2Sunny

So just got back from some intense discussions of motor technology and one of the guys asked an interesting question:

Does the increased mass of an outrunner provide increased inertia that helps stabilize the nose of the plane? If inertia is the product of mass and velocity then it makes sense that an outrunner would have significantly greater inertial stability than an inrunner. Heck . . . for that matter the rules prohibit gyroscopic stabilizers right? And here we have a 1 lb+ gyroscopic stabilizer built in?

So what say all you geeky engineering types?


Joe P.

can773

My Feedback: (1)

The vast majority of the mass of the outrunner is stationary. The prop is by far and wide a bigger contributor to any gyroscopic forces I would think.

Jetdesign

My Feedback: (8)

I was asking about this in another thread. If I understand you correctly, you're wondering about the overall extra weight in the nose? Or the spinning mass of the motor?

If it's the increased weight in the nose, the result would appear to be a more stable plane; it would take more energy to get the plane to change it's attitude, and take more energy to stop it. While this would appear smoother, it would be more difficult to control with precision as the plane would resist the changes you are telling it to make with the sticks.

I had a hard time ending loops on a perfect horizontal line with my Venus; it had a heavy motor (for that size plane) and 3 full size servos mounted in the tail. I originally thought the issue was the speed of the elevator servos, but after talking about this I think it was the inertia of the plane.

If you're talking about the spinning mass of the motor, I'd have to think about that one...

lsjpeng

Don't guess, let's calculate.

Outer runner: the rotor can be treated as a cylinder. So the momentum of inertia is mr^2 in good approximation
The rotor is ~200g, radius ~3cm ==> I=200x3^2=1800 g-cm^2

For propeller, we can use the stick approximation. So the momentum of inertia is (mL^2)/12
20" prop is ~50.8cm, ~100g ==> I=(100x50.8^2)/12=21,505 g-cm^2

Therefore, momentum of inertia of rotor is ~8% of the propeller. Is it significant?

Luke

patternflyer1

My Feedback: (11)

Luke, you are my fatha.. But I have no idea what you are talking about.. lol

C

ExFokkerFlyer

My Feedback: (4)

Don't worry about it Chris, I'm pretty sure that was Greek...

T

riot3d

Wax on, wax off, Grasshopper ...

2Sunny

Luke you have the force for sure dude Thanks for bringing us right to the heart of the matter!



. . . . but the original question was whether or not there is a significant difference between an outrunner and an inrunner as all planes have a prop so they all benefit equally from its stabilizing force when we try to hold a line or its destabilizng force when we want to move that momentous spinning stick out there. The real question I'm driving at is whether or not the outrunner has an as yet to be discussed additional advantage, and I would argue that the spinning ring of magnets outside the windings on the Pletty has an approximate radius of 3 cm whilst the much smaller and lighter ring of spinning magnets inside the windings on the C50 has a radius of about 1.5 cm so the approximate comparison of the moments of inertia might look more like:



Of course . . . . as much fun as all that is to think about . . . I promise this is one grasshopper who will continue to "Wax on; wax off" once all the snow melts. Problem is there's only so much simulator time I can take so I find way too much idle time to ponder silly questions


Snow Bound in NY - Joe P





EDIT: For anyone who read my initial post I fixed my mistake of using d instead of r

J Lachowski

My Feedback: (46)

My head hurts.

2Sunny

WARNING: More fun ramblings related to physics below . . . This is NOT for everyone and will NOT make you a better pilot . . .




. . . . . of course one might argue that adding moment of inertia to the nose of an airplane is not helpful but rather quite the contrary since a gyroscope will oppose changes in the position of it's axis of rotation with an equal force in a direction at 90 degrees to the change. So although the added moment of the outrunner may help stabilize the plane in level flight, once one attempts to pull that nose around in a loop it also increases the need to use right rudder to counteract that gyroscopic effect, and that is definitely NOT helpful for me since I have a hard enough time simply making a symmetrical loop while using the elevator, let alone worrying about the darn rudder. Anyways just more fun thoughts . . .

J Lachowski

My Feedback: (46)

On second thought, it is pounding

lsjpeng

Hi Joe:

It is not an issue for model airplane, at least not for my fingers.... may be those top F3A pilots can tell the difference.

