Was Marchaj having us on?

Al
Normalish keelboat number for upwind sideforce, around 450 kg for 6,000 kg yacht. Foward force around 120 kg

 

Can you put that into the context of your 5Kg/sqm or my conclusion that would generate 5,000 Newtons of aerodynamic force please?

 

...roughly 4,500 Newtons of sideforce and 100 m2 of sail area...

 

Sorry, but I'm still not sure what point you are making.

 

These are real world numbers for a (say) 38 foot irc keelboat. The forward force is around 25% of the sideforce, expressed in newtons or kgs....for optimum upwind approximation.

 

I simply pointed that that was quite a lot of force: around 5,000 Newtons.
You are now introducing another interesting fact:

Again, I don't know where you are getting those numbers, they may be right, they may be wrong. What was your source?

I still don't know what point you are making.

 

Point 1 was that at these pressure differentials, air compression is a non issue.

The estimates are derived from CE, CLR and RM calculations per ORC methodology; accepted viscous drag equations and VPP outputs from leading contemporary design firms, which are tested in actual 'sailing operations'.... as in top level competition.....

 

But your figure of 5 kg/sqm generates 5000 N of force. Why do you say that's a non issue?

 

It is a factual estimate, that you could back calculate just using the CE height, the vertical arm to the CLR and righting moment.

For RM we can use typically light ship RM20 x 1.2 for fully crewed racing RM.

 

Are you claiming that the forward force is around 25% is a "factual estimate"?
You may be right, I'm just asking for a source, that's all.

 

An original source was some data released by North Sails from wind tunnel and CFD work.

But if you work through an equilibrium calculation with known drag and sideforce, this fraction can also be derived.

 

Can you provide a reference please?
But please, not from CFD, since CFD relies on the inapplicable principles of fluid dynamics. See Q11 in my theory

 

Cmon!
Do the equilibrium maths.... quite straightforward

Even if you use the original Froude equation for resistance, you will get close enough. Froude's estimate up to speed of 1.2 x root LWL just used his constant, wetted area and speed: Resistance = (Constant*wetted area*speed) ^1.82

 

But you said "An original source was some data released by North Sails from wind tunnel and CFD work.", now you are asking me to to derive this "factual estimate" of 25% myself. It's your number, all I'm asking for is a reference to where you found it.

(BTW, I don't think North Sails has ever published any of their experimental data from wind tunnel tests. I may be wrong, but I haven't see any.)

 

Well!
If you do the maths for a real boat, is this likely to be more or less accurate than an extrapolation of a simulation?

So we can do this with an actual 6,000 kg yacht, using resistance, heel etc for 1.2*root L.

There are plenty of published numbers for side force:forward force upwind in the range 23 to 27% from wind tunnel and other sources.

But the full size numbers could well be more accurate.