I do not know what "blue cell which looks like cracked glass" you are talking about, but I have seen cells on sale rated up to 173 W/m2 nominal power - these where best quality, selected cells. The ratings are however usually given as open circuit voltage times short circuit current, and even with the best cells, the most power you are ever going to get even with the cell facing directly the sun, is 70% of that, which is about 120 W/m2. But that nominal power you are only going to get at full sunlight with the cell facing exactly in the direction of the sun. Yesterday I was making tests with some small cell around 14:00, and when the cell was in horizontal position, I was able to get about 67% of the maximum power it had facing the sun. However when the cell was in shade, the output was only about 6% of the maximum power.
If you are interested in getting 12 V & 40 A only at clear sky, if you buy the best quality cell of rated nominal power 173 W/m2 (real maximum power 120 W/m2), the rough calculation would be:
$$12 \mbox{ V} \times 40 \mbox{ A} = 480 \mbox{ W}$$
$$\frac{480 \mbox{ W}}{120 \mbox{ W}/\mbox{ m}^2 \times 67\%} \approx 6 \mbox{ m}^2$$
That might fit on a van if it is a moderately large one, but if you want it to provide that power also when there are clouds, or any shade, you will need a panel at least 10 times as large. If you don't buy the highest quality selected cells, but go for most common panels, they have only 114 W/m2 nominal rated power, the required area will be 50% more.
If you would want to supply that 480 W round the clock, you will need batteries, account for the loss of power while charging (10% - 50% depending on what batteries and charge controllers you use), and take into account that sun only shines a small part of the day.
I read that from a panel in horizontal position, in London, UK, in December, you get on average 0.434 Wh of energy per each nominal W of panel power per day. In July the value is about 10 times as much Wh per day.
That means to provide 480 W continuously even in December, you would need a panel of nominal power 26544 W, which would be at least 154 m2 in size. And that's not taking into account loss on charging the battery.
Battery University is a good starting point. A vast amount here Battery University
Particularly:
Charging Lead Acid
Based on the following, it would SOUND wise to charge fully for "a while" under trickle and then to open circuit the charger until Vbattery fell to approaching 2.1V/cell. A controller to implement this could be extremely simple.
Dangerous in isolation, but, they say:
Most stationary batteries are kept on float charge. To reduce stress, the so-called hysteresis charge disconnects the float current when the battery is full. As the terminal voltage drops due to self-discharge, an occasional topping charge replenishes the lost energy. In essence, the battery is only “borrowed” from time to time for brief moments. This mode works well for installations that do not draw a load when on standby.
Lead acid batteries must always be stored in a charged state. A topping charge should be applied every six months to prevent the voltage from dropping below 2.10V/cell. With AGM, these requirements can be somewhat relaxed.
Measuring the open circuit voltage (OCV) while in storage provides a reliable indication as to the state-of-charge of the battery. A voltage of 2.10V at room temperature reveals a charge of about 90 percent. Such a battery is in good condition and needs only a brief full charge prior to use. If the voltage drops below 2.10V, the battery must be charged to prevent sulfation.
Observe the storage temperature when measuring the open circuit voltage. A cool battery lowers the voltage slightly and a warm one increases it. Using OCV to estimate state-of-charge works best when the battery has rested for a few hours, because a charge or discharge agitates the battery and distorts the voltage.
Best Answer
Extracting power from such a low voltage is tricky because even otherwise small voltage drops are significant fractions of the input.
Fortunately, you have the 12 V battery available to power the control electronics of a boost converter. At its core, a basic boost converter will be a inductor, switch, and diode:
The switch closes, which makes the inductor current rise linearly with time. Once the inductor current gets to a decent level and a useful amount of energy is stored in it, the switch opens. At that point the inductor current can only go thru the diode, dumping current onto the output. The reverse voltage now on the inductor causes the current to decrease linearly over time, transferring the energy stored in the inductor to the output.
Since your input is 500-750 mV, the most critical part of this circuit is a switch that will only drop a small fraction of that. A bipolar transistor with maybe 200 mV saturation is not appropriate here. You want a FET with low on resistance (Rdson). Let's say you want to keep the drop accross the switch to 50 mV or less at 8 A. By Ohm's law, that means you need 6.3 mΩ Rdson or less. There are FETs that can do that, but for something like this paralleling a couple of them may be quite appropriate.
Fortunately, the drop accross the diode is relative to the output voltage, not the input voltage. A regular Schottky diode rated for the current should be good enough here. Let's say it drops 500 mV when dumping current onto the 12 V output. That's a 4% power loss, which in this case, especially considering you're asking here, I'd just swallow. To get around that you'd need to implement synchronous rectification, which is not a beginner concept unless you can find it included in a chip.
Since the battery is there, you can run the control circuit that turns the FET on and off from it. Many microcontrollers can do this. The micro would measure the input and output voltages, and produce the switch signal via built-in PWM hardware. The micro can power itself down when there is insufficient voltage from the solar cell to make it worth it to run the boost converter. With the right circuit, the micro can take less current than the internal leakage of the battery when powered down. It can then wake up every few seconds to check whether sufficient input voltage is available to run the converter, and go back to sleep when not.
You have to pay attention to low current design with the micro, and make sure things like the battery voltage measurement circuit don't take any current when off. Fortunately this is something a beginner can do, although it requires paying attention.