|Subject: RE: [harryproa] forces on the rudder|
|Date: 5/5/2007, 6:51 AM|
ps.com.au [mailto:harryproa@ yahoogroups. com.au] On Behalf Of Doug Haines
Sent: Saturday, 5 May 2007 5:08 PM
Subject: [harryproa] forces on the rudder
I would like to hear some explaination on the rudder balance.
How is it, that a shaft that exits the top of a rudder a third the length back - should this then be tending to turn the shaft around because of the large surface area back from the shaft?
I understand the shaft is at the thickest , which is a third from the forward edge.
sigurd grung <nosupersnail@
yahoo.no>wrote:rob said to me, regarding my tornado conversion:
< Make it as long as you can within handling
I am not sure why.
1: I am expecting a bow-up attitude, since there is no
mast to push it down. This will be draggy, and I think
that extra length will not be enough to prevent it to
Thus I will need a T rudder or something anyway, so
hobbyhorsing will be much less of a nuisance.
2: speed, I was looking at this michlet calculation of
the tornado drag, and it appears that wave drag is
only a tiny part of the total drag, even as fast as
20kt. Thus I guess a longer hull will be draggier,
except maybe at *very* high speed, since WSA must be
inet.fi/private/ muu/torodrag. htm
Let me know if I'm wrong, please!
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