12 volts to 48 volts for an electric bicycle

[[email protected]]

On Friday, September 21, 2012 2:26:23 PM UTC-7, amdx wrote:

....

I put a suggestion on the DIY electric car forum to mount a

motor/generator on a trailer that you can tow behind your electric car

for long distance trips. If it's a short trip say 35 miles just take the

car, for a longer trip connect the trailer and let it charge your

batteries while you drive.

Oh, neato!

I'll my video again, some of you have seen it and probably tired of

it, This is an electric gokart my 16 yr old son and I put together

about 3 years ago. Since then I overheated the original motor, we

installed 4 deep cycle group 27 batteries and a new motor. The

overheating was at a connection between one #4 wire and 7 #14 wires.

The motor windings were ok.

Nice!

Just wondering, why did you put more voltage on the motor than it was designed for?

Oh, and where did you get the batteries? Sounds like a deal to get used ones!

 

[[email protected]]

how are you charging these batteries, and are they balanced correctly?

Hmm, I'm pretty sure the four were new (two from Home Depot, two from Fry's). I returned them all though (well, except for one from Fry's, which fell out and had scuff marks on it).

12V battery maintainer from Wal-Mart. Schumacher brand, I think.

when they're dead from use with your bike, check the voltages across each

battery, if they're not close, something is wrong with one or more of

them, of they're just too chinese to even string up and get a nice 48 volt

string with. One off battery or even just one cell in one battery will go

into voltage reversal if it drains before the others, and that will really

kill performance as a whole as well as that battery.

Oh ok. I'll measure the voltage when I take the batts for a spin next week!

I use a gas engine on my bicycle and know from use starting and stopping

aka "city use" is the real killer with gas use. If I never stopped, way

over 200miles per gallon, if not more would be easy.

Ooh, a 30cc 2-stroke? How many mpg do you get with city driving?

Is it tolerably loud?

 

[amdx]

On Friday, September 21, 2012 2:26:23 PM UTC-7, amdx wrote:

...



Oh, neato!




Nice!

Just wondering, why did you put more voltage on the motor than it was designed for?

MORE POWER!!
The motor can take the amperage for certain amount of time.
Time at high amperage equals heat, to much time at high amperage
equals new motor
At start with the pedal down it will draw over 250 amps at 48 Volts.
At 90% efficiency that's 14 hp. It gets up and goes.

Oh, and where did you get the batteries? Sounds like a deal to get used ones!

I have a wrecker service that has a lot for their junk cars, some of
the wrecks still have good batteries, so I got some cheap. I only got
cheap ones until I was sure I had a working model. I spent $300 on
batteries once everything was working.

Mikek

 

[P E Schoen]

"amdx" wrote in message

MORE POWER!!
The motor can take the amperage for certain amount of time.
Time at high amperage equals heat, to much time at high amperage
equals new motor
At start with the pedal down it will draw over 250 amps at 48 Volts.
At 90% efficiency that's 14 hp. It gets up and goes.

The motor may have 90% efficiency at rated current and voltage, but at twice
rated current the I^2R losses will be four times, so you might get 40%
losses and 60% efficiency. You will get more power output, but more like
12kW * 0.6 = 7.2kW = 9.6 HP.

At some point you will get less power out with more amps. You can get more
torque, but if it is under nearly locked rotor conditions all the power will
be used up as heat. There is no output power if the vehicle is not moving.
HP = Torque * RPM / 5252.

Spinning the tires and doing burn-outs seem impressive but take less power
than maximum acceleration just before losing traction.

Paul
www.muttleydog.com

 

[amdx]

"amdx" wrote in message

The motor may have 90% efficiency at rated current and voltage, but at
twice rated current the I^2R losses will be four times, so you might get
40% losses and 60% efficiency. You will get more power output, but more
like 12kW * 0.6 = 7.2kW = 9.6 HP.

At some point you will get less power out with more amps. You can get
more torque, but if it is under nearly locked rotor conditions all the
power will be used up as heat. There is no output power if the vehicle
is not moving. HP = Torque * RPM / 5252.

Spinning the tires and doing burn-outs seem impressive but take less
power than maximum acceleration just before losing traction.

Paul
www.muttleydog.com

Ahh, shit! and I thought I was havin fun.

Mikek

 

[Cydrome Leader]

Hmm, I'm pretty sure the four were new (two from Home Depot, two from Fry's). I returned them all though (well, except for one from Fry's, which fell out and had scuff marks on it).

12V battery maintainer from Wal-Mart. Schumacher brand, I think.

They're probablhy all from the same grade school in china, but you never
know.

Oh ok. I'll measure the voltage when I take the batts for a spin next week!

Maybe it just a weak battery.

Ooh, a 30cc 2-stroke? How many mpg do you get with city driving?

Is it tolerably loud?

It's really not bad, it just sort of buzzes, and I sit in front of it, so
sounds really doesn't matter. It has a 25 or 30cc 2 stroke Redmax engine.
The thing starts no matter what, and has only needed the fuel tank gasket
replaced as the ethanol crap in gas eats that stuff up.

As for powered bikes, the real killer on gas or electric is pulling out of
a stop. Mine can't since it has a leafblower clutch, so I have to pedal to
get going. If you're lazy with that, the milage drops like crazy per tank
of gas. Once you're moving, the thing might as well be idling, unless
you're going up hill.

 

[[email protected]]

"amdx" wrote in message

The motor may have 90% efficiency at rated current and voltage, but at twice
rated current the I^2R losses will be four times, so you might get 40%
losses and 60% efficiency. You will get more power output, but more like
12kW * 0.6 = 7.2kW = 9.6 HP.

Power is power. If your losses are 4x, so is your useful power (assuming
nothing lost in the wiring to the motor - a design issue). Unless you have
nonlinear resistance in there somewhere.

At some point you will get less power out with more amps. You can get more
torque, but if it is under nearly locked rotor conditions all the power will
be used up as heat. There is no output power if the vehicle is not moving.
HP = Torque * RPM / 5252.

That's a different issue and true at any current. Higher is better, since it
won't stay locked (as long).

Spinning the tires and doing burn-outs seem impressive but take less power
than maximum acceleration just before losing traction.

Less *useful* power, for sure.