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I went to that site and saw that the Crosley is rated at only 25 hp at 3900 rpm, and weighs 100 lb with the starter. For that kind of power I would use the Rotax 277 2-stroke (28 hp), or the single cylinder Hirth at 30 hp. The Crosley was tiny at only 44 cubic inches, and kind of puny. How the Mooney Mite was able to fly on that was a miracle in airplane design.

Torque! Much better measure of prop swinging power than hp. 44 cubic inches is 721 cc, which is about three times the volume of a Rotax 277.

 
Dick O'Reilly

I see people go round and round about horsepower versus torque. 'Torque' sounds cool. But only horsepower is power. Torque is pressure. On the other hand, I could be wrong.....

Mooney Mite is the cutest plane with forward sweeps i seen in a long time. always did wonder about less draggy properties of the blanik glider's rudder. though for that kind of hp, a swissauto250 engine may be better at 35hp for 33lbs.

And then, the high wing high lift E-MBN714 (Eppler-Mark Beierle . .) airfoil is known as the hardest working wing in the industry for a very important reason :)

read once that hp = (torque*rpm) / 5250; interesting equation that's indicates that, in internal combustion engines 'with crankshafts', torque curves up and horsepower curves down until they are equal at 5250rpm; which is why gearboxes are required to turn the propellor slower, i think; some effect of the crankshaft mechanics more than internal combustion, it feels , but a discussion about it would be great. now, with electric (and steam) seem to be constant maximum torque from the get go...

(attached a dove's photo for Maishalabe and the group, made a nest for her recently, thought i'd share)

Mooney Mite is the cutest plane with forward sweeps i seen in a long time. always did wonder about less draggy properties of the blanik glider's rudder. though for that kind of hp, a swissauto250 engine may be better at 35hp for 33lbs.

And then, the high wing high lift E-MBN714 (Eppler-Mark Beierle . .) airfoil is known as the hardest working wing in the industry for a very important reason :)

read once that hp = (torque*rpm) / 5250; interesting equation that's indicates that, in internal combustion engines 'with crankshafts', torque curves up and horsepower curves down until they are equal at 5250rpm; which is why gearboxes are required to turn the propellor slower, i think; some effect of the crankshaft mechanics more than internal combustion, it feels , but a discussion about it would be great. now, with electric (and steam) seem to be constant maximum torque from the get go...

(attached a dove's photo for Maishalabe and the group, made a nest for her recently, thought i'd share)

Gents –

Torque is the twisting force put out by the engine; think about a torque wrench. It is measured in a force (lbs) at a given distance (feet) from the center of rotation. If you have a 1 foot wrench and apply 15 lbs of force that is 15 Foot-Lbs of torque. If you have a 2 foot wrench and apply 15 lbs of force that is 30 Foot-Lbs of torque. So how hard the engine has to twist the shaft at the speed you are operating is its torque.



Speed is pretty self-explanatory, we like to use Revolutions Per Minute, or RPM.



Power is the product of Torque and Speed, ie Torque x Speed = Power



Horsepower = Torque (in Ft-Lbs) x Rotation Speed (in RPM)/5252. The 5252 isn’t some magic crossover speed or anything, it’s just a number to get the units correct and to accommodate the definition of what a Horsepower is. Will derive it below for those that don’t mind equations.



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So if an engine puts out 100 Ft-Lbs of torque at 2500 RPM, that’s 100 x 2500/5252 = 47.6 HP.



If, for example, you have a geared down engine, say 2:1 ratio to make it the math easy, and neglecting losses in the gear, in order to

produce the 100 Ft-Lbs or Torque at a prop RPM of 2500, the engine will obviously need to run twice as fast, or 5000 RPM.

But the torque it needs to put out is cut in half since it is geared down, so it only needs to put out 50 Ft-Lbs of torque.



As a check, given that we are ignoring the gearbox losses, the engine is doing 50 Ft-Lbs x 5000 RPM/5252 = the same 47.6 HP of power.



The power is the same. You can trade RPM and torque around (inversely) with belts and gears, but the power always is the same.



You don’t have to read the following if you don ‘t want to; it’s how the 5252 gets in there:



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Imagine a pulley (with Radius R) on a shaft like in a water well. Say the bucket we are lifting weighs B pounds. If we turn the pulley

one revolution we have raised the bucket by a distance of 2 x (Pi) x R.



Work is force times distance. So the work we have done to raise the bucket is its weight B, times the distance we raised it, 2 x Pi x R.

So the work done is B x 2 x Pi x R



Power is time rate at which work is performed. Say we lift the bucket 10 feet in 10 seconds, it took a given amount of Power to do that.

