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Sep 22, 2019, 03:42 PM
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Extreme Electric Power - The Math


We've been actively analyzing electric power for the past 8 years on the RCG Power Systems for Extreme Flight Fanatics! thread, and I posted this about a month ago to recap some of the mathematical concepts that interrelate propeller specifications and load ratings with motor Kv, voltage, power-in, power-out, torque, and current.

For the benefit of anyone that may come by this blog, a basic understanding of algebra is all that's needed to understand these concepts. Those of us that have posted online calculators understand that the mathematical concepts that relate with each other are as follows:

Concept 1 - Efficiency:

We start off with an assumption about the efficiency of our outrunner motors and go from there. Testing will tell us exactly how efficient each motor is by comparing unloaded RPM/v against RPM/v with a recommended load. This can be affected by the ESC programming settings since it is, after all, the 'speed controller'. But if we test each example with the same controller and settings, and stick with the recommended settings, we should end up with a fair and relative comparison of motor and propeller performance.

As an example of the process of determining the Kv efficiency between loaded and unloaded (KvE), I will use a common 50cc power systems. If we have a 24x10 PJN with a typical 980 to 1180 gram 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.

In real testing, I have found this efficiency to be anywhere from 76.5% to 83.8% at nominal voltage with outrunners weighing from 152g to 1520g turning the recommended propellers at nominal 3.7 volts per cell with LiPO batteries.

This is not too bad considering that a gas engine runs at about a 35% efficiency from the fuel to the shaft.

Therefore, a 190Kv motor that runs at about an 80% Kv efficiency (KvE) will take a 12S battery at nominal voltage (44.4v) and turn a recommended prop to about 6748.8 RPM.

RPM = Kv * volts * KvE

Concept 2 - Power-In vs. Power-Out:

We can calculate that a 24x10 PJN spinning 6748.8 RPM is going to need 4250.39 Watts at the shaft. This is a very typical formula for Watts-out (Wo):

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

So we take the propeller diameter in inches converted to feet and raised to the 4th power. We then take the RPM and divide it by 1000 and raise it to the 3rd power. This tells us that the prop diameter and how fast it is spun is the most important factor in determining the power requirements.

The prop load constant will increase the product of these two values by as much as 2 to 6 percent, on average.

The pitch of the prop, like the prop diameter, is also divided by 12 inches. This means that the prop pitch has no significance to the product of the formula at 12, can reduce the product of the three significantly if lower than 12, or raise it significantly if higher than 12.

If I take off the 8-pitch and replace it with a 12-pitch, I will require 50% more power to turn the prop to the same RPM.

Watts-out = (((24" / 12")^4) * ((6748.8 RPM / 1000)^3) * 1.0372 pK * (8" / 12"))

Watts-out = 3400.32

Watts-out = (((24" / 12")^4) * ((6748.8 RPM / 1000)^3) * 1.0372 pK * (12" / 12"))

Watts-out = 5100.47

(5100.47Wo - 3400.32Wo) / 3400.32Wo = 50%

At this point it is important to remember that the difference between the unloaded RPM/v and the real RPM/v is 20%, or the same difference between 100% of the product of the voltage times Kv when unloaded and 80% of the product of the voltage times Kv when unloaded if 80% is the real tested difference. This value will change from one motor brand to another and as I mentioned before, has ranged from 76% to 84%.

The unloaded RPM/v is within the statistical proximity of Volts times Kv.
The real RPM/v is within the statistical proximity Volts times Kv times the efficiency difference.

This means that the difference between Watts-in and Watt-out is also the same as the difference between unloaded and loaded RPM.

So in the example above with the 24x12 to 6748.8 RPM, 5100.47 Watts-out divided by 0.80 = 6375.6 Watts-in.

So when you spin the 24x12 on a 190Kv motor that operates at an 80% efficiency, you should see about 6749 RPM on the tach and about 6375.6W on your meter at 44.4v.

As you can see, this also means that 6375.6W divided by 44.4v equals about 143.59A.

Concept 3 - Watts, HP, and Torque:

Now this is where most people really start to understand the important 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 6375.6 Watts into the system in order to generate 5100.47 Watts at the shaft.

What is 5100.47 Wo in terms of Horsepower, or lb-ft/min?

