So.. I tried the fluro thing, it works, however with a 15W lamp it seems to be banding rapidly, I upped it to a 40W lamp and it seemed OK, then a 60W lamp and it started sputtering.. so I'm guessing i must remain between the 15W to 40W range?? is 18W like you mentioned earlier the ideal wattage Medved?
The exact ballast lamp specification should be based on some calculation guess, but then it should be corrected based on real world measurement. There are effects, which are very hard to cover by the calculations.
What I gave were very quick, but still safe (should not lead to quick blow up due to an overcurrent) guesses - I took the rated arc current and take an incandescent leading at full voltage to 50% higher than the fluorescent rated operating current (usually the minimum specified preheating current), as you may see below, such quick guess could be still quite far off.
But lets calculate it more accurately, yet still neglecting errors like reignition delay:
The exact calculation is a bit complex, I will try to explain it:
You need to reach the current the fluorescent is rated for. So for an F6T5 or F8T5 that is about 0.155A, for the F15T8 it is 0.22A.
In the circuit you have a 120Vrms (so 170V peak) sinewave voltage source (the mains), a square wave voltage source (the fluorescent; 57V for F8T5, 43V for F6T5 and 75V for the F15T5; the actual voltages may differ a bit, but it won't be too far away) and a resistor.
The voltage across the resistor is, what remains from the mains sinewave when the fluorescent voltage is subtracted from the mains.
So you just "cut" a part of the mains sinewave (e.g. 75V for the F15T8; lets use this lamp as an example here) from the zero, so what remains are just the peaks above the 75V and zero in the meantime.
For our example we get nearly half sine pulses with 95V amplitude, but with gaps corresponding to 1/6 of the mains periode, so 2/3 of the time we have the half-sines, 1/3 we have zero.
You then calculate the rms voltage from that: Vincrms = 95V*0.707*sqrt(2/3)=55V (for our example). That is then the rms voltage the incandescent will see. You find somewhere a graph (e.g.
here) depicting what the incandescent resistance does when the voltage falls such low (about 75% to normal resistance).
At the same time you calculate the real resistance the ballast have to have, in order to pass the required power to the fluorescent.
Because the fluorescent act as a constant voltage source of 75V, the dissipated power corresponds to a "rectified average" current (Pfluoro=75V*Iavg = 75V * average(abs(i(t))) )
The current (i(t)) are the 95V pulses, so the averaged value is 95*2/Pi()*2/3/R = 15W/75V
And the R is the resistor we want (=200Ohm).
And for a 120V incandescent we know this is 75% of the nominal resistance (at 120V), so the corresponding incandescent will have to have 270Ohm (I'm rounding a bit, but to stay within few percent tolerance) at 120V.
And given power and Ohms law equations we get (120V)^2/270Ohm=53W.
Because the rms current will be higher than the tube is rated for (the pulses have higher crest factor than the sinewave), better to underpower the system a bit, so go for lower wattage, so e.g. 40W incandescent.
You may see, the initial guess based only on the preheating current could be really quite off.
And even this neglects some effects, so the result should be adjusted to keep both preheat, as well as operating currents in check.