Found this Ebay Listing for 35/55w SOX gear, just by browsing, but it includes a power factor capacitor. The value is not mentioned.

I got the same Philips ballast & ignitor. I have made 2 fixtures which I have sold in the past using these two Philips ballast+ignitor parts, but without power factor capacitors. There is no mention of capacitor on the ballast or ignitor.

What is the value required for the ballast & ignitor pictured? I intend to use a 120V-to-240V step up transformer 60hz for N.America. Would it be different for a 35w SOX lamp compared to a 55w lamp?

There are related topics Here & Here, but it's not stated for this watt, or not clear to me.

At first this ballast is designed for 50Hz operation. As the mains frequency is directly in the "ballast equations", it will not work properly on 60Hz. The current level may be corrected by the voltage, but it would mean the ballast core losses will increase by something between 20 to 40%. The core losses use to be small portion of the total ballast losses, but the small difference may be killing for the ballast temperature.
Second is, you would indeed need a transformer to step up the voltage. The problem is, no one would tell you, to which voltage - because there is a lamp in series, it is not directly "60/50*240", but something a bit less. So you will need to try out.
Regarding the power factor correction capacitor, the series reactor ballast does not need it, it does not influence it's working in any way. It's presence (assume correct value) only reduces the input current from the supply source. But because you will be using a the step up transformer, the presence of this capacitor will reduce it's load (without the transformer would have to be rated at 0.6A, with the cap just for about 0.3 to 0.4A on the secondary).

In any way the original capacitor will be useless for you: It is calculated to compensate the ballast at 240V/50Hz feed, while you will be feeding it at about (guess) 280V/60Hz. So the capoacitor for you would have to be of lower capacitance, but rated at higher voltage.
Quick calculatiopns:
Assume the ballast losses are dominantly the wire resistance and present 30V drop.
Assume 55W LPS arc voltage ~103V (= 55/0.6/0.9)
For 240V/50Hz I got about 6.6uF (= 0.6A * sqrt(240V^2-133V^2) / 240V^2 / 2*pi()*50Hz ) for complete phase compensation; probably will be a bit lower, expect about 6uF (it is safer to have the mains slightly undercompensated than overcompensated - otherwise a running induction motor may turn into a generator with voltage runaway, so high voltage surge when the mains get suddenly interrupted)
Assume the 280V would lead to the correct current:
For 280V/60Hz I got about 5uF (= 0.6A * sqrt(280V^2-133V^2) / 280V^2 / 2*pi()*60Hz ). Because you will have local transformer, I would keep that on this border - the required slight inductive component will come from the transformer.

Ok thanks Medved. I have understood that the ballast will overheat assuming it is operated for let's say typical dusk-to-dawn conditions using a 120v to 240v transformer @ 60Hz. The transformer I've used in the past, and will be using again, is rated for 100W power max.

But what do you mean "no one would tell you, to which voltage - because there is a lamp in series"? I would assume this ballast/ignitor will NEED to see 220 to 240 Volts (ok, at 50hz, but is 60hz that much off?), otherwise it will not even ignite the lamp. Why are you saying I will "be feeding it at about (guess) 280V/60Hz"?? Your statements are a little cotradictory to me as you point out this series reactor ballast requires a voltage "it is not directly "60/50*240", but something a bit less"??

I ALSO understood, please correct me, that operating this ballast/ignitor with a step up transformer (for ½ to ¾ hour) would be BETTER WITHOUT a power factor capacitor, mainly because the correct value of capacitance is difficult to pinpoint because of the step up transformer?

I have been running 50 Hz ballast on 60 Hz here no problem. As i been running Fizzy light pack for over 2 years now.

Anyway here your solution to your problem as this ballast is dual voltage and it a leak style HX ballast

http://m.ebay.com/itm/281577916037?_mwBanner=1&ul_noapp=true

The ballast is an an impedance connected in series with the lamp to limit the current to the desired value.
For the moment for the moment let's forget the reignition effects after a current zero cross (when the AC current changes it's polarity, the arc effectively extinguishes and then has to ignite again for the opposite polarity).
With that simplification, anything that poses the impedance may do the job, so it could be just a resistor, capacitor and/or the inductor.
The resistor is the only one, which has no inertia, so does not depend on the frequency. But it is very lossy.
Both of the other have the impedance directly linked to the frequency:
So the capacitor impedance is
Xc = 1/(2*Pi()*Freq*C)
That means increasing the frequency, the impedance decreases. Now as the discharge operation (mainly to handle the reignitions) requires the ballast to act as close as possible to a current source, so have high impedance, the capacitor does not work well - when the voltage changes rapidly, it means high frequency content, where the impedance is low.

