

May 22, 2012, 12:45 PM  

RPM is the big variable in the formula for gyroscopic effect and the prop radius is the big variable for the torque formula, so the questions is this... can the 8500 RPMs be the tipping point for the 14 inch prop since we know the prop size seems to be fine on lower Kv 2820's?
From my data chart, it seems that all 14 inch prop/motor/cell combinations that I have eliminated are all 14 inch props that are potentially spinning greater than 8500 RPMs at WOT. If there is some calculated validity to this, using the formulas for angular acceleration, with moment of inertia and angular momentum, then we may also be able to calculate/predict the durability of the DM2820 with a 13 inch in a setup whose potential is 9500 RPMs, which is what we're getting on 5S with the 600Kv and the 638Kv. Hmmmmmmmmmmm........... 
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May 22, 2012, 01:52 PM  

Crap!
I just did some checking and this is going to be nearly impossible to calculate because there are just too many variables that can't be predicted. The elements of torque effects vary too much with changes in flight situations. But the bottom line is that 3D flight, and the changing angle of attack with high RPM and prop diameters will significantly apply vectored torque on the rotor at amounts relative to prop diameter, RPMs, and airspeed, which is directly proportional to pitch and RPMs. I'm sure it can be done... calculating the maximum torque of a 14x7 prop at 8500 RPMs throughout the scope of angles of attack, at the maximum airspeed possible for the pitch and RPM, and compare them to the effects of a 13x6.5 prop at 9500 RPMs at the airspeed possible with this pitch and RPM. I'm wondering if we can just use the maximum values possible and compare them? Did you see the last video of Tom flying the Edge? He does a levelwing turn into a lomcevak... three or four tumbles at full throttle across the field. I do this tumble now with the Edge... no video yet but I do have one from last year with my SHP. The tumble rate is just about the same and I'll try to get some video of the Edge doing it the next time out as a comparison. Sorry for the quality but it is a phone camera:


May 22, 2012, 02:10 PM  

Quote:
stgdz... here's the 48 Edge video... high winds, 14x7 APC, 683Kv, 4S:


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May 22, 2012, 02:46 PM  

A little googling filled in some of the gaps in my memory...
Angular momentum: L = Iω I is the moment of inertia of the object and ω is the angular frequency. This is the angular equivalent of the 'normal' momentum(or inertia) that you experience when accelerating any object. So by using this formula it's easy to work out the equivalent mass that would produce the same inertia reaction acting on the prop driver when the plane changes direction. Moment of inertia has to be calculated from the geometry of the spinning object. For a disc spinning about its central axis, I = 1/12 * m * l^2 this is not strictly accurate for a prop because a prop has most of it’s weight toward the hub but using this formula is conservative and is fine for comparison of one prop to another The angular frequency can be calculated from the RPMs by: > Converting RPM to Rev/s (divide by 60), which gives you frequency; > multiplying frequency by 2π, which gives you angular frequency. Looking at the APC 14x7 (356mm dia) spinning at 8800RPM
So basically an APC 14 x7 spinning at 8800RPM is similar to having a weight of 0.36Kg (0.8lb) fastened to the end of your motor shaft, it’s the inertia of that equivalent mass that’s putting all the stress into the motor. It’s easy to compare other props against the APC 14x7, for instance a 12x6 APC spinning at 10000RPM is equivilat to a 0.23 kg mass I’ll calculate the APC 13x6.5 at 9500 RPM when take the prop off the Edge to weight it. ****Edit****.. I changed the moment of inertia calc from a disk to a rod rotating about it's centre.. this is more realistic for a prop and brings the angular momentum down a bit. 
May 22, 2012, 02:54 PM  

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May 22, 2012, 07:09 PM  

Quote:
Jim 

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May 22, 2012, 07:34 PM  
Omaha Nebraska
Joined Apr 2008
326 Posts

We are now drilling in to the bearing tube where the set screws go to prevent this from happenin. Some are fine with out, but there have been about 3 backplates that have come off recently so it will be standard now to do the modification on all motors.
Ken 
May 22, 2012, 07:41 PM  
zürich
Joined Jun 2008
1,074 Posts

Quote:
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May 22, 2012, 07:49 PM  

Quote:
will that be your setups only or will it come out of the factory like this? 

May 22, 2012, 08:17 PM  

Quote:
Jim 

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May 22, 2012, 09:07 PM  

It may help you more than you think, but here is an explantion…
If my prop weighs 37g and the radius is 178mm and I cut the radius into ten 17.8mm segments, and weighed each individually, and found that more weight is within the first two of ten segments, and then less for the next two, then more for the middle two, then it diminishes as we go toward the prop tip, then we can set up the formula like this: I = mr^2 + mr^2 + mr^2 + etc I = (.0045*.0178^2 + .0045*.0356^2 + .0033*.0564^2 + .0033*.0712^2 + .0044*.0890^2 + .0044*.1068^2 + .0033*.1246^2 + .0032*.1424^2 + .0032*.1602^2 + .0029*.1778^2) This would equal 0.000409, and when I multiply by Omega: Omega = 2 x 3.1416 x (8800 RPMs / 60 minutes) = 921.54 I get 0.38kg/m^2 I multiplied pie by 2 above because we have two blades in the 360 degree disk, which means a 2pie radiant. The idea now is to run the same calculation for the 13x6.5 APC at 9500 RPMs and see how the Angular Momentum compares. If this is all correct, and 0.8 pounds of Angular Momentum is causing the stator and the rotor to impact each other, then maybe we can predict how many RPMs are needed to duplicate the Angular Momentum on the 13 inch prop. I mean, what the heck. I think that 9500 RPMs will do it if the 13 comes in at a proportional weight, but I don't think so because the tips are going to be where the least amount of mass is. 
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