Mosaic
Well-Known Member
Ok guys, I am confused about the values that my BT100 CCA tester displays.
http://www.amazon.com/Schumacher-BT...BOI0/ref=sr_1_2?ie=UTF8&qid=1338663317&sr=8-2
As is typical of this direct load tester it applies a fixed .13 Ohm high wattage load (about 100A) @ a batt V of 13V and then correlates a CCA value based on the voltage drop.
Well here's the data from the voltmeter gauge markings on it:
1000CCA @ >=10.6v after 10sec load.
800CCA @ >=10.4V
600CCA @ >=10.2V
400CCA @ >=10.0V
200CCA @>= 9.6V
using V=I*R.
Now @ a 100A load, to get the 800CCA display value we have a drop of 13-10.4 = 2.6V => a battery IR of about .026 ohm. Now this might actually mean a higher Int.R as the terminal voltage of 10.4 will work out to an actual current draw of 10.4/.13 = 80Amps.
So we actually have 2.6/80 = .0325 ohm in resistance.
I cannot see how this can correlate to 800 CCA! Even based on the standard that permits the largest voltage drop to 7.2V , we have a drop of 13-7.2 = 5.8V, @ 800CCA => 5.8/800 = .00725 ohms Int Resistance MAXIMUM!.
An Int Resistance of .0325 ohm is going to give 800 x .0325 = 26V drop at 800CCA...which is impossible.![Confused :confused: :confused:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I cite this site as reference:
http://www.smartgauge.co.uk/nosurge2.html
Can anyone clear this up for me please? I was planning to make an electronic version of the load test but this is too confusing.![Frown :( :(](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
http://www.amazon.com/Schumacher-BT...BOI0/ref=sr_1_2?ie=UTF8&qid=1338663317&sr=8-2
As is typical of this direct load tester it applies a fixed .13 Ohm high wattage load (about 100A) @ a batt V of 13V and then correlates a CCA value based on the voltage drop.
Well here's the data from the voltmeter gauge markings on it:
1000CCA @ >=10.6v after 10sec load.
800CCA @ >=10.4V
600CCA @ >=10.2V
400CCA @ >=10.0V
200CCA @>= 9.6V
using V=I*R.
Now @ a 100A load, to get the 800CCA display value we have a drop of 13-10.4 = 2.6V => a battery IR of about .026 ohm. Now this might actually mean a higher Int.R as the terminal voltage of 10.4 will work out to an actual current draw of 10.4/.13 = 80Amps.
So we actually have 2.6/80 = .0325 ohm in resistance.
I cannot see how this can correlate to 800 CCA! Even based on the standard that permits the largest voltage drop to 7.2V , we have a drop of 13-7.2 = 5.8V, @ 800CCA => 5.8/800 = .00725 ohms Int Resistance MAXIMUM!.
An Int Resistance of .0325 ohm is going to give 800 x .0325 = 26V drop at 800CCA...which is impossible.
I cite this site as reference:
http://www.smartgauge.co.uk/nosurge2.html
Can anyone clear this up for me please? I was planning to make an electronic version of the load test but this is too confusing.