That is an excellent post [repeated below]! Heat transfer is a big part of the equation when considering appropriate wattage ratings. My calculations were based on ideal conditions.

Volumes of material exists on heat transfer in industrial electronic equipment, especially the equipment that does not employ fans.

John

Coulomb Magician wrote:

The power rating of a resistor is based on how much heat can be removed under normal operating conditions such as the maximum operating temperature and given maximum ambient temperature, the transfer mechanism of heat out of the resistor such as still air convection, and the size of the physical package. That's why the rating of an 0805 resistor is the same whether it's 1k or 100k. If you solder one SMD on top of another the effective area for dissipating heat has not really increased very much so your trying to dissipate twice as much heat through about the same area. If the resistors are physically separated you have about twice the area and twice the power dissipation so about the same temperature rise and you're still within rating.

The wirewound example has to do with the fact that with a pot where all of the leg terminal current is flowing through the wiper and none is flowing through the other leg terminal all the power is dissipated in the small area between the first leg and the wiper and so you have the same problem of smaller area through which to transfer the power. Take a 100W 5 Ohm and set the wiper for 1 Ohm wiper-leg and try to dissipate 100Watt in that wiper-leg connection, you'll get the picture and go from theoretical to practical very quickly!

The discussion up to here has involved lumped circuit parameters and has ignored the distributed nature of current density and heat dissipation, that's my point.