Subject: Re: [harryproa] Sail to windward in a gale?
From: Gardner Pomper
Date: 3/28/2011, 12:39 PM
To: harryproa@yahoogroups.com.au
Reply-to:
harryproa@yahoogroups.com.au

 

Arto,
 
That website looks like it would be really helpful, if I could follow it. I am hoping you can clear up whatever my misunderstanding is. I scrolled down to the Windward Ability section, and in particular, these formulas:
 

Cdp = CP x AP / SA ..............8

CdTOT = Cdp + Cd ...............9

  • Cdp = coefficient of drag of parasitic element related to sail area
  • Cp = parasitic drag coefficient of each element
  • Ap = area of parasitic element
  • CdTOT = total coefficient of aerodynamic drag for whole boat + rig
He mentions "slab sided bridgedeck saloon cat with a total coefficient of drag (Cdtot) of 0.392", but I can't see how he can get to that kind of number. I must have a problem with my assumptions.

 
For a first approximation, the area of a boat, with the wind coming from 45 degrees off the bow, should be the length * width * avg height * sqrt(2). For my old Maine Cat, that would be 30 * 18 * 5 * 0.7 = 1890 ft^2. The sail area was 500 ft^2. Plugging these numbers in, I get Cdp = 1.0 * 1890 / 500 = 3.78, which is an order of magnitude greater than his "slab sided bridgedeck cat" and the Maine Cat was open bridgedeck and pretty streamlined.
 
What am I misunderstanding?
 
- Gardner
On Wed, Mar 23, 2011 at 4:03 AM, Arto Hakkarainen <ahakkara@yahoo.com> wrote:
 

Your assumption is correct. For better analysis see John Shuttleworth's article Multihull Design Considerations for Seaworthiness on his website here: http://www.john-shuttleworth.com/Articles/NESTalk.html and scroll down to the part "windward ability" to see how it could be calculated. You need to find out quite many variables to the equation first though.
 
Arto

--- On Wed, 3/23/11, Gardner Pomper <gardner@networknow.org> wrote:

From: Gardner Pomper <gardner@networknow.org>
Subject: Re: [harryproa] Sail to windward in a gale?
To: harryproa@yahoogroups.com.au
Date: Wednesday, March 23, 2011, 3:33 AM


 
I would assume that there must be some relationship between windage, righting moment, steering and leeway resistance that would define if a boat can go upwind in any weather. It is the windage that I would assume becomes the major issue as the wind goes up to gale force or higher. If the driving force of the sail does not exceed the wind drag of the boat, it won't go upwind. So, should I be calculating the maximum sail area for the righting moment at 40 knots of wind, then compare that to the force of the wind on the cross-section of the boat? I don't know the formulas for that.

Your description is nice and shows the boat as safe and comfortable, but I don't see how it can be used to predict if it will actually sail upwind.

- Gardner

On Tue, Mar 22, 2011 at 9:28 PM, Rob Denney <harryproa@gmail.com> wrote:
 
On mine, I said

UPWIND IN 40 KNOTS
The simple, forgiving rig, very high righting moment, shelter on the windward side and lack of things to worry about should make it as easy as possible for the crew.  The large, twin rudders and schooner rig give excellent, low stress control at low speed with no flogging sails or sheets.  Windows in the cuddy allow a seamanlike lookout to be kept without exposing the crew.  Shunting really comes into it's own in these conditions, particularly if there are big waves as well.

Use what you like, but probably diplomatic not to use an exact copy.  I explained about simple forgiving rig, lack of worry, etc  elsewhere.

rob


On Tue, Mar 22, 2011 at 10:29 PM, Gardner Pomper <gardner@networknow.org> wrote:
 
Part of the wooden boat design competition is the ability for the design to sail to windward in a gale. How can you prove that from a drawing? I am pretty sure that my design will float, and it should be fast, but I don't know how to calculate upwind performance in heavy weather. Can anyone tell me what criteria might be used to judge that?
 
- Gardner
 




__._,_.___
Recent Activity:
Visit Your Group
.

__,_._,___