Aha! I think part 25.107(a) (1) & (2) linked before explains that part.
So if I’m a performance engineer at Learjet doing takeoff calcs and figuring out takeoff performance charts I’d be governed by the following:
So we start out with the assumption that VEF is greater than or equal to Vmcg from 25.107(a)(1), and then add the requirements from 25.107(a)(2).
Learjet and the other manufacturers must have some empirical reaction time number (based on pilots doing V1 cuts in sims maybe?) of the delay time between the engine failure and how long it takes the pilot to notice and initiate the first action…
They should also know the takeoff acceleration provided by the remaining good engine…
This is one of the derivations of the classical mechanics kinematics equations from physics:
View attachment 63376
If you’re not into calculus, it’s basically say that because acceleration is change in velocity per a given change in time, you can multiply that change in time on both sides and take the integral (calculus for area under the acceleration curve) and you get the equation [1] at the bottom.
Equation [1] is saying your final velocity (v) = your initial velocity (v0) + that known acceleration (a) times the change in time (delta t).
For our purposes v0 is VEF which is Vmcg or slightly greater, a is the acceleration provided by your remaining good engine and delta t is the pilot’s reaction time to initiate action after the engine failure, and the resulting v is going to be our V1.
In conclusion if we interpret the FAR part 25 definition of what V1 is and do some physics, it makes sense why V1 > Vmcg. Hope this helps!