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Old 09-15-2019, 04:03 PM
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Motor for new plane

I just bought a new Skyline Raven 30cc 74"overall weight 9.9-11.2lbs. I have a Rimfire 1.60 I would like to use so I dont have to buy a new motor and esc but im afraid it will be under powered. I'm new to electrics and just need to be pointed in the right direction.
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Old 09-15-2019, 05:23 PM
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That will be fine it , Will swing an 18x8 well, your setup should be very light so will be a rocket Just get the correct Esc and Battery to suit and will be fine.
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Old 09-18-2019, 10:44 AM
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Thank you!!
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Old 09-18-2019, 02:08 PM
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Quote:
Originally Posted by grizzly2279 View Post
... I'm new to electrics ...
Plane type and mass are leading when selecting motor power (in watt). Rules of thumb for electric power

Note that Kv (in rpm/volt) says absolutely nothing about motor max.power, max.current, torque, efficiency.
Kv only depends on desired rpm and battery voltage.
It does effect current and power drawn greatly though.

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Vriendeljjke groeten Ron
• Without a watt-meter you are in the dark ... until something starts to glow •
e-flight calculatorswatt-metersdiy motor tips&tricksCumulus MFC
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Old 09-18-2019, 02:10 PM
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Below an excellent quote about motor selection.
From
brushless motors Kv?.
Quote:
Originally Posted by scirocco View Post
While an absolutely critical part of the system ...
... Kv is actually the item one should choose last.
  1. Decide your peak power requirement based on the weight of the model and how you want to fly it.
  2. Pick a preferred cell count (voltage) and pack capacity for how to deliver the power.
  3. Pick a prop that will a) fit on the model and b) fly the model how you want - often as big as will fit is a good choice, but if high speed is the goal, a smaller diameter higher pitch prop will be more appropriate.
  4. Look for a size class of motors that will handle the peak power - a very conservative guide is to allow 1 gram motor weight for every 3 watts peak power.
  5. Then, look for a motor in that weight range that has the Kv to achieve the power desired with the props you can use - a calculator such as eCalc allows very quick trial and error zooming in on a decent choice. For a desired power and prop, you'd need higher Kv if using a 3 cell pack compared to a 4 cell pack. Or for a desired power and cell count, you'd need higher Kv if driving a smaller diameter high speed prop compared to a larger prop for a slow model.
The reason I suggest picking Kv last, is that prop choices have bounds - the diameter that will physically fit and the minimum size that can absorb the power you want. On the other hand, combinations of voltage and Kv are much less constrained - at least before you purchase the components.

So Kv is not a figure of merit, in that higher or lower is better, it is simply a motor characteristic that you exploit to make your power system do what you want, within the constraints you have, e.g. limited prop diameter, if it's a pusher configuration, or if you already have a bunch of 3S packs and don't want to buy more, and so on.


Minor lay-out changes by RvS
Vriendeljjke groeten Ron
• Without a watt-meter you are in the dark ... until something starts to glow •
e-flight calculatorswatt-metersdiy motor tips&tricksCumulus MFC
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Old 09-20-2019, 08:45 PM
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We've been actively analyzing electric power for the past 8 years on the RCG power thread, and I posted this about a month ago to recap some of the math.

This is extreme overkill for answering your question, but I do recognize a few people here that may like to review these concepts in one place. Plus I always do the math first before I buy motors and props and only need the meter to confirm the math.

For the benefit of everyone that may come by, those of us that have posted online calculators understand that the math looks something like this....

Concept 1 - Efficiency:

We start off with the assumption about the efficiency of our outrunner motors and go from there. Testing will tell us exactly how efficient each motor is, and this can be affected by the ESC since it is, after all, the 'controller'.

If we have a 24x10 PJN with a typical outrunner that has an efficiency difference of 80%, it means that the power we put into the system results in about 80% of that power at the shaft. This is not too bad considering that a gas engine runs at about a 35% efficiency from the fuel to the shaft.

The next thing we know is that a 190Kv motor that runs at about an 80% efficiency will take a 12S battery at nominal voltage (44.4v) and turn a recommended prop to about 6748.8 RPM... RPM = Kv * volts * 0.8.

