@A_lights: Do not mix up the Ballast factor (that youaredescribing) and Power factor.
Ballast factor say, what would be the lamp output on a ballast under question, relative to a reference ballast. SoBF=1.0 mean lamp run atit's rated power.
But the "Power factor" mean, how the ballast utilize the upstream wiring. It is defined as PF=RealPower/(Vrms*Arms).
It become unity with only exactly resistive load (soe.g. a heater and/or incandescent lamp).
For other real load (ballast, computer, motor,...), where the current waveform isoff phase and/or a different shape than the voltage, the PF is below one. E.g. the mains current is 4A at 120V, the real power input is 300W, so the power factor of such device would be 300W/(4A*120V)=0.625.
For the "CWA without cap" the first thing it mean, it won't be a CWA anymore, but something like HX autotransformer.
The CWA isbased on a resonance effect together with core (magnetic shunt) satruration to keep the lamp current constant over wide range of external conditions, mainly the input mains (so it's main task isto compensate out the mains variation).
When you remove the cap, the "regulation" effect will be in fact reversed (so higher mains voltage would cause even higher lamp current) and it is a question, how the result will behave.
I would guess the ballast is designed so, under such condition it would run at lowwer than ratedcurrent, it won't saturate, as capacitor short circuit is a failure mode most likely anticipated bythe ballast design (assume quality ballast).
But with some ElCheapo ballast this assumption does not have to be the case and so such configuration may overload both the lamp, as well as the ballast coil.
