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Daox · 04-29-11, 01:59 PM

Great. The calculation is pretty simple.

103Ah / 1.9A = 54 hours.

Of course, you don't want to drain a lead acid battery down that far because it'll reduce the number of cycles you get out of the battery. For optimal life, 50% is used for lead acid batteries. So, you really only have 51.5 Ah available to use, or 27 hrs.

If you were drawing more power you'd also have to take into effect the Peukert effect which basically says the faster you drain a battery, the less capacity it able to be used. However, the amperage you're pulling is so small compared to the battery size that it won't matter.

04-29-11, 01:59 PM	#5
Daox Administrator Join Date: Aug 2008 Location: Germantown, WI Posts: 5,525 Thanks: 1,162 Thanked 374 Times in 305 Posts	Great. The calculation is pretty simple. 103Ah / 1.9A = 54 hours. Of course, you don't want to drain a lead acid battery down that far because it'll reduce the number of cycles you get out of the battery. For optimal life, 50% is used for lead acid batteries. So, you really only have 51.5 Ah available to use, or 27 hrs. If you were drawing more power you'd also have to take into effect the Peukert effect which basically says the faster you drain a battery, the less capacity it able to be used. However, the amperage you're pulling is so small compared to the battery size that it won't matter. __________________ Current project - To view links or images in signatures your post count must be 0 or greater. You currently have 0 posts. To view links or images in signatures your post count must be 0 or greater. You currently have 0 posts. & To view links or images in signatures your post count must be 0 or greater. You currently have 0 posts. Last edited by Daox; 04-29-11 at 02:04 PM..