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Help with determining battery draw

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Old 02-04-2018, 09:15 AM
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Default Help with determining battery draw

I am fixing to install an LED lighting kit on my 2017 FLHXS, and I am trying to determine how long I can run this kit and not have to worry about battery going dead to where I can't start it. This kit has a total of 12 strips and 14 pods. Pods pull .05 amp/hrs and strips pull .1 amp/hrs. Thanks for the help
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Last edited by Nismogsxr; 02-04-2018 at 09:18 AM.
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Old 02-04-2018, 09:23 AM
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It depends on how you are connecting the pods and strips either in parallel or in series.

12 strips at a current draw of 0.1 amp per strip = 1.2 amps.

Plus 14 pods at a current draw of 0.05 amp per pod = 0.7 amps.

Total draw 1.9 amps.

What is the ampere hour rating of the battery you have installed on the bike?

With a new totally fully charged battery divide the ampere hour rating of your installed battery by 1.9 and that will give you the total time you can run the LEDs without running the engine. Do not expect your bike to crank afterwards. If you are looking for a few hours say less than 4-5 hours you should have no worries.
 

Last edited by K9F; 02-04-2018 at 09:27 AM.
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Old 02-04-2018, 11:43 AM
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It's the OEM battery from Harley. I tried to locate the amp/hrs of the battery's but all I saw was the CCA
 
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Old 02-04-2018, 11:58 AM
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I just found that the Drag Specialties premium batter is rated at 30 amp/hrs
 
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Old 02-04-2018, 01:16 PM
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A 30 ampere hour battery over 6 months old and in a temperature of between 0 and 85 degrees Fahrenheit will discharge down to approximately 50% on the load of 1.9 amps after 6 hours.

Have a play with the calculator below. Hope it works for you.

https://www.batterystuff.com/kb/tool...to-a-load.html
 
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Old 02-05-2018, 09:32 AM
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leds are very complicated and can not be likened to batteries and incandescent lighting. each type of connections have issues whether series or parallel.
a whole lot of info is needed such as how is the current limited, aka, one resistor, multiple resistors or a sum of leds (each dropping a portion) and what type of power supplies.
leds have a IV curve and is not always consistent and vary greatly with temperature. ever seen a string of leds with one not as bright or dimmer than the others? ex: if your first string drops voltage, the next string will increase in current, good/bad?, leds can survive sometimes almost double the current draw as advertised but will not be long lived.
the best way to drive leds is with a power supply feeding a driver which then feeds the leds, the driver regulates the attached string and in case of multiple strings, driver for each.
as far as power consumption and battery, leds can have a wide margin of current draw and therefore, will be only a ballpark figure(remember temp affects draw).
what i would do if you want to show off the bike is to find a place for a secondary battery and check it with diodes so it does not back flow (isolates from other battery) and you will have a battery for starting yet both are charge-able from the alternator system. the other option is a bigger battery or hook up a voltage cut out system so the leds are dropped out when the battery voltage is pull to set point (tricky business here).


 

Last edited by bustert; 02-05-2018 at 09:35 AM.
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Old 02-13-2018, 09:43 AM
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The easiest way to know for sure is to:

1. Install everything.
2. Charge the battery to 100%
3. Put a volt meter on the battery.
4. Turn on the LED lights.
5. Record the voltage drop over time.

For an AGM type battery:

12.8v+ is 100% charged.
12.7v is 88% charged.
12.6v is 75% charged.
12.5v is 62% charged - probably your last chance to start the cold engine.
12.3v is 50% charged.
12.0v is 25% charged.
Less than 12v is a dead battery.

If you want to know how low you can go before you need a jump, drain to 12.7, then try to start...if it starts, charge back to 100%, then drain to 12.6...keep doing that until you hit a voltage where the bike won't start...note the voltage, and the time it took to discharge there,... and you know exactly how far not to push it...

Also, if your bike has a cig-lighter plug, you can buy a plug in voltmeter to make it easy to keep an eye on that voltage drop when you're away from home and don't want to mess with a volt meter.
 

Last edited by HDSlimJim; 02-13-2018 at 12:35 PM.
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Old 02-13-2018, 03:02 PM
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There's your answer ^
 

Last edited by Campy Roadie; 02-13-2018 at 03:59 PM.
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Old 02-14-2018, 01:35 PM
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while that sounds nice and fuzzy and warm and cozy, it only would work IF the system was consistent.
1. leds have a CURRENT/VOLTAGE curve and can change quite a bit
2. operational temperature has effects
3. inrush currents can demand way more from a battery with diff start senarios
4. batteries are in a constant decay process
5. many more one could consider

so yes if you want that approach, do install a voltage cut out relay but do not expect consistency.

when i drove an 18 wheeler, i ran a refrigerator so i set up the battery bank in a buck/boost configuration. at static, one battery was isolated from the bank and ran the refrigerator but when i started the truck, a relay used it to boost the others and all were together. not a very good option here but a secondary battery may/may not be the answer.

how about installing a manual foot if it can be done.
 
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Old 02-14-2018, 01:48 PM
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Originally Posted by bustert
while that sounds nice and fuzzy and warm and cozy, it only would work IF the system was consistent.
1. leds have a CURRENT/VOLTAGE curve and can change quite a bit
2. operational temperature has effects
3. inrush currents can demand way more from a battery with diff start senarios
4. batteries are in a constant decay process
5. many more one could consider
Trying to account for all of that with math is futile; you'll get much closer to the actual run time number by running actual run time tests.

Once he knows at what voltage his bike won't start, he can easily monitor that with a cig plug voltmeter and be sure to stay above that "won't start" voltage. I think he wants a simple solution...that's as simple as it can get. Second batteries, voltage cut-out, ect., I think is way more than he wants to get involved with.
 
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