2.7mph down
1.8mph up
Assume 5.4 miles up and 5.4 miles back.
3 hours, 2 hours back.
Real average speed = (5.4+5.4) / 5 = 2.16mph

If your average speed was 2.25 in a non current swim, your time would be 10.8/2.25 = 4.8 units of time.

That's a four percent saving in a calm swim.

Unless my maths have failed me.

I know in an out and back run, a tailwind seems to not give as much assistance as the headwind hampered me...

Harmonic Mean (2 numbers):

H = 2xy/x+y

Plug in your speeds and it is 2.16

---

Formerly TriBrad02

Remember speed is a function over time ~ you spend a greater duration at the slower speed vs the faster speed over the same distance, so the slower leg carries more 'weight'; i.e., the average speed is not an even split in the middle, but skewed towards the slower pole. So, the greater the current, the more the upstream leg will drag down the net average.

But who is going to remember that obscure formula??? The way Luke approached the problem is better IMO b/c he did not have to remember and/or look up that formula. His approach is much simpler.

---

"Anyone can be who they want to be IF they have the HUNGER and the DRIVE."