Upwind-DownCurrent Tug of War Brainteaser

[leong]

A Winter Brainteaser

Almost daily I paddle through the Palm Beach inlet going to and from the ocean (no, I haven’t met Trump yet and I keep away from Mar-a-Lago when the Coast Guard is patrolling).

The current in the inlet is generally about 2-knots and sometimes the wind is in the opposite direction. In fact, sometimes the wind exactly cancels out any drifting of my kayak; i.e., the current drifts me west at 2-knots and the wind pushes me east at 2-knots. So here’s a brainteaser.

Under the non-drift conditions above, which is easier (less exertion) to paddle: upwind/down-current or vice versa? Give reasons for your answer.

-Leon

 

[Brian Nystrom]

Since both wind resistance and hull drag are exponential functions, there's going to be little or no difference initially. Going against the current, you will approach your hull speed limit quicker, so it will become harder, sooner. Going with the current, you'll reach a higher "ground speed" before you hit the hull speed limit.

[leong]

On 12/19/2016 at 6:52 AM, Brian Nystrom said:

Since both wind resistance and hull drag are exponential functions, there's going to be little or no difference initially. Going against the current, you will approach your hull speed limit quicker, so it will become harder, sooner. Going with the current, you'll reach a higher "ground speed" before you hit the hull speed limit.

Brian,

You might want to change your answer based on the following hints:

A wind speed of approximately 9 knots against a kayaker has a force of roughly 1 pound and that cancels the 1 pound water drag on a kayak in a 2-knot current in the opposite direction.

Case 1: Say you paddle with the current (against the wind) with a speed over ground (SOG) of 4 knots. Then the relative water speed is 2 knots; (4 -2). Also, the relative air speed is 13 knots (9 + 4). Again, at a relative water speed of 2 knots the water drag is 1.00 pounds. Also, the air drag at 13 knots is 2.29 pounds. Thus the total drag you have to overcome is 3.29 pounds (1 + 2.29).

Case 2. Instead, say you paddle against the current with the same SOG of 4 knots. Then the relative water speed is 6 knots (4 + 2). Also, the relative air speed pushing you is 5 knots (9 – 4). At a relative water speed of 6 knots the water drag is 11.91 pounds. And the force of the relative air speed of 5 knots pushing you is 0.34 pounds. So the total drag you have to overcome is 11.57 (11.91 – 0.34).

Therefore, unless I made a mistake somewhere, I believe the example demonstrates that it’s easier to paddle downcurrent/upwind then vice versa, at any speed (not just as you approach hull speed). My actual paddling experience is consistent with this answer. Also, as you approach hull speed the exponent in the approximate exponential formula for water drag grows larger than 2.0. Also note that it’s less than 2.0 at slow paddling speeds. I validated this using John Winter’s drag program, named Kaper. Note, in a program I wrote I got answers consistent to the two cases at lower speeds. The examples were just easier to present.

One mistake I think you made is assuming that both wind resistance and hull drag are the same exponential functions. Yes, to a first approximation, the water drag function is sort of like Dw = W*s^2 and the air drag function is exactly Da = A*s^2 (where s is speed and A and W are constants). But, because water is much denser than air, W >>A so a small change in s changes water drag much more than it changes air drag.

The water drag values are from keelhauler.org/khcc/seakayak.htm and the air drag values were calculated from the standard air drag formula like the one published here https://www.grc.nasa.gov/www/k-12/airplane/drageq.html

-Leon

 

[Phil Allen]

Leon

I still need to think more about your details, but I'm suspicious of your 9 knots of wind in the face being equal to 2knots of contrary current.  Practical experience and numerous publications would suggest that it would take close to 20-25 knots of wind to get a 2knot boat drift.

Happy holidays!

Phil

 

[leong]

2 hours ago, Phil Allen said:

Leon

I still need to think more about your details, but I'm suspicious of your 9 knots of wind in the face being equal to 2knots of contrary current.  Practical experience and numerous publications would suggest that it would take close to 20-25 knots of wind to get a 2knot boat drift.

Happy holidays!

Phil

 

Phil,

There is a lot of uncertainty in the necessary wind speed because of what to use for the frontal area in the air drag formula. I used a 2 square-foot area for the paddler and kayak. At 9 knots, the wind pressure is ½ pound per square-foot. So the force of the wind is 1 pound and, according to KAPER, that’s the total drag on a kayak going at 2 knots. But let that go.

Suppose it does takes about a 20-knot wind to overcome the 2-knot current. Then, for case # 1, the drag increases to about 6 pounds and for case # 2, the drag decreases to about 9.5 pounds. So the qualitative result is the same; i.e. it’s still easier to paddle downcurrent/upwind than vice versa.

In my experience in the inlet, it does seem like a 10-knot or so wind is enough to overcome the current. But, I don’t know the exact current speed or wind speed opposite to the current. Also, I don’t account for wind generated waves on the surface that might be moving opposite to the current flow.

I think a key point is that a change in current speed costs a paddler more than the same change in relative wind speed. 

-Leon

 

Edited December 21, 2016 by leong

[Brian Nystrom]

I see two issues with your examples. First 2 square feet is a low estimate of frontal area given that your chest alone is at least 1 square foot. Second, your 4 knots SOG results in extremely high hull drag when going against the2 knot current. Most sea kayaks hit their max hull speed well below that and cannot be paddled at 6 knots by a normal human being. Run the test again at a more realistic speed and the results will be much closer.

[leong]

14 hours ago, Brian Nystrom said:

I see two issues with your examples. First 2 square feet is a low estimate of frontal area given that your chest alone is at least 1 square foot. Second, your 4 knots SOG results in extremely high hull drag when going against the2 knot current. Most sea kayaks hit their max hull speed well below that and cannot be paddled at 6 knots by a normal human being. Run the test again at a more realistic speed and the results will be much closer.

Brian and Phil,

I used 2 square feet because I used a large coefficient of drag. The extremely high hull speed (as you say) is not that high. I get my QCC700X to 6 knots (in still air and water) for short sprints. Max hull speed is just a number, not a hard speed limit. The Blackburn Challenge course is almost exactly 18 Nautical miles. The winning Epic 18’s and similar kayaks finish the course in 3 hours, or so. That’s an average speed of 6 knots for a long course. But so-called “max hull speed” of these kayaks is about 5.6 knots (1.34 time the square root of the water-line length of ~ 17.5’)

But let all this go. Nothing good on TV so, instead of rerunning the test, I decided to prove my result analytically so there can be no questions on the values that I used (it’s good that I’m retired ). I also gave it a whirl on the water but couldn’t get the proper conditions today.

See attached Mathcad worksheet for my analysis. Mathcad - DownCurrent-UpWind and vice a versa.pdf

-Leon