If we lift the bucket the same 10 feet, but in 5 seconds instead, we needed twice the power to do it, although the work we did was the same.

The work was the same because even though we used twice the power, it only took us half as long to do the work.



So power is Work [B x 2 x (Pi) x R] divided by Time, so Power = [B x 2 x Pi x R/Time]

Where B is the weight of our bucket, and R is the radius of the pulley,



1 Horsepower is DEFINED as 550 Ft-Lbs/Sec. Somebody a long time ago figured out how much work a horse could do and defined it.



So we have to get our power equation above to come out with the units of Ft-Lbs/Sec. We can use any units we want, ie the weight of

our bucket it ounces or pounds, the radius or our pulley in inches, feet, or miles, and the time in seconds, minutes, or hours. But as we do that,

we will end up with a conversion number that we have to use for our chosen units. Nothing magic, the power to lift a given number of ounces will

be one sixteenth of the power to lift the same number of pounds.



So if we want to have our inputs, Torque in Ft-Lbs, and Speed in RPM, here is what we do:



Power = [B x 2 x Pi x R/Time]



Power = 1 HP = 550 Ft-Lb/Sec = (B lbs x 2 x Pi x R ft) / Time minutes) x (1 Min/60 sec)



Power = 1 HP = (B lbs x R Ft / Time minutes) x (2 x Pi) / (550 x 60)



Power = 1 HP = (B lbs x R Ft/ Time minutes) / 5252.



Power = 1 HP = Torque (Ft-Lb) x Speed (Rev/Min) / 5252.



The 5252 isn’t some magic thing, it’s the 550 for the definition of HP times the 60 to get minutes to seconds, divided by 2 x Pi , the circumference of the pulley.



I hope this helps….



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Thanks, Bruce; this was easier to understand than a terse howto
I glazed over earlier

James Watt defined it to sell his steam engine to customers who had only used horses earlier, instead of a royalty of a third of coal savings compared to an engine he improved upon. He used a pony first, then a horse which prompted a discussion on calibration, finally settled on a brewery horse

then how come electric motors start with high nearly constant torque until a few thousand which then reduces steadily till eleven thousand rpm? (some even have a small pull-up torque in the constant region)

Rahul –

After posting the Horsepower thing below, I was afraid somebody was going to ask the electric question…



Some pre-discussion notes:



In an attempt to not get massively flamed by others, the following discussion applies to DC motors with magnets inside. There is a whole lot of other stuff going on with AC motors designed to work at a CONSTANT AC frequency (ie 60 Hz or 400 Hz sine wave driven) that typically self induce rotating magnetic fields, but can also have magnets as well. There are probably also some obscure classes of DC and/or magnet motors to which the following do not totally or exactly apply.



In its simplest form, a DC motor has magnets, windings, and brushes. The power goes to a winding, and that energy goes around and around in the winding and creates a magnetic field. The power to the windings are switched on and off (or commutated) by the brushes so that the electrical energy goes to the correct winding with the correct polarity so that the magnetic field it creates reacts with the permanent magnets to either attract and /or repel the permanent magnets and cause a torque or rotation to occur. You can see how the electric motor does not need speed to create torque, even stopped the forces are there.



The inrunner and outrunner motors like the Zero and Joby are really DC motors with magnets inside, not true AC motors. The only difference is that the brushes have been replaced with some circuitry that switch the coils on and off at the proper times (through a variety of methods not discussed here) without having to resort to brushes. They are called Brushless DC motors. All of the following also apply to them as well.



All of the power, horsepower, speed, and torque discussion below equally applies to electric motors. Yes, I know that people say that an electric horsepower and an engine horsepower are not the same, BUT THEY ARE. Yes, if you pull a gas engine off of a machine and you put an electric motor with the same HP rating as the gas motor had it will likely work better, but this has to do with how the gas engine HP is rated, and the fact that the gas engine may not have been operating at its peak power point, and that most gas engines are really not designed to operate at their peak rated power for continuous operation, whereas most electric motors are happy running all day at their rated HP.



Basically, to produce a power curve for a gas engine, you run the thing at full throttle and note the no load max RPM, then you slowly add friction to the shaft (with a machine called a dynamometer) and note the torque and RPM as the load is increased, all the way until the engine stops running. You now have a curve of torque as a function of speed. You multiply the torque and the RPM at each point and you get a curve of the power (remember torque x speed = power). The available torque from the engine will change as the speed changes and they are related to each other. At the min and max speed endpoints, the torque available is greatly reduced (or even zero), so that the power you can get is really small, and somewhere in the middle of the band there is a place where the product of speed and torque (ie power) is at a maximum. This is the peak power point, and you will have a given RPM and a given torque available at that point. If you operate the engine at any place other than this peak power point, you will get LESS POWER.



