Electronic – My Solar PV system correctly shuts off power during a blackout – I want to safely override that

This is dependent on what you are willing to pay. A remote controlled car pulls Amps of power normally, you will not be able to sustain this with a solar panel you can afford and is not too large.

If you want to use a solar panel for recharging the battery during down time this is very possible.

Solar panels are based on a watts per square inch they can achieve. They are still normally sold with a wattage rating.

The amount of available power is quite variable. Most marketing give you an absolute best case. Things that cause great variance are...

1. Alignment to light source. Yes, being off-angle from the sun is bad.
2. Amount of light available. Small variance of conditions can have a large affect.
3. Light source. florescent and the sun are very different, there are panels for each. There can be many more variables, but these alone will account for you getting less than 10% of what it is rated in most cases.

If I get a chance tomorrow I will look up more information for you.

Disclaimer: I may have made a mistake in my working's in which case I would appreciate if someone could correct me. OP: Take what's below with a pinch of salt.

To be honest, If I try to make a perfect calculation on what you will and won't get, this will just be too complex for me to answer. Instead, I'm going to make very rough approximations.

"The system will be located near Narbonne where the insolation varies from 1.18 (winter) up to 6.44 kWh/m²/day (summer)."

What we need to do first is establish the average amount of energy per day you'll get. After doing a basic mean average of the values in your table, I got:

44.47 / 12 = 3.706 kWh/m²/day

Your solar panel has the following dimensions:

160 mm x 138 mm

We'll ignore the uncertainty and so we have a surface area of(standard form converts to meters):

160*10\$^{-3}\$ m * 138*10\$^{-3}\$ m = 0.02208 m²

Hence we can now see that per day, assuming (I have absolutely no idea of the efficiency of your solar panel) a rather optimistic 20 % efficiency, you'd get:

0.02208 * 3.706 * 0.2 = 0.0163 kWh/day

0.0163 * 1000 * 3600 = 58913 J/day

Continuous Wattage Possible: 58913 / (3600 * 24) = 0.6819 W

Now, at this point we haven't even addressed the fact that in winter you're getting on average 6 times less power from the solar panel than in summer. Potentially this means you have to store a great deal of power from summer and utilise it all throughout winter. Assuming you could run on 0.68W (including the efficiency of the Arduino, etc), The main problem as I see it is that some days you'll have virtually no power whatsoever. Additionally, you may need to step up or step down the power produced from the solar panel, which, in itself will incur efficiency penalties.

I advise that you total up the exact power consumption of your Arduino, then actually conduct real world testing and procure a maximum power consumption under peak load. At the minute, you may need to either add more solar panels(increasing the area) or use solar power in conjunction with grid power.

Edit(Assuming the purported 50 mA current consumption at 7 V):

P = VI

7 * 50x10\$^{-3}\$ = 0.35 W

Now, based on this, you'll be fine if you're perfectly storing every ounce of energy you get from your solar panels and there aren't weeks where there is barely any solar power available. So let's assume that we want the Arduino to function even with the worst winter insolation values.

Taking your minimum of 1.18, redoing the calculations:

1.18 * 0.02208 * 0.2 = 0.00521 kWh / day

0.00521 * 1000 * 3600 = 18759 J / day

Continuous Wattage Possible: 18759 / (3600 * 24) = 0.217 W

So in winter months you'll have around 0.217 W available, but the reality could be worse then that as is the case with all weather based power sources. What does this mean? It means that realistically in order to A. Power the Arduino and B. Have a decent safety margin, you will need (Assuming a safety margin of 2X that required):

0.35 W * 2 = 0.7 W

0.7 / 0.217 = 3.22

Hence you need to increase the surface area of your solar array by 3.22 times. In other word's, you'll want four solar panels, presumably connected in parallel (don't quote me on this :D) in order to power your Arduino through thick or thin.

Final note: Your power booster will incur efficiency penalties as well as the act of energy storage, which is why I took such a large safety margin. Hope this helps.