# Question for all you wind and current

geniuses ?

Paddling on flat water against what MPH wind speed would be equivalent to paddling against a 4 MPH river current? -(assuming that there is no wind against you on the river)

Jack L

I’ve often heard wind to current at 10:1
So a 10 mph headwind would be roughly like paddling into a 1 mph current. I’m sure the boat’s windage and such all affect this so this is just a very rough rule of thumb.

10:1 seems high
I’d hazard a guess of around 20. If you’re going to count the wind driven waves that are going to slow you down too.

Just sort of a guess based on what I think I could do for “speed made good”.

Rough Guess

– Last Updated: Nov-09-10 8:09 PM EST –

I'd guess 1mph current for every 5mph of wind just based on what keeps me from making any forward progress. I ain't know jeenyus but that would be about a 20 mph headwind for your 4mph current.

hmmm
I’ve never seen a clean formula for this. Most of the sea kayak nav books will tell you how much resistance you will feel in given wind speeds.

Shelley Johnson’s book says that a 25 knot wind will produce 6 pounds of force and “slow you to a crawl at your normal cruising effort”. Assuming a normal cruising effort produces a 3 to 4 mph pace, then a 25 knot wind might be close to paddling into a 4 mph current.

From personal experience i’d say that pretty much jives. I might put it closer to 30 knots.

I think we can do better than this

– Last Updated: Nov-10-10 12:38 AM EST –

http://www.ckapco.com/Kayak/KayakManual/WindWavesandCurrent/tabid/129/Default.aspx

Indicates that over 25 mph you are pretty well done. Wind effect is not linear it seems.

Do you want to add drag(sinking) factor in shallow water?

Over 25 mph and a hammock looks better. Did my 43 mph wind spur a question about if it was possible..??

We had a diagonal course so it was more of a wind ferry..
to the hammock.

Less than genius …
is probably the reason I was out paddling in a gale this

autumn. Anyway, the water was definitely not flat.

Shape would matter
If you were paddling something like a whitewater boat, the 4mph river current might have you stationary, while on a ski or something you could make plenty of way.

I’m sure a simple formula doesn’t exist, but for my Prospector canoe solo I can’t move forward in much more than 30km/h, and then I don’t really want to.

Can’t say exactly

– Last Updated: Nov-09-10 9:44 PM EST –

since I'm not a genious of any sort, but paddling against about 22-25 mph I can go at about 2 mph. Against 15-20mph I seem to be doing b/w 2-3 mph (with some waves against me).

However, that comes at a level of effort that would take me into the 5+ mph speed without wind in the same boat. And that was not a "fast" boat, to give you an idea about the effort needed. I'd be doing near 6mph in my "fast" boat at the same effort...

At around 30mph sustained I've only paddled once or twice and there I'm below 2mph, just barely making any progress.

The above speeds in the 20+ mph winds were maintained for short periods of time with breaks, e.g. I'm paddling upwind for 10-15 minutes, then surfing back. Did not feel like paddling more than a few miles this way would be fun... After paddling about 5-6 miles this way over less than a couple of hours feels like paddling probably 15 miles at near my top speed on flat water (would be close to 3 hours for me).

Perhaps with a highly feathered paddle I'd do better than with my unfeathered paddles - switching to a canoe paddle makes a notable difference in wind resistance but not so much in overall speed due to the need to switch.

Drag calculations

– Last Updated: Nov-09-10 10:04 PM EST –

This is long, and has equations so I apologize in advance. What can I say, I'm a fluid mechanics nerd:

I did a similar calculation a while back to judge the drag on my kayaks while mounted on the roof rack. The order of magnitude approximation is based on using the concept of drag coefficient.

Drag coefficient (Cd) is an experimentally determined dimensionless quantity that is used as a proxy for the complicated dependence of drag on body shape and fluid velocity. Briefly, if you know Cd, then drag (D) is given by:

D = Cd * 0.5 * rho * V^2 * A

rho is fluid density, V is velocity, A is frontal area (perpendicular to the fluid stream). For a streamlined body like a hull, Cd is perhaps 0.2. For a bluff body like a person crosswise to the wind, it's about 1.

For my original calculation, I assumed that the frontal area of the kayak on top of the car was twice the frontal area of a kayak in water, because both the hull and deck encounter wind resistance. In water, obviously just the hull encounters water resistance. I also assumed that Cd for the hull is about the same in air and water, which is a pretty good assumption for a streamlined body like a kayak. Equating drag forces in the two fluids gives a simple relation:

V(air) = V(water) * SQRT( rho(water)/rho(air) * A(water)/A(air) )

or more simply:

V(air) = V(water) * SQRT( density ratio * Area ratio)

For water and air, the density ratio is about 850, and the area ratio as written is 1/2. So the Velocity ratio is the square root of 425, which is about 20.

