# How many 45ft luxury motor coaches can you park on one (1) acre of land (with enough room to drive them on and off)?

You have an acre, with maximum use going to the RVs, how many can you park on one acre of land giving each the ability to drive in and out of a place to park?

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## 12 Answers

Is the acre square? Shape is important.

To allow for room to get in the RV, assume ten feet wide. An acre is 43,560 sq feet, so if you had a lot that was 45 feet wide, it would be 968 long, and you could fit 96 RVs with room to get in and out of them and fit the side mirrors.

So, **96**.

@zenvelo That’s a reasonable approach. One long strip of vehicles packed side-to-side is probably optimum. The spaces might work at 10 feet apart, but I wonder if 12 feet is necessary for easy entry / exit from the widest RVs. The length of each parking space, equal to the short dimension of the one-acre strip, must be almost two vehicle lengths long, so the RVs can be backed out and turned 90 degrees. So I think 50 or 60 feet may be necessary. At 60 feet only **72** RVs will fit.

I am thinking of an acre squared because it is easier to deal with. I am not sure how many access roads or driveways one would need because I do not know how many rows there would have to be. For the sake of argument, let’s say the spaces were at a diagonal in the rows and the driveways were one-way; you drive in on the south you had to back up and leave on the north side.

@Hypocrisy_Central You just made the problem much harder and not at all an optimal way of parking RVs to the maximum. My design was for an RV dealership to see a lot of them, just back them into the space.

A square of area one acre = 43,560 sq ft has a side of about 207 feet.

Since the RVs are 45 feet in length (a detail of the OP I originally missed) they’d have to be arranged in rows with an access road running along each row. It works out that three rows of RV’s can share two 36’-wide access roads between them. Total length of the parking lot is thus 45 + 36 + 45 + 36 + 45 = 207 feet.

What is the width of a 45’ RV? Apparently it’s between 8 and 8.5 feet. Parking RVs in 10’-wide spaces allows 1.5 to 2 feet of door clearance for entry & exit. 207 ft / 10 ft/RV = 20.7 RVs – rounded down to 20 – that will fit across the 207’ width of the acre.

Three rows of 20 fits about **60** RVs. I say “about” because you might squeeze 21 RVs into each row, while a few parking spaces might need to be removed where access roads connect up.

@gasman Just curious, where did you come up with a 36 foot wide access road? And why no access road on the side to get from the front to the back?

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Are these RVs to be used or just stored? If you are considering “In use” then you must consider the slide outs and awnings.

I just checked my friend’s Crescendo Coach parked on my property as we speak. His awning extends 8 ft. The pop-out on the opposite side extends 34”. Also he said he likes to park facing north. (We live in the northern hemisphere.) That prevents the sunlight from coming directly into the front windshield. He said 25’ wide space is reasonable for camping. 30’ is much better. He has been in a place where it was only 20 and that was very crowded.

@gasman Couldn’t two rows of RVs share one access in if one row angles east and the other west? Then they just back up and head out north when time to leave or move them out of the acre lot? How wide would the access lane have to be just to make a 45 deg turn in and out?

@LuckyGuy *Are these RVs to be used or just stored?* Stored, parked there and not used, any sliders, awnings, etc retracted and in place.

@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.

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