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<p>[quote user="ROBERT PETRICK"]</p> <p> </p> <div class="quote-header"> </div> <blockquote class="quote"> <div class="quote-user">gregc</div> <div class="quote-content"> <p><a href="http://www.w8ji.com/track_resistance.htm">Nickle-silver track has a resistance of 1.45 millohm</a>/in (0.057 ohm/m) which is the <a href="http://www.daycounter.com/Calculators/AWG.phtml">equivalent of #22g</a> wire.</p> <p>32" radius amounts to 17' of track per loop.    The track length fed from one feeder/loop is 8.5', reached from two separate points and ~0.15 ohm or 0.15 V drop at 1A.</p> <p> </p> </div> </blockquote> <div class="quote-footer"> </div> <p> </p> <p>Just a quick math question . . .</p> <p>If the feeder spacing is 8.5 feet, doesn't that mean that the load (engine) is always less than 4.25 feet from the feeder? So, why isn't the resistance of 0.00145 ohms per inch multiplied by 4.25 feet (51 inches)?</p> <p>Just asking.</p> <p>Robert </p> <div style="clear:both;"> </div> <p>[/quote]</p> <p>Greg,</p> <p>The postulated scenario states one set of feeders in 17' feet of track.  => Furthest distance a locomotive can be is 8.5 feet. </p> <p>Sheldon,</p> <p>Most locomotives now draw 1/3 amp or so.  Is the OP running modern production equipment...  </p> <p>That 1/3 amp everyone throws around is the steady state draw of the locomotive measure current draw on a 30" curve with a 3% grade (or whatever this ends up being).</p> <p>You are also assuming one locomotive will pull the train up this grade.</p>
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