount and so on along the length of the wire. This results in a very small temperature gradient across the wire with the middle slightly warmer than the ends.
Copper has a positive temperature coefficient of about 0.4 percent per degree C. This means that the warmer the wire the higher the resistance.
The middle of the wire is hotter which means its resistance increases. From the above equations this means more power is dissipated in the middle of the wire than in the ends.
More power means more heating in the middle than the ends and you get a positive feedback effect. The middle is hotter which means it has a higher resistance and more power is dissipated there which means it gets hotter...
This continues until almost all the power is dissipated in the middle of the wire, you never get all of the power in a single point because the heat conduction along the wire means that the sections near the middle also have reasonably high resistance. Eventually you reach an equilibrium where the thermal conductivity spreads the energy enough to balance the positive feedback effect.
The best example of a positive temperature coefficient is an old style incandescent light bulb. If you measure the resistance when cold it will be a fraction of the value you would expect for its power rating, they operate at about 3000 degrees and so the cold resistance is about 1/10th of the normal operating resistance when on. They are made of tungsten not copper, copper would be a liquid at those temperatures, but the thermal coefficient is about the same.