# Electric Airplane - Glider Tow



## JAKiii (Nov 20, 2008)

Hello DIY Brain Trust:

I am curious about what it would take to convert a 235hp Piper Pawnee. This airplane is used around the world for glider towing. As a tow plane, it will fly for 5-10 minutes and spend 30 or more minutes on the ground before flying again. Additionally, Pawnees were designed for agricultural use (crop dusting) and have a large hopper bin between the cabin and firewall, directly above the center of gravity. This bin can carry more than 1,000 lbs of weight.

So my question is this: What is the basic calculation for converting the airplane while maintaining 235 hp at 10-minute flight times? Yes, there are many, many considerations for electric flight and for conversions, but I want to understand the basic power calculation, motor, and battery pack requirements. For now I won't even consider charging time.

Thanks for any input,
JAKiii


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## dougingraham (Jul 26, 2011)

What a fun project this would be.

I dont have a good feel for powering something like this so my estimates could be all wet.

If you are talking about full throttle for 10 minutes meaning 235HP for the whole time I suspect cooling the motor will be the problem. 235hp is 176kw. Assuming 70% efficiency that means you would need to feed the motor with 250kw. At 1000 amps this would require 250 volts so you could probably do it with a WarP11 HV if you could keep it from melting down. Any controller that could do 1000 amps could do this. As for batteries you would need to do a 98 cell pack of LiFePO4 cells. For a 10 minute flight at 1000 amps you would need a minumum of 167AH. So 180AH cells would actually work with a small reserve. A 98 cell pack of 180AH calbs would weigh 1215 lbs. So that is one problem right there. I also suspect that if you put 1000 lbs of additional weight in the Piper it would not be a very good tow plane anymore.

Charging is not an issue. You can pretty much fully charge these batteries in 20 minutes if you have enough power. Your 10 minutes at full power would consume 42kwh. With a 30 amp dryer outlet it would take 5.8 hours to recharge. A Tesla supercharge station could charge it in about 21 minutes.

I doubt this is practical yet but really not too far off. For a big bucket of money you could make a larger LiPo pack like Assault & Battery (drag racer) uses and that would weigh less for the same capacity. Cooling the motor and the weight of the pack are the real issues as I see it. There are probably FAA regulations as well that might be difficult to meet. Aren't you required to have a 20 minute reserve?


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## JAKiii (Nov 20, 2008)

Thanks Doug,

FAA Regulations for Daytime VFR (clear weather) call for nearest airport plus 30 minutes fuel. So that is one problem, but possibly there are ways around it. The glider itself doesn't have 30 minutes of fuel, it flies under a different classification. I need to research that. As for alterations to an FAA certified aircraft, it would require taking the airplane into the Experimental category where you have a lot of freedom to experiment, not sure if you're allowed to pull a glider with an Experimental class airplane. Like I said, there are many issues...

As for horsepower, one thing I'm trying to understand is how much of the rated power makes it to the prop. The Pawnee's engine is rated at 235 hp but I don't know how much of that gets to the prop or if an electric motor would allow a lower horsepower rating for the same output. Horsepower versus torque might change the equation.

Motor cooling does seem like an issue, and just as with combustion engines in airplanes, it's difficult to air-cool while climbing because the nose is pointed up and speeds are slower. Does it help cooling if you oversize the motor and run it at less than 100% power?

I think the first thing I need to clarify is how much energy the tow plane is actually using. By Doug's math, 42 kWh is required for 10 minutes at full power. My instinct is that this seems high. The Pawnee is burning about 10 gallons of AvGas for that 10 minute flight, it seems like 42 kWh properly applied would go farther than 10 gallons of gas on fire.

Jamie


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## dougingraham (Jul 26, 2011)

With an aircraft engine and a prop directly connected to the crankshaft I think all of it makes it to the prop. There isn't anyplace else for it to go. In a car some of it is absorbed by the gears in the transmission and differential and turned to heat and sound there. If you have the prop mounted to the motor shaft the only significant losses are going to be as heat in the motor. Iron and copper losses. There are also some losses in the motor controller but they are usually dwarfed by the motor losses. I was guessing about 70% of the energy coming out of the batteries will make it to the prop. This could be low but I would be surprised if you were able to get a lot better than that with a motor you are willing to carry in an airplane at those power levels.

