To get back a meaningful result for the car example, c=1 does not really help. It is much easier to plug in u and v directly into the simpler equation that con

To get back a meaningful result for the car example, c=1 does not really help. It is much easier to plug in u and v directly into the simpler equation that contains c^2. Otherwise (the indirect method), to get fractions, first you do a substitution a=u/c, b=v/c, arrive at: c * (a * b) / (1 + a * b) Then you convert the car speeds to fractions, only to multiply back by c=299792458. Curiously enough, this sort of thing rarely occurs in the original publications, so they are usually easier to read than obfuscated textbooks or blog examples.