Math's problem - Page 1
Forum section: electronics / general

author: PICmonster
author_rank: Regular Contributor
timestamp: January 14, 2014, 12:39:30 pm
content: Greetings GuysI am new to thius forum so please excuse me if I miss something or I am not on the right forum, I call upon thee engineering geeezers for help in confirming if my calculation is wright or wrong, I want to see how you all approach the math's problem and if you get similar figures to me or not.As in apollo 13 when individual engineers used slide rules to get the same figure for the stricken crew problem, this is an indication that the most common answer was accurate and arrived at by individuals using most probably different techniques to solve the same problem.So if we all concur then it must be right is my thinking,this is the basis of scientific truth is it not.Here what its about, I am attempting to calculate the life span of the batteries we are using, the particulars are as follows.You will need this formula, Ah = I(Amps) * t(Hours)Parameters(1) 5Ah battery that we have to consider has 80% start capacity, so that 4Ah you need to use in the calculation (5*0.8 = 4).(2) Circuit 1 that consumes 0.3A for 4 Minutes/Month, calculate total Ah in one month please.(3) Circuit 2 that is consumes 4.7mA for 100mS, every 3.9 Seconds continuously, calculate total Ah in one month please.(4) Sum the Ah from above for total Ah per month.(5) Divide 4 Ah by total Ah per month to give number of months life, now convert this to years life.I get 3.09 yearsPlease let me know if you require any further details.RegardsJim

author: AndreasF
author_rank: Frequent Contributor
timestamp: January 14, 2014, 12:52:27 pm
content: Calculations look good to me (mathematically).I can't agree with this statement though:Quotescience is not a democracy!

author: grumpydoc
author_rank: Super Contributor
timestamp: January 14, 2014, 01:00:34 pm
content: This sounds a little too like a homework question, and one which hardly needs the combined minds of the eevblog forum members but what the heck.QuoteUh, yes - I hope that most forum members could figure that out unaided.FWIW I get a similar, but not quite equal, answer so you are in the ballpark.The difference could be what you define as "a month" - for simplicity I used 31 days, but the average month length is actually closer to 30.4 days.For a real application you need to consider self-discharge which will depend on the battery chemistry - which you haven't told us.Quote from: AndreasFThe majority isn't always right and as Andreas points out science is not about some consensus view being correct. If it were then the Earth would, indeed, be flat with the sun revolving around it since that was the majority, consensus view for many, many years.Edit: BTW it's a "Math problem" or possibly a "Maths problem" but definitely not a "Math's problem" - I realise your first language might not be English which is why it is so useful to set a country in your profile to avoid grammar pedants (like me)

author: Excavatoree
author_rank: Frequent Contributor
timestamp: January 14, 2014, 01:10:24 pm
content: Another problem is that battery discharge is not perfectly described by one capacity rating, it will vary according to load. A battery may have a higher number of ampere hours at a low discharge than it will at a high discharge. A 10 ampere hour batter, for example, may not supply 10 amperes for an hour - it will give up before then. Likewise, it may be able to supply 100 milliampers for more than 100 hours. Battery discharge is a complicated topic.Self discharge is significant, as grumpydoc said. In many cases, a battery will go flat after a few weeks with no load.

author: grumpydoc
author_rank: Super Contributor
timestamp: January 14, 2014, 01:54:01 pm
content: QuoteThis is very true - the only reason I didn't mention it was that the discharge rates from PICmonster's problem will mean he'll run into self-discharge but not into de-rating because of heavy current loads.Temperature also plays a part, generally batteries have lower useful capacity when they are cold. In fact get cold enough and they can stop working altogether which is why a car battery which was marginal can start the car just fine in the summer but then fail when the weather gets colder.

author: KJDS
author_rank: Super Contributor
timestamp: January 14, 2014, 02:11:28 pm
content: You've got the answer wrong, though the arithmetic is right.Self discharge, temperature variations and most importantly someone borrowing the batteries out of the camera to put in the TV remote have been omitted.C- could do better.

author: GEuser
author_rank: Frequent Contributor
timestamp: January 14, 2014, 02:27:25 pm
content: Quote from: grumpydoc on January 14, 2014, 01:00:34 pmCould you please set your country to another one then please lets say usa for instance , there are other pedantics <(more than 1) around that say Math is singular and goes nowhere as in 9 multiplied by ,, to do non singular mathematics whilst being mathematic would be 9 multiplied by 24 going by my maths ...dost thou or thee follow .

author: PICmonster
author_rank: Regular Contributor
timestamp: January 14, 2014, 02:35:10 pm
content: Thank you to everyone who has answerd my post, I realise that there are there aspects such as temperature and self discharge amonst others mentioned, I was mearly looking to see if other members arraived at the same or approximate figure so at least I know they will have worked this out in a similar way to myself.Thanks againTo try and answer some questions raised in the post the battery is Litium Manganese Dioxide 3V 5Ah, its datasheet states a continuous discharge rate of 1500mA, this is non-rechargable technology and generally considered fully discharged at ~2V, I am aware the capacity is load dependant.