5100.47W divided by the constant 745.7 gives you HP, or 6.84 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 6.84 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 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 it comes to delivering low end power to quickly accelerate large propellers, electric motors have the advantage.

In high performance automobiles, Formula-1 racing teams use large batteries to power two important electric motors. One assists the turbo at low engine RPM, and one assists the drive train by providing the much-needed torque power that helps accelerate the cars out of slow turns during periods of low engine RPM.

Concept 4 - Amps & Torque:

We also know that just as Watts and HP are related, and HP and torque are related, Amps and torque must also be related too. But since Wi and Wo at a specific RPM under load is a function of the Kv efficiency at a specific voltage, there must be a relationship between the Wi, Amps, torque, and Kv as well. This also means a direct relationship between Amps and Kv.

Since Torque = ((HP * 5252 RPM) / real RPM) then 6.84 HP at the shaft equals 5.3234 lb-ft at 6748.8 RPM.

And now that we know the torque, we can also calculate for Amps and check the 143.59A we calculated above from Watts by dividing by the volts.

The torque constant used to relate torque with current is 1352.4, and this is the formula used:

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

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

Amps = ((5.3234 lb-ft * 192) / 1352.4) * 190Kv

Amps = 143.59

So when you run your 24x12 to about 6750 RPM with your meter on the system and see about 143.6A and 6376W 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 propeller driven aircraft using rotational energy are related mathematically as well.

Concept 5 - Prop Load Constant:

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

then....

pK = (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, angular momentum, and disk area.

pK = (((((1 / 12) * (Prop weight in grams / 1000)) * ((((Propeller Diameter * 25.4) / 1000)^2)) * (2 * 3.1416 * ((Kv * volts * Kv Efficiency) / 60)) / 0.95) / 100) + (1.01 + ((1 / 12) * (Propeller Weight / 1000) * (((Propeller Diameter * 25.4) / 1000)^2))))

Concept 6 - Measured Thrust:

The thrust that I refer to here is not static thrust since static thrust requires the advance ratio element to estimate thrust at a specific airspeed. The thrust I refer to is instead the amount of pull the power system will have on a weight scale. Although it is another mathematical element that can be confirmed with a scale, it seems important to me to use a formula to produce a relative thrust value that is void of any other dynamic environmental variation, like air temp and altitude.

I developed a formula for this and have been able to confirm the results here in Florida as close to sea level and a constant 85 to 90 degrees temp Fahrenheit or 30 to 32 degrees Celsius.

This formula is a function of torque and disk area.

Thrust = (Torque (lb-ft) * ((600 (k) - (25 (inches) - Propeller Diameter (inches)) * 10) / (Propeller Diameter (inches) * 3.14159)))

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.

(All propeller dimensions are in inches)

Amps = (((((((((Prop Diameter / 12)^4) * ((RPM / 1000)^3) * pK * (Prop Pitch / 12)) / 745.7) * 5252 RPM) / RPM) * 192 (oz-in conversion)) / 1352.4 (k)) * Kv)

Thrust = ((((((Prop Diameter / 12)^4) * ((RPM / 1000)^3) * pK * (Propeller Pitch / 12)) / 745.7) * 5252 RPM) / RPM) * (((600 - (25 - Propeller Diameter) * 10) / (Propeller Diameter * 3.14159)))


You can also substitute the pK formula for the pK above and make one big formula, and perhaps the next step would be to condense this formula further by combining redundant elements.

Concept 7 - Flight Time:

Calculating for a safe flight time for electric RC airplanes has become less of an issue now that telemetry has become more widely used. But for most pilots, flight time is relative to three factors -- the pilot's personal and subjective power average (throttle control), the maximum current of the power system at WOT, and the capacity of the batteries being used.

The pilot's personal power average is a component of flight time that seems to rarely change for each pilot from one plane to another regardless of the size. In most cases, this also remains the same over the years. Once a person feels comfortable with a plane, they generally tend to use similar power inputs. This, of course, is a sweeping generalization, but for myself, my son, and my flying buddies, it seems to be the case.

The math behind this is as follows:

Flight Time:

Take a fully charged pack and fly normally for about 3 minutes.
Recharge the battery and record the capacity replaced to the battery.
Let's say I'm using a 12S 3300mAh main pack and replace 1320mAh after a safe 3 minute flight to test my capacity usage.
1320mAh divided by 3 minutes is 440mAh per minute.
3300mAh times a safe 80% capacity is 2640mAh of usable capacity.
2640mAh divided by 440mAh per minute equals 6 minutes of a safe flying time to 20% capacity.

