# Need help calculating headwind drag in watts from force in newtons

Discussion in 'Hydrodynamics and Aerodynamics' started by Jacques_clue_no, Jul 20, 2012.

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### Jacques_clue_noJunior Member

Just like the title says.

This is for an ericson 27. I estimated the force due to various headwinds here:

http://alain.fraysse.free.fr/sail/rode/forces/forces.htm

I used 1/2 of what the calculator output due to comments lower on the page.

10kt - 310N (aka 31 DecaNewtons)
15kt - 700N
20kt - 1240N
25kt - 1940N

So my question is how much power will it take to overcome the headwind? I understand that the force * the speed gives you watts, but the numbers get very large at 25kt and do not seem right at first glance.

eg.

25kt headwind is ~13 m/s, so 13 m/s * 1940N = 25,000 watts or 33hp!

Is this right?? (other than adding the speed of the boat to the speed of the headwind?).

Also, attached is a graph of the power required push an ericson 27 using a solidnav drive unit. I got the power data from an old solidnav promotional video. I assume this is for calm conditions.

Specs:

Ericson 27
Motor - Mars me0201013001 brushless
Reduction 1 : 2.2222
Sevcon millipak 4kw controller

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### Jacques_clue_noJunior Member

Ok, forgot to divide values from the calculator page by 2.

Now I get:

10kt headwind - 829w / 1.1hp
15kt headwind - 2811w / 3.74hp
20kt headwind - 6634w / 8.84hp
25kt headwind - 12978w / 17hp

Does this seem reasonable?

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### latestarterSenior Member

Well to get the discussion started, you are right that you have to add the speed of the boat to the wind speed to calculate the apparent wind speed and thus find the drag force.

Were this a car, to work out the power requirement you would multiply the drag force by the speed of the car not the speed of the wind.

However boats are different, you could be using the engine to keep the boat stationary into the strong headwind but the engine power used is obviously not zero.

It is similar to doing a hill start in a car as you release the brake and feed in the clutch the force times speed for the car is zero but the engine is doing a lot of work.

Hopefully someone who knows will be along soon.

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### Jacques_clue_noJunior Member

I think with a car you would want to factor in the speed of the wind too. It's just not as important since the windspeed is usually small compared to the speed of the vehicle and most cars have gobs of excess power.

I'm guessing I wouldn't need to add the speed of the boat when adding the power lost due to a headwind to the graph I posted. The graph is based on empirical data and the aerodynamic drag is already factored in for calm conditions. So I think I could just shift the curve up by the power lost for a given headwind.

Also, looking more carefully at the rode forces page I think the calculator output would be divided by 3 like the guy said, not two (I misinterpreted one of the graphs).

So divided by three (to better match empirical data) the power lost due to a headwind would look like this:

10kt headwind - 553w / 0.73hp
15kt headwind - 1872w / 2.5hp
20kt headwind - 4423w / 5.9hp
25kt headwind - 8651w / 11.3hp

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### Jacques_clue_noJunior Member

Here's the graph showing the efficiency of the solidnav on an ericson 27, for anyone who is interested. Quality is not so hot because the only place I could find it was in a youtube video.

Again, I believe this is what they were using:

-Motor - Mars me0201013001 brushless (aka Motenergy ME0907)
-Sevcon millipak 4kw controller
-Reduction 1 : 2.2222 (18 and 40T cogs, saw this on a store display - may change depending on application)
-Gates powergrip 360H100 belt

Solidnav is out of business, so I feel it's ok to do a little industrial espionage. The more electric boats out there the better

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### gonzoSenior Member

A 27 foot sailboat with a 15HP engine will go into a 25kt wind with no problem. You need to re-calculate the resistance. Somehow you must be either accounting for the rig and hull to have more than they do or doing the math wrong.

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### philSweetSenior Member

Agree with Gonzo. My old Cal 28 had a 15hp merc in a well and got me around just fine. This in the Keys with the trades blowing 15-30 all winter.

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### mydauphinSenior Member

Wish you would break this down to sq ft of windage. I have to figure hp to push boats sideways, like for thrusters. The aerodynamics have a lot to with it. This why I would rather have formula based on that.