So the last remaining is the inductor, where the impedance is
Xl = 2*Pi()*Freq*L
You see the impedance increases with the frequency, so whatever steep voltage step imposed by the lamp, the current won't change that fast. That is the main reason, why most series reactor ballasts are either inductor based (or at least contain significant inductance).

Now for the lamp to operate correctly, you have to keep the arc current (otherwise the lamp would be under or overpowered). So if you want a 240V mains ballast to feed a 133V resistive load (arc plus wire resistance drop) by 0.6A, the voltage across the inductance would be 200V, so you need an 333.3Ohm reactance.
For 50Hz it means an inductance of 1.06H (for 60Hz it would mean 0.883H).
Now what you have is an 1.06H inductor and you want to feed using that 0.6A into the 133V load at 60Hz. The inductance is the same, so it's reactance will be 2*Pi*60*1.06 = 400Ohm (at first order approach; assuming all losses coming from the copper losses and none from the core).
So for the same 240V, but 60Hz it would mean the lamp current just abut 0.5A. Fluorescents may tolerate this, but LPS will loose quite a lot from it's efficacy (because of the steep dependency of the radiated light on the arctube temperature, there is practically fixed amount of power required to keep the thing at the operating temperature, only the rest is "available" for the light, so reducing the power you reduce just the part generating the light output, the "heating power" component remains the same).

Now you want to still feed the 0.6A, but compare to 50Hz, your ballast has now 400Ohm (instead of 333Ohm). So the only thing you may do is to select the feeding voltage so, the current will be where you need it, so at 0.6A. So:
Vsupply = sqrt(133^2 + (0.6A * 400)) = 274V. So you have to design your step up transformer for this voltage.
(I didn't do such detailed calculation before, the "280V" was just a guess, as simple 240 * 60/50 = 288 and reducing that a bit; it was intended as a rough estimation, just to illustrate the effects)
And because part of the ballast losses is related to the core, which is (mostly) acting as a resistor parallel to the coil, the power dissipation here will increase, increasing the ballast operating temperature (or making more demand on the heat management of the fixture). At the same time this current component does not follow the frequency rule, so you will have to really try it out and adjust the voltage according to the current you get. That is the risk you are facing.

I plugged a 35 watt 240/50 SOX light into 240/60 and it lit and didn't seem that dim.
Is using a buck/boost transformer to bump it up to 262/60 to compensate for the frequency difference a good idea, or no?

Boosting the voltage is, what you have to do, so yes, you need such transformer. The only open question is, to which voltage to go, there you will likely need some room for some tuning - o keep ballast temperature reasonable and  still maintain the arc current close enough to the specification.
So yes, in that the 265V is better than the 240V...

Regarding the brightness: When you do not have any side-to-side compare with a properly ballasted lamp, you won't nottice even 50% drop in brightness. With many sources this drop of the power means shift in the color, but if the LPS already evaporate sufficient amount of sodium, the color will be all the time the same Sodium D-line pair, so no difference at all.

What percentage of brightness drop does running a 240/50 ballast on 240/60 produce? Is it a linear drop with LPS underdriving it until it no longer lights?

What percentage of brightness drop does running a 240/50 ballast on 240/60 produce? Is it a linear drop with LPS underdriving it until it no longer lights?

The light output is (roughly) following an expression:
Output = (Power - A) * B
where the A and B are parameters.
A corresponds to the power required to just keep the arctube at temperature, it uses to be in the range of 50..70% (decreases with increasing lamp power rating) of the rated power input.
B corresponds to the efficacy of the light generation from the power exceeding the level required to keep the arctube hot. It uses to be in the range of 300..500lm/W.
So power reduction to ~80% (corresponds to operating 50Hz rated ballast at 60Hz) would yield about 50..60% of the rated light output (my guess, if it will be just 20% or so, or if the lamp does not warm up at all, don't blame me).

The only quick and rather reliable option you have is to try it out...

What I have is a standard buck/boost transformer.

Would it be better to wire it at 252 volts or 264 volts? I'm assuming a slightly higher line voltage won't hurt the gear like too much power would (like running a 60 hz ballast on 50hz). Or so my father, a retired electrical engineer, told me.