In an airplane the power is purely a function of the RPM of the prop which also depends on temperature, humidity, and pressure. It doesn't matter if it is an ICE turning the prop or an electric motor, if they turn the same RPM they are producing the same power given the same prop and conditions.

A gallon of gasoline has 33.7kwh of energy. The overall efficiency of a magnetic drive system in a car is around 8 times better than the ICE. If you actually burn 10 gallons in that 10 minutes that would be 337kwh consumed. 8 times more efficient would be about 42kwh which is surprisingly close to to my 41kwh estimate.


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## lithiumlogic (Aug 24, 2011)

dougingraham said:


> With an aircraft engine and a prop directly connected to the crankshaft I think all of it makes it to the prop. There isn't anyplace else for it to go. In a car some of it is absorbed by the gears in the transmission and differential and turned to heat and sound there.
> 
> *In an airplane the power is purely a function of the RPM of the prop which also depends on temperature, humidity, and pressure*.


Does the Pawnee have a variable pitch prop? I understand this is something of a high end feature, so I'm guessing it doesn't. Which means that although the Pawnee makes 235HP at peak power RPM it isn't necessarily going to do so for the whole duration of the aerotow. At the start of the takeoff run, with airspeed and altitude low, the engine may be operating a bit below max power RPM. As speed builds and you climb to thinner air, you may go a bit above it. Obviously as the air gets thinner you loose a bit of power there too.

The big one seems to be battery weight. What % increase in the takeoff weight of the Pawnee itself are we going to see? The takeoff speed will increase with the square root of the weight gain, although i'm sure the airplane manufacturer lists takeoff speeds for different weights already. You want to check the gliders can handle the increased speed, though i understand their limits are surprisingly high.

The extra weight is going to slow the climb rate as well.

Let's say the ICE Pawnee weights 1900lb with pilot and fuel.

EV Pawnee 2900lb with 1000lb batteries.

Glider weighs 800lb

So instead of hauling 2700lb (tug and glider) to 2000" we are hauling 3700lb (tug , glider, batteries)

That's a 37% increase. Count on a 14 minute climb to altitude not 10 minutes.

Mind you Wikipedia lists the rate of climb of the Pawnee as over 600ft/minute @ 2900lb. So 10 minutes is surely a very pessimistic estimate.

As for the 20 minute reserve thing, I imagine you're using a heck of a lot less than 235hp just maintaining level flight on an unhitched airplane..


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## aeroscott (Jan 5, 2008)

max power is only good for 1 to 3 minutes . then it's something like 60% . And the ice prop/engine is shaking everything including it's coupling to the air much more then a electric prop/ motor would .I think of it like beating the prop with a sledge hammer( many little explosions) More losses


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## JAKiii (Nov 20, 2008)

Most Pawnees have fixed pitch props.

The 6-cylinder Lycoming O-540 weighs 438 lbs. I suspect (uneducated guess) another 150 lbs could come out of the airplane. This leaves an empty weight of 850 lbs before adding back the electric motor and batteries. Keep in mind that full fuel (100LL AvGas) in this airplane is 150 gallons = 900 lbs. It seems like an electric conversion could net out close to the current ICE weight.

Another interesting thing about airplanes is that they require regular engine overhauls. This is a $20,000 maintenance hit and makes any Pawnee that's due for an overhaul a possible candidate for ICE subtraction.

I need to understand how much energy is actually used by the tow airplane. If I can get to a kWh number for towing plus 30-minute reserve then we could understand the motor/battery requirements. So this is where I'm currently stuck, it seems like more accurate math is required -

Thanks for the input so far,
Jamie


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## PStechPaul (May 1, 2012)

I don't know much about propeller or aircraft design, but I found some information that may be helpful:
http://www.experimentalaircraft.info/articles/aircraft-propeller.php
http://www.pilotfriend.com/training/flight_training/fxd_wing/props.htm
http://theknowledgeworld.com/world-of-aerospace/Cal-Power-Require-Propeller-Airplane.htm

From this I learn that, for a given prop design, there is generally one optimum airspeed and prop RPM for any condition. Since an ICE has a rather narrow peak power and peak torque curve, variable pitch props may be more efficient. But since an electric motor has relatively constant available torque at all speeds, it might be able to spin at a much lower optimum speed during take-off, and not require anywhere near as much HP as an ICE (which is usually rated at peak HP). So I think a 50 to 100 HP electric motor might be sufficient, and actual energy usage during take-off may be considerably less than has been calculated. 