Power Average:

Using a Watts meter, check the Amps on your power system at wide open throttle (WOT) in a ground test.
Let's say the WOT current is 70A.
3300mAh is the same as 3.3Ah.
70A divided by 3.3Ah is a 21.21 A/Ah discharge rate.
21.21 A/Ah is the same as 21.21C.
60 minutes divided by 21.21 A/Ah equals 2.829 minutes to zero capacity at WOT or 70A.
2.829 minutes to zero capacity times 0.80 equals 2.236 minutes to a safe 20% capacity.

In order to achieve a safe 6 minute flight time from a system that will discharge to 20% capacity in 2.236 minutes, then the current must be lower than 70A by some percentage.

The average power required by our throttle control must be 2.236 minutes divided by 6 minutes.... 37.27%.

So if I know that my power average is approximately 37%:

70A / 3.3Ah = 21.21C
60 mins / 21.21C = 2.829 minutes
2.829 minutes * 80% capacity = 2.263 mins at WOT
2.263 minutes / 0.3727 = a 6 minute normal flight

My power average is about 35% and my son's is about 40% and this does not change much from 48" planes flown with 4S packs all the way up to 102" planes flow with 14S packs -- mainly because we seem to fly mostly aerobatic 3D planes and generally fly them all the same way.

Of course, this concept topic is highly subjective and when practicing slow maneuvers that require much less power, such as Elevators, flight times will increase a lot. This is a solid argument for using telemetry to monitor battery capacity as it enables us to gain the most time from our battery packs in a very safe way.

The variables you will need to know to calculate flight time are as follows:

Your personal power average (typically between 33% and 40% for low and slow 3D).
Peaks Amps of the power system (A slow and steady throttle-up to WOT is the best way to read with from a meter or your telemetry).
Total battery capacity being used.
The safe capacity you will be leaving in the batteries at the end of the flight (20% being the safe minimum).

For our example, let's use the following:

Personal power average = 40%
Peaks Amps of the power system = 125A
Total battery capacity being used = 5.5Ah (5500mAh)
The safe capacity you will be leaving in the batteries at the end of the flight = 20%

With this information, we calculate the A/Ah, which is the C discharge.

125A / (5500mAh /1000) = 22.727C

We then divide 60 minutes to an hour by A/Ah to get the minutes of flight time at WOT to zero capacity.

60 minutes / 22.727C = 2.64 minutes to zero capacity.

We then multiply the minutes to zero capacity by the 80% we will be using.

2.64 minutes * 0.80 = 2.112 minutes to a 20% remaining capacity.

If our personal power average is 40%, then we divide the minutes to 20% capacity by 40%.

2.112 minutes / 0.40 = 5.28 minutes to 20% capacity.

The complete formula looks like this:

Flight Time = (((60 mins / (Peak Amps / (total mAh /1000))) * Total Capacity % being used) / Personal Power % Average)

Flight Time = (((60 mins / (125A / (5500mAh / 1000))) * 0.80 capacity) / 0.40 power average)

Flight Time = 5.28 minutes to 20% capacity

Concept 8 - Propulsive Force:

Electric powered propeller-driven systems have been defined many times and for many years by the Watts product of Amps and voltage.

The issue I have with this is that measured Watts is the total power required to drive the propeller load at the shaft and not the actual power developed by the propulsion system as a fractional result of the total input power. This means that a less efficient system that requires 5000 Watts-in to develop 3800 Watts-out has the same propulsive power as a more efficient system that requires only 4200 Watts-in to develop the same 3800 Watts-out. I would rather not represent the less efficient motor as generating more propulsive force than the more efficient motor.

For this reason, the combination of thrust, speed, and Watts-out, is a more accurate representation of propulsive power.

For example:

A gear reduction drive system of 3.33:1 turning a 178g 28x12 beechwood propeller with 44.4 volts with a 525Kv motor operates at a 90.09% efficiency and will therefore develop 6300 RPM.

The Watts-in will be the product of the 198.17A and the 44.4v or about 8800 Watts-in. But the power at the shaft does not consider the entirety of the energy being used or dissipated by other circuit components, such as heat, resistance, capacitance, impedance, magnetic field, noise, etc.