Should I use the old 1980s? Thorn Gear or the modern Philips Gear (the same as in the picture), or doesn't it make a difference?

The 265V would be definitely better.
And if both ballasts are of a series series inductor type, both would be OK, they will do exactly the same...

Would 277 volts be better, or is that approaching the risk of damaging something because of the extra voltage.
If I lose 20% of the power going from 50-60, would I want to compensate by adding 20% to the voltage?
240*1.2=288 would be the "ideal" voltage.
Or is there something conceptual I'm not getting?
 

Would 277 volts be better, or is that approaching the risk of damaging something because of the extra voltage.
If I lose 20% of the power going from 50-60, would I want to compensate by adding 20% to the voltage?
240*1.2=288 would be the "ideal" voltage.
Or is there something conceptual I'm not getting?
 

The extra voltage may damage just the power factor capacitor, if that would remain there. As it will overcompensate the circuit, the one rated for 50Hz is anyway unusable.
For the coil, the main factor is the operating temperature and current (how close it is towards the saturation). As we want to reach exactly the same current as with the original 240V/50Hz setup and still are below the target, there won't be any problem.
The ignitor sees just the arc voltage and that is 100% given by the lamp, so no difference from the original setup at all.

For the voltage calculation you are really very close:
What get's multiplied by the "60/50" factor is the voltage across the inductive component of the ballast.
The mains is a vector sum of the inductive, resistive (losses) and lamp voltages. As the resistive and lamp voltages are in phase and the inductive 90deg phase shifted, a Pythagoras equation works there:
Vmains^2 = Vind^2 + (Rlosses*Iballast + Vlamp)^2.
The inductive part could be then stated like:
Vind = Iballast*L*2*Pi()*Freq
The ballast losses are (just an estimate based on similar current fluorescent ballasts; it does not play that much role) 10W, so the related voltage drop is about 30V.
The lamp arc voltage is about:
Vlamp = Plamp / Iballast / 0.9
The 0.9 is the power factor of the lamp: It is fed by a sinewave, but the voltage is rectangular. The resulting power factor of the shape mismatch is about 0.9. So the voltage drop for the 18W is about 56V

Now these equation are valid for any supply yielding the rated Iballast current (on other currents the L may change due to magnetic nonlinearity). As this is where we aim for, you may form that into a set of two equations, one for 50Hz and second for 60Hz:
240^2 = (0.36A*L*6.28*50Hz)^2 + (30 + 56)^2
Vwanted^2 = (0.36A*L*6.28*60Hz)^2 + (30 + 56)^2

Now if you will play a bit with the "lamp" and "losses" voltages, you will find out the thing is pretty insensitive to these (if they stay below half of the mains). Putting zero there then leads your equation and even with that is not that far (282 vs yours 288V).

So an naswer to the original voltage question: Yes, the 277V is the voltage closest to the calculated one (from the numbers you said are available for you)...
So if you have that, just try it out.
Measure the lamp arc current and after a while check the ballast temperature.
For the temperature you may measure just the winding resistance when it is cold and then just after shutting it down and calculate the temperature from the resistance change - there is just copper and nothing else. It may not look so, but this method is actually more accurate in estimating the temperature inside of the winding (that is, what matter for the ballast reliability) than measuring the surface temperature. Just don't forget to disconnect the thing from mains before connecting the Ohm-meter.
If you have some IR thermometer or so, you may monitor the temperature while the ballast is running.
The expected temperature difference when on free air is printed on the ballast ("dtXXX" means the expected difference is XXX degC), so compare the measurement on bench with that figure. It should not be higher. Then when installed, the ballast should not exceed a temperature marked as "twXXX". These rated values are to be measured on the ballast top (iron body, where it does not touches the eventual heat sink or so).

Thanks a lot for the help with this thing, it is a 35 watt SOX.
I found a thread from 2011 where you said 277 would be too much for a 240/50 ballast, which is the only reason why I'm questioning it
http://www.lighting-gallery.net/index.php?topic=2043.0
So the summary is
1) Leave it underdriven since it does light, or
2) Disconnect the capacitor (if I use the one that has one), and make sure the ballast doesn't get to hot.
I don't have 277 available at my house, but 120/277 transformers being extremely common it's one thing I thought of.

If you want to run it on standard 120v OR 240v 60hz, I found this advance ballast for 35/55w LPS bulbs. The thing here is you need to buy your own 14uf correction capacitor, the seller did kind of a favor by not including the plastic cased crapacitor.