Check My Math Please

Yes. Based on your math, your battery would be dead at sunrise. And the battery will lose a small amount of it's power every hour. By the 9th hour, your battery would be too low to power the charge. If you are designing a system, start with overkill, and delete when you are sure it produces no use. IMHO, I would use a group 31 battery designed for solar charging. good luck.

Hi RR,

I approached it this way:

The draw is 3A firing 0.5s/5s or 10% of the time.

The draw is 0.3A the rest (4.5s/5s = 90%) of the time. This becomes (3)(0.1) + (0.3)(0.9) = 0.57A draw average.

So, for 10 hrs you need a 5.7 A-hr. battery. (I think).

Mike

RDGRNR,

Yes, you badly missed something. You did the average over an hour (36000 seconds) but you didn't count the total number of 3A cycles throughout the hour. You can do average ampere calculation using either an hour or 5 seconds period. Here the calculations are for both cases:

(1) Over the hour (36000 seconds) period:

Total number of 3A pulse in an hour = 36000/5= 7200

Average current = (3*0.5*7200)/36000 + (0.3*32400)/36000 = 0.3 + 0.27 = 0.57 A

(2) Over the 5 seconds period:

There will be only one 3A pulse in 5 seconds and 0.3A for rest of 4.5 seconds

Average current = (3*0.5)/5 + (0.3*4.5)/5 = 0.3 + 0.27 = 0.57 A

Ampere-Hour for 10 hours = 0.57 * 10 = 5.7 AH

For both cases, the average ampere is same. The final unit should be AH (Ampere-Hour), not Amp/Hrs. This is the AH the flare ignition system will consume during the dark time. When you select the battery capacity in AH, select at least 50% more of this requirement.

- MS

I just pulled out my calculator and it says there is 3600 seconds in an hour.

Yes, you are right. In fact, the calculation (1) has been done for 10 hours period (36000 seconds). The calculation can be done using any time interval which is multiples of 5 seconds. However, 5 seconds calculation is simpler though. The result will be same all the times.

- MS

Sorry, I was drinking and feeling really smart. Thanks for taking it easy on me.

Also... the number of sun hours varies greatly by location. Make sure you have enough capacity for the worst case scenario... and that there will be enough sun hours to charge the batteries during the time available to do so.

Others have worked the math.

I have an observation: "Assuming 10 Hrs. of darkness..." As my pessimistic nature kicks in, I wonder if that might be better as ten hours of usable light? Maybe less.

[edit] I see North of 60 voiced a similar opinion. Didn't mean to step on you.

I would approach the problem more conservatively:

8 [hours] good sun
16 [hours] not enough sun
Use only 50% of battery capacity to extend life.
Assume 75% efficiency for all other conversions.

0.3 [A] x 0.90 + 3 [A] x 0.10 = 0.57 [A] average current

0.57 [A] x 16 [hours] / 0.50 = 18.2 [A-hours] battery capacity needed

Solar panels must be large enough to supply charge current AND ignitor current during sun hours.

Assuming a 12 [Volt] system...

Solar power needed to charge battery

12 [V] x 0.57 [A] x 16 [hours] / 8 [hours] / 0.75 = 18.2 [W]

Solar power needed to run ignitor

12 [V] x 0.57 [A] / 0.75 = 9.1 [W]

Average solar panel output during an 8 [hour] sunny day should be at least 30 [W].

Good Luck!

You have had many suggestions in the maths and etc.

In all cases, when you reach a conclusion about the exact required AH for the next 14 hrs or 16 hrs of darkness, double the battery capacity (at least) and make sure that the solar panel has enough power to run the Gizmo and charge the battery to full.

You can only draw 1/2 what you put into the battery (normally).

Mikerho has provided the easy way to look at the total current drain: it's 3A for a tenth of the time and .3 Amps for the rest of the time, about 0.6 Amps average (ignoring the sort of overlap which he has accounted for).

Lots of people have pointed out the foibles of batteries.

I'm more worried about your short estimate of night time. The best scenario from your point of view is having the equipment at the equator: 12 hours daylight, and 12 hours night. Any other part of the world you might get longer days in the summer, but, you also get shorter days/longer nights in the winter.

Thanks for the attention! FYI; I had already chosen a 30W BP Solar, a 26 AH Powersonic sealed LA Batt. and an ASC charge contoller. Test phase underway.

Would the light from the flare produce any output from a solar panel?

Just asking,,,,,

I believe you are on the right track and some empirical testing should confirm operation at your location.

Probably too costly and not worth the effort (you have to run the numbers on your end), but what about using the waste heat and a thermoelectric generator to augment the solar panel and battery backup?