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post #1 of 17 Old 09-10-2010 Thread Starter
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Jeff_h or other Math whiz...

I put this in the design forum because it seemed the logical place... so here goes:

The debate came up on wing keel versus fin keel (yeah, like we have never had that before!!!). Anyways, my question was how much a fin keeled boat would have to heel in order for it to be or equivalent draft to a wing keel.

The numbers in this excersize are based upon the world famous, internationally recognized, infinitely superior Catalina 400!!

So if you have a C400 wing with a 5'4 draft, and the fin is 7 feet, how much would the fin keeled boat have to heel over just to equal the 5'4 draft?

Brian

Anyone?

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post #2 of 17 Old 09-10-2010
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if you would put enough BBQ equipment on the boat to sink the wing keel 1' 8" everything would be equal.

You can thank me later, Happy to help!!!!!!


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post #3 of 17 Old 09-10-2010
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Quote:
Originally Posted by bubb2 View Post
if you would put enough BBQ equipment on the boat to sink the wing keel 1' 8" everything would be equal.

You can thank me later, Happy to help!!!!!!
Nicely played old fart... One brick BBQ pit would do the trick I think...

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post #4 of 17 Old 09-10-2010
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Quote:
Originally Posted by Cruisingdad View Post
I put this in the design forum because it seemed the logical place... so here goes:

The debate came up on wing keel versus fin keel (yeah, like we have never had that before!!!). Anyways, my question was how much a fin keeled boat would have to heel in order for it to be or equivalent draft to a wing keel.

The numbers in this excersize are based upon the world famous, internationally recognized, infinitely superior Catalina 400!!

So if you have a C400 wing with a 5'4 draft, and the fin is 7 feet, how much would the fin keeled boat have to heel over just to equal the 5'4 draft?

Brian

Anyone?

42.....

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post #5 of 17 Old 09-10-2010
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Brian--

It's not a difficult computation if you know the distance between the metacenter (the point of rotation) and the bottom of the keel. If, for the sake of argument, one assumes that the yacht rotates about a point roughly 1 foot below the water's surface, such that the radius of rotation of the keel is 6 feet (7 foot draft less 1 foot) and the wing has a total draft of 5.34' then the necessary angle of rotation (heel) = arccos(4.34/6) or, roughly, 42.83 degrees.

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post #6 of 17 Old 09-10-2010
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CD, you dont want to heel too much in that fantastic plastic thing. Leave it tied to the dock or it may get some salt water on it.
& pls dont strain our brains - its easy to do

cheers mate


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post #7 of 17 Old 09-10-2010
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Well, I will give you the method:

You have just to pick the drawing of the boat with the fin keel version, put it on a cad, scale it and incline the drawing till the draft is equal to 5.4.

I would say something like 40ş.

why do you want to know that?

Regards

Paulo
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post #8 of 17 Old 09-10-2010
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Don't forget the corollary question:

How much will the draft of a wing keeled boat INCREASE when heeled?

We'll need the "wingspan" of your keel, CD.


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post #9 of 17 Old 09-10-2010
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So if I need to cross a shoal with a fin keel, I just have to sheet the sails in too far and heal 45 degrees or more while I cross?

Calculating the draft at a given heel angle is easy:

upright draft*cos(heel angle)=heeling draft

Last edited by casioqv; 09-10-2010 at 04:01 PM.
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post #10 of 17 Old 09-10-2010
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And don't forget to ask at what degree of heel does the wing keel maximize its draft...

Quote:
Originally Posted by JohnRPollard View Post
Don't forget the corollary question:

How much will the draft of a wing keeled boat INCREASE when heeled?

We'll need the "wingspan" of your keel, CD.

Sailingdog

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