[Spaceships] Conventional and EM Gun ranges in atmosphere?

[vicky_molokh]

Originally Posted by Anthony

Kinetic energy is proportional to v^2. Drag is proportional to v^2. Loss of energy from drag is equal to (drag force) * (distance). Thus, dE/dd = kE. This is a well known differential equation, and means E ~= e^(kd) (where E is energy, d is distance, and e is the base of the natural logarithm) and half-energy range is ln(2) / k (note: doing this quick by memory, so some signs may be off).

I have trouble parsing that - my math is very rusty (the equation-related parts, anyway). Sorry.
What is 'k'?

What I was trying to say was that more powerful throws do result in greater range, so if range is proportional to calibre only, something seems missing.