Liz Caldwell wrote:We're doing shuttle a little differently by going to the put-in first, unloading, then running shuttle to the low-water bridge take-out, and getting back to the put-in by 3.
DISCLAIMER: WHAT FOLLOWS IS AN OBNOXIOUS SHUTTLE CALCULATION! YOU HAVE BEEN WARNED! 
I was surprised to find such differing opinions on running shuttle, so I decided to run the numbers to see if good ol' mathematics could shed some light on this perplexing question.
In both scenarios, we’ll assume we have 4 cars with paddlers all going to the river. Everyone will be driving through Murfreesboro. If you go straight to the put in on 19 from Murfreesboro it’s 6 miles. If you go to the put in from the take out using the short cut, it’s 5.8 miles, but for simplicity, we’ll just assume any trip between the put in and take out is 6 miles, or from Murfreesboro to the put in is 6 miles.
Scenario #1:
All four vehicles meet at the take out. They load all boats and paddlers in 1 car, leaving the other 3 cars at the take out.
They drive the 1 loaded car to the put in: 1 car x 6 miles to put in.
They paddle down the river, having lots of fun.
At the end of the day, 1 vehicle is needed to go get the single car up at the put in: 1 car x 6 miles to put in, PLUS 2 cars x 6 miles back to take out.
Everyone loads up and goes home.
Total shuttling miles driven = 24 (4 vehicle trips x 6 miles)
If everyone can load up to shuttle to the top in a reasonable time, you could
meet at 2:30 and be ready for the 3:00 siren.
Scenario #2:
All four vehicles meet at the put in. Since coming from Murfreesboro this is 6 miles further than the take out, this is 4 vehicles x 6 miles.
Everyone unloads their boats, and then all 4 cars drive to the take out (4 vehicles x 6 miles). Hopefully there’s an extra paddler to watch everyone’s boats and gear.
Once at the take out, they leave 3 vehicles, and 1 vehicle brings all drivers back to the put in (1 vehicle x 6 miles).
They paddle down the river, having lots of fun.
At the end of the day, 1 vehicle is needed to go get the single car up at the put in: 1 car x 6 miles to put in, PLUS 2 cars x 6 miles back to take out.
Everyone loads up and goes home.
Total shuttling miles driven = 72 (12 vehicle trips x 6 miles)
To give everyone time to unload, drive the take out and back, and then get on the river, you’d better plan to
meet at the put in at 2:00.
Total shuttle miles saved in Scenario #1 vs #2 = 48 miles. If everyone’s driving a Prius (we’re not) that’s maybe 1 gallon of gas = $3.50. If we’re more likely driving trucks or SUV's (not all of us, but a lot of us), that’s maybe more like 4 gallons = $14. Let’s split the difference and say we’re all driving Subarus, and that it’s 2 gallons = $7. Anyway, how many beers will $14 or $7 buy at dinner?
In my hypothetical scenario we’re only talking about 4 cars (could have been more paddlers than 4). Last Friday we had 25 people on the river…so instead of shuttling a handful of cars, say that was 2x or 3x as much shuttling. So instead of 48 more miles driven, with a big group that could be as much as 100, maybe 150 extra miles driven…and since everyone has to drive the shuttle both ways, it’s more time driving as well as more miles.
See, isn’t math fun? Math is our friend! 
I'm sorry - where are we meeting again, and at what time?
