# Hub motors for EV project- why buy?



## Jason Lattimer (Dec 27, 2008)

It depends on what you are building. Hub motors for bikes motorcycles and reverse trikes would be great. You can even buy 6kw 13" hub motors from kelly controllers that are rated to run 130kph. 

I agree that a full sized car wheel hub would be super heavy by car standards.


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## Anaerin (Feb 4, 2009)

Jason Lattimer said:


> I agree that a full sized car wheel hub would be super heavy by car standards.


Not necessarily. After all, on a car you would use a pair (Or a quad) set of motors, so they wouldn't have to be THAT beefy, and the outside edge of the motor could be used as the tire rim. As the "hub motor" would be in the airflow, cooling wouldn't be so much of an issue. With Regen systems to slow you down to 0.5mph, and a very small mechanical brake for parking, a lot of the weight in discs, pads and calipers could be recovered (Again, re-using the inside of the motor rim as a braking surface, perhaps). Finally, the weight of drive linkages, bearings, differential, transmission and gears would all be removed from the vehicle, giving more carrying capacity (for more batteries, say).

So, at the end of the day, all you have is a minor (2-5Kg per wheel) increase in unsprung weight as a "Disadvantage".


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## arftist (Mar 20, 2008)

Both Mitsubishi and BMW think they are worth trying, as do I. Regen braking is simplified, overall # of parts is reduced, traction is improved, the downsides seem to be a potential increase in unsprung weight, and cost. It is possible that the unsprung weight issue could be engineered around, depending upon the cost issue. There is little doubt that someday wheel motors will be common.


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## rudy (Feb 15, 2009)

arftist said:


> Both Mitsubishi and BMW think they are worth trying, as do I. Regen braking is simplified, overall # of parts is reduced, traction is improved, the downsides seem to be a potential increase in unsprung weight, and cost. It is possible that the unsprung weight issue could be engineered around, depending upon the cost issue. There is little doubt that someday wheel motors will be common.


Cost is the biggest issue for cars with in wheel motors. 2 yrs ago enough people were prepared to pay $110,000 for an electric Elise (tesla) but not in 2009.


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## John (Sep 11, 2007)

Anaerin said:


> Not necessarily. After all, on a car you would use a pair (Or a quad) set of motors, so they wouldn't have to be THAT beefy, and the outside edge of the motor could be used as the tire rim. As the "hub motor" would be in the airflow, cooling wouldn't be so much of an issue. With Regen systems to slow you down to 0.5mph, and a very small mechanical brake for parking, a lot of the weight in discs, pads and calipers could be recovered (Again, re-using the inside of the motor rim as a braking surface, perhaps).


When people talk about 100% regen systems I think they under estimate the amount of power such a system must be capable of absorbing from a vehicle at speed to provide reasonable retardation so that the vehicle will slow as quickly as one with conventional brakes.

Power = Force x speed where Force is in Newtons and speed is meters per second. If the car weight was say 1000kg and could achieve 1.2g deceleration (some with ABS can) then Force = 1000kg x 9.81 N/kg x 1.2g = 11772N. At 130 kph (36.11m/s) Power = 11772N x 36.11m/s = 425100W or 425.1kW. Given that the quoted hub motor is capable of 6kW or 24kW for a set of four this leaves a short fall of braking power of around 400kW.


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## Anaerin (Feb 4, 2009)

John said:


> When people talk about 100% regen systems I think they under estimate the amount of power such a system must be capable of absorbing from a vehicle at speed to provide reasonable retardation so that the vehicle will slow as quickly as one with conventional brakes.
> 
> Power = Force x speed where Force is in Newtons and speed is meters per second. If the car weight was say 1000kg and could achieve 1.2g deceleration (some with ABS can) then Force = 1000kg x 9.81 N/kg x 1.2g = 11772N. At 130 kph (36.11m/s) Power = 11772N x 36.11m/s = 425100W or 425.1kW. Given that the quoted hub motor is capable of 6kW or 24kW for a set of four this leaves a short fall of braking power of around 400kW.


