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Discussion Starter · #1 ·
Hi All,

Is the voltage vs speed relationship basically a straight line? I am
currently running a 48 volt nominal pack and have a top speed of 32 MPH.
The controller is not in current limit at only 120 amps with a constant
current rating of 325 amps. If I up the pack voltage to 72 volts, could I
expect the top speed to increase to approximately 48 MPH? This is assuming
I don't change the ratios and the higher voltage controller doesn't end up
in current limit. The expected controller is a Curtis 1209B with a 1 hour
rating of 175 amps.

I am contemplating going from the current pack of 8 T145s which is 48V 265
AH to possibly a pack of 6 Genesis XE60 which would be 72V and 62 AH. I
realise this would cut my range down to 25-30% of what I have now. I would
look at the XE90s but I am told the lead time is 22 weeks.

respectfully,
John

The Skunk
http://www.austinev.org/evalbum/751

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Discussion Starter · #3 ·
No. For many motor technologies (brushless DC, permenant magnet DC,
synchronous AC), doubling voltage doubles speed. For wound-field DC
motors, though, speed is only limited by the output torque and input
power. Lowering output torque makes it go faster (lower rolling
resistance and better aerodynamics). Increasing input power will also
make it go faster (increasing current/voltage).

The power required to drive goes up according to speed cubed (plus a
linear term), though, so doubling voltage won't double speed.

There's equations you can use to calculate the approximate power
required to drive at a certain speed, and from there you can use
voltage and current limits to find how fast you can go on a certain
power. You may have to make educated guesses on some of the numbers,
though.

Max power your motor puts out:
A=max amps
V=battery volts
E=efficiency (0.6-0.8 for DC)
P=A*V*E

Power required to drive in 0 wind:
Af=frontal area of your car (google it)
Cd=coefficient of drag for your car (google it)
V=speed the car is traveling
Crr=coefficient of rolling resistance (depends on your tires, roughly
0.01 to 0.03, much much bigger on hills)
P=0.5*Af*Cd*V^2+Crr*V

So you can find out how much power your motor can put out and how much
power your car needs to drive, and this will give you an approximate
top speed.

-Morgan LaMoore

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Discussion Starter · #4 ·
> Hi All,
>
> Is the voltage vs speed relationship basically a straight line?

It would be if the load (torque output) remained constant. Unfortunately
that isn't the case here.
The torque needed to overcome aerodynaic drag goes up as the square of the
speed (or is it the cube, I always get that messed up).
Anyway, the higher torque means that the motor will slow down from a
linear relationship to a curved one.

I am
> currently running a 48 volt nominal pack and have a top speed of 32 MPH.
> The controller is not in current limit at only 120 amps with a constant
> current rating of 325 amps. If I up the pack voltage to 72 volts, could I
> expect the top speed to increase to approximately 48 MPH? This is
> assuming
> I don't change the ratios and the higher voltage controller doesn't end up
> in current limit. The expected controller is a Curtis 1209B with a 1 hour
> rating of 175 amps.
>
> I am contemplating going from the current pack of 8 T145s which is 48V 265
> AH to possibly a pack of 6 Genesis XE60 which would be 72V and 62 AH. I
> realise this would cut my range down to 25-30% of what I have now. I
> would
> look at the XE90s but I am told the lead time is 22 weeks.
>
> respectfully,
> John
>
> The Skunk
> http://www.austinev.org/evalbum/751
>
> _______________________________________________
> For subscription options, see
> http://lists.sjsu.edu/mailman/listinfo/ev
>


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Discussion Starter · #5 ·
Peter VanDerWal wrote:
> > Hi All,
> >
> > Is the voltage vs speed relationship basically a straight line?
>
> It would be if the load (torque output) remained constant. Unfortunately
> that isn't the case here.
> The torque needed to overcome aerodynaic drag goes up as the square of the
> speed (or is it the cube, I always get that messed up).

Fortunately, it is the square.
--Steve

> Anyway, the higher torque means that the motor will slow down from a
> linear relationship to a curved one.


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