Hi All,

This is an update regarding the school project, originally designed around the etek motor. Sourcing, cost, time, and a couple other reasons persuaded us to go with the Leeson motor that was previously mentioned. BTW it is 3hp 3000 rpm 86 amp, duty: INT, with 84 in/lb of torque @ 36V. It is a bit powerful but it was free...

For the time being we plan to run it at 36v. We may step down to 24v so battery mounting can be symetrical, but that would require a level of amperage that is not feasable and will affect my torque...right? (Also is the voltage constant simply: rated rpm/rated voltage)? i.e. 3000/36 = 83.3rpm/volt???

With help from UsaTracy and many other I crunched some numbers and came up w/ the following regarding batteries..I'm pretty sure I messed up...lol)

3hp = 2238W = 2500W needed for 1 hour...1250W for .5h

65% needs to go to motor(Please tell me where 65% comes from...is it the chemistry?)..therefore thats approx 815W/h need to supply in capability

815/36 = 23 Amps average draw

which is 46 Amp draw over 30 min

Thats how far i got...THE THEORETICAL APPROACH

Then theres the issue of discharge rates...ahhh...lol

now...the more practical LETS RIDE & TEST APPROACH

Power sonic has some very cool batteries, I was recommended to them. A particular 35AH battery will provide a peak draw of 100Amps for less than or equal to 7min. And guess what since i will only need to have to "operate/navigate w/" the scooter for 30min theres no chance I will peak out at 86amps for 7min continous. So technically this battery will do the job i think.

Presenting this problem to many technical individuals, the latter approach is the more practical common way.

the link...http://www.power-sonic.com/site/doc/prod/99.pdf

Eek thats all for now, I hope im not stressing too much..lol

Many Thanks for your continued support!

I'm sorry should the required power (at full draw) be 3096W. 86A X 36V?????

Yes, unfortunately the full power needed is 3096 W. Actually it's more, because there are losses in the controller (and wiring, contactor, connectors, ...).

I haven't followed the discussion but I guess the 65% might be a way of describing how much energy you can expect to get out of the system, compared to the nominal capacity. If you are planning to run the motor at 2500 W for half an hour, you would at least need 2500*0.5/0.65 = 1923 Wh of nominal battery capacity. Or more. Only perhaps 65% of the nominal battery capacity will actually find its way to the motor.

I am also confused about the 3 hp, 36V/86A specification. For that to be true, the efficiency of the motor is 3*741/3096 = 72%. That doesn't make sense, you'd better check the figures again.

But anyway, let's say you plan for 2000 Wh battery capacity. At 36 V, that's 2000/36 = 56 Ah. So you buy three 12 V / 56 Ah batteries. Batteries are rated at 20 h current draw, but when you pull more current, capacity will be lower. Much lower. With some luck, you might get 70% out of nominal, i.e. 0.7 * 56 Ah = 39 Ah. Another few percent will be lost in the controller and wiring. So you get maybe 0.95 * 39 = 37 Ah to the motor. That means the motor can pull 1332 Wh, or 2500 W during 32 minutes.

Remember, this is the absolute minimum of needed battery capacity (in the example). Lead acid batteries age quickly, so soon you will only have 70-08% of the capacity left. I'm not sure if it matters in a school project, but in a real world application you want to design so that you normally use only perhaps 50% of the (lead acid) battery capacity.

Very interesting school project btw. I wish I could have done things like that at school.