hydrofoil formula

sorry guys I just wondering if there is an STUPID EASY formula for calculate

the area of a foil for a certain speed and how much lift related to this area...

can we make fly 300kg at 5 knots of speed??? is this possibile?

(can a 470 foil?)

 

I fear there is no simple formula available because drag and power are both factors. Given a big enough foil and sufficient power a 300kg boat will foil just fine at 5 knots of speed. However a 470 may be unable to generate sufficient power to push the drag of that large foil through the water at 5 knots in any reasonable wind conditions.

 

==============
On page 58 of the book "Icarus The cat that Flies" the author tells the story of Claude Tisserand in about 1974 where he added retractable surface piercing foils to a 470(two forward and a rudder T foil). He achieved an unofficial timed run of 20 knots and until approximately 1999 was probably the fastest and one of only a very few monofoiler dinghies.
----
There is a book you should buy called "Hydrofoils Design Build Fly" by Ray Vellinga and available on Amazon(I think). Ray is a hydrofoil pioneer and an evangelist in helping people interested in foiler design.
He published this simplified (but very accurate) formula:
L=V^2 X S X CL where
L= lift in pounds
V= velocity,ft per second(1 mph =1.47 ft/sec.)
S= surface area of the top side of the foil projected vertically, in sq.ft.
CL= Coefficient of lift, is a dimensionless number found in tables of wing profiles in the Theory of Wing Sections by Abbott and Doenhoff .
---
You can also solve the equation for CL to get an idea:
CL= L/ (v^2 X S)
-----
Good Luck and have fun!
===============

 

Doug,

the water density, usually indicated with "ρ" (Greek "r"), is required in that formula in order to have correct units and in order to use the Cl from the book Theory of Wing Sections by A&D.
So the correct formula should read:

L = 0.5*rho x V^2 x S x Cl,​

where ρ = 62.4 lb/cu.ft.

If one doesn't use the density, and implicitly (but dimensionally wrongly) assumes that it's effects are contained in the lift coefficient, it equals to saying that the lift coefficient of the hydrofoil is 0.5*rho (31.2) times smaller than the lift coefficient of the airfoil, which is way too off the ballpark.

However, the above formula is good just for a very first estimate of what a hydrofoil could ideally give. It doesn't take into account the influence of the finite span of the foil, water surface proximity, interference with other foils or with the hull, etc. A much more complete set of formula and general technical info can be found in Groper's thread: http://www.boatdesign.net/forums/boat-design/foil-assisted-multihull-design-48462.html.

Cheers

 

=======================
Vellinga says" We can calculate the value of 1/2 p, where p(rho)= density of the fluid. The density of seawater is 1.99 slugs and when multiplied by 1/2 the result rounds off to 1.00(fresh water is 3.5% less).
The formula this is based on is: 1/2p X v^2 X S X CL
See Chapter 4 in the book. Vellinga is a well known hydrofoil designer with many boats to his credit. The difference between what you say and what he says is probably explained by the units used(difference between imperial and metric) as pointed out by Olav in post 5 below:
See post #5 here: http://www.boatdesign.net/forums/hy...ler-design-foil-assist-full-flying-40894.html
====
I've compared this formula's results to two of my boats and to the Moth and it is very accurate. It's also very simple along the lines of what was requested by the OP in post #1. There is much more that can be learned-this formula is on one page and the book is 251 pages. But it is a start and without a doubt answers a most frequently asked question.

 

I was wondering if some factor was needed for end effects. CL from books and 2d analysis is for infinite length. Is there some formula for losses at the edges? Home builders are limited to rectangles or at best trapezoid plan-forms.

 

===============
Vellinga's modified formula is fine for homebuilders but no homebuilder should use it without getting the book which is the best book on small hydrofoils I've ever run into.
Anyone considering designing and/or building should learn all they can and this book is an excellent place to start.

 

No, actually the difference is imo due to the confusion which exists in imperial system regarding the unit "pound":

http://en.wikipedia.org/wiki/Pound_(mass)
http://en.wikipedia.org/wiki/Pound_(force)
​

Cheers