Thread: Splitting The Positive View Single Post

12-27-2017, 08:14 AM
 BroMikey Platinum Member Join Date: Jan 2013 Posts: 5,266
Fresh Charge Resting 12:90v All day

Keep an eye on this page this morning as I am updating right now.

NEW CONTROL TEST with inverter. Time Out 11:45pm

12.90v resting 6 hours after last night and this mornings bulk charging
at the C/20 rate on ALUM batteries. Second stage of charging was done
at half the C/20 rate for 5hrs then the amps were raised to top off
well under the C/20 rate holding at 16v for 10minutes and then 16.4v
for another 5 minutes. This was all done using a transformer (toroid) to
bridge rectifier and smoothing caps. RAW DC or RAW DIRECT CURRENT.

The inverter was loaded and the drive energy is as follows.
Feeding the the drive booster at 1600ma and a differential of13.64v at
the beginning portion of the 1 hour run and the ending differential 13.72v
which averaged is 13.68v.

Drive energy was 13.68v X 1600ma = 21.888w

Over the 1 hr run the battery went from 12.90v then stop to rest 1 hr
the battery was 12.77v. This is a 13 point drop. The run loaded volts
were 12.65v down to 12.55v at the end of 1 hr running. This is a 10
point drop.

The Control test was stopped at 12:45am central and at 1:45am the
final reading was took. Using the smaller of the two point drops we see
the 10 points was lost of battery voltage.

Joules are watt/seconds and for each hour we have 3600 seconds this is

21.888watts X 3600 seconds = 78,796 joules that were used up over
a 10 point drop. So we may find the amount of joules used up for each
.01v drop on the meter. This is

78,796J / 10 = 7,879.6J per point drop. or 7,880J rounded off.

Each .01v drop

.01v = 7,880J
.02v = 7,880J
.03v = 7,880J
.04v = 7,880J
.05v = 7,880J
.06v = 7,880J
.07v = 7,880J
.08v = 7,880J
.09v = 7,880J
.010v = 7,880J

The inverter drive booster was set at 14.5v while the run batt recharge
booster stayed as always at the 26.5v setting. The recharging booster
ran at 3.60amps during the entire control test.

This is only relevant when you consider efficiency losses. 2 Boosters
passing a calculable amount of power at an average of 90%. This is
another subject completely.

This is our control test standard for conventional watt burning.

__________________

Last edited by BroMikey; 12-27-2017 at 08:42 AM.