Subject: Flexifoils: how do they work?
Message-ID: <1992Nov19.132019.12311@ohm.york.ac.uk>
From: jkn@ohm.york.ac.uk (John Nicoll)
Date: 19 Nov 92 13:20:19 GMT
Organization: Electronics Department, University of York, UK

Having built my flexifoil (see recnt postings), I've been
discussing with colleagues how they actually fly. The
side view of a Flexi- is a sort of inverted wing shape,
and we have speculated that it the downward `lift`
levers around the leading spar to align the kite
at an angle to the wind - almost 90 degrees, as it happens.
But I'm not sure where all that `forward' velocity comes from,
nor how the bending of the spar affects things.

Anyone else have any thoughts on these matters?

	john N
-- 
john nicoll (jkn@ohm.york.ac.uk)
Department of Electronics  University of York
Heslington   YORK YO1 5DD U.K
NEW PHONE! Tel +44 (904) 433221   Fax +44 (904) 432335



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From: spaceman@kuhub.cc.ukans.edu
Subject: Re: Flexifoils: how do they work?
Message-ID: <1992Nov23.114109.45046@kuhub.cc.ukans.edu>
Date: 23 Nov 92 17:40:58 GMT
Organization: University of Kansas Academic Computing Services

In article <1992Nov19.132019.12311@ohm.york.ac.uk>, jkn@ohm.york.ac.uk (John Nicoll) writes:
> Having built my flexifoil (see recnt postings), I've been
> discussing with colleagues how they actually fly. The
> side view of a Flexi- is a sort of inverted wing shape,
> and we have speculated that it the downward `lift`
> levers around the leading spar to align the kite
> at an angle to the wind - almost 90 degrees, as it happens.

The Flexi uses an inverted airfoil cross section, which will pivot the
foil about the l.e. spar so that the net moment of the forces (tension in
string, lift on the curved bottom ('downward lift'), and drag about the
center of gravity (very close to the spar) are
zero. Think of it this way:

Sum of the moments about the c.g. = T*Xt + L*Xl + D*Xd +Lt*Xlt= 0
where:  T is tension in the line, Xt is the distance from c.g. to the tension
	force
	L is lift, Xl is the distance from the c.g. to the lift force
	on the cambered surface (bottom of the flexi)

	Lt is lift at the top of the flexi (see 'forward velocity' question)
	Xlt is the distance between c.g. and the Lt force

	Note that the moment arm Xl for Lift is much greater than moment
	arm for the lift produced above  the mesh

	D is drag, Xd is the distance from the c.g. to the drag force.
	There is also drag at the top of the flexi!

        Note that Xlt << Xl, so not much pivoting will occur due to
	the lift produded at the top of the flexi

These moments will position the flexifoil at a certain angle with respect to
the relative wind (aka angle of attack) since kite is moving.


> But I'm not sure where all that `forward' velocity comes from,
> nor how the bending of the spar affects things.

Here comes the interesting part (and the part that made me pull out a 
picture of the flexi to answer this question)

If you notice the position of the spar relative to the leading edge of the
kite- it is below the mesh.  But, there is a curved part ( the leading edge
radius) above the mesh.  This curved radius will generate lift also (decreases
the pressure near the top-front of the flexi).

(Note that lift is perpendicular to the relative wind)
(Also note that even though the airfoil is at a large negative angle of attack,
 i.e. the airfoil is inverted, lift can still be produced on the non-cambered
 side- like a flat plate at some angle to the wind)

It is this lift (in conjunction with the moments) which gives the flexi 
the forward velocity.

The bending of the flexi is for control (and probably a little stability).
By pulling on one string, you effectively decrease the angle of attack on one
side, which produces less lift on that side (L and Lt) and the kite 
pivots about the other string.
 
> 	john N
> -- 
> john nicoll (jkn@ohm.york.ac.uk)

Paul
spaceman@kuhub.cc.ukans.edu



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