Power Systems for Extreme Flight Fanatics! - Page 102 - RC Groups
 May 22, 2012, 11:22 AM Volts>amps So is it coming down to rpms or prop size or a combination of both? The 638 drops the rpms by about 900. I ran 14x7 on 4s and really didn't like it. I think I was at about 7800 rpm. The 683 drives it to 8400. I understand about the black loctite. I was more concerned about the grub screws but that brings up another question. If I had a problem I would have seen some play on the stator to the grey mount plate when both grub screws were removed, correct? I didn't feel any. I think I may just stay rite where I am. I talked to ken about the setups we run and he agrees we are on the hairy edge. I offered to send the motors in at the end of the flying season for a checkup and he said sure as long as I pay for shipping Its nice to have good customer service
 May 22, 2012, 11:45 AM Registered User 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...........
 May 22, 2012, 12:34 PM Volts>amps You would also need the speed of the airframe. Also consider 10mph tailwind as worst case scenario. Figure a wall is easiest to calculate as you are going from 55 to zero in under a second. I believe doc a recorded 16g on his mxs with telemetry. What is the 14" weight difference between xoar and apce? One other variable to consider is that people like me and steve probably tumble way more than tom as its easier for us.
May 22, 2012, 12:52 PM
Registered User
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 level-wing 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:

 SHP Pop Top Tumble.wmv (0 min 49 sec)
Last edited by Aeroplayin; May 22, 2012 at 01:01 PM.
May 22, 2012, 12:57 PM
Grumpy old git.. Who me?
Quote:
 Originally Posted by stgdz You would also need the speed of the airframe. Also consider 10mph tailwind as worst case scenario.
Headwind tailwind makes no difference, that only effects the speed of the plane over the ground which is only an issue if you come into contact with the ground!. It's rate of direction change that's the issue.

along with rate of direction change moment of inertia of the prop(effected by weight and diameter) and RPM are the factors effecting gyroscopic procession.

Steve
May 22, 2012, 01:10 PM
Registered User
Quote:
 Originally Posted by JetPlaneFlyer Headwind tailwind makes no difference, that only effects the speed of the plane over the ground which is only an issue if you come into contact with the ground!. It's rate of direction change that's the issue. along with rate of direction change moment of inertia of the prop(effected by weight and diameter) and RPM are the factors effecting gyroscopic procession. Steve
Steve -- do you think we can calculate the relative procession comparison between a 14x7 at 8500 RPMs and a 13x6.5 at 9500 using airspeed calculated by pitch and RPM at the maximum AOA? Same mass.

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

 Tom Kitt - Extreme Flight Edge 540T EXP__Wind Therapy (9 min 6 sec)
 May 22, 2012, 01:46 PM Grumpy old git.. Who me? 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 8800RPMI = 1/12 * 0.037 * 0.356^2 = 0.000391 Kg m2 ω = 8800 min-1 * (1 min)/(60 s) * 2π = 921 s-1 L = 0.000391 Kg m2 * 921 s-1 = 0.36 kg m2 s-1 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. Last edited by JetPlaneFlyer; May 22, 2012 at 04:30 PM.
May 22, 2012, 01:54 PM
Registered User
Quote:
 Originally Posted by JetPlaneFlyer So basically an APC 14 x7 spinning at 8800RPM is similar to having a weight of 0.54Kg (1.2lb) 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 0.34 kg mass I’ll calculate the APC 13x6.5 at 9500 RPM when take the prop off the Edge to weight it.
Thanks for doing the legwork on this one Steve, but I think it may be valid in our comparisons and observations. If we can get a comparison between the props we use, and how fast we're accelerating them, we may at least bring the props, the cell count, the motor Kv, and the RPMs, into a relevant perspective.
May 22, 2012, 06:09 PM
Registered User
Quote:
 Originally Posted by JetPlaneFlyer ****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.
I saw that and it brings up another question... Since this is a 2-pie radiant, and since the mass is evenly distributed along the radius, I was thinking that I=2mr^2 (I = mr^2 + mr^2 = 2mr^2). There is more mass at the end of a spinning baton, but at least in this formula, the two rotating masses of equal radii are taken into consideration. I have to think about this because 2mr^2 means twice the original angular momentum.

Jim
 May 22, 2012, 06:34 PM Registered User 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, 06:41 PM
Quote:
 Originally Posted by Aeroplayin I saw that and it brings up another question... Since this is a 2-pie radiant, and since the mass is evenly distributed along the radius, I was thinking that I=2mr^2 (I = mr^2 + mr^2 = 2mr^2). There is more mass at the end of a spinning baton, but at least in this formula, the two rotating masses of equal radii are taken into consideration. I have to think about this because 2mr^2 means twice the original angular momentum. Jim
Quote:
 I = 1/12 * 0.037 * 0.356^2 = 0.000391 Kg m2 ω = 8800 min-1 * (1 min)/(60 s) * 2π = 921 s-1 L = 0.000391 Kg m2 * 921 s-1 = 0.36 kg m2 s-1
will help me to find my next setup
May 22, 2012, 06:49 PM
Quote:
 Originally Posted by Subsonic Ken 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
Ken,
will that be your setups only or will it come out of the factory like this?
May 22, 2012, 07:17 PM
Registered User
Quote:
 Originally Posted by Subsonic Ken 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
Thanks for stopping by Ken. By the way, we know that the 13x6.5 is okay for the 5S setup on the DM2028 638Kv and 600Kv, but the question is if these modifications are made, with that mean that the 14x7 APC on 4S with the 750Kv will now be okay?

Jim
May 22, 2012, 08:07 PM
Registered User
Quote:
 Originally Posted by long leng will help me to find my next setup
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 2-pie 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.
Last edited by Aeroplayin; May 22, 2012 at 11:09 PM. Reason: added all 10 segments
 May 22, 2012, 09:08 PM JKasprak ... that's sprocket not socket ... Seriously I'm glad you guys are doing all the calculations and groundwork. I'm in tune as I'm looking for a more efficient set up for cooler temps and longer flight times. My learning curve is just vertical at the moment!