If an average of 100 HP (75 kW) over the 10 minute climb is sufficient, then the total energy will be something like 75/6 = 12.5 kWh. Once at altitude the power requirement is probably much less, maybe 10-20 kW, so for 20 minutes additional reserve a battery pack of 25-35 kWh might be more than enough. If the lower number proves sufficient, the battery pack might weigh only about 600 pounds. And a 60 HP three phase motor (using 3x overclocking) might be about 300 pounds (20 HP frame size).


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## JAKiii (Nov 20, 2008)

I think PStechPaul's estimate is closer to the actual required energy. I can drive my Ford Focus EV up a mountain 3000' virtically and cover 20 miles laterally using 15kWh of electricity, and that vehicle weighs 3500 lbs. Apples and Oranges I know, but still...


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## PStechPaul (May 1, 2012)

I installed the Aximer program and I tried it with some WAGs for a small plane as being considered:










From one source I found that the conversion of ft.lb/s to HP is a factor of about 0.0018 so 50,000 ft.lb/sec is about 90 HP. 
http://answers.yahoo.com/question/index?qid=20121020150722AANw0gh

Here are results with better guesses (especially the propeller efficiency). 90 times unity is a really fantastic device! The program needs better reality checking, and I found that the charts do not update unless you close and reopen. 










I think there is still something not quite right, because 3000 ft/min seems pretty high. Maybe 250 HP in a 2000 lb plane is overkill? And the power goes from 40,000 to 20,000 ft.lb/sec which is like 72 to 36 HP. Maybe that's all that's needed? 

Here is another link I found which has numerous calculations for a Fairchild F-22 monoplane (1936), showing a weight of 1467 lb and a 95 HP engine. 
http://thesis.library.caltech.edu/951/









http://en.wikipedia.org/wiki/Fairchild_F-22


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## dedlast (Aug 17, 2013)

I wonder if looking at what the scale RC guys are using would be of any use. A lot of them are using brushless motors and LiPo these days and you should be able to extrapolate something from the energy usage they are getting.


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## PStechPaul (May 1, 2012)

I wonder if you could launch gliders with a giant slingshot or crossbow type contraption? You could pull it back with a big winch or even a tractor. Isn't that what they use on aircraft carriers?

http://en.wikipedia.org/wiki/Aircraft_catapult
http://en.wikipedia.org/wiki/Electromagnetic_Aircraft_Launch_System (like a rail gun or linear motor)

or...


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## mizlplix (May 1, 2011)

Back in the 1960's, in west Texas, we used a big Skadgit winch. It had a diesel engine and a 48" diameter drum and could get close to 80 MPH line speed.


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## dougingraham (Jul 26, 2011)

dedlast said:


> I wonder if looking at what the scale RC guys are using would be of any use. A lot of them are using brushless motors and LiPo these days and you should be able to extrapolate something from the energy usage they are getting.


The rule of thumb the electric RC fliers use for power planes is 50watts per lb to be able to take off and fly comfortably. Ive been flying electric RC exclusively since 1991 and I have a motor glider that will maintain altitude in dead air on 6 watts which worked out to 4.6watts per lb. In the dense air near the ground you can go vertical on 200 watts per lb and your plane is quite aerobatic on 100 watts per lb.

Aircraft are not operated like automobiles. They use full power for several minutes at takeoff and generally cruise at 75% power often for hours. Full scale sailplanes are most likely more efficient than my little motor glider. Full scale power aircraft seem to follow the wh/lb rules of thumb the modelers use pretty well when operated within the norms. The gallon per minute at full throttle and 12.5% efficiency correlation of 42kwh for 10 minutes makes sense. Of course if the Pawnee is not towing a sailplane or a cargo load of crop dusting materials it probably feels like a rocketship to the pilot. That 10 minute full throttle burn wasted 295kwh as heat out the back of the aircraft.

Could it be done, I think so, barely. Could it be done under the safety rules in place today, doubtful because you would not have a long enough power reserve. I am pretty sure you would want to use the LiPo pouch cells because they have 1.3 to 1.5 times higher wh/per unit weight energy density. The first step would be to convert an already efficient plane to magnetic drive and see if you could get an hour of flight time. Worry about towing gliders up later.


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## major (Apr 4, 2008)

JAKiii said:


> I am curious about what it would take to convert a 235hp Piper Pawnee.


