@Hypocrisy_Central I was thinking 3 rows abc with roads ab and bc. But you’re right! Four rows abcd with roads ab and cd. Rows b and c can be adjacent—I picture it as vehicles parked nose to nose. But 45×4 = 180 feet of vehicle length, leaving (207 – 180) / 2 = less than 14 feet per road, probably not sufficient if parked perpendicularly.
Angling the parking spaces is necessary. Let Z be the angle, where Z=0 degrees means the RVs are parked perpendicular to the access road. Then the vehicles have an effective length of (45 feet)*cos(Z). If Z = 45 degrees, then effective length is 32 feet.
So four rows is 4*32 = 128 feet allowance for the rows of parking spaces, leaving (207 – 128) / 2 = 39-foot-wide roads. More than enough. The trade-off, however, is that by angling the RVs you reduce the number that will fit side to side. The last one sticks out across the property line by 32 feet. You lose around 3 or 4 vehicles per row, or 12–16 total.
Say we make the angle only 30 degrees. Now effective length = 39 feet, leaving 25-foot wide roads. That sounds about right. And now you only lose 2 per row, 8 total.
Estimating 20 vehicles per perpendicular row (see earlier post), with four rows of vehicles, less 2 per row, makes (20 – 2)*4 = 72 RVs, minus
I’m assuming the problem is to cram as many parked vehicles into a one-acre square as possible, moving them just one at a time. The only “real world” concerns in this analysis are providing room to maneuver between parking space to access road, plus sufficient side clearance for the driver to enter & exit the vehicle. @LuckyGuy I think the vehicle itself is less than 9 feet wide & I assume all side extensions are retracted except for side mirrors, which hopefully don’t prevent doors from opening.