So I own similar to pic related and am thinking about using for

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>>2771517
do you have a multimeter ?
wire it in parallel to measure voltage and measure when it drops past the cutoff of your load
to measure amperage , wire it in series

but thats the cheapest way to do it unless you want to drop mad dosh on a prophoto one

>>2771517
if it would even work, you will need a DC to AC converter, once it is all hooked up you can probably expect 10-50 strobes depending on your lights draw.

Q1 - trial and error

Q2 - Profoto makes a decent battery pack that will give you 150 strobes at full power, using their 600w monoblocks, it has a regular AC out so you can use any lights but results may very.
Dynalite makes a good kit for their lights, I almost bought into their system, it is quite nice and compact.
Speedotron has an ancient system that you could probably pick up for nothing, you will need to rebuild their NI-CAD batteries (battery specialists can still do this) and you get about 25 strobes at 300 wat seconds

Battery capacity in Ah (amp-hours) divided by the current draw of the load. In practice it will be about 30% worse than the calculation. If there isn't any Ah rating on the product you could open it to check the battery itself, or just test it.

>>2772753
AC plugs are built in.

I have multimeter but will trial and error it as there are so many factors, ( modeling lights, power, etc)

>>2774745
built in or not, AC to DC is fairly straight forward, but DC to AC is really incredibly inefficient.

what lights do you have? my guess is 25-30 pops of the flash at 600ws

>>2772753

Always really sucked at electrical engineering stuff but are my calculations correct for the following product:

http://www.ebay.com.au/itm/Godox-RS-600P-600W-Wireless-Studio-Flash-Strobe-Outdoor-Light-Battery-Trigger-/201477408857?hash=item2ee8fd4859:g:h~MAAOSwAKxWXraY

600 Ws Light from a 12.8V 8000 mAh battery.

600 Ws / 12.8V = 46.875 Ah = 46875 mAh

1 full power burst is 1/800 so does that mean we divide our total by this figure?

46875 mAh / 800 = 58.6

Given the capacity of the battery this gives us very approximately 150 shots at full power. Now I have a feeling I am getting something wrong here, but I cannot see how the advertising material arrives at 450 shots at full power for the battery pack given those figures.

Can someone please educate me what I am doing wrong? Are they advertising more than the battery pack can do. By my simple and probably wrong calculations the light would have to only be drawing 17 mAh per discharge.