# how is a controller supposed to work?



## frodus (Apr 12, 2008)

> Assuming that your normal commute speed is closer to 50mph does the controller holds motor voltage below pack voltage and the motor is running slower than 3600rpm? These different speeds mean different amps draw from pack right? ie. slower means less power is used.


not neccessarily. You can be drawing the same power at lower RPM because of the load on the motor. You could be drawing twice the amps at half the RPM (half the voltage) and get the same power.



> But if the speed stays at 50mph and you have gear shifting transmission you also can controll the motor to drive shaft ratio and make the motor turn slower or faster. How would this effect the controller and the motor? If the speed and road conditions are the same would the power CONSUMED be different at all if the motor is turning faster or slower and why? Does the gearing actually change the power consumed at speed?


well, shift into a low gear and you have high RPM, low amp draw, but you don't go that fast. If its in a higher gear, then you put more load on the motor when its at that lower RPM, causing a higher current draw. It does help though, because motors have an RPM limit. The power could be the same in a low gear and higher gear. It all depends on the load and how you're accelerating.




> Does the volatage of the pack effect the range / efficiency and why? If the motor is designed for 24V and 3600rpm and uses 15Kw? What would a 72V motor that is also rated for 3600rpm and 15Kw make to the system?


it can.

At 24V and 15kw, you have 625A draw on the batteries. If they're 100Ah batteries with a 3C (300A) continuous rating, you're stressing them. which will cause range to drop significantly.

at 71V and 15kw, thats 208A draw on the batteries. With the same 100Ah batteries, they're well under their continuous rating, so you're not really stressing them.

Research peukert effect and look at some lithium discharge curves. You'll see what I mean. The more amps you draw continuous out of a battery, the less overall energy you'll get out of the pack. Energy determines range.


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## dmac257 (Jun 30, 2010)

frodus said:


> not neccessarily. You can be drawing the same power at lower RPM because of the load on the motor. You could be drawing twice the amps at half the RPM (half the voltage) and get the same power.
> 
> 
> 
> ...


OK .. assuming the same flat road, driver, vehicle, state of charge, steady speed, all things the same .. slower speed means less power drawn from the traction pack correct??

and assuming the above, would it matter which gear you are in as far as power consumed if you are not changing the ground speed??

and if I understand you correctly, whatever gearing you use, the higher the voltage the lower the current for the same power draw and batteries last longer if low current draw and there is less of a drop in voltage on the pack as well. But in order to get that 72V 100Ah I would need 3times the batteries. All in series would still be 100Ah but not stressing the pack. If three times the batteries at 24V 100Ah three parallel pairs WOULD the current be 208a and no stressed pack also just much higher MOTOR current?? So same number of batteries but different voltage motors. Higher voltage motor would be ABLE to turn faster with lower MOTOR current.

If I understand correctly, then I am ready to start making my preliminary plans for my first built/convert

dmac257


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## frodus (Apr 12, 2008)

> OK .. assuming the same flat road, driver, vehicle, state of charge, steady speed, all things the same .. slower speed means less power drawn from the traction pack correct??


it depends on what gear you're in. It all depends on the LOAD. If you're loading the motor down by putting it in too high of a gear, you draw lots of amps, even at lower voltage, and it might be the same power drawn as a higher voltage and lower current.


> and assuming the above, would it matter which gear you are in as far as power consumed if you are not changing the ground speed??


Ever shift a car from 1st to 5th? what does it do? It bogs down the engine. Same thing here, if you load the engine too much, amps go up, power goes up even though voltage may stay the same.



> and if I understand you correctly, whatever gearing you use, the higher the voltage the lower the current for the same power draw and batteries last longer if low current draw and there is less of a drop in voltage on the pack as well. But in order to get that 72V 100Ah I would need 3times the batteries. All in series would still be 100Ah but not stressing the pack. If three times the batteries at 24V 100Ah three parallel pairs WOULD the current be 208a and no stressed pack also just much higher MOTOR current?? So same number of batteries but different voltage motors. Higher voltage motor would be ABLE to turn faster with lower MOTOR current.


motor current is NOT battery current, almost never is.

for an EV, i would NOT use a 24V motor. For a car, you should do at least 96V. For a motorcycle at least 48V. 

you keep thinking that everything is linear, its not. There's a torque curve. As RPM increases, there's not neccessarily an associated current draw. Current draw is all dependant on LOAD. If you try to accelerate heavily at low RPM, you may draw the same amount as you would trying to accelerate heavily at high RPM. Its all dependant on the motor you use.