So the drag on a kayak on a roof rack at 60 mph is equal to the drag in the water at 3 mph, which is only a few pounds. That's straight-line drag, doesn't count the cockpit opening, etc., just an estimate. If the hull strays from straight, much higher drag forces will come into play.

For the situation originally proposed above, drag due to the wind is contributed mostly by the paddler's body, not the deck of the boat. An equivalent analysis gives a similar expression with additional terms:

V(air) = V(water)* SQRT( density ratio * area ratio * Cd ratio)

Again, density ratio is 850, I assumed the hull to paddler area ratio is about 1/5 (frontal area of the person is larger than the frontal area of the hull in the water) and the Cd ratio is about 1/5 as mentioned previously. So:

V(air) = V(water) * SQRT(850/25) = 6 * V(water)

This is a velocity ratio of about 6, e.g. for 4 knot paddling in water, a 25 knot headwind gives about the same drag, which is in line with the experience of others mentioned above. Clearly, one can tweak this result by assuming different Area and Cd ratios, but the result is fairly robust due to the square root.

No, it had nothing to do with that
I’m training in my QCC-700 for a river race where we go up river for five miles against a current, and then turn around and come back down.

I am training on a lake, and as much as I hate racing into a wind, right now it is my friend, since I can’t similate a current on a lake.

By the way did you get the e-mail I sent with all the info on the southern Everglades. (never heard back)

Maybe I should get Scupper Frank in touch with you.

If he reads this, he will know what I mean !

Jack L

Hey Carldelo
How would the Cd affect the boat when it’s going with the current? I’ve always heard folks say that they were paddling in a, let’s say, 3 mph current and simply subtracted that from their overall speed to get how fast they were paddling. But, is that exactly true? Due to the Cd from the rear, do you really get the full force of the current as you’re paddling?

Depends on whether you are Ichabod
Crane or Batman.

Relative velocity
Yes, it’s correct to just subtract a following current. Your hull only cares about the relative velocity between the hull and the water. Going into a current, you add it on to get the higher relative velocity in that situation.

Well, at least if I’m a paddling…

– Last Updated: Nov-10-10 1:24 PM EST –

...Ichabod pushin' my nag (let's say an Old Town Disco 159, cuz I've often thought DougD looked like a strugglin' Ichabod) into a 25 MPH headwind, I may be able to take a wee tad of solace in the knowledge that any headless SOB closin' on my stern is go'n to have themselves one hell of a heave with their pumpkin prosthesis into my upwind direction.

"Quick, Robin, to the BatCanoe!"

And, in other exorcisms of mathematical exercises in futility, or nearly so, I do know that it once took me 4-1/2 hours to go 5 miles into a continuous 20-25 MPH headwind whilst double-blading an eighty-pound canoe filled with 400-lbs. of me and gear and firewood. With only one, five-minute stop with an arm wrapped around a duckblind ruin's pile so that the other arm might tip an open can of Guinness that had been teasin' me like some sultry sip of Irish siren ever since I left the dock two hours prior!

River racing
My bet is that those who can read the current and guess where the suckwater is will be most successful. Unfortunately, can’t train for that in a lake…

4mph current seems like a lot - there’s got to be some eddies and back currents to take advantage of upstream and then go with the fastest current downstream, all the way optimizing speed/distance/depth of water/current speed…

Sounds like fun…

No, there is no suck water in this river
I know it and have paddled it.

Yes I know how to use the current going both directions - been racing a long time !

On the current; I might be estimating it on the high side.

Jack L

Let’s imagine that our power paddler manages to stay put paddling against 4mph current, or his power output is equivalent to boat drag at 4mph. Of course, no wind is blowing

Then we can ask ourselves - would paddling down, say 2mph stream, result in 6mph? Again, no wind.

The ratio in your eqns is 34.

So, Vmax=4mph, Vw=2mph, and Vp - is how fast the paddler is going.

0=Vp^2*(1+1/34)-2VpVw+Vw^2-Vmax^2,

and the relevant root is 5.87 mph.

Now, most people would probably laugh at .13mph lost, but a serious racer might start doubting his manliness and loose sleep.

can’t give no mathmatical ratios but …

– Last Updated: Nov-11-10 1:36 PM EST –

...... I know I've been on a mountain river heading downstream and the wind picked up something fearce . At one point as we approached a bend we had to give all we (two guys) could muster just to gain an inch at a time , and then at one point giving it all we had we actually were going backwards upstream ... I guesstamated the wind was somewhere in the 25 mph. range (??) , but who knows , and we know how to dig in and push hard when required .

I've also rowed a small boat before in heavy winds and staying near shore have noticed that the shore and I weren't changing relative position ... all I had was only holding me still in comparison to shore .

And I have other similar wind and current memories that make me think of the words "very hard work" . So all I know is if you want to keep moving forward against wind and or current , you got to dig in and push it to the max sometimes . I can do that for quite some time , but sooner or later something , me or the wind/current has to give .

Good luck!
Sounds like fun!