Which quoted hub motor? 6kW would not be a very effective motor for a car-sized vehicle. I was thinking more of 30kW per wheel, which gives a 120kW power input. Considering that Instantaneous peak ratings are typically 3x typical running rates for most motors, you would have a total braking power of 360kW. So not quite to your 400kW limit, but close enough to be within the standard "Stopping distance" measurements, even without mechanical assistance.

Typically, if your electric motor has enough power to get you moving, it has enough power to stop you.

What is more difficult is doing something with the power you're pulling out of the motor. Dumping it into a capacitor bank to seep back into the batteries is one option (Though you will need a large and empty capacitor bank, as calculated above), or you can dump it into a resistor bank (Good for when your batteries and capacitors are full already).

[edit]
Ah, I see where you got the 6kW figure from. If you read that post more carefully:


Jason Lattimer said:


> It depends on what you are building. Hub motors for bikes motorcycles and reverse trikes would be great. You can even buy 6kw 13" hub motors from kelly controllers that are rated to run 130kph.


(Emphasis mine)
You'll see that the 6kW wheelmotor (technically, hub motor) mentioned was meant for bicycles and motorbikes, not for cars.
[/edit]


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## John (Sep 11, 2007)

http://www.kellycontroller.com/shop/?mod=product&cat_id=16&product_id=474

Sorry I wanted to use an actually available motor in my example. The motor in question has a through axle and is intended for some sort of fork mount though obviously not a bicycle. 130 kph is not very fast and 24kW should be capable of propelling a small light aero car to this speed. 120kW should be capable of propelling a vehicle to a much higher speed. As the speed increases so does the power absorption requirement. For instance if your hypothetical vehicle is capable of 210kph (like the Tesla) the power absorption requirement would rise to 687kW. If your car was a bit heavier at 1400kg this would become 962kW and you pursued a slightly more aggressive 1.3g deceleration it becomes 1042kW or over 1mW. If you artificially limited your top speed to 110kph 360kW would be sufficient for a 100% regen braking system in a 1000kg car otherwise you will need to keep the mechanical braking system.


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## Greenflight (Sep 13, 2007)

360kw would be roughly equivalent to 480hp with instantaneous torque... You'd have a pretty quick car... Your continuous rating is 160hp!!!!  

Very few cars can actually accelerate as quickly as they can stop...

Also I do believe the effectiveness of a motor as a brake is decreased at lower RPM, which would mean that say the last 5 or 10 mph of braking would need to be done largely by a mechanical brake. So eliminating the mechanical brake or having a parking brake only is probably not a good idea.

A hub motor with a built in rim and disc brake assembly makes a lot of sense. Personally, I think it would be an awesome, well-integrated solution, but you're still talking about a lot of unsprung weight and I don't know what that would do to ride quality. Maybe heavier shocks etc.? With a properly designed suspension I doubt it would be much of a problem...


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## Anaerin (Feb 4, 2009)

John said:


> http://www.kellycontroller.com/shop/?mod=product&cat_id=16&product_id=474
> 
> The motor in question has a through axle and is intended for some sort of fork mount though obviously not a bicycle.


No, I'm thinking that motor is meant more for a motorcycle than a car.



John said:


> 130 kph is not very fast and 24kW should be capable of propelling a small light aero car to this speed.


130KPH is more than fast enough (80MPH, which is above most legal speed limits in most places in the world, Germany's unrestricted Autobahn sections nonwithstanding).



John said:


> 120kW should be capable of propelling a vehicle to a much higher speed. As the speed increases so does the power absorption requirement. For instance if your hypothetical vehicle is capable of 210kph (like the Tesla) the power absorption requirement would rise to 687kW.


Splitting hairs slightly, but the Tesla has an electronically-limited top speed of 125MPH, or 201KPH. Your point does still stand, however.



John said:


> If your car was a bit heavier at 1400kg this would become 962kW and you pursued a slightly more aggressive 1.3g deceleration it becomes 1042kW or over 1mW. If you artificially limited your top speed to 110kph 360kW would be sufficient for a 100% regen braking system in a 1000kg car otherwise you will need to keep the mechanical braking system.


