That graph shows the life expectancy as well. The life on the 60% voltage gous out of the graph, so the real life will depend more on the manufacturing quality (failures are likely to be of the "early fails" nature, not from a normal wear mechanisms, but mainly from the defects propagating through the different materials)
With quality lamps very likely you will suffer practically no wear.
Even the starting should be of not much a problem, the voltage drop of a cold arc is around 40V, so you would be already at about 200V supply, so at levels where the extrapolated life is still "way high above the roof"

By the way when writing the previous, I got a bit lost in the graph, the lumen output at 60% voltage (that is about the 137V) is barely 17% of the rating (and not the 25% I have stated; that would correspond to operation at around about 60% of the power and not voltage)

And by the way:
I just looked more in detail at the DuroTest rating and did some calculations:
Rated voltage is 208V
That means with a standard 132V arctube (250W 2.1A) the filament will see just 94V.
With 2.1A that means the 200W in the filament
Viola, you have 200W in filament and 250W in the arc tube.
So in my eyes no "special design for high efficacy", but just plain "use just standard components to cut costs as much as possible".
Yes, good marketing guys there, when they were able to sell primarily a cost cutting measure as a "desire to get the best performance"...

You are working with 230V mains supply, but the DuroTest are working with 208V supply.
The arc voltages are about the same
The lower mains means first there is less peak voltage across the filament and seconds there is lower conduction angle (i.e. the current flows for shorter time, leaving longer gap without any current around the zero cross).
So when you have to drop 137V across the filament, the filment power will be about 137/230*TotalRating. So with a 415W total, you will be at about 247W at filament.
When DuraTest is dropping just 94V on a 208V mains lamp, their filament power is roughly 94/208*TotalRating=94/208*450W=200W.

If DuraTest would like to use the same arctube at 230V, they would end up with similar 135V across the filament (a bit lower than you, because the 250W MV has a higher voltage drop than the 175W one, but just a bit as the difference is small), so their total power would end up at nearly 500W and filament power at about 250W.

There is one error I made n these calculations and that error is in what rms current the arctube need for the given power and arc voltage. I took all the time the factor of 0.9, but that is valid only with a sinewave current onl. But as the relative arc voltage gets larger,the longer are the gaps without any current, so the lower the conduction angle, so the arctube power factor won't be the 0.9 but will get lower and I have neglected this.

For more accurate analysis you would have to resort to more detailed calculus and use e.g. some numerical method. Just an ordinary table calculator (Excell, KCalc,...) is good enough, once you know what you are doing, but it can not be explained in such condensed way as I tried here, so I decided to neglect this for the moment.

There are other errors as well:
The arc ionisation, so its resistance has some inertia, so it is not as constant voltage either. The main discrepancy is in the reignition overshoot (although suppressed by the auxiliary probe, it is there), but I would not dare to say it is all.
As these errors are leading to lower power than originally calculated, I deemed neglecting them won't hurt that much.

When you account for these and for some reignition delay (=increase the equivalent arc voltage by about 10%) and finetune the filament resistance (= the ballasting lamp combination) to get the exact power dissipated in the arc (175W for your case, 250W for DuroTest, arc voltage for calculation increased by 10% both to account for the reignition delay), I got following figures (I've added a case when you would like to incandescent ballast a 250W MV):

Configuration: Your175W/Durotest/Your250W

Mains voltage: 230/208/230
Arctube power: 175/250/250
Ballast power: 238/202/248 (real dissipated power, not the ballast filament rating)
Total power: 413/452/498
Ballast resistance: 69/38/47.6
Circuit current: 1.85/2.3/2.28
Power factor: 0.965/0.935/0.95

Pretty close, the mains voltage difference makes really very significant contribution to the "ballast efficiency", so the lower mains means way lower power dissipated on the filament...

I understand this, but wouldn't the arc tube have lower RMS power (more time the arc does not have power after the zero crossing) so current would have to be increased accordingly?

When you say power factor are you also referring to cosine phi? I could be wrong but I don't think conduction time in a sine wave vs none conduction time in a sine wave is the same as leading or lagging current in relation to the voltage. Forgive me if I am misreading.