The Watts-out is in fact 7928 Watts and 10.63 HP of usable lb-ft/min and 8.86 lb-ft of torque.

7928 Watts-out will also be the same power at the shaft as a non-geared 157.5Kv outrunner operating at only about 80% efficiency at 44.4v and 223.2A for 9908.73 Watts-in.

Therefore, comparing the two systems that are turning the same prop to the same RPM and developing the same thrust and airspeed by implying that the less efficient outrunner is developing more propulsive force seems counter-intuitive. They are both developing the same propulsive force, and one is doing so with less input energy requirement.

Therefore, the propulsive force should be the product of thrust, and airspeed, making both systems equal and relative to the force being developed.

It has been widely accepted that Propulsive Force is the product of Thrust and Airspeed.

Thrust * Airspeed = Propulsive Force

So in this specific case 38.414 m/s * 28.78 kg = 1106 kg-m/s

Note here that Airspeed conversion from the calculated pitch speed is 1.2003:1 which to me is an arbitrary value because it reflects an advance ratio that is never constant due to the angle of attack and air conditions. Airspeed is always a value that requires measurement and I use it only to be consistent to the units involved in the formula.

I would personally prefer it if pitch speed was the standard unit since there is nothing arbitrary about it. Ground testing with zero advance ratio is normally the way we compare systems anyway, and we should expect that these comparisons will translate to in-air performance in a very relative way.

Now this is where the interrelationship of all the formulas and all the calculated results steps up to prove to us that we are using the correct methodology. All values throughout all formulas must be consistent. Consider the following:

The total energy of the system is supposed to be calculated by using the ratio between the Wi and Wo values. In this example, it's the margin between the 90% efficiency (really 90.09%) of the geared system and the 80.00% efficiency (really 79.03%) of the non-geared outrunner.

This efficiency comparison would for one, be valuable in determining how much more flight time I would get from the more efficient power system using the same battery capacity. It would also perhaps provide insight into how the added stress of the less efficient systems may affect the battery cycle life. But first, it must be correct.

The actual straight through physical conversion of Watts to kg-m/s is a constant 0.1019716213.

7927.8W * 0.1019716213 = 808.408 kg-m/s

In order to bring the original 1106 kg-m/s calculated above using Airspeed down to 808.408 kg-m/s converted normally from 7927.8W, I find that I must use the pitch speed instead.

The calculated pitch speed of 32.004 m/s * 28.8 kg of thrust = 921.465 kg-m/s.

The difference between 921.465 kg-m/s and 808.408 kg-m/s is the 12.3% difference between 90.09% KvE and the 79.03% KvE.

1 - (0.7903 / 0.9009) * 100 = 12.296%

Therefore, for the math to successfully quantify the interrelationship of all the power values, the pitch speed must be used in place of the Airspeed.

Propulsive Force (kg-m/s) = ( (Thrust in lbs * 2.2046) * (Pitch Speed in MPH * 0.4704))

Concept 9 - Tip Speed:

Propeller efficiency will be affected by rotational speed the same way any airfoil's efficiency will be affected by airspeed. Compressibility and the vortex produced at the propeller tips are both phenomena that will quickly reduce the operational efficiency of the airfoil. Since the tips of the propeller rotate at a speed faster than the rest of the propeller, we monitor the tip speed to determine if we are approaching the point where resulting vortexes and compressibility will eliminate portions of the propeller from the overall maximum performance.

It is widely accepted that an efficiency threshold is approached at a Mach 0.70 tip speeds. So when calculating the prop size and maximum RPM we can use to maximize Propulsive Force without diminishing the system's efficiency, I personally use the 0.68 Mach number.

In the example from Concept 8, 6300 RPM with a 28" propeller is a tip speed of Mach 0.682.