Let's work the other way around. Given a 1400kg vehicle, 4 30kW motors (something like the e-traction SM/350 http://www.e-traction.nl/content_wheel_SM350.html) and a travelling speed of 130kph ('cause just because the posted limit is 120kph, doesn't mean you're going to go that fast, right?), how long would it take to stop the vehicle?

1400kg * 33.33ms = 46,662N of inertia.
30,000 * 4 = 120,000W of braking power

...And at this point my hazy memory of A-Level physics fails me. Anyone care to help?

[edit]
Apologies, Jason. When I hit "Quote" for the message, it had your post as the first quote, so I took that as gospel. Attribution corrected.
[/edit]


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## Jason Lattimer (Dec 27, 2008)

Hang on there speedy,

You need to edit that last post. If reread this thread you will find most those quotes are not mine.


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## Anaerin (Feb 4, 2009)

Jason Lattimer said:


> Hang on there speedy,
> 
> You need to edit that last post. If reread this thread you will find most those quotes are not mine.


Edited. The "Advanced" post editor came up with your name as the quoted party. That really needs to be fixed (Along with the inserting of a png image at the top of every post in font tags).


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## rudy (Feb 15, 2009)

Interesting reading. So a performance car should not rely on hubs alone for brakes. And if not using a hub motor its still possible to regen using an ac motor? I saw these http://www.greenmotorsport.com/green_motorsport/products_and_services/3,1,388,17,12958.html


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## Jason Lattimer (Dec 27, 2008)

Actually, after emailing Kelly Controllers about their 6kw hub motor, I found out that it is 6kw continuous. It can produce 28 horsepower peak for short bursts, which means even in a reverse trike it is capable of 84 horsepower if used one on each of three wheels like I intend to use them.

Regen on this will be great. Even on this application it will put a little power back into the batteries and save the mechanical braking system at the same time.

This is the only consumer hub motor I can find anywhere. The rest are all WAY to expensive or still in trial stages.

I wonder if it is possible to build your own, say with a pancake motor and a 14" rim or so?


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## John (Sep 11, 2007)

Anaerin said:


> Let's work the other way around. Given a 1400kg vehicle, 4 30kW motors (something like the e-traction SM/350 http://www.e-traction.nl/content_wheel_SM350.html) and a travelling speed of 130kph ('cause just because the posted limit is 120kph, doesn't mean you're going to go that fast, right?), how long would it take to stop the vehicle?
> 
> 1400kg * 33.33ms = 46,662N of inertia.
> 30,000 * 4 = 120,000W of braking power


There would be two phases to stoping the vehicle in this scenario. the first phase would be power absorption limited and the second phase would be friction limited as with mechanical brakes. Using a purely friction limited calculation the weight of the vehicle is irrelavent to the calculation which is simply average speed (m/s) divided by deceleration (m/s/s) = 36.11 / 9.81m/s/s / 1.2g = 3.06s. Distance = 36.11/2 x 3.06 = 55m. This is your bench mark for friction brakes.

Transposing the original formula for initial retardation in gravities = 120000W / (9.81N/kg x 1400kg x 36.11m/s) = 0.242g. Transposing to find speed at the end of the first phase - speed = 120000W / (1.2g x 9.81N/kg x 1400kg) = 7.28m/s. Solving for second phase time = 7.28 / 9.81m/s/s / 1.2g = 0.62s Energy absorbed in the second phase = average power x time = 120000W/2 x 0.62s = 37200W.s Total energy needing to be absorbed = 1/2 x m x v^2 = 1/2 x 1400 x 36.11^2 = 912809W.s. Energy absorbed in first phase = 912809 - 37200 = 875609W.s. Divide by power for time = 875609 / 120000 = 7.3s giving a total time of 7.91 seconds


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## rudy (Feb 15, 2009)

Jason Lattimer said:


> Actually, after emailing Kelly Controllers about their 6kw hub motor, I found out that it is 6kw continuous. It can produce 28 horsepower peak for short bursts, which means even in a reverse trike it is capable of 84 horsepower if used one on each of three wheels like I intend to use them.
> 
> Regen on this will be great. Even on this application it will put a little power back into the batteries and save the mechanical braking system at the same time.
> 
> ...


would these work for a Tuk Tuk?? 2 motors at the back say..


