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Old 02-09-2014, 03:11 PM   #31
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Run the furnace and drive an inverter is gonna be hard on a single battery. Even with two (depending on other draws - like incandescent lights), you may find you won't get through the night depending on the state of charge prior to quiet hours at the campground).

Knowing the actual % of capacity remaining is critical when there are quiet hours posted at the campground and you need the heater at night.
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Old 02-09-2014, 03:14 PM   #32
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If you are going to be operating on battery/inverter for any significant amount of your camping time, a good battery monitor system is a must.
Yes, and now would probably be the right time to mention that I also bought a Trimetric, just like Herk.

The FR panel, while merely ok for monitoring tank usage, is not an accurate portrayal of battery charge.

Also since herk mentioned incandescent lights, if you havent thought about swapping all your lights for LEDs, now would be the time.
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Old 02-09-2014, 03:15 PM   #33
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Yes, and now would probably be the right time to mention that I also bought a Trimetric, just like Herk.

The FR panel, while merely ok for monitoring tank usage, is not an accurate portrayal of battery charge.

Also since herk mentioned incandescent lights, if you havent thought about swapping all your lights for LEDs, now would be the time.
And not very good at "Tank Usage" either!

has anyone actually tested the current draw of their interior LED bulbs?
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Old 02-09-2014, 05:34 PM   #34
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Nope. I see we have similar tastes in LED panels as well.
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Old 02-14-2014, 07:48 PM   #35
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OR, you could just multiply AC amps by 10 to get DC amps. (120/12 = 10)

You meant divide as in the example correct?
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No multiply: 1 amp at 120 VAC is 10 amps at 12VDC, thus multiply AC amps by 10 to get DC amps.
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Old 02-14-2014, 08:08 PM   #36
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No multiply: 1 amp at 120 VAC is 10 amps at 12VDC, thus multiply AC amps by 10 to get DC amps.
A factor of 10 to 1 : DC amps to AC amps.

However, the ratio does not work when the DC voltage provided by the battery fluctuates from fully charged 12.7 volts to depleted 10.5 volts).

Appliances to be powered by your inverter are "amperage rated" (watts) at a defining voltage AC. For example 1000 watts at 120 volts. The inverter will manage battery draw to deliver (as best its controller can) 120 volts AC.

In this case the amperage would be 8.33 amps. Using the above ratio, you would draw 83.3 amps from a 12 volt battery (1000 watt demand).

However, a battery delivering 12 volts is actually 50% depleted. If the battery is fully charged, the amperage drawn by the inverter would be 1000 watts / 12.7 volts or 78.7 amps.

As the battery passes 50%, say down to 11 volts, The amperage draw will increase to 90.9 amps (1000 watts / 11 volts) which will accelerate battery depletion to inverter cut out voltage.
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Old 02-14-2014, 10:52 PM   #37
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83.3 amps vs 78.7 amps is only about a 5% difference.

I guess we could start talking about the accuracy of everyone's ammeter and see what changes that makes. Also, is it really 110 VAC or maybe it's 120 VAC. Or maybe less when it's 100 degrees out and everybody in the campground is running their AC units.

The 10 to 1 ratio is a "rule of thumb."
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Old 02-14-2014, 11:13 PM   #38
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Also, is it really 110 VAC or maybe it's 120 VAC. Or maybe less when it's 100 degrees out and everybody in the campground is running their AC units.
Then we'll all need to carry some capacitors to shift the phase angle back from all those inductive loads!

Sorry! A blast from the past.
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Old 02-14-2014, 11:21 PM   #39
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Then we'll all need to carry some capacitors to shift the phase angle back from all those inductive loads!

Sorry! A blast from the past.
I am sorry; I thought we were talking about inverter loads.
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Old 02-15-2014, 10:45 AM   #40
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Okay, somewhere on this forum somebody posted a list of typical current draws for various appliances, etc. Should have saved it when I saw it. Now I can't find it. Has anyone else seen this and, if so, where?

Thanks.

Barbara
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