Well
Hells Bells, even I am shocked. My last article on the “Heart
of Gold” ezy-build precision F2B
stunter covered mods necessary to
control gyroscopic precession. In my usual “cut and paste”
style of modelling, I made no attempt to calculate the gyro
force. Like I don’t know how to.
So I just experimented with the “Supercool Slash” and fixed
the problem.
If only I had turned the
page. Page 50 of Frank Zaics
“1951-52 Model Aeronautic Year Book” has the gyro formula:
even a worked example! Just in case the formula is wrong,
lets give credit to the original presenter of this formula,
D. J. Cameron, from the 1938 Zaic year book.
Lets not
waste time. Here is the formula. Skip the maths if you want. But be prepared for a shock!
With W = mass of the prop
in pounds (lbs)
N = RPM of prop
V = speed of
flight in MPH
r = radius of
prop in feet
X = distance
of prop from the C/G in feet
R = radius of
turn in feet (say, square corner)
Then the gyro force F (in
pounds) is given by:
F = .00043 x W x N x V x r x r / ( X
x R )
where
x means multiply.
This is all a bit messy,
still in the old (but gold) Imperial units: we might need
some conversion factors. Here they are:
1 pound (lb)
= 0.45359 kilograms
1 MPH
= 0.44704 metres per second
1 foot
= 0.3048 metres
Now lets try a calculation for my glorious “Heart of
Gold” precision F2B flying wing.
W = 22 grams
mass for a Supercool 11X8.5. We gotta make this into pounds.
So W = 22
/ 1000 x 0.4539
N = 8000
RPM for my mighty Enya 45 6002
V = 60
MPH
r = 11 / 12 /
2 prop radius in feet
X = 6.5 / 12
for Heart of Gold prop-to-C/G distance in feet
R =
5 feet for F2B square turn rules
Then
F = .00043 x (22 / (1000
x .4539)) x 8000 x 60 x (11 / (12 x 2)) ^ 2 /( 6.2
/ 12 x 5 )
=
0.776 lbs
= 12.4 ozs
This means that when you
give full-up to do a square turn, there is effectively a lump
of lead on the nose weighing 12.4 ozs
trying to yaw the model out, with the same amount trying to yaw it in
on down-elevator. Yikes! The “Heart of Gold” weighs 43 ozs
and here we have a force corresponding to 29% of the weight
of the model trying to yaw it about in the square turns! Can’t
be true!
Well, that’s OK, it isn’t
true. The gyro force is there alright, but it is to some degree
counterbalanced by the “P” effect (thrustline
offset during turns) mentioned in my previous article on curing
gyro effect (the self-called Supercool Slash).
Nevertheless, it would appear
that high RPM and heavy props could be a problem. On my Eather “Firecracker”, the model requires heavy use of elevator
when using massive props, like the injection moulded APC and
Bolly props. These latter props are
twice the weight of my Supercool “petal” style F2B props.
Perhaps this is why I see
so many experts using wooden props or lightweight Eather carbon props. One day they will wake up and use the
Supercool F2B props!
So for all you thinkers and tinkeres out there, here
are some sleepless nights for you!
feedback:
Hi Stuart,
I've just finished reading your article on Gyroscopic force
calculations and have a few questions.
First... does the formula given relate to a prop or to a
solid disc like a flywheel? From the little I understand the
gyroscopic force is dependent on the distance from the centre
of rotation of the mean angular momentum (for want of a better
way to say it). In other words, for a prop wouldn't it be
the CofG of one blade x the radius of that CofG from the crankshaft?
I'd guess a typical blade has it's CofG at maybe 40% of the
blade length. Maybe "center of mass" is a better
way to say it than CofG.
Second...building on the above question, a flywheel has a
centre of mass that's distributed evenly around the entire
flywheel. A 2 bladed prop though has 2 distinct centres of
mass, one for each blade. It seems to me that precession would
vary from zero to maximum in a hard turn depending on blade
position, zero when the blades are vertical to a maximum when
the blades are horizontal. This would cycle precession twice
per rev and I'd assume give a very high frequency vibration.
I've read on SSW that a 3 bladed prop distributes each blade's
centre of mass evenly no matter what position the blades are
in and this gives a smooth precession similar to a flywheel.
This of course would mean that a 3 bladed prop would have
a non-cyclic but higher precession force than a 2 bladed prop.
Third...is that .00043 coefficient intended to make the formula
apply to a 2 bladed prop as compared to a flywheel and cover
my second question?
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