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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 1A - Known Efficiency:

If we have the motor and propeller in-hand, the actually testing and collection of the power and current variables is simple to do. We can also test for the motor's efficiency so that we can calculate the power and current variables with different size props without having to buy expensive propellers that may not provide the performance we have in mind.

To do this, we simply follow this procedure with the one of the recommended propeller sizes we do have in-hand along with a volt meter and tach.
  1. With no prop, run the motor slowly to wide open throttle (WOT)
  2. Don't worry about running a motor to WOT unloaded. I've done this no fewer than 140 times and never lost a motor.
  3. Record the voltage and RPM at WOT
  4. This will be your unloaded RPM/v (URPM/v)
  5. Run the motor with the recommended prop size slowly to WOT
  6. Record the voltage and RPM at WOT
  7. This will be your loaded RPM/v (LRPM/v)
  8. Divide the LRPM/v by the URPM/v
  9. This will be your Kv efficiency
Calculate the Watts-out using this formula:
Watts-out = (((Prop Diameter / 12)^4) * ((RPM / 1000)^3) * prop constant * (Prop Pitch / 12))

Calculate the Watts-in using this formula:
Watts-in = Watts-out / Kv efficiency

Calculate the Amps using this formula:
Amps = Watts-in / recorded Voltage

To check this using the relationship between torque, current, and Kv, first calculate HP at the shaft using this formula:
HP = Watts-out / 745.7

Then calculate Torque using HP:
Torque in lb-ft = (HP * 5252) / LRPM

From torque and KV, we can calculate the Amps:
Amps = ((Torque * 192) / 1352.4) * URPM/v

In the Amps formula we are using above, the 192 will convert lb-ft to oz-in which is required for correct results with this formula. The 1352.4 value is the conversion constant that is the relationship between current and torque of the prop at the loaded RPM compared to the unloaded RPM/v.

Now that we have the Kv efficiency of this specific motor, we can use it to calculate current and power for any other viable propeller we may want to consider without having to buy it.

The only missing variable for these calculations is the prop constant (pK). This is also a calculated field by I have addressed that formula separately in Concept 5 below.

This testing and the resulting calculations 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.

Concept 1B -Unknown Efficiency:

If we do not have the motor and propeller, but want to estimate what motor size, motor Kv, and propeller size, will provide us with the power we want, we must start off with an assumption about the efficiency of our outrunner motors.

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 and only advance fuel formula can boast near 50% efficiency.

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

When estimating KvE for outrunners, selecting from the efficiency rage of 76% to 82% will require some subjectivity, but it is fair to say that both smaller and larger motors will typically be on the low end of this range while mid-sized motors in the 950 to 1200 gram range will perform at the top end of the range. I have seen some motors run to 70% when heavily loaded, and I have seen mid and larger motors run to 86% efficiency when very lightly loaded. But heavily loading motors will make it less probable that you will hold nominal voltage (3.7 volts per cell) for more than a few second, and underloading motors will result in higher voltage for longer amounts of time. More volts means more RPM, and more RPM with the same prop means more power.

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))


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.

If you do not yet know your RPM for this formula, you can accurately -- or at least effectively -- predict that too if your Kv is accurate relative to RPM/v. If so, then you can use this formula:

Amps = (((((((((Prop Diameter in inches / 12)^4) * (((Kv * 3.7 * LiPO cells * 0.803) / 1000)^3) * prop constant * (Prop Pitch in inches / 12)) / 745.7) * 5252) / (Kv * 3.7 * LiPO cells * 0.803)) * 192) / 1352.4) * Kv)

If you are using a gear reduction unit with your motor, then the motor Kv will have to be adjusted to the reduced Kv, and the 0.803 in the formula will need to be changed to 0.901.

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; Feb 07, 2020 at 02:39 PM.
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Oct 05, 2019, 10:57 AM
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Mechanical Advantage and Machine Efficiency

In physics, it is important to have our terminology defined because outside of physics, words are easily and often confused. As with any professional discipline, how we define the language we use is important. I cannot emphasize enough how the independent definition of Force, Work, Energy, and Power, should be understood and not confused.

For this specific topic about the advantage of gear reduction, there are some concepts that need to be defined first.

The first law of thermodynamics is about the conservation of energy. There is a finite amount of Energy in the universe and it cannot be created or destroyed. But Energy can be transferred, and as humans, we have always been looking for ways to transfer energy as efficiently as possible. Tools we design and develop also help us enhance our own strength to do things, like move big or heavy objects we would otherwise not be able to without these tools.

