ORIGINAL: lsjpeng
. . . . I agree the out runner has higher momentum of inertia than the inrunner, but it's still small compare to the prop's that I can not feel the difference.
Luke
Well now in fairness I would point out that I=mr^2 is for a cylinder with an infinitely thin wall whereas a more accurate estimate for a thick walled cylinder of constant density would be I=m(r1^2+r2^2), and the RASA prop would be closer to 70g so I could fudge a little and bring the comparison closer to:
200g x (3cm^2 - 2cm^2) = 2600 g-cm^2
versus
70g x 50.8^2 = 15053 g-cm^2
so now we're looking at an increase of around 2000 g-cm^2 which is more than a 10% increase.
BUT way more interesting is this formula worked out in the '50s which directly estimates the forces felt on the nose:
F = .00043 x W x N x V x r x r / ( X x R )
where
W = mass of the prop
N = RPM of the prop
V = speed of flight
r = prop radius
X = distance of prop from CG
R = the radius of the turn your trying to make
unfortunately the equation was designed with a factor for lbs so we need to first convert our numbers and we get a set like this:
W = 70g = .154 lbs
N = 5000
V = 80 mph
r = 25.4 cm = 0.833 feet
X = 50 cm = 1.64
R = 2m = 6.6 feet
so throwing in some rough numbers for the prop:
F = (.00043 x .154 lbs x 5000 x 80 x .833^2)/ (1.64 x 6.6) = 1.8 lbs
now the motor:
W = 200g = 0.44 lbs
N = 5000
V = 80
r = 3 cm = 0.098
X = 47 cm = 1.54 feet
R = 6.6
so:
F = (0.00043 x 0.44 x 5000 x 80 x 0.098^2)/(1.54 x 6.6) = 0.7 lbs
The motor pushes the nose left with nearly a pound of force which is nearly half as much as the prop!
EDIT: I made a math mistake in the original estimate so the number is more like 0.07 lbs
Hmmmmm . . . . . maybe the motor does make a difference
I think the trick here is the fact the I=mr^2 is the moment of a body at rest and the rotational speed imparts a far greater added angular momentum not taken into account by the static equation.
Joe P