Concept 2 - Power In vs. Power Out:

The next thing we know is that a 24x10 PJN spinning 6748.8 RPM is going to need 4250.39 Watts at the shaft in order to do this. This is the same formula you've seen here forever for Watts-out:

Watts-out =(((Prop Diameter / 12)^4) * ((RPM / 1000)^3) * prop constant * (Prop Pitch / 12))
Watts-out =(((24 / 12)^4) * ((6748.8 / 1000)^3) * 1.0372 * (10 / 12))
Watts-out = 4250.39

At this point it is important to realize that the difference between the unloaded RPM/v and the real RPM/v is 20%, or the same difference between 100% unloaded and 80% loaded.

The unloaded RPM/v is Volts times Kv.
The real RPM/v is Volts times Kv times the efficiency difference.
This also means that the difference between Watts-in and Watt-out is also the same as the difference between unloaded and loaded RPM.
4250.39 Watts-out divided by 0.80 = 5313.5 Watts-in.

So when you spin the 24x10 on a 190Kv motor that operates at an 80% efficiency, you should see 5313.5W on your meter at 44.4v.

As you can see, this also means that 5313.5W divided by 44.4v equals about 119.66A.

Concept 3 - Watts, HP, and Torque:

Now this is where most people really start to understand the relationship between Kv, power, and torque, and it's not that difficult if you can follow this logic between power in and power out.

We put 5313.5 Watts into the system in order to generate 4250.39 Watts at the shaft.
What is 4250 Watts in terms of Horsepower, or lb-ft/min?
4250W divided by the constant 745.7 gives you HP -- 5.70 HP in this case.
But this is the power at the shaft, and not the power we need to put into the system to produce this power at the shaft.

So what is 5.7 HP at the shaft in terms of torque?
Well HP is lb-ft/min and torque is just lb-ft, and torque power and horsepower actually are the same at 5252 RPM.

This is important to us in 3D because below 5252 RPM, torque power is higher than HP, and above 5252 RPM, HP is more than torque power.
A motor can accelerate a large prop through the lower RPM range faster than most gas engines, so when we need low end power to quickly accelerate large masses (propellers), we use motors.

Concept 4 - Amps & Torque:

We also know that just as HP and torque are related, Amps and torque are related too.

Since Torque = ((HP * 5252 RPM) / real RPM) then 5.7 HP at the shaft equals 4.436 lb-ft at 6748.8 RPM.

And now that we know the torque, we can also calculate for Amps and check the 119.66A we calculated above by calculating Watts and dividing by volts.

I mentioned the constant 1352.4 above and this is where that constant comes into play...

Amps = ((Torque in oz-in) / 1352.4) * Kv

To change lb-ft to oz-in, we just multiply lb-ft by 192. So....

Amps = ((4.436 lb-ft * 192) / 1352.4) * 190Kv

Amps = 119.66

So when you run your 24x10 to about 6750 RPM with your meter on the system and see about 120A and 5313W at 44.4v, you know the motor is also a true 190Kv and all is still right with the world. As with all things in our physical space, all things in our niche of rotational energy are related mathematically as well.

Concept 5 - Prop Load:

For years. the debate about the prop load constant (pK) was hot. In the end, I feel the mathematical evidence that pK is not a constant as all, but a variable dependent on RPM for the same reason that all the physical forces of a rotating mass are RPM dependent.... centripetal, centrifugal, angular momentum, torque, etc, etc.

I never really understood why the debate raged while all the time the formula was staring us in the face.

Since Watts-out equal (((Prop Diameter / 12)^4) * ((RPM / 1000)^3) * prop constant * (Prop Pitch / 12))

then....

pK equals (Watts-out / ((((Prop Diameter / 12)^4) * ((RPM / 1000)^3) * (Prop Pitch / 12)))).

And of course, Wo and Wi are related by the same efficiency as the motor's loaded RPM/v versus its unloaded RPM/v.

From testing over 160 prop and motor combinations over the past decade, I can confirm that pK is a function of torque and disk area.