You might check out the numbers and gear Chip is using. http://www.wired.com/autopia/2013/10/yates-world-records/ Similar power.


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## PStechPaul (May 1, 2012)

Also check out the entries in the Wiki:
http://en.wikipedia.org/wiki/Electric_aircraft

The E-Genius weighs 2100 pounds at take-off, has a 58 kW (100 kW peak) PM synchronous motor which weighs only 59 pounds, and has a 56 kWh battery pack. It achieved a flight of over 211 miles at an average 100 MPH.


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## Caps18 (Jun 8, 2008)

http://www.solarimpulse.com/en/airplane/hb-sia/

I'm not an airplane guy, but would it be a crazy idea to mount some motors to the wings of a glider? Basically, you would use the motors to get up to altitude and then turn them off and glide. It might classify as an ultralight then, but you would only need one pilot to fly it too.

Would the numbers work with all the added weight?


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## lithiumlogic (Aug 24, 2011)

Using a (front-engined, non-retractable propeller) motor glider might be a better basis for the conversion to get past the 20 minute reserve rule, since something with a >30:1 glide ratio will take a lot less power to keep in the air 20 minutes than a 10:1 Pawnee. 

TBH, if you're hitting the cells with a 4C discharge on takeoff and climbout, you probably want to finish the tow job with at 20-25% SoC remaining anyway... routinely hitting them for that much power down to empty is going to be brutal. 

It's a brutal application to start with TBH. LiFePO4 may lack the energy density, but LiPo cells won't last that long under a combination of fast charge and discharge and deep cycling. Perhaps using NMC cells, and routinely skipping the CV part of the charge curve so they run 80%-15% each flight is a compromise that'll give reasonably quick charge times, flight times and battery longevity.


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## PhantomPholly (Aug 20, 2008)

The airplane in the photo has a fixed pitch prop. That kind of airplane (similar to a crop sprayer) would have the pitch set for "climb" performance (think of being stuck in 2nd gear - not quite optimal from a stop, great for hills, limited top speed).

Airplanes at full power burn about .45 pounds of fuel per hour per horsepower. If we assume that the average horsepower being generated on a hot day and a few thousand feet is 200, we are talking about 90 pounds of gas per hour, or 45 pounds for 30 minutes of flight.

In a previous thread we determined that batteries (including installation, bracing, etc.) are roughly 30 times heavier than gasoline for a given amount of power produced. using this rule of thumb the battery pack would have to weigh 1,350 lbs for 30 minutes at full throttle - but that would run the pack dead and would require high C rates.

Short answer: The technology is not yet practical in this application.


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## major (Apr 4, 2008)

PhantomPholly said:


> the battery pack would have to weigh 1,350 lbs for 30 minutes at full throttle - but that would run the pack dead and would require high C rates.


 Drain the battery in 30 minutes. That is 2C. Not high IMO


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## PhantomPholly (Aug 20, 2008)

major said:


> Drain the battery in 30 minutes. That is 2C. Not high IMO


Ok, but dead has got to be a serious consideration if you hope to recharge, yes?


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## major (Apr 4, 2008)

PhantomPholly said:


> Ok, but dead has got to be a serious consideration if you hope to recharge, yes?


Full discharge is not down to zero volts. Battery cycle life is rated for "full" discharge of useable energy stored. It is true you can extend cycle life with shorter discharges and shorter charges. But if you have a battery rated for 100 kWh at 2C, you can use the full 100 kWh, at first. You will lose capacity over the life of the battery. Maybe to the tune of 20% over 3000 deep (full) cycles, depending on the type.


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## PhantomPholly (Aug 20, 2008)

major said:


> Full discharge is not down to zero volts. Battery cycle life is rated for "full" discharge of useable energy stored. It is true you can extend cycle life with shorter discharges and shorter charges. But if you have a battery rated for 100 kWh at 2C, you can use the full 100 kWh, at first. You will lose capacity over the life of the battery. Maybe to the tune of 20% over 3000 deep (full) cycles, depending on the type.


Well, I stand corrected.

Provided the pilot needs only fly to altitude over the field, it may be feasible - especially if full power is only needed for 10 minutes.

And, 3,000 20 minute flights is a LOT. With AvGas at $6.00 per gallon and electricity around $0.30 per gallon-equivalent, you might well save money.


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