The design of a self-steering device using the wind
entails devising a means to convert the light air forces
on a wind vane to an output line force capable of steering the boat.
This is easily done by extracting power from the water flowing
past
the hull. This type of device is known as a type of servo system where
the small input force from the wind vane is amplified to a
line
load
large enough to control the boat.
This magnification of load is referred to as " gain " and
can be
as much as several hundred times the input load from the wind vane.
Getting
this amplification is not difficult but getting it
with out over steering is more difficult.
The linkage action shown on the" Basic Action" page depicts some
of the mechanical motion used to provide smooth control.
The use of over counter weight of the wind vane and or
an elastic line arranged to pull the vane to a neutral position
can provide additional means of avoiding erratic steering.
The actual loads involved are surprisingly small
and sample calculations below verify this.
The device needs no more output than is required to steer
the
boat by hand.
Check for errors in calculation and typing
Please report any you may find.
Theoretical Loading Calculations for Wind and Water Vanes
Wind Vanes
Loading for a sample blade at a wind speed of 10 knots
Size (
Sv), 6 x 30 inches mounted such
that the center of pressure is
16 inches above the rotation axis. 180 in2 /144 = 1.25 Ft2
Metric 1.25*0.0929 = 0.1161 M2
Angle
of attack ( Alpha ) of 5.0 degrees. This assumes a 5 degree course
error.
Gravitational constant ( G ) , 9.8 meters/sec/sec
Coefficient of lift ( Cl) , For
ALPHA = 5.0 use Cl =0.5 ( Use this low value to
help compensate for the low Reynolds number of
such a small surface. Even this may be optimistic.
Air Density (
Da) = 1.215 Kg/M3
Velocity (
Va )= Wind speed in Knots , 0.5148 M/sec x Knots
Dynamic
Pressure (Qa) = ( Density x Velocity2)/(2xG)
At 10 Knots, Qa= 1.215 * ((0.5148)2
* (10)2)/(2*9.8)
Lift Force on Foil (
Lv) = Cl *Qa*Sv
= 0.5*1.64*0.1161
= 0.095 Kg
= 0.21 Lbs
Comment---
This low incidence angle on the vane implies that
most of the torque is being caused by the lifting effect such as
an airplane wing would show. For a wind vane situation the
drag loading becomes more of an influence at angles where the
lift begins to stall. However, the result is that the torque actually
continues to increase with more incidence even when the flow
is in a stall condition. The incidence angle is determined by the
off course error, the vane tilt, the boats heeling and any yawing
or rolling so the calculation of the vane's output requires all these
factors to be considered.
Oar loads
Oar loading at 7 knots boat speed in salt water
Size ( So)
= 5 x 24 inches submerged blade = 120 in2
120 x0.0006452 = 0.077 M2
Cl=
0.7, for Alpha of 10 degrees
Dw
=64 lbs/ft3 , 64x 16.02 = 1025.28 Kg/M3
Vo
=(0.5148 x Knots) M/sec
Qw =
1025.28 x ((0.5148 x 0.5148 ) x (7 x 7 ))/(2 x 9.807)
= 679 Kg/M2
Force on oar blade = 0.7 x 679 x 0.077
= 36.6 Kg , 36 X 2.205 = 81 Pounds
Insert sketch of push rod and oar crank here
Loads on the push rod and oar linkage.
Referring to the "Basic Action " page the calculated
loading of the push rod and associated parts are:
For a force of 0.21 Lbs acting normal to the wind blade and having
an effective center of pressure 15 inches above the vane's rotation
axis with a moment arm of 15 inches produces a torque of
(0.21 X 15 ) = 3.15in-lbs
With an output crank arm of 1.5 inches ( A )
The push rod loading is :
Pr= 3.15/1.5 , = 2.1 Pounds ( per the example
above ).
This push rod in turn acts on the crankshaft with an arm of 1.5 inches
resulting in an input torque of 3.15 also neglecting the minor friction
in the link fittings The output side of the crankshaft which
is twisting
the oar about it's vertical axis is likewise causing 3.15 inch-lbs of
torque on the oar axis.
For the rod that is bent to 45 degrees, Rotation of the oar for the full
wind vane travel of 45 degrees is approximately 35 degrees.
A bend of 30 degrees will rotate the oar some 22 degrees.
This should help explain the relationship of the crank
to the oar rotation. Click on the thumbnail to enlarge.
The geometry of the oar carrier and the slotted oar head will cause
the oar to rotate back to zero angle of attack and align with the
stream flow past the hull.
The power of the oar to steer the boat will require it to have enough
force on it at the travel required to balance the load being
produced by the boat's rudder.
For the case calculated above the angle of 10degrees angle of
attack of the oar occurs when the 45 bent crank has rotated the
carrier over about 25 degrees.
For a 30 degree bend it would only swing over about 12 degrees
to have this attack angle and equal load.
Reynolds Number
The Reynolds number of the wind vane is calculated as follows:
Reynolds number equals velocity times the Chord length
all
divided by the Kinematic viscosity in feet per sec
Velocity = 10 (6070/3600) ft/
X= 0.5 Feet of chord length
v=0.0001567 ft2/sec
R = V (X/v)
R = 5.38x104
At
these low values the flow will be all lamina and simple flat plate
sections will be almost as good as more curved foils.
PVC and sunlight.
This is a question that comes up quite often and
from my personal experience and searching the net has
found no problem with the physical properties of PVC
piping and fittings made to USA ASTM D 1784/5
specifications. However ,there is no doubt that many
mixes referred to as " PVC " that will not qualify.
In countries where the quality is not very good the builder
should look for local knowledge and paint any exposed
pipe.
Jan Data_1
These files were posted on the forum
Showing (-) 30 degree tilt axis
Showing wake running straight downwind
Normal vane geometry
USD vane geometry
Jan's oar working
Jan's rudder
Heeled
Jan's wire rig
Auto Vs Vane system recording
Z bar crank Vs oar angles
Jan July 2003
Jan's boat with USD and RHM
