Re: Re: [indoor construction] VP or VD

From: Bill Gowen <b.gowen_at_earthlink.net>
Date: Wed, 25 Oct 2006 10:21:01 -0400

I can't vouch for the aero theory but the results line up very well with what seems to work.

In response to Slobodan's thought that we would be seeing more VD's - I'm not so sure. This isn't a new idea and most of the people who've tried it went back to VP's. I spent a lot of time drawing folding mechanisms in AutoCAD and could never come up with a linkage that worked on paper - much less in real life. I think that the hubs that work (which are in reality just the ones built by Jim Richmond) there must be some flexing going on that enables the folding process to occur.

Both types of hubs are pretty aerodynamically dirty when folded.

  ----- Original Message -----
  From: John Barker
  To: Indoor_Construction_at_yahoogroups.com
  Sent: Wednesday, October 25, 2006 7:55 AM
  Subject: RE: [Indoor_Construction] Re: [indoor construction] VP or VD


  I have no practical experience of VD propellers on high class indoor
  aeroplanes but I think the following basic aerodynamic stuff may be a help.

  The standard expression for propeller torque is Q = kq. rho. n^2. D^5 where
  kq is the torque coefficient, rho is the air density, n is the rotational
  speed and D is the diameter. If the torque coefficient and the air density
  are taken as constant for our purposes it can be said that Q is proportional
  to n^2.D^5 or that n^2 is proportional to Q/D^5.

  Now assume suffix h is applied for the high torque and large diameter and
  suffix b for the low torque and diameter then:
  n^2 = Qh/Dh^5 and n^2 = Qb/Db^5
  Now as we want n to be constant through the flight we can equate the right
  hand sides and rearrange to give:

  Db = Dh.fifth root(Qb/Qh)

  A typical torque curve has a peak about 5 times higher than the cruise so
  that Qb/Qh is 0.2 and the fifth root of this is 0.725. If the propeller
  diameter required at high torque is 20" then the diameter required at low
  torque is 20 x 0.725 which is 14.5".

  I think in practice, and certainly with a two position movement, you would
  probably ignore the first few second of excess torque and work on a high/low
  torque difference perhaps 2.5. In this case, if the high diameter is still
  assumed to be 20" then the low diameter would be 16.7"

  John Barker - England

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Received on Wed Oct 25 2006 - 07:24:06 CEST