Buddy,

Do you have a Lithium pack? If so, you are being way too conservative
limiting regent to only when your pack is below 80% SOC. You would be
better served to limit regen based on battery terminal voltage than on
SOC.

On Thu, Mar 20, 2014 at 12:04 AM, Buddy Mills <[email protected]> wrote:
> So much math...it hurts my head.   Regen good. It kicks in, the amp gauge
> goes positive.
> As for the setup;  I have a switch on the dash.  Position 1, it only works
> when the brake is applied.  Second position. it work when I let off the
> accelerator.  Third position it is turned off.  I also have a disengage
> switch on the clutch to keep the motor from coming to a dead stop when it is
> working off the accelerator and trying to shift gears.  And last but not
> least my BMS turns off regen if my SOC is above 80% as to not overcharge the
> battery pack.
>
> Buddy Mills
> [email protected]
>
> Look mom, no gas. http://www.evalbum.com/2887
>
> Disclaimer:  No animals were harmed or killed in the process of writing this
> email.  Any stories to the contrary are, for the most part, either fictional
> or greatly exaggerated.
>
>
> Cor van de Water wrote:
>
>> you have a total efficiency of:
>> 90%(batt) x 95%(controller) x 85%(motor) x 90%(drivetrain) = 65%
>
> Excluding the battery efficiency for simplicity and considering the
> situation from the point of view of energy provided by or received by the
> battery:
>
> Consider a 1000kg (approx. 2200lb) vehicle travelling 50km/h (approx.
> 30mph); it has a kinetic energy of 96450.6J.
>
> Using your numbers above, 96450.6J/(0.95 x 0.85 x 0.90) = 132715J were
> provided by the battery to achieve this.
>
> Regenning to a complete stop provides 96450.6J * (0.95 x 0.85 x 0.90) =
> 70095.5J to the battery.
>
> So, (0.95 x 0.85 x0.90) = 72.7% of the available kinetic energy is provided
> to the battery, however only 70095.5J returned / 132715J consumed = 0.528,
> or 52.8% of the battery energy consumed in achieving that kinetic energy
> level is recovered.
>
> Joules aren't a unit that means a lot to some of us (though SI units are
> easy to work this example in ;^), so let's convert it to something more
> familiar:
>
> 1J = 1W-second, so 70095.5J = 19.47Wh.
>
> If we assume that you accelerate to 50km/h and then regen to a stop every
> 500m (so you start from a stop at 0m and are again stopped at 500m), then
> out of every 500m travelled you recover 19.47Wh.
>
> So, how much does this extend your range?
>
> If your vehicle consumes 200Wh/mi (=200Wh/1609m) at 50km/h, then every 500m
> you recover enough energy to travel a further 156.6m: 156.6m/500m = 31%
> increase in range.
>
> Another way to look at it is that accelerating to speed after the stop
> consumes 132715J from the battery; this is 36.86Wh, or enough energy to
> travel 296.6m at constant speed.
>
> So, regenning to a stop allows you to recover enough energy to travel
> another 156.6m, but avoiding the stop in the first place saves you enough
> energy to travel 296.6m further.
>
> Obviously, the shorter the distance between each start and stop, the greater
> the % range increase appears.  How high can we push it?  Let's assume it
> takes us 5s to accelerate to 30mph (50km/h) and 5s to regen back to a stop.
> Let's assume that we accelerate to speed and then immediately regen to a
> stop, and continuously repeat this pattern.
>
> During each acceleration we travel 34.7m and during each deceleration we
> travel another 34.7m.  So, for every 69.4m travelled, we consume 132715J to
> accelerate, plus about 31055J (69.4m @ 200Wh/mi) = 163770J.  We recover
> 70095.5J, which is enough to travel (70095.5/163770) = 42.8% further in this
> inefficient driving pattern.
>
>
>
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-- 
David D. Nelson
http://evalbum.com/1328
http://www.levforum.com
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