@Chris "(L1+L2)/(L1/v1+L2/v2)=2v1v2/(v1+v2)=/=(v1+v2)/2." Which yields, as required, c for v1=c(1+q) ; v2=c(1-q) . For your suggestion v1=c/(1+q); v2=c/(1-q) it would yield c/(1-q^2) for the two way speed of light

@Thomas Are you seriously trying to tell me you plugged those in to 2v1v2/(v1+v2) and got c? Try it with q=.5 and L=3 light seconds, so v1=1.5c and .5c using your math. Then one way takes 6 seconds and the other way takes 2 seconds. This yields a round trip speed of (6 light seconds/8 seconds)=3c/4.

@Chris Sorry, I misread your notation. I though the last bit was the result of your calculation. Note though that your round trip assumption does not apply here. In bartleby.com/173/a1.html it is about two different light signals travelling in opposite directions, and any speed anisotropy added here should be of the form c(1+q), c(1-q) (analogously to the case of a swimmer swimming with and against the flow, the flow should have speed c).

@Chris In any case, even with c/(1+q), c/(1-q) you can not get a generalized version of the LT without violating the relativity principle for the motion between the two frames (the unprimed frame would have to move with a different speed relatively to the primed frame than vice versa unless q=0). Just try it.