but might 36v from a pair of panels damage the actuator circuitry?
So here's the deal. Lead-acid batteries look electrically like a voltage source/sink with a small series resistance, with the voltage level a function of state of charge. 2V/cell (there are 6 cells in series in a 12V battery) is nominal, and if I remember right, their open circuit voltage is something like 1.9V empty, 2.1V full. That covers 90% of their behavior.
Considering that, the "1W@18V" spec of the solar panel isn't going to be able to "win" against the battery, and the solar panel's voltage will be pulled down to battery voltage, delivering probably 0.055A (=1W/18V) at whatever the battery voltage is.
When a battery gets completely full, however, its series resistance goes up dramatically, and the voltage goes up, until there's enough voltage to start electrolysis of the fluid and you get H2 and O2 generation at the terminals and loss of the electrolyte. A lead-acid battery, depending on the type + manufacturer, has a certain recombination rate of H2 + O2 => electrolyte that it can handle; if you electrolyze at a higher current than that, it leads to permanent electrolyte loss (+hence capacity loss)
So there is a safe current that can be delivered to a lead-acid battery continuously, where its own self discharge due to electrolysis balances the charging current. It depends on the manufacture + construction. I wouldn't feel worried about a C/10 or C/20 rate of charge (where C = the current needed to discharge a battery in 1 hour). Garage door batteries are probably > 1Ah capacity so you should be safe with 55mA charging current.
HOWEVER -- I would probably put a (zener diode and resistor in series) in parallel with each battery, the zener diode being about 14V and resistor being maybe 10 ohms or so, so that it keeps the battery terminals from getting charged too far.
Also: if you can, wire each solar panel to each battery (and keep the diodes), rather than the pair of panels in series wired to the batteries in series -- i.e. try to connect the center taps. By doing so, you'll charge each battery independently. Otherwise, what can ruin battery life is if the battery voltages diverge -- the one with the higher voltage will tend to get overcharged, while the other one will tend to get overdischarged and not completely charged.
Both do the same, the LiPo Rider Pro just has everything on board. The 1n4001 diode used in the second instructable is used to prevent battery discharging though the solar panel at night, prevents reverse current through the solar panel when the usb charging (powering the lipo charger through usb) is used, and prevents the solar panel from affecting the charger when both light and usb charging is used. The LiPo rider has the same type of protection.
Best Answer
That charger takes any voltage between about 2V and less than 5V and boosts it to 5V.
You could use that circuit with a 2V to say 4.5V solar panel BUT you can instead just use a solar panel directly.
A solar panel with full sun loaded voltage of 6V or more will charge a USB cellphone either directly or with the addition of a series diode. (Diode needed ONLY if cellphone battery voltage appears on charging input. )
For 'safety' you should also connect a 5V1 zener-diode (5.1 Volt) across the solar panel output so that Vout can never be > 5.1V.
Zener Wattage = Wmp panel = max panel power in full sun.
Charge rate is related to panel Wattage.
Panel peak watts in full sun = Wmp.
The panel wattage for maximum charge needs be no more than :
Vpanel_loaded_at_full_sun / Ah_of_cellphone_battery
eg a panel with Vmp=6V and charging a 1000 mAh = 1 Ah cellphone battery need have a Wattage of no more than 6V/1Ah = 6 Watts. This would charge the battery at maximum possible rate. Higher wattage panels will heat the zener and lower wattage panels take longer to charge.