Jim
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Old 09-20-2019, 08:56 PM
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So Kv is a lot more than stated in post #5 above.... unless Amps and torque are of no significance as well.... which I'm sure we can agree is not the case. It tells me what the Amps will be with any prop at any RPM. A lot of merit there, IMO.
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Old 09-21-2019, 09:15 AM
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To put the math into practical use, we can use the 30cc power we may be looking for to power the 74" Raven. This is the process I always follow for aerobatic planes, and my power systems are typically extreme --- minimum 275 W/lb with a prop no less than 27% of the wingspan.

The 27% is something I do for myself as a result of testing an is a personal preference. I feel it transitions wing loading to disk loading better when approaching stall speed at a high angle of attack for 3D maneuvers. It also develops a bigger thrust cone for better control surface authority at the same 3D stall speeds.

One of my other personal preferences is to develop a pitch speed no less than 58 mph for my 3D planes since it translates to a 70 mph level flight airspeed. I also like to keep the tip speed under Mach 0.7 in ground testing. This keeps the prop within the efficiency curve as it releases slightly in the air... motors won't release much but the batteries do hold nominal voltage better and longer.

So for a 74" aerobatic 3D/XA plane projected to come in at 11 lbs, I would be looking for a motor that can spin a 20" prop to about 3000 Watts.

A 20" prop will approach Mach 0.7 at 9000 RPM so this is not going to be a factor because I already know that this is about 1000 RPM more than what a 20" electric prop needs to generate 3000W.

If anyone has ever visited our thread on RCG, they will see that we transitioned to 12S years ago on planes from 70" and up. So when I consider powering a 74" plane, I typically don't consider anything else, but the same math applies regardless of individual preferences.

So for now, I will consider a 20" prop on 44.4v.

Since most motors in this size have tested from 79% to 82% efficiency with a few outliers on both sides, I will estimate the prop pitch and motor Kv using 81%.

At this point, you can make it easy on yourself by taking these variables to an online calculator, including my PAR Calc.

But the math is manageable as well.

Watts-out = (((Prop Diameter / 12)^4) * ((RPM / 1000)^3) * prop constant * (Prop Pitch / 12))

Watts-in = (((Prop Diameter / 12)^4) * ((RPM / 1000)^3) * prop constant * (Prop Pitch / 12)) / .81

Using a 20x8 prop, we find that at 44.4v, we need 7732 RPM to develop 3000W.

Kv = 7732 RPM / 44.4v / 0.81 KvE
Kv = 215

So a 215Kv on 44.4v at an 81% efficiency will turn a 20x8 to 7732 RPM with a pitch speed of 59 mph and a tip speed of Mach 0.60.

This is why there are so many 215Kv motors in the 600g class.

Now to check the Amps and confirm the Kv, we know that:

Watts-out = (((20 / 12)^4) * ((7732 / 1000)^3) * 1.024 * (8 / 12))
or 2434Wo

2434W / 0.81 = 3005Wi

3005Wi / 44.4v = 67.69A

But we now also know that a 20x8 spinning 7732 RPM develops 2.217 lb-ft or torque. So.....

67.69A / ((2.217 * 192) / 1352.4) = 215Kv

Finally from all of this, we can prove the interrelationship between Amps, Torque, and Kv by developing a formula for Amps using Kv and torque.

Amps = (((((((((Prop Diam / 12)^4) * ((RPM / 1000)^3) * pK * (Prop Pitch / 12)) / 745.7) * 5252) / RPM) * 192) / 1352.4) * Kv)

The product here is 67.69A.
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Old 09-21-2019, 09:48 AM
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Sorry to keep going here but another important math element is flight time and capacity now that we know everything else.

As a rule, I always try to keep my discharge rate under 25C and try hard to get below 22C to keep everything cool and batteries healthy as long as possible under the extreme conditions of 3D and XA..

Using common battery capacities, 67.69A is a 20.5C with a 3300mAh pack.

67.69A / 3.3Ah = 20.5C

60 minutes / 20.5C = 2.927 minute flight time at WOT to zero capacity.
2.927 mins times 80% capacity is 2.341 minutes at WOT.
2.341 minutes at a 40% power average is almost 6 minutes.

My personal power average is about 35-37 percent.
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