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## dmac257 (Jun 30, 2010)

frodus said:


> it depends on what gear you're in. It all depends on the LOAD. If you're loading the motor down by putting it in too high of a gear, you draw lots of amps, even at lower voltage, and it might be the same power drawn as a higher voltage and lower current.


I think my problem is that what I think I am asking is not what is actually what is typed. (or I am just too dense)

Are you saying that the LOAD on a motor is not the same thing as the POWER consumed by the motor.

I read that you can calculate the power required to push a vehicle at a given speed. I thought that made sense at the time, but now I am confused. Does the interposition of a gearing mean that if you drop from 5th gear on a flat road to 4th gear to maintain the same speed would require a different amount of power? Untill I understand how the motor speed effects how much power is drawn I dont care about the batter load part. If the motor is only capable of 3600rpm and that rated speed will give road speed of 60mph, but you are not driving as fast as you COULD go, are you using less power? What I am trying to grasp is if slower road speed means less power consumed and that the motor will turn slower AND if gearing to make the motor turn the rated speed at that slower road speed would consume more power or the same.

These are different situations.. I origionally asked too many questions and got too much information all at once. 

Thanks,
Don


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## Duncan (Dec 8, 2008)

Hi Dmac

Forget the controller and the electric motor, for a minute

It takes FORCE to maintain your vehicles speed (think of somebody pushing it)(measured in Newtons)

WORK = FORCE x DISTANCE (this is engineering work or energy) (measured in Joules) (1 Joule = 1 Newton x I meter)

POWER = rate of doing WORK = FORCE x DISTANCE per second (measured in Watts)
1 Newton x I meter / second = 1 watt

As your car goes faster the FORCE required increases - try putting your hand out of the window at 60mph

But there is a double hit because you are going faster the FORCE increases and the DISTANCE per second increases

Back to your car
It takes about (Dependant on the car) 20,000 Watts - 20Kw to drive a car at 60 mph

You will need the 20Kw - It can come from a small motor running at 10,000 rpm or a monster running at 1000 rpm 

The gearbox does not change the POWER but it can enable the motor to run at an rpm where it can deliver the power needed

You can normally get more power by increasing the revs of a motor until it blows up 

Now electric motors - DC Series Motors

Your battery pack provides the voltage to drive current into the motor 
Motor torque (PUSH) is roughly proportional to current 

The motor provide a back voltage proportional to motor rpm that fights against the battery voltage - so if you have too low a battery voltage you are limited by the motor speed below the blow up speed

Current provides torque - but it is also a limiting factor as it heats everything up and most batteries prefer lower current

So you can drive in top gear with lower rpm and increased current
or in a lower gear with higher rpm and reduced current

Does that help??


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## Franky.EV (Feb 27, 2010)

You can see the effect of gear shifting in this ev calculator : http://www.diyelectriccar.com/forums/showthread.php/yes-another-ev-calculator-45278.html

The best is to start with single gear reduction, and you can see you have to optimize the ratio to go fast, say 200 km/h or to get a good acceleration in the 0-100km/h range.

You can't have both, unless you get a big motor.

So you have to use gearbox, and you see ICE gearbox are not optimized for ev, with an ev you can use only 2 ratios.


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## dmac257 (Jun 30, 2010)

Franky.EV said:


> You can see the effect of gear shifting in this ev calculator : http://www.diyelectriccar.com/forums/showthread.php/yes-another-ev-calculator-45278.html
> 
> The best is to start with single gear reduction, and you can see you have to optimize the ratio to go fast, say 200 km/h or to get a good acceleration in the 0-100km/h range.
> 
> ...


I downloaded that calculator.. and then installed openoffice.org so i could look at it... uh .. is there a textbook to explain all those numbers


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## Franky.EV (Feb 27, 2010)

I have commented most cells (place the mouse over the cell with a red dot).

You can enter your values in any blue text cell.

Concerning gearbox, change the value in the right part (I give as an example Lotus Elise, Mitsubishi and Tesla gearbox).

And choose your gearbox in the selection list, the values should update.


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