<html><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; ">How about using your Wattmeter? Take readings of voltage and current at a number of throttle settings. Divide the change in voltage by the change in current between any two settings to get the resistance. This is easier to do on a spread sheet if you have a lot of settings and makes it easy to plot a curve of resistance vs. current. You will be able to see if it is linear as well as get a reading at the current of interest. Do it again with your pack after it is discharged.<div><br class="webkit-block-placeholder"></div><div>Save yourself $20 and 4 hours. Send contributions to Jim O so that we may continue to provide you with this great entertainment. Hope you are all smiling. </div><div><div><br class="webkit-block-placeholder"></div><div>Jim O</div><div><br class="webkit-block-placeholder"></div><div><br></div><div><br class="webkit-block-placeholder"></div><div><br><div><div>On Mar 16, 2008, at 5:42 PM, Lance Van Nostrand wrote:</div><br class="Apple-interchange-newline"><blockquote type="cite"><span class="Apple-style-span" style="border-collapse: separate; color: rgb(0, 0, 0); font-family: Helvetica; font-size: 18px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: 2; text-align: auto; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-border-horizontal-spacing: 0px; -webkit-border-vertical-spacing: 0px; -webkit-text-decorations-in-effect: none; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0; "><div bgcolor="#ffffff"><div><font face="Arial" size="2">Thanks for all the posts. I understand the piece of the puzzle I was missing. Actually pieces. bob's point is important but the basic approach is to start with a resistor of a known resistance that will give you the current where you want internal impedance measured. Let's say I have a fresh 2S pack measuring 8v unloaded. I buy a heavy duty resistor of 1 ohm that actually measures exactly 1 ohm (for example only). Under load it measures 7.2v across the resistor and is therefore pulling 7.2A (I = V/R)</font></div><div><font face="Arial" size="2">The voltage drop is due to internal resistance so 7.2a = 8v / (1 + internal impedance)</font></div><div><font face="Arial" size="2">internal impedance = .11 ohm</font></div><div><font face="Arial" size="2"></font> </div><div><font face="Arial" size="2">Is the idea of measuring at 1 ohm and 10 ohm a way to find an average number? If so, I think just finding a resistor that is in the neighborhood of your worst case situation may actually be more representative.</font></div><div><font face="Arial" size="2"></font> </div><div><font face="Arial" size="2">--Lance</font></div><div> </div><blockquote style="padding-right: 0px; padding-left: 5px; margin-left: 5px; border-left-color: rgb(0, 0, 0); border-left-width: 2px; border-left-style: solid; margin-right: 0px; "><div style="font: normal normal normal 10pt/normal arial; ">----- Original Message -----</div><div style="background-image: initial; background-repeat: initial; background-attachment: initial; -webkit-background-clip: initial; -webkit-background-origin: initial; background-color: rgb(228, 228, 228); font: normal normal normal 10pt/normal arial; background-position: initial initial; "><b>From:</b><span class="Apple-converted-space"> </span><a title="bob@toprudder.com" href="mailto:bob@toprudder.com">Bob Richards</a></div><div style="font: normal normal normal 10pt/normal arial; "><b>To:</b><span class="Apple-converted-space"> </span><a title="nsrca-discussion@lists.nsrca.org" href="mailto:nsrca-discussion@lists.nsrca.org">NSRCA Mailing List</a></div><div style="font: normal normal normal 10pt/normal arial; "><b>Sent:</b><span class="Apple-converted-space"> </span>Sunday, March 16, 2008 6:54 PM</div><div style="font: normal normal normal 10pt/normal arial; "><b>Subject:</b><span class="Apple-converted-space"> </span>Re: [NSRCA-discussion] Internal battery impedance</div><div><br></div><div>The resistor does not really need to be a precision resistor, you just need to know precisely what the resistance is. :-) And, if you are comparing or matching cells, the measurement only needs to be relative and the accuracy may not important.</div><div> </div><div>I do agree that the measurement will probably not be linear, IOW the cell impedance may vary with large changes in current. If I normally run 30 amps, I would want to make a measurement at 30 amps and probably 25 and 35 amps as well, comparing the differences in voltage at the three currents. This would be more meaningful to me than comparing the unloaded voltage with the voltage at 1 amp.</div><div> </div><div>Just M2CW.</div><div> </div><div>Bob R.</div><div><br><br> </div><div><br class="webkit-block-placeholder"></div><hr><div><br class="webkit-block-placeholder"></div>_______________________________________________<br>NSRCA-discussion mailing list<br><a href="mailto:NSRCA-discussion@lists.nsrca.org">NSRCA-discussion@lists.nsrca.org</a><br><a href="http://lists.nsrca.org/mailman/listinfo/nsrca-discussion">http://lists.nsrca.org/mailman/listinfo/nsrca-discussion</a></blockquote>_______________________________________________<br>NSRCA-discussion mailing list<br><a href="mailto:NSRCA-discussion@lists.nsrca.org">NSRCA-discussion@lists.nsrca.org</a><br><a href="http://lists.nsrca.org/mailman/listinfo/nsrca-discussion">http://lists.nsrca.org/mailman/listinfo/nsrca-discussion</a></div></span></blockquote></div><br></div></div></body></html>