Dauntless said:
swbluto said:
You seriously don't have a cheap little voltmeter?
It's pretty simple measuring the capacity. (energy)capacity = INTEGRAL( v(t)*i(t),dt) [(amphour)capacity = INTEGRAL(i(t),dt)]
A simple Riemann sum is good enough, IMO. Don't really need to come up with a polynomial approximating curve for the v(t),I(t) curves. (For the less mathematically trained, you just sum up the voltage*current*time_step values over time, with measurements every 1-60 minutes(The time step); more frequent measurements makes for more accuracy)
Now, if you're saying you lack a suitable load (Power resistors, bulb, nichrome wire, etc.) and a couple of alligator clips, I feel for you.
Those 200w 10ohm variable resistors you can get from China via ebay for $30 are awfully nice for this kind of thing.
If only I had a functioning lab power supply (For charging), I might offer to do some free testing for the community.
We're back to my being the home tinkerer rather than the engineer. Of course I have the voltmeters, but your description is still vague. I'm to assume that I stand there and keep the voltage level for every x minutes of my choice? I have an AA in math, when I see that 'i' I think imaginary number. Is this a definite integral? I don't see no c.
Is the Riemann Sum going to work similar to the 200 day moving average on Wall Street? Specifically for the trending downward, of course.
I'm assuming you're more concerned about amp-hour capacity.
So, hook up two voltmeters, one for voltage and one for current.
The one for voltage is just to make sure you don't over-discharge. (Or, if you're trying to caculate watthours(energy capacity), you'll use it in the calculations.)
Hook up the load, and every 10 minutes, record the current. Stop recording when fully discharged.
Then, do this calculation.
(10/60)*I_0+(10/60)*I_1+(10/60)*I_2+...
(Where I_x is the current you recorded at every 10 minutes; I_0 is starting current (In amps), I_1 is the current 10 minutes later, I_2 is the current 20 minutes later,...)
That sum will give you the battery capacity in amp-hours.
Easy peasy lemon squeezy
(Note, you can do more frequent measurements for higher accuracy, taking care to change the '10' number you see in the formula appropriately. Conversely, you can do less frequent measurements, reducing accuracy.)
Note, the keen observer will notice that 10/60 is common to all the terms in the sum, which can be factored out, leaving...
(10/60) * (I_0+I_1+I_2+...)
Implying it's as easy as adding all the currents you measured every 10 minutes together, and multiplying that number by 10/60(Which is the time step, in units of hours) to get Amp-Hours.
EDIT:
The correct formula is
(10/60) * (I_1+I_2+I_3...)
That is, don't count the initial current at time 0. If you do want to account the initial current, you should remove the last current measurement from the sum that marks the exact end of the test, otherwise, an artificially high amphour count will result.
Measurements taken over time should look like
I_0 .......... I_1..........I_2............
Or
............I_1...............I_2..............I_3
Measurements that look like
I_0..............I_1............I_2...........I_3
with measurements at both ends of the test, if both are added to the sum, it's wrong.