The type of tools that help us do this provide something we call mechanical advantage. They are tools like levers, pulleys, and gear reducers. They are applied in common tools that enhance strength, like wrenches, hammers, scissors, golf clubs, hand cranks, pliers, bicycle gears, car transmission, etc.

But if we only have a certain amount of strength and energy, and we cannot create new energy to help us, then how do these tools enhance our strength to move the object we never could without them?

The answer to this question defines how the mechanical advantage these tools apply works. But we must define the words that we are using and define how they are different.

It is Energy that cannot be created from nothing or destroyed.
In physics, Energy is defined as the capacity to do Work.
Work is defined as the transfer of energy.
Power is the amount of Work done per unit if time.
Force is mass times acceleration and represents mass in motion.

The most important concepts of these definitions have to do with the fact that we are not creating energy -- only transferring it in the form of Work. But since Work is the transfer of energy, and we cannot create any new energy, then how are we able to move a big object without increasing energy?

The answer to this question is where Force comes in. Force represents mass in motion and is the product of mass times acceleration. So, for Force to be a value greater than zero, it must have motion.

As it turns out, Work is also defined as the product of Force times distance. If we think about this, what we are really saying is that:

Work is the transfer of energy.
Energy cannot be created during this transfer.
The same Energy applied to the system must equal the resulting Energy.
This also means that the same Work applied to the system MUST equal the resulting Work.
Work is the product of Force times distance.

With this in mind, let's think about what is happening when we apply energy to a common tool, like a lever. Levers help us move masses easier than if we tried to move the mass without it. Some objects that just won't budge when we try to lift them ourselves, seem easy to move when we apply the use of this simple tool. In effect, they give us seemingly superhuman strength when we need it. But how does this lever really do this if I'm not gaining energy from someone?

The answer lies in the definition of the terms above, and if we look closely at the image below, we will see the specific things that changes Force without changing Energy or Work.

In the lever image below:
I am pushing down on the left edge of the lever with a specific amount of Force.
The lever edge moves a specific distance at a specific amount of time.
This defines the Work that is being applied on the left side of the diagram, and let's call this Input.
On the right side, and let's call that the Output, the edge of the lever is rising a specific distance that is half that of the Input side, over the same amount of time.
With this in mind, let's consider our definition of Work.

Work (W) equals Force (F) times distance (d): W = F x d

On the Input side, Work is the Force I am applying times the downward distance the edge of the lever arm is moved.
On the Output side, Work is the resulting Force applied to the object mass times the shorter distance the edge of the lever arm moves upward.

For Work and Energy to be equal on both sides, the upward Force applied on the Output side must be twice the amount on the Input side because of the difference in the distance.

Work Input = 1 Force * 2 distance
Work Output = 2 Force * 1 distance

So with the first law of thermodynamics intact because Energy and Work are the same on both the Input and Output side, we can see how the Force must be increased on the Output side for this to happen. In practical terms, the magnitude of Force is enhanced with distance.

Of course, managing the fulcrum location can also change the magnitude of the Output Force as well because it changes the distance ratio of both sides. So there is an opportunity to control the amount of Force we need while using the same Input Force.

By understanding the concept of mechanical advantage using a simple lever, we can now see how Force extended over distance can increase.

Gear Reduction:

With gear reduction, the difference is that the motion is rotational (angular) which means that the distance (d) can continue indefinitely around in a circle as if there is no floor in the image below, and Input distance continues to drive a shorter Output distance with greater Force.

With rotating gears and rotating propellers, the Force we are talking about is torque (t). Since Force causes acceleration and torque causes angular acceleration, then torque is the force needed to accelerate or decelerate propellers. In 3D and aerobatic flight, there is infrequent episodes where the propeller is spinning at a constant speed, so infrequent episodes of no accelerate or decelerate, or rotational equilibrium. So we must consider that torque is the most important vector applied to the propeller to gain the desired result.

The StinGR utilizes a synchronous or timing belt drive system, where we have a small pinion gear on a shaft that is rotating at a very high speed. It is connected to a large main gear with a drive belt. The distance in our Work equation for the Input side is represented by the high revolutions per minute (RPM) of the smaller pinion gear. The distance of the Output side is represented by the low RPM of the bigger main gear.

As with the lever in the image below, if there can be no net gain in Work or Energy from the Input to the Output side, then the Force MUST me greater on the Output side.