Tip Speed in Mach values = ((((((Prop Diameter * 3.1416) / 12") * RPM) / 5280 ft/mile) * 60 mins) / 770 mph)

Jim Kitt
Last edited by Aeroplayin; Oct 16, 2019 at 07:28 AM.
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Sep 23, 2019, 01:35 AM
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Regarding efficiency I wish to point out that, even if it is true that for a combustion motor is 35% and for electric motor 80%, if you produce electricity with a fuel (carbon, oil or gas) the efficiency of the transformation is at the best 50%, than you have to consider the losses for transport, voltage transformers, charging and discharging process of batteries, only after the one of the motor, so at the end I do not know which one is best ( for the total system and entropy increase). And this is valid also for electric cars, that everybody believe not to pollute; that is true only if charging is made with electricity produced with sun, wind and water (and also in this case you must consider the energy used and pollution created to produce the necessary devices versus the energy you get in their life). I was forgetting the huge pollution involved in the production and disposal of the batteries.
Sep 23, 2019, 09:51 AM
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Yes, there is a lot to consider with regard to where carbon based fuel and electrical energy comes from and then what happens after the fact -- the clean-up. Just as carbon fuels are needed to charge the batteries, and lithium based batteries will need a special way to be processed when disposed of, the raw material for fossil fuels also need to be found, drilled, transported, and processed as well, then stored and distributed of course. It's a much bigger picture for both.

Even with solar energy charging batteries, imagine the lithium waste if every form of energy throughout the world required this type of energy storage.

One of my degrees is in marine and environmental science, but I always suggest that people try to get their science-based information about environmental impact from peer reviewed journals as opposed to biased media based outlets. We should use the real science that is peer reviewed to make decisions on the future of energy. Peer reviewed data means that the results of the study have been duplicated by others who are both professionally and academically dedicated to the discipline.
Sep 24, 2019, 07:24 AM
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Do not forget lobbies that duplicate data bending them to their needs (profits)
Sep 24, 2019, 09:27 AM
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I really didn't want this blog to meander into the political realm and I prefer to stick with the mathematical relativity between volts, Amps, power-in, power-out, thrust, torque, angular momentum, load, disk area, and the Kv efficiency. But I guess any time electrical power as a substitute to engines using fossil fuels is discussed or suggested, we may have to address the two even though it may tend to invite others to debate the legitimacy of environmental impact of each or both. If it does digress in that direction, I will have to monitor and manage posts that are not related to mathematical relevancy.

With that said, I was part of several post graduate programs and research funded by government grants, and I know for a fact that it is easier to receive grant funding if the study's hypothesis is in line with the political agenda. But in the end, published findings must still pass the smell test.

What I have seen frequently, that bothers me a lot, is when pseudo scientists write an article for mainstream media and take charts, graphs, and quotes, from peer reviewed studies, and use them to reinforce their personal slant and bias to the conclusions being drawn by the article. This can happen on both sides of the debate. Sighting peer reviewed studies gives their conclusions legitimacy. But when you read the actual study in the journals, the study's results by no means come to the same conclusions. This is no different than Selection Bias, which is what most reviewers look for in scientific studies to begin with.

For me, electric power as a substitute to gas engines in my high performance 3D RC aircraft was a consideration because of two reasons -- convenience and power. No cleanup is needed, the airframes and components don't take a beating from the vibration of pulsing pistons, and it's safer for those that don't have help at the fields they fly from since we never have to stand anywhere near the spinning propeller.

When stall speeds are approached in 3D maneuvers, transitioning from wing area for lift to disk area for lift is a requirement. A prop of at least 25% of the wingspan is usually the minimum requirement to provide thrust over the ailerons, elevators, and rudder, to maintain control of the plane in stall speeds. Most 3D pilots would agree that 26% to 28% is even better, and this requires more low end torque.

If we are going to use a power system with similar weight in order to keep the center of gravity and static margin the same, we typically have to trade off some of the top end horsepower and airspeed with low end torque power. Accelerating a big prop disk with less pitch from zero RPM through a function RPM zone on an RC airplane instead of accelerating a smaller disk with a deeper pitch to a very high RPM means more power and control for low and slow 3D maneuvers. This requires more torque.

For this reason, bigger props and electrical energy go hand-in-hand.

For years, people have debates the limits of electric motor power for RC airplanes. But I have been working with many other pilots recently to help power their 28 to 36 pound airplanes to 300 Watts per pound and a 2:1 thrust to weight ratio with very good air speeds exceeding 90 mph. In some cases, 9 to 10 minute flight times are achieved -- which reminds me that a Concept 7 should be added for the relationship between flight time, Amps, pilot power average, and capacity. I will add that soon.

Edit: Flight Time has been added.
Last edited by Aeroplayin; Sep 24, 2019 at 11:27 AM.
Oct 05, 2019, 10:57 AM
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Two more concepts have been added to the first post.


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