Run time reality check needed

BatteryMooch

10 kW
Joined
Dec 13, 2008
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New York City, USA
Don't know why I'm having so much trouble getting my head around this calculation. It's not for an e-bike but I'm hoping you guys can help. :)

The question is...If I have a 300uA constant load on a pack, how long will it take for 1.8Ah of capacity to be used by the load?
My calcs were as follows...
300uA constant = 300uAh of capacity drawn each hour
1.8Ah / 300uAh = 6000

So it would take 6000 hours to drain 1.8Ah at that rate?

Thanks for the reality check!
 
Not kidding. I've had 10 hours sleep since Friday and I'm beat and perhaps not thinking clearly when it comes to this stuff. Don't have time to waste with a silly post. :)

Need a reality check regarding this calculation.
 
Other than self-discharge of the pack, then that sounds right (then again, I can't add two and two and get five each time...er...three, I mean). If it is a NiMH pack, for instance, it'll kill *itself* by self-discharge *loooooong* before your load ever would. :lol: OTOH, a lithium pack would probably last at least 90% of that 6000h, maybe more. :)
 
1.8Ah/0.0003A = 6,000hrs.

Sleep deprivation happens.
 
I remember in school, after a certain point at night you have to start using scientific notation, becuase counting all the 00000000s is no longer possible :twisted:
 
amberwolf said:
Other than self-discharge of the pack, then that sounds right (then again, I can't add two and two and get five each time...er...three, I mean).


2.47 + 2.47 = 4.94

(|| 2.47 || = 2) + (|| 2.47 || = 2) = (|| 4.94 || = 5)

2 + 2 = 5

The same sort of beer logic could make 2+2=3
 
I don't even know what the symbology in that statement means...but I can presume from context that it means to chop off all but the integers, rounding up or down as needed. Math, programming, musical notation...can't really wrap my head around any of them. Intuitively I do pretty well, but if I try a logical progression especially written out formally with math, I'm doomed.
 
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