The SSPRR Run Room is a good place to visit. Good information that is well organized and easy to navigate.
In the Layout Wiring section under Load Calculations I noticed the abbreviation "CM" and some values listed next to different wire sizes. Larger sizes have larger CM values. Some research shows that CM = Circular Mils and that there is a simple formula that can be used to calculate CM for a given application.
CM = (2500*I*L) / (%drop*V) where I is amps, L is wire length in feet, V is supply voltage and %drop is the amount of voltage drop you are willing to tolerate.
The calculated value for CM is an indication of the minimum wire size needed.
For example, on my small layout I want to run 5 feet of wire drop for 2.5 amps while providing 16 volts from the power supply with only a 1% drop. This is currently a pretty big load for me.
So, CM = (2500*2.5*5) / (1*16) = 1953 which suggests that an 18 awg wire is a pretty good choice (CM = 1624).
In this case the selection of a 1% drop was arbitrary. My question is: what is acceptable and expected for good track performance? Why wouldn't a 5% or 10% drop be ok over a 5 ft length allowing me to use a smaller wire? My wiring is all 14/16 and 18 as needed but why wouldn't 20 awg solid core be ok? Does % drop effect DCS performance or how well speed control works?
Thanks