Since the angular acceleration Force is torque, then the torque force must be higher on the Output side than the Input side. With the change of distance (difference in RPM), and the fact that Work must be the same on both sides, the torque on the Output side must be greater.
Last edited by Aeroplayin; Jun 23, 2020 at 02:41 PM.
Oct 22, 2019, 01:38 PM
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Getting a bit more technical

A final concept to know about these systems is that torque (t), Force (F), and power (in Watts) are directly related.

Force is measured in Newtons (or pounds)
Torque is measured in Newton-meters (or pound-feet)
Watts is measured in Newton-meter per second (or pound-feet/second)

When we speak of rotation or angular Force (N), we usually speak in terms of torque (N-m or lb-ft or oz-in) because of the rotational distance element.

Torque is the angular force that tends to cause rotation around an axis. Mathematically, torque can be calculated as the Force times the radius of the rotating object times the sin of the angle of rotation (known as angle theta). So simply put, torque is nothing more than the Force needed to twisting or rotating a mass with a radius around an axis. So if Force is increased on the Output side, the ability to rotate bigger objects over a specific distance is increased.

1 Watt divided by 745.7 times 5252 RPM divided by the shaft RPM equals 1 lb-ft of torque.

And 1 lb-ft of torque times 192 will give us the torque in oz-in.

Torque & Amps

One of the reasons that we may want to make this conversion is because we can predict the current that an electrical system will draw when using a motor to power the system to a specific level of torque.

We begin by taking the torque value at the Output shaft in oz-in and dividing it by the constant 1352.4. We then take this value and multiply it by the motor's Kv. This will give us the predicted current in Amps. And this will be true regardless of if the electric power system is using a direct drive or a gear reduction to increase power and torque.

Getting even more technical:

Since we now know that we are gaining power and torque though a mechanical system driven by an electric motor, and can predict the Amps needed to drive the system, and that Amps is directly related to the motor Kv, we can now explore the physical efficiency of gear reduction.

If we have a non-geared motor with a 180Kv turning a 28x12 propeller to 6400 RPM at 44.4 volts (12 LiPO source at nominal voltage), we can calculate the Watts needed at the shaft to do this.

Watts at the shaft (Wo) = (Prop Diameter / 12)^4) * ((RPM / 1000)^3) * Prop Load Constant * (Prop Pitch / 12)
Watts at the shaft (Wo) = (28 / 12)^4) * ((6400 / 1000)^3) * 1.0698 * (12 / 12)
Watts at the shaft (Wo) = 8311.7

HP = 8311.7 Wo / 745.7
HP = 11.147 lb-ft/sec
Torque = (11.147 HP * 5252 RPM) / 6400 RPM
Torque = 9.148 lb-ft
Torque = 1756.4 oz-in
Amps = ((Torque {oz-in} / 1352.4) * Motor Kv
Amps = ((1756.4 oz-in / 1352.4) * 180 Kv
Amps = 233.77A

Since we are using a 12S LiPO source, the Watts-in must be:

Watts = 233.77A * 44.4v
Watts = 10.38 kW

The difference between 8311.7 Wo and 10,380 Wi is 80.1%

This means that 10.38 kW is applied to the overall non-geared 180Kv motor to result in 8311.7 Watts of power at the shaft.

To test the StinGR's ability to increase efficiency, we can spin the 28x12 on a 3.33:1 StinGR gear reduction and find that a 533Kv motor at 44.4v will turn this propeller to the same 6400 RPM.

This means that the Output production as far as Force and Power are the same. But didn't we say that the StinGR will develop more power than a non-geared outrunner?

For this example, we chose to see what the efficiency was, and not the 'relative' power difference was. So let's calculate the difference in current first.

With a 3.333:1 reduction, the 533Kv is actually 160Kv at the power shaft. (505 / 3.33) = 160Kv

So with the same HP and torque, the different Kv means the following:

Amps = ((1756.4 oz-in / 1352.4) * 160 Kv
Amps = 207.80A
Watts = 207.80A * 44.4v
Watts = 9226W

The result is that the StinGR is developing that same power at the shaft with 25.97 fewer Amps.
This also means that we needed 1154 fewer Watts-in (or power-in) to develop the same power at the shaft.

The difference in current and power is 11.1%, so the StinGR can:

- Develop the same power as the non-geared system at 11.1% more efficiency.
- Develop 11.1% more power at the shaft with the same current (Amps).