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### Jacques_clue_noJunior Member

PhilSweet and gonzo - Thanks for the info. Would you say you need full throttle into a 30kt headwind? Or would you tack into the wind?

Running the numbers again I get 20hp for a 30kt headwind (up from ~11.3 for 25kt). So it sounds like I'm at least 50% high if 15hp is good enough for a 30kt headwind in the real world.

mydauphin - I didn't use square footage since I have no idea what the drag coefficient is. (or the square footage for that matter). I'm really just trying trying to get a ballpark figure here.

The page I linked to above calculates anchor rode forces for a generic monohull of a given length for a given windspeed. The author mentioned the figures were very conservative and empirical measurements are typically 1/3 of the calculator output.

It's also possible they are accounting for the angle of the rode which would make the calculated force higher than the force needed to overcome the wind.

But I think this is good enough to understand the tradeoffs in choosing different sized motors, which is all I was trying to do.

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### philSweetSenior Member

Yes, you would want full throttle. My arrangement with the prop behind the rudder meant I didn't have any positive steering unless I was moving. I would need more than two knots to hold course in 30 knots.

You would go dead into wind. Windage from the 30-45 degrees is usually quite a bit larger than straight ahead. I might also need to fall off to lower than a beam reach to get enough speed to round up. A forward prop location would behave better. The Cal had a lot of windage forward. It didn't anchor well. It needed to be kept moving to keep the keel hooked up.

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### DCockeySenior Member

To estimate the aerodynamic force on the boat use the apparent speed of the wind relative to the boat. For going directly upwind this is the sum of the wind speed relative to the bottom and the boat speed over the bottom.

For the useful work done in moving the boat though the water against the force of the wind use:
Useful Work = Aerodynamic Force * Boat speed relative to the water
Important - This is equal to the portion of the effective power, not shaft power, required to overcome the aerodynamic force.

Shaft power can be found from the effective power divided by the propulsive efficiency.

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### Jacques_clue_noJunior Member

Ok, so with the best estimate of forces according to the link (rode forces/3) are as follows:

10kt - 103N
15kt - 233N
20kt - 413N
25kt - 647N
30kt - 943N

Making the power required at 4kts (~2.14 m/s) as follows:

10kt - 103N - 220W / 0.29HP
15kt - 233N - 500W / 0.66HP
20kt - 413N - 884W / 1.18HP
25kt - 647N - 1385W / 1.85HP
30kt - 943N - 2018W / 2.69HP

Then say the propulsive efficiency is 60%......

10kt - 0.48HP
15kt - 1.1HP
20kt - 1.96HP
25kt - 3.08HP
30kt - 4.48HP

Now the numbers seem low

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### latestarterSenior Member

Using a generic formula and then dividing by an arbitrary number is never going to produce a reliable figure.

You have been focussing on the headwind, you will need to add the power to move the boat through the water at 4 knots plus punching into waves.

Commenting that the previous estimate was too high and the new one is too low suggests you already have a rough idea what it should be.

I am surprised no one has picked up that you are proposing a battery powered electric motor on this size of boat. The consensus of opinion is that is not practical with the present technology.

I am a fan of this system and will be using it on an easily driven canoe but never on a 27 footer.

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### DCockeySenior Member

At 4 kt into a 10 knot wind would be the same as anchored with a 14 kt wind, into a 15 kt wind the same as anchored with a 19 kt wind, etc.

Actually it's more complicated than that because natural wind has a profile with wind speed slower close to the water while the "apparent" wind from the boat moving is the same at all heights. So the wind force when moving at 4 kt into a 10 kt head wind would be between the wind force of a 14 kt wind when anchored and moving at 14 kt without any wind. But the difference due to this will probably be less than the uncertainty of the estimates being used of wind drag vs wind speed.

As latestarter mentioned the drag of the water and waves on the boat needs to be added to the aerodynamic drag.

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### gonzoSenior Member

You have to add skin friction and wave making resistance to the wind drag while anchored.

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