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From: Michael Edelman <mje_at_spamcop.net>
subject: [Paddlewise] Newtonian Physics
Date: Mon, 29 Jul 2002 15:21:25 -0400
> From: "Jeffrey Steck" <jeffrey_steck_at_hotmail.com>

>.. The equation for power is simply:
> 
> POWER = FORCE * VELOCITY
> 
> The FORCE is just the resistance of the boat at that
> speed, becasue of the constant speed assumption.
> You can look up the resistance numbers in Sea Kayaker. (I am
> ignoring the effects of "coasting" between strokes.)
> 
> The VELOCITY is the velocity of the paddle relative to the
> paddler, which is the velocity of the boat through the
> water PLUS the velociy of the paddle through the water, i.e.
> the speed of paddle slippage.

> Thus, the power that a paddler has to put in to keep
> a constant boat speed is equal to:
> 
>      BOAT_RESISTANCE * (BOAT_SPEED + PADDLE_SLIPPAGE_SPEED)

Let's work some numbers. Say we want to go 4 mph (5.87 fps), and say 
Matt's table says that takes a force of 8 lbs (It's a heavy boat).

Say our big fat WW paddle moves through the water twice as fast as the 
boat to go 4mph:

        8lbs * (5.87fps + 11.74 fps) = 140 ft-lb/sec = 0.256 HP.

  Now let's try it with a narrow Greenland paddle that requires paddling 
four times as fast as the boat:

        8 lbs * (5.87fps + 23.48 fps) = 234.8 ft-lb/sec = 0.427 HP

I sped up the paddle and (by the model) put in twice as much energy for 
the same speed. The model says the Greenland paddle is only half as 
efficient, as I'm putting in twice as much power to go the same speed.

But that doesn't make sense, because paddles with less area have less 
resistance to being pulled through the water, and hence do less work on 
each stroke.

So if we pull a paddle twice as fast to get the same speed, less work 
must be getting done. But the model says paddle *resistence* isn't 
important. Only paddle velocity relative to the boat. Ergo there's an 
error in the model.

-- mike
-----------------------------------
Michael Edelman   medelman_at_ameritech.net
http://www.foldingkayaks.org
http://www.findascope.com



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From: Shawn Baker <shawnkayak_at_yahoo.com>
subject: Re: [Paddlewise] Newtonian Physics
Date: Mon, 29 Jul 2002 13:09:19 -0700 (PDT)
Only if the paddle and boat have the same amount of resistance in the
water, be it any combination of lift, drag, wavemaking resistance, or
skin friction.

Shawn

"Jeffrey Steck" <jeffrey_steck_at_hotmail.com> wrote:
>Efficiency of 5% would mean that for a boat speed of 3knots,
>the paddle is slipping backwards at 57 knots, something none
>of us have experienced.


Yahoo! Health - Feel better, live better
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From: Michael Edelman <mje_at_spamcop.net>
subject: Re: [Paddlewise] Newtonian Physics
Date: Wed, 31 Jul 2002 12:54:12 -0400
> From: "Peter Chopelas" <pac_at_premier1.net>
<I wrote>
>>Say our big fat WW paddle moves through the water twice as fast as the
>>boat to go 4mph:
> 
> 
>         8lbs * (5.87fps + 11.74 fps) = 140 ft-lb/sec = 0.256 HP.
><snip>


> Your model is wrong because you are comparing the wrong forces.  

Actually, it was a misstaement of Jeff's model. I was trying to show the 
errors.

Jeff's model actually stated Force = hull drag * (hull speed + slippage)

>...In both
> cases you used the resistance of the drag of the boat, 8 pounds, and then
> multiplied that by an assumed paddle speed.  You have to multiply the hull
> drag by the HULL speed and you get the power OUT-put. 

Right. Jeff multiplied drag * (hull speed + slippage) and called that 
force needed to propel the boat at V.

-- mike
-----------------------------------
Michael Edelman   medelman_at_ameritech.net
http://www.foldingkayaks.org
http://www.findascope.com



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