The major gyro effect comes from the prop. It was corrected or smooth out by pattern plane design..... the longitudinal stability. For typical patternship, the tail moment (CG to elev. hinge) and nose moment (prop hub to CG) ratio is 2 or higher, huge side area etc. that's why our patternship flies so smooth and so stable compare with those hotdog 3D planes. Theoretically, the out runner rotor has higher momentum of inertia than the inrunner, but it is over shadowed by the prop's contribution.

You can do an experiment:
Remove the prop from your planes. Hold your plane, start motor to full throttle. pitch up from horizontal to vertical quickly. Do you feel the head tends to pull left? Repeat for both of out runner and inrunner, compare the difference.
Now is the fun part; put your prop back, repeat the same steps (watch your fingers.... I do not take any responsibility if you cut your fingers.. ), how about the side pull this time, much stronger, right? Do you think w/o prop is still an issue?

Theoretically (Chris, it's Greek again....), if prop tips wobble during the transition (quick pitch moment), the transition is supposed not so smooth as a rigid prop. If the prop rotation plane is somewhat flexible (blade is rigid, but the spinning prop "disk" is flexible) the transition will be even more smooth. Again, as a sandbag in Master class for more than 11 years, I can not feel the difference.[&o]

I guess, whole this discussion was originated from the WWI rotary engine. It was in a totally different scale: Mass ratio of rotary engine to the prop is way much higher than in our case. Diameter of rotary engine and prop size is almost 1:2, much smaller than ours. The contribution from the rotating engine is around 30-50% of prop's (OK, Chris, no more physics calculation ).

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. Don't think too much, just enjoy and fly!

Luke

DaveL322

I think Luke is on target with this one. I've flown the same plane with the same prop at the same RPM with both geared inrunner and direct drive outrunner, and couldn't tell the difference in terms of trim. The same plane with a larger prop does make a difference - increased gyroscopics (90 degree reaction to input commands) are noticeable. All other things equal, reducing the peak gyroscopic load is a good thing (reduced prop mass, or hinged hub, or reduced diameter at same weight), but the benefit is not easy to find on many planes/setups, especially if flying in wind.

Regards,

Dave

2Sunny

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:



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


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

lsjpeng

Dear Joe:

I teach physics in college for years. This calculation is meaning nothing to me. 8% to 10% even 12% doesn't matter since they are all approximation at very begining. Do the experiment then come back we can disucss more. Thanks!

Luke

2Sunny

And I suppose if I'm really gonna take this all the way we should make the calculation for an inrunner. So here it goes:


F = .00043 x W x N x V x r x r / ( X x R )

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


W = 150g = 0.33 lbs
N = 5000
V = 80
r = 2 cm = 0.066
X = 47 cm = 1.54
R = 2 m = 6.6 feet

F = (0.00043 x 0.33 x 5000 x 80 x 0.066^2)/(1.54 x 6.6) = 0.0245 lbs or less than an ounce!





But then that's the opposite of my opening premise because I would say the fact that the outrunner makes it harder to hold a nice radius is a detriment NOT a plus!

2Sunny

Luke,


a) I'm just having fun so please don't take it personally.
b) You obviously did not read the whole post because you missed the entire discussion of the actual angular force exerted being 1.8 lbs for the prop versus .7lbs for the motor


Joe


EDIT: Obviously if you read below the .7 lbs turned out to be wrong so I offer my apologies for dragging this out, and simply throw myself at the mercy of the crowd.

Scott Smith

My Feedback: (4)

I'm coming up with .07 for the Outrunner...

And don't forget the inrunner is turning 6.7 * prop rpm (for the C50)

can773

My Feedback: (1)

I think you need to increase your R to about 40 or 50 feet.....I don't see too many 6' radii in a sequence.

2Sunny

Ohh drat and double drat


Caught in a common math error!

I liked the answer better when it was .7lbs not .07 lbs!


I guess I stand corrected and offer up my head on a silver platter for public humiliation


BUT I sure had fun gettin' there!





Joe

lsjpeng

Hi Joe:

I did read through every detail. I know you enjoy doing this.
In my flying skill level, I don't feel the difference. That's what I want to say. If you feel it affects, change your trim to correct it. Not a big deal. Sorry for making you feel personal.
Have fun!

Luke

2Sunny

Scott,

Thanks for keepin' me honest. I shoulda used a spread sheet too instead of the Mac wiki calculator


Joe

Tall Paul

I'd like to see a good test using the common inrunners and outrunners on the large pattern planes.. using the good old Prony Brake to check the rotational forces between the two.
My motors are piddling 400s, and I will never get the motor$ used in these planes, but surely someone has both.