So, here goes:



DC motors create speed and torque just like any other motor. The core difference between them and a gas motor is that the speed and the torque of a DC electric motor are basically INDEPENDENT of each other. The DC motor is really quite happy providing its maximum torque at zero speed, whereas a gas motor is producing zero torque at zero speed.



The Speed of the DC motor is proportional to the voltage applied. There is a number associated with it, called the Kv, which is a function of how the motor is designed, that tells you how fast the motor will turn for a given applied voltage. For example, a Joby motor with a Kv = 54 RPM/Volt will spin at 54 RPM for every volt applied to it. So give it 100 volts and it will turn at 5400 RPM. [Volts x Kv = RPM]



The Torque of the DC motor is proportional to the current applied. There is a number associated with it, called the Ka, which is a function of how the motor is designed, that tells you how much torque the motor will output for a given applied current. For example, the Joby motor above has a Ka = 0.117 Ft-Lb/Amp (more on how I got that Ka later) will torque the shaft at 0.117 Ft-Lbs for every Amp applied to it. So give it 100 amps and it will twist the shaft with 11.7 Ft-Lbs or torque. [Amps x Ka = Torque]



So this is the really great thing about DC electric motors – You control the speed by the applied voltage, and the current provides the necessary torque. There are, of course, limits. The max limit on voltage (and therefore speed) is caused by how much insulation is on the wires (if you go higher they start to arc and short each other out), the bearings, and the balance of the rotating parts. The max limit on current (and therefore torque) is caused by the gauge (ie diameter) of the wires and the cooling characteristics of the motor, keep putting in more current and the motor wires (and or the insulation) will start to melt and the heat in the wires will cause even more losses, heat, and lost efficiency.



But, IN GENERAL, you can treat the DC electric motor’s speed and torque (which together make power) as completely independent. This is just like flying where the elevator position controls you speed and the power controls your climb, essentially independent of each other (steady state). You can design an electric motor to provide the torque and the speed you want and not have to use gearboxes, and you don’t have to operate at a certain RPM to get peak power like a gas engine.



That said, if you take an existing electric motor, (like the Zero), it has a limit on the maximum current based mostly on its wire diameter, and Mark could not provide the torque necessary (due to the current limit) with it to direct drive the prop, so he had to use a reduction gear to get enough torque out of the engine. Had the motor been designed specifically for the airplane application, the wires inside would have had less turns and have been larger in diameter (ie a lower Kv and a higher Ka) and it would have been able to direct drive the prop.



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Some helpful equations (and example):



Speed x Torque = Power (Speed in RPM x Torque in Ft-Lbs / 5252) = Horsepower

Voltage x Current = Power (Volts x Amps = Watts)

[please see correlation above between (speed and voltage) and (torque and current)!!!]



1 Horsepower = 550 Ft-Lb/Sec = 746 Watts



Volts x Kv = RPM

[Example: 100 Volts x 54 RPM/V = 5,400 RPM]



Amps x Ka = Torque

[Example: 100 Amps x 0.117 Ft-Lb/Amp = 11.7 Ft-Lbs]



Voltage x Current = Power (input)

[Example: 100 Volts x 100 Amps = 10,000 Watts]



Speed x Torque = Power (output)

[Example: (5,400 RPM x 11.7 Ft-Lbs / 5252) = 12.029 HP [Convert: 12.029 HP x 746 Watts/HP =8,974 Watts]



Output Power (watts) / Input Power (watts) = motor efficiency

[Example: 8,974 Watts Output / 10,000 Watts Input = 89.74% Efficiency]



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So, how did I get the unlisted Ka for the Joby Motor – I assumed a given efficiency and derived it from the above equations.



Here is an example for two different windings of the JM1:



Item Winding B Winding D Units

Nominal RPM 6,000 6,000 RPM

Cont. Shaft Power at Nom RPM 13,200 13,200 Watts

Continuous Torque (metric units) 21 21 N-M

Continuous Torque (English Units) 15.49 15.49 Ft-Lbs (N-M x (1 Lb/4.448 N) x (1 M/3.28 ft)

Motor Kv 18 54 RPM/Volt



Calculations:

Voltage for Nominal RPM 333.33 111.11 Volts (RPM/Kv)

Assumed Motor Efficiency 90% 90% %

Input Power 14,667 14,667 Watts (Output Power / Efficiency)

Input Current 44 132 Amps (Input Power / Voltage)

Motor Ka 0.352 0.117 Ft-Lb/Amp (Torque / Amps)



I hope not to get flamed badly by anyone over generalizations, just trying to help with the concepts…



Regards

Bruce Markle

Great calculations, and just watt I thought.