Comparing battery power/achievable mileage question

I was encouraged to see someone say recently that there's no such thing as a stupid question, so I'll push the boundaries of that theory with this one

 

Take a hypothetical situation where you have a bike that has, say a 36V, 250W 6 Ah battery. And if you rode that bike on a certain day, let's say you got 14 miles out of the battery using pedal assist over rolling hills.

 

Now, if you could roll back time and take exactly the same journey, apply exactly the same amount of leg power to it - but this time you'd swapped out the battery for a 9 Amp Hour one, would you get 50% more miles (i.e. because of the 6 to 9 Amp upgrade)?

 

 

Basically, I'm trying to calculate whether a bike I'm thinking of buying has the potential to give the mileage on one charge that I need. Currently, my wife lugs around a second battery for any journey over 12 miles, which isn't ideal.

 

Thanks for any help.

Hi YES Amp hours is the amount off amps X volts storage for one hour so the amount off electricity stored

 

Frank

OP, use WattHours for the basis of your calculation. Wh is the amount of energy available in the battery. Wh is the Ah multiplied by the voltage so 36v x 6Ah = 216 Wh

 

If you are getting 14miles from 216Wh then your consumption is 15.43 Wh / Mile. Quite a typical figure for many.

 

So base your distance calcs on an assumption that you will use 15.5Wh / mile, that means your 9Ah battery at the same voltage would give you a range of approx. 21 miles....

 

I should add if you are using SLA batteries then you must reduce the stated battery Ah by 30% due to a strange effect called the peukert effect

 

http://en.wikipedia.org/wiki/Peukert's_law

Edited August 24, 201114 yr by NRG

In practice you actually gain a little more with an increase in battery capacity than the arithmetical calculation indicates.

 

This is because batteries work by chemical reaction and the stress of that affects the true capacity. The higher the nominal capacity, the lower the stress of a given current draw rate. The lower stress results in greater efficiency and thus less heat loss. Since heat loss is wasted energy, you gain that reduction in slightly extra range and slightly extra performance. The latter is sufficient to even be noticeable to the rider.

OP, use WattHours for the basis of your calculation. Wh is the amount of energy available in the battery. Wh is the Ah multiplied by the voltage so 36v x 6Ah = 216 Wh

 

If you are getting 14miles from 216Wh then your consumption is 15.43 Wh / Mile. Quite a typical figure for many.

 

So base your distance calcs on an assumption that you will use 15.5Wh / mile, that means your 9Ah battery at the same voltage would give you a range of approx. 21 miles....

 

I should add if you are using SLA batteries then you must reduce the stated battery Ah by 30% due to a strange effect called the peukert effect

 

Peukert's law - Wikipedia, the free encyclopedia

 

Thanks for that clarification. I'm guessing that because Peukert's Law affects Lead batteries, it won't be a factor with LiIon or LiFePO4s?

 

I have a follow up question. If I'm considering buying Bike A that has a LiIOn battery and comparing it with buying Bike B that has a LiFePO4 battery, then presumably I can take it that the WattHours rule is the same? It's just that the LiFePO4 battery is expected to have a longer life, is it?

 

Ged

Yes, it can be more or less ignored for Lithium batteries although take note of what Flecc said re Ah size....the Wh calculation is the same.

 

 

 

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Lifepo4 batteries are lower nominal voltages. So with that in mind - to reach the same Watt's the physical size of the battery will be larger.