Many of us have done experiments ballasting various fluorescent lamps using an incandescent lamp, but while almost every regular fluorescent lamp has been tried (and I have even tried a mercury vapor lamp) I have yet to find any evidence of anyone ever having tried an HO or VHO lamp. (Or a T17, but I reckon one wouldn't want to make one into a guinea pig for such an endeavor.)

Based on my calculations, an F24T12/HO, F30T12/HO, F36T12/HO, and F42T12/HO lamp should run well on an incandescent ballast. Possibly an F48T12/HO and F48T12/VHO would work, but they would likely be difficult to start, at the very least. As for starters, you could use the same ones that you would use for a regular lamp of similar arc voltage. For example, the F24T12/HO has a similar arc voltage to an F15T12, so an FS-2 starter would work. However, I don't know if the increased current would wreak havoc on the starter or not. Also, as HO and VHO lamps are rapid start, I have no idea if the cathodes could handle the amount of current they would get during preheating. (Although if not, it would certainly be a simple fix to insert a resistor in series with the starter. That would reduce wear on the starter, too.)

If anyone would be interested to try one of the above-mentioned lamps on such a setup (I don't have my own to do so), I would love to hear about it, and see a picture! Based on my calculations, a 150 watt incandescent lamp would work for the F24T12/HO, and a 500 watt lamp would work for the F48T12/VHO.

Trent

I have tried lighting a 48" VHO tube with a halogen ballast before and it worked for about 20-30 seconds. I have a case of 48" 110w VHO lamps (6000K color) that ive got no use for so I connected one tube in series with a 500w halogen lamp and used the driver out of a plasma ball for ignition. It lit for about half a minute but appeared only about half brightness. These lamps can also take a surprising amount of power, I have a 4-lamp 110w VHO electronic ballast (instant start) that I wired for overdrive (all leads into a single lamp) about 450w power and the tube took forever for the ends to blacken, but never actually failed (was incredibly bright).

Cool!

I'm just repeating your experiment with one of these old MV lamps.
The lamp is 220V/50W, incandescent is 230V/100W. Lamp voltage should be: 95V +/- 15%, is: 74V. And it works - now burning about an hour. (But the arc in another piece of these died within 10 minutes.)
I don't find it easy to calculate the right ballast exactly as the filament resistance is varying with its temperature.

Btw. I can remember a heater-ballasted sun-taner for home use. It was called "mixed IR/UV radiator".

A 100 watt lamp isn't enough. The MV lamp is only doing 25 watts or so, which explains why it's only running at 74V (it's not running up all the way). According to my calculations a 200 watt lamp will be best. I've attached a formula I came up with which estimates the proper wattage of incandescent lamp. I is the desired current (0.6A for a 50 watt mercury vapor), Vs is the source voltage (240V for you), Va is the arc voltage (95V for a 50 watt mercury vapor lamp), and 0.83 is a constant I came up with, based on my own experimentation, to correct for an incandescent light bulb's resistance when underdriven.

@TheMaritimeMan:
The correct formula isn't that simple, even when assuming an ideal resistor (so not dependent on the voltage).
The reason is, proper operating power is essential to keep not the rms current, but the average of an absolute value.
And because the arc conduct current only part of the sinewave (in an ideal case the part, when the mains exceed the arc voltage, in real case even less, as the voltage has to first rise to the reignition level and only then the current reappear).
I would guess that is included in your

So the rated 0.6A rms (the rating is made for a sinewave) actually means 0.54A of the "average-absolute" value (for the sinewave the exact ratio is sqrt(2)/(Pi/2) = ~1.11).
With a 230V mains and about 95V arc voltage the current will be present for just 81% of the time (acos(95/325) /(Pi/2)).
Treating the current shape as a half-sine wave (it is actually not, but the error is very small) it means the rms current would have to be 1/sqrt(0.81) times larger than the sinewave rating, so about 0.65Arms for a "0.6A" rated lamp. All this with still neglected reignition overshoots, they make the difference just greater.

And it is the overshoots, what cause the rms voltage to become larger than the rating: With the rated inductive ballast the spikes are very short in time duration, so they do not practically contribute to the average, nor rms value of the voltage across the lamp, so (as it is a 95V rectangle) on a rated ballast it reads the rated 95V.
But as with the resistor the overshoots are larger, it cause higher rms voltage reading of the essentially equal arc voltage (105V i.s.o. 95V is inside what I would expect).

Nice equation, Trent

@Medved: I am assuming that Trent (TheMaritimeMan)'s formula is for the DC circuit.  Basically we need to add the AC voltage sinewave equation to the ratio that he has given to us?

Nice equation, Trent

@Medved: I am assuming that Trent (TheMaritimeMan)'s formula is for the DC circuit.  Basically we need to add the AC voltage sinewave equation to the ratio that he has given to us?

When exactly described:
You have to go through the input voltage sinewave and calculate, what would be the current doing when the lamp would be fed via given resistor. And then integrate the product over the complete period and then divide the result by the time duration of that period.
Quite tedious, but when done in e.g. Excell (each column represents one variable, as it evolves over the time), it could give reasonable results. But it expects the correct formulas are there (include the need of a reignition,...)