Lawrence, At 4.5 miles per kWh when driving continuous 55 MPH, you drive 55 miles in 1 hour. In that time you consume 55/4.5 = 12.2 kWh of energy. If you want to generate that same energy while driving, to keep your battery at the same level, you need to be generating an average of 12.2 kW into the battery. Since you will likely not generate power at the exact battery voltage, you need to convert, which can mean anything from 80 - 95% efficiency depending on money and choices in system components, so let's say that you need to generate 13 to 14 kW from the panels before conversion to battery voltage. But due to sub-optimal conditions (you cannot point the panel to the sun) you get a varying amount of power from the panels, depending on the angle that the sun hits them and the occasional shading that you encounter from trees, buildings, signs and so forth. So, you probably need to start with a panel of about 20kWp in order to end up with an average stream of 12.2kW into your pack. Now do the math on the size of that amount of solar: at roughly 15% panel efficiency (which is extremely high, 12% used to be the norm for many years) you would need 20/0.15 = 133 sqm which is over 1400 sqft. I believe that 12 ft is the absolute max width of a vehicle on the road, so this would mean a trailer with a length of 120 ft. I don't think it will be roadworth or legal or still get 4.5 mi/kWh.
BTW, others have suggested, many years ago and in context of charging trailers, that you do not need to maintain battery level, just slow the drain to the point that with stops included, you replenish enough that your battery is drained at the end of the trip. Say you take a 330 miles trip. That means 6 hours driving at 55 MPH. But you also will add a stop for a meal, and possibly a restroom break with a coffee, say that you have a total of 7 hours trip. You start with full battery and can drain safely 20 kWh from the battery by your arrival. The 330 mi trip takes an amount of energy of 330/4.5 = 73.3 kWh. You only need to generate 53.3 kWh and have 7 hours to do it. Now you only need 7.6 kW average into the pack. If you produce less, you need to take longer breaks, or plug in for fast charge. You get the idea. Cor. -----Original Message----- From: EV [mailto:ev-boun...@lists.evdl.org] On Behalf Of Lawrence Rhodes via EV Sent: Wednesday, December 27, 2017 2:48 PM To: email@example.com Cc: Lawrence Rhodes Subject: [EVDL] Solar trailer calculation Hi all, My math is good enough to figure out I get about my 30kw Leaf gets 4.5 miles per kw at 55mph more or less depending on wind. What I'd like to know is what size solar panel would be needed to support that speed. Now don't tell me it can't be done. I just want to know the formula for figuring it out. My puny brain exploded when I thought of all the variables. I do know that a 3000 watt system would charge my car in 10 hours. I would just like to know the numbers of what would be needed for continuous power at speed. I guesstimated it would be about 10000 watts. I'd just like the exact number. Thanks, Lawrence Rhodes...the point is to build a teardrop solar range extender/RV that could charge while driving. -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.evdl.org/private.cgi/ev-evdl.org/attachments/20171227/7cfe6581/attachment.html> _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org Please discuss EV drag racing at NEDRA (http://groups.yahoo.com/group/NEDRA) _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org Please discuss EV drag racing at NEDRA (http://groups.yahoo.com/group/NEDRA)