Calculating inverter inefficiencies?

I have a need of tools in remote areas without power. (I use Milwaukee M18 "Fuel" brushless tools.)

I have a need to charge the batteries in those tools.

Idling my truck (Ford Expedition with a 5.4L V8 gasoline engine) is both inefficient, wasteful, and not environmentally friendly.

The rooftop seems to be good for a pair of 200 watt panels, so about 400 watts of solar panels.

I can charge a 12 volt battery bank on that... After that, I have a decision to make.

I can convert the 12 v DC to 110 v AC by way of inverter. Or I can purchase chargers that will direct convert 12 v DC to 18 v DC to charge the batteries. (Milwaukee 48-59-1810)

Converting energy twice (DC to AC to DC) seems less effective than doing it once. (DC to DC)

How can I calculate the efficiency of cheap square wave/Modded sine wave inverters that don't publish the spec?