# Derivation of Formula

This is the method for calculating the State of Charge (SOC) of the 12V lead acid vehicle battery. Based on the formula for battery capacity (C) formula

$$C = (\text{Current}) \cdot (\text{Time of Discharge})$$

and my approximation that under loaded conditions (even 1 amp) if the battery voltage reaches 10.5 the SOC is 0%.

So, if the present battery voltage is say $V_1$ and on applying a load of say $I_1$ the voltage has fallen to $V_2$ and the present SOC is $\text{SOC}_{\text{present}}$. Then the formula for the new SOC is

$$\text{SOC}_{\text{new}} = (V_1 - V_2) \cdot \dfrac{\text{SOC}_{\text{present}}}{V_1 - 10.5}$$

This is the final equation I derived after intermediate calculations. But my main concern is when you crank the vehicle the voltage falls sometimes below 10.5 and hence my SOC is 0% in fact less than 0% which is not practical and returns back to normal SOC after vehicle starts though.

So, my question is can I omit that one condition and still say the formula is valid? If required I can show the intermediate calculations. For simplification I made lot of approximations like neglecting peukart equation, temperature affects which in fact will be reflected in the voltage is my assumption.

• What? "In science, a formula is a concise way of expressing information symbolically" - Wikipedia. So, can it be defined this way? – Eugene Sh. Jul 28 '15 at 17:10
• This question needs more details. – K. Rmth Jul 28 '15 at 17:53
• I'm voting to close this question as off-topic because lacks all sort of details and, as it stands, the only "question" is about math terminology (when a formula can be called a formula). – Lorenzo Donati Jul 28 '15 at 18:25
• Sorry for not providing additional details. Is the thread closed or can I give further details. – rajesh Jul 29 '15 at 2:33
• Rajesh, you can always add more details, even if the question is on hold. If it is on hold (it's not, yet, but it's close to being put on hold), your edit may help open it again. So, please, feel free to edit it and add the details. – Ricardo Jul 29 '15 at 11:32

## 2 Answers

So your formula does not work in all cases. You can change your formula to a conditional formula like this:

$y = \left\{ \begin{matrix} \text{(your formula)} & , & \text{if } cranking=0 \\ \text{unknown} & , & \text{otherwise}\end{matrix} \right.$

The range (set of possible outputs) of this formula is no longer just real numbers, but real numbers plus "unknown". Whatever you use this formula for will need to account for this. For example, if you will be averaging the values, you could just ignore "unknown" values unless a certain number of them accumulate. Assuming the "cranking" case is short lived, the resulting averaged/filtered output may be acceptable.

Another option would be to make the formula explicitly time-dependent. For example, if you know that cranking always lasts less than 5 seconds and reduces the state of charge by 2%:

$y(t) = \left\{ \begin{matrix} \text{(your formula)} & , & \text{if } cranking=0 \\ y(t-5\text{ seconds}) - 0.02 & , & \text{otherwise}\end{matrix} \right.$

You can keep adding more conditions. Perhaps the above 2% reduction only applies above 50% and below that it is a 1% reduction. You'll have to measure and refine until your model matches reality well enough for your purposes.

• Thank you very much for that reply. Do you agree with my formula during non cranking conditions or it is flawed one, because till now I could not find out whether it is correct or wrong.Can you provide any comments. – rajesh Jul 31 '15 at 15:34

Your formula suffers from several serious flaws.

1. It doesn't take into account voltage drop due to current flowing through the internal resistance of the battery.

Even a small load can cause significant voltage drop. Surface charge on the plates only keeps internal resistance low for a short time. When current is drawn continuously the resistance increases as ions move through the bulk electrolyte, so voltage drops even lower.

1. You assume that voltage drops linearly as the battery is drained.

In reality the output voltage of a Lead-Acid battery drops slowly over most of the discharge, then steepens dramatically towards the end. So your formula will find the point at which it goes 'flat', but not accurately determine SOC at intermediate points.

1. Capacity reduces at higher current. A battery that delivers 60Ah at 3A ('20 hour' discharge rate) may only get 30Ah at 60A ('1 hour' rate).

Considering the nonlinear voltage curve, variation with current draw, and capacity reduction at high current, your formula is almost useless. To get a more accurate result you need to measure (or at least estimate) current draw. With voltage and current measured simultaneously you should be able to develop a formula that provides meaningful SOC figures throughout the discharge cycle.