I have a motorcycle battery rated as 8.4 Ah (20 HR). It also says 8 Ah (10 HR) and I am not sure what the difference is.

Anyway, I have a GPS tracker fitted with a rated consumption of 5 mA. According to this formula I have to divide battery capacity by consumption:

8.4 Ah from battery / 0.005 A consumption = 1680 hours (70 days)

Is this correct? The reason why I am posting this is because the battery says 8.4 Ah, which is only double as many mobile phone batteries:

Why is this motorbike battery so heavy then if it's only double the capacity? Is it because it's a 12 V one? My knowledge in electronics is zero as you can see.

## Best Answer

The capacity decreases at higher discharge rates. The figures are telling you that discharge current of \$ \frac {8.4 \ \text {Ah}}{20 \ \text h} = 0.42 \ \text A \$ will last 20 h and \$ \frac {8.0 \ \text {Ah}}{10 \ \text h} = 0.8 \ \text A \$ will last 10 h. Given that doubling of discharge rate the total energy out is actually surprisingly close.

Correct method. (I didn't check your numbers but they look right.)

Two reasons:

energystored in the battery is given by \$ V \times I \times t \$ so a 12 V battery will have three times the energy storage of a 4 V battery. Your 12 V battery has a capacity of \$ 12 \times 0.42 \times 20 = 100 \ \text {VAh} = 100 \ \text {Wh}\$.Figure 1. Energy densities for various technologies. Source: EPEC.