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## rudy (Feb 15, 2009)

thai electric tuk tuk- look at the power rating...


http://www.pcd.go.th/info_serv/en_air_tuktukth.html


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## Anaerin (Feb 4, 2009)

John said:


> There would be two phases to stoping the vehicle in this scenario. the first phase would be power absorption limited and the second phase would be friction limited as with mechanical brakes. Using a purely friction limited calculation the weight of the vehicle is irrelavent to the calculation which is simply average speed (m/s) divided by deceleration (m/s/s) = 36.11/2 / 9.81m/s/s / 1.2g = 3.06s. Distance = 36.11/2 x 3.06 = 55m. This is your bench mark for friction brakes.
> 
> Transposing the original formula for initial retardation in gravities = 120000W / (9.81N/kg x 1400kg x 36.11m/s) = 0.242g. Transposing to find speed at the end of the first phase - speed = 120000W / (1.2g x 9.81N/kg x 1400kg) = 7.28m/s. Solving for second phase time = 7.28/2 / 9.81m/s/s / 1.2g = 0.31s Energy absorbed in the second phase = average power x time = 120000W/2 x 0.31s = 18600W.s Total energy needing to be absorbed = 1/2 x m x v^2 = 1/2 x 1400 x 36.11^2 = 912809W.s. Energy absorbed in first phase = 912809 - 18600 = 894209W.s. Divide by power for time = 894209 / 120000 = 7.45s giving a total time of 7.76 seconds


And I'm an idiot. Sorry, I gave you 4x the constant power figure, not the momentary impulse figure.

So, 4*120kW (Peak power out/input) = 480,000W



Windows Calculator said:


> Initial Retardation = 480000 / (9.81 * 1400 * 36.11) = 0.9679g
> Speed at end of first phase = 480000 / (1.2 * 9.81 * 1400) = 29.1248m/s
> 
> Solving for second phase time = 29.1248 / 2 / 9.81 / 1.2 = 1.237s
> ...


Though I'm a little uncertain here. You've thrown in a 1.2g, which I think is supposed to be the initial retardation value, correct? If so, this comes out to:


Windows Calculator said:


> Initial Retardation = 480000/ (9.81 * 1400 * 36.11) = 0.9679g
> Speed at end of first phase = 480000 / (0.9679 * 9.81 * 1400) = 36.109m/s
> 
> Solving for second phase time = 29.1248 / 2 / 9.81 / 0.9679 = 1.901s
> ...


Still, not too shabby a time, no matter which way you slice it.

Of course, my math could be well off (I'm using Windows 7's calculator in scientific mode, and as it's beta software anything could be happening!), and I'm sure someone who knows more about this stuff than me could jump in and correct me.

And yes, I know regenerative braking tapers off effectiveness at lower speeds, which is why I'm not advocating a complete removal of mechanical brakes. However, they would not need to be nearly so substantial as they are now, seeing as they would mostly be serving to slow the vehicle from low speeds (10kph and under) and holding the vehicle at park.


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## John (Sep 11, 2007)

Sorry my mistake solving for second phase time shouldn't have a divide by two in it. I have corrected the calculation below and my previous post.



Anaerin said:


> So, 4*120kW (Peak power out/input) = 480,000W
> 
> Originally Posted by *Windows Calculator*
> _Initial Retardation = 480000 / (9.81 * 1400 * 36.11) = 0.9679g_
> ...


1.2g is the retardation reached at the end of the first phase and throughout the second phase. The initial retardation isn't used in the rest of the calculation but just gives a sense of the initial decelerative force. You are right that the time is not much worse than friction brakes but it is with a top speed of just 130kph and a rather fat 480kW power absorption.

_Edit _PS. The e-traction SM350 30kW direct drive motors (not hub motors) weigh in at 85kg each so four would add 340kg to the vehicle. Even the 6kW kelly controllers motors weigh over 20kg each.


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