Could I correct it by simply over-sizing the filaments a bit say by 5%? Honestly to my understanding MV bulbs are hearty creatures that will eat anything so to speak with minimal effect on health. Unlike metal halide and HPS lamps with their trapezoids, MV can tolerate wide range of voltages, start at voltages well below their OCV, and tolerate being over and under driven without issue. In fact I know of some MV ballasts such a regent that under drove most lamps, ie 175 watt yard blasters had 125 watt ballasts.

Same ball game for 400 and 1000, correct? 

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Power factor is a ratio of the average transferred power to the apparent power. And the apparent power is a product of rms voltage and current.
It in real life means how well you are utilizing the power transfer devices (wiring, transformers,...) in transferring the real power.
With a pure resistive load, the power factor is always a unity, regardless of the shape of the waveforms.
With some few pages math you may prove the maximum power factor is, when both current and voltage waveforms match in shape, frequency and phase (obviously they will have different scale, as one is current and second voltage).
Mathematically the power factor is in fact a correlation factor between these two functions (one is the voltage as function of time, the second current as function of time)
If the power factor is lower than unity, it means there is some mismatch between the voltage and current waveforms.
In AC circuits the mismatch could be of linear distortion, so a phase shift. In such case both voltages and currents remains sinewave, both keep the frequency, but there is a phase shift between those. This phase shift (Phi) is then enough to determine the power factor, as the PF=cos(Phi) (again some lines of calculus and you may get a math proof of that)
But the non-unity power factor may come also from other reasons. The phase may remain the same, but one of the values contains e.g. higher harmonics due to distortion (e.g. a pulsing input current of a rectifier with sine voltage feed, or a rectangular voltage across a discharge with sine current feed). Because one value (e.g. the arc current in a series reactor ballast circuit) does not contain any higher harmonics, the harmoonics from the second value (e.g. the arc voltage) do not transfer any power to the lamp (sinewaves of different frequencies have zero correlation), but they are contributing into the overall rms voltage.
So (after some math) you may get the PF=2/Pi()/sqrt(2)=~0.9 when a sinewave feeds a load that leaves squarewave in-phase voltage across it (as the discharges do, therefore the discharges with sinewave current exhibit PF=0.9 even when there is no phase shift)

Now with resistive ballast the current becomes nonsinusoidal too, so the power factor equation becomes way more complex. And in fact it becomes not much practical to express it analytically, when a brute force correlation could be calculated numerically in e.g. the Excell (or so), we have the machines usable for that at hand...

But wouldn't apparent power (as measured via true RMS devices) = average power even if the device is only drawing current in various parts of the sine wave? Voltage and current are still in phase as you say- just stopping and starting at various times. I personally would not call this power factor, rather average power consumed over a period of time. Much like a diode in series with a bulb or a triac dimmer. Less work is done per second. To lower power factor (shift current in relation to voltage) you need either capacitance or inductance, devices which briefly "store" electricity and then "send" it back to the source after having "taken" it from the source. For example a capacitor charging (taking) and then discharging (sending back) electrons creates added current on the line in addition to the current already present actually going to do work. Because of I2R losses and amps being amps in terms of thermal limits, we restrict equipment to VA instead of watts. The more VA is doing work relative to "bouncing" back and forth, the more we consider assets to be well utilized.

Cheesy explanation, but that is how I was taught it- I'm sure you can word it better and by the looks of it you did.   

Correct me if I am wrong.


 

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At the end you will be limited in what incandescent wattages are made to select from, so you would have to deviate from the calculated value anyway.
Plus nor the mains, nor MV's parameters are exact, both have their tolerances.
And indeed, MV's are very tolerant species, mainly towards slight unnderdriving. It was the reason, why I neglected certain aspect in a way the result will yield a bit lower loading than aimed for, so you won't have troubles with an overloaded lamp (with overload it starts to age way faster)

   
Ahhh, thank you- almost did not consider that. But, I have to ask- at what point is a Mercury arc tube the most efficient (lumens per watt)? If we are dealing with a ballasted roadway or tunnel fixture longevity would be the primary goal, but because in this case we are using filaments (for the sake of the discussion we will assume they are an integral part of the bulb inside the envelope) I am less concerned about arc tube longevity as the filament will probably break before the tube fails even if over driven. At least it is my thin belief that over driving does not shorten life that much.

Also deluxe phosphor- how would that play with the filament's light?