Since current and flight time are related, this also means the StinGR can provide 11.1% more flight time with the same capacity as the non-geared motor system or provide 11.1% more power at the same flight time.
Last edited by Aeroplayin; Jun 17, 2020 at 03:14 PM.
Jul 04, 2020, 08:16 AM
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Horsepower vs Torque Force

An important concept regarding gear reduction is the difference between Horsepower and torque and how much torque force is valued over Horsepower in our hobby. Especially for 3D pilots.

Horsepower does not have a linear relationship to torque and is virtually meaningless to us in this hobby unless we're flying pylon races -- in which case electric power is the wrong choice anyway considering that a 25C discharge will bring a battery to 20% capacity in under 2 minutes at full throttle.

On the other hand, prop sizes and motor Kv can be manipulated to use one specific power level at the same voltage to develop a wide range of torque options. Here are just a few examples how HP and torque are quite different:

Spinning a 29x12 to 5490 RPM is about 8 HP and 7.6 lb-ft of torque. It is also a pitch speed of 63 mph and 54 lbs of thrust and ~162A to 165A.
A 28x12 to 5757 RPM is also 8 HP and 7.3 lb-ft of torque. It is also a pitch speed of 65 mph and 52 lbs of thrust and about the same 165A, of course.
A 27x12 to 6045 RPM is also 8 HP and 6.9 lb-ft of torque. It is also 69 mph and 51 lbs of thrust. 165A.
A 26x12 to 6365 RPM is also 8 HP and 6.6 lb-ft of torque. It is also 72 mph and 49 lbs of thrust. 165A.
A 25x12 to 6705 RPM is also 8 HP and 6.3 lb-ft of torque. It is also 76 mph and 48 lbs of thrust. 165A.
A 24x12 to 7100 RPM is also 8 HP and 5.9 lb-ft of torque. It is also 81 mph and 47 lbs of thrust. 165A.

So you can see how HP is exactly the same while we can manipulate torque at the expense of speed.

Since torque and Amps are related, you may be wondering why the current stays aligned with HP instead of increasing and decreasing with torque. The answer is because torque and Amps are related by Kv in motor power, and to get the same HP from motors at the same voltage with different props, we must change the Kv. And this change in Kv brings the Amps in line with the power at the shaft (Watts-out).

For example, the 24x12 with a motor of 200Kv at an 80% Kv efficiency will do 7100 RPM and 5.9 lb-ft.
Amps = ((5.9 lb-ft*192)/1352.4)*200Kv
So 165A

The 28x12 with a motor of 162Kv at 80% Kv efficiency will do 5757 RPM and 7.3 lb-ft.
Amps = ((7.3 lb-ft*192)/1352.4)*162Kv
And ~165A

The StinGR is a gear reducer that amplifies torque by utilizing the concept of mechanical advantage, defined in the previous post.

Torque = (HP*5252)/RPM

5252 represents the RPM at which the torque force and Horsepower curves intersect. This means that below 5252 RPM, torque RULES.
Thrust is also related to torque and the area of the propeller disk, so large props and higher torque develops more thrust at lower RPM.
In 3D, thrust helps us accelerate to operation speeds faster, and pulls plane from near stall speeds harder.
Propeller sizes between 25 inches and 28 inches have an efficiency range between 6340 and 7050 RPM in ground testing.
This is true because the speed of the rotating propeller tips releasing load in the air will approach the speed of sound and the resulting vortex will affect the efficiency (fluid mechanics).
Therefore torque force is essential in 3D flying and our hobby in general since zero to 5252 RPM is 82% of this RPM range.
And 5252 RPM is where torque begins to give way to Horsepower.

This is why gear reduction can out-perform non-geared outrunners starting at about 6000W and 91" planes.

With this in mind, a gear reduction system will amplify torque at the rate of the gear ratio the same way a lever amplifies torque by moving the fulcrum, or the same way a wrench with a long handle can provide greater torque force on a stubborn nut.

Additionally, because of the combination of transferring torque force at the rate of 3.333 to 1, the StinGR also improves the efficiency of electrical energy by narrowing the margin between the Watts-in and the net Watts-out. This means that the same net Watts-out, or power at the shaft, can be developed with less circuit current in spite of the fact that some electrical power is slightly reduced by losses caused by things like friction and the result of added rotational/angular forces.
Last edited by Aeroplayin; Jul 04, 2020 at 08:30 AM.

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