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OK, let me try again. Let's say that you have one flashlight that you want to recharge along the trail, and it has an ordinary home-style charger that you use when you have AC power. Let's say that the home-style charger is marked with its output specs: "5 Volts, 100 milliamperes." That means it puts out (Voltage in volts times Current in amperes) 5x0.1=0.5 watts output power. If you use that charger to charge up your flashlight for 1 hour (and if that gets it done), then that is 0.5 watt-hour that is required. It it is ten hours, then that would be 5 watt-hours that is required. So, the first step is to find the normal charge rate that your flashlight device expects. Your own numbers might be quite different.
If you have a solar panel, it also has output specs in watt-hours that may be similar to the home charger. If those specs are similar, then the solar panel should be able to charge the flashlight in a similar time as the home charger. My experience with small solar panels is that the manufacturers seldom advertise a full set of specs. If they do (rare), they advertise specs for "full sun." Your guess is as good as mine as to what that means. I guess it means something like full, direct, Southern California sun at noontime. As you get away from noontime by a few hours, often the sun intensity falls off dramatically. Also, as you get clouds in the sky, it falls off dramatically. With some solar panels, a small shadow across the panel will ruin its output significantly, and that could be from something as dinky as a pack strap hanging across it. In defense of the manufacturers, I will say that they don't really know how good the user will be at keeping the solar panel oriented or at keeping it in good sun, so they just omit a few specs to protect themselves.
Also, there is the sun angle. If you were located on the Equator, the sun would pass almost directly over your head at noon, so you would want to have your solar panel relatively flat to the Earth to collect that good noon sun. If you were at the North Pole, the sun would not be anywhere near overhead. Instead, it would be lower toward the horizon, so you would need to point the solar panel much lower. I remember that the correct angle is Latitude (degrees)+ 23 degrees for optimal winter operation in the Northern Hemisphere. For summer, I don't remember, but it is less.
--B.G.--
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