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- Path: sparky!uunet!zaphod.mps.ohio-state.edu!caen!kuhub.cc.ukans.edu!spaceman
- From: spaceman@kuhub.cc.ukans.edu
- Newsgroups: rec.kites
- Subject: Re: Flexifoils: how do they work?
- Message-ID: <1992Nov23.114109.45046@kuhub.cc.ukans.edu>
- Date: 23 Nov 92 11:40:58 CST
- References: <1992Nov19.132019.12311@ohm.york.ac.uk>
- Organization: University of Kansas Academic Computing Services
- Lines: 67
-
- 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
-
-