----- Original Message ----- From: Horace Heffner <[email protected]> Date: Thursday, August 13, 2009 11:59 am Subject: [Vo]:BB motor filament heating test for Harry Veeder
> Maybe there was a bit of an anomaly here? > > This is in regards to and an addendum to the Marinoff Ball Bearing > Motor experiments documented here: > > http://www.mtaonline.net/~hheffner/HullMotor.pdf > > At the request of vort Harry Veeder, I did a test to show how much > time it takes to heat up the resistor when the motor is running vs > when stopped. I added a little green LED just below the filament > so > you can see exactly when the current comes on, and also provided a > clock to see the time. Here it is: > > http://www.youtube.com/watch?v=PWlVn-uqxig hey that's great. When I asked the question I had in the back of mind that some of the electrical force which could have gone into heating the coil was dissipated as mechanical energy when the motor was running. > Looks to me like about 8 seconds when the motor is running, and > about > 5 when stopped. This at first glance appears to be yet another > indication this is an ordinary magnetic effect. The reduction in > final current can be attributed to a back-emf. To some degree it > might also be attributed to non-conduction time when the motor is > running, but the scope traces have indicated pretty much full time > current conduction in all runs since the bearings were cleaned. > > However, the current sense resistor voltage drop doesn't look like > what I'd expect. > > http://www.mtaonline.net/~hheffner/HullRunningTrace.jpg > > http://www.mtaonline.net/~hheffner/HullStoppedTrace.jpg > > The traces show: (1) motor running, takes about 5 seconds to go > from > 7 V to 9.6 V, but about 9 seconds to heat orange, (2) motor > stopped, > takes about 5 seconds to go from about 6 V to 10.8 V, and to heat > orange. > > The difference in peak voltage makes sense in that the running > motor > peaks at 1.2 V less, so the back emf must be about 1.2 V. > > However, the traces don't make sense with regards to how the > filament > heats. > > P = I^2 R = V^2/R > > The resistance R = 0.0631 ohms cold. > > So, at startup the running resistor heating power Prun and stopped > power Pstop are: > > Prun = (7 V)^2/(0.0631 ohms) = 777 watts > > Pstop = (6 V)^2/(0.0631 ohms) = 571 watts > > The ratio is 1.36, with the running motor circuit initially > producing > more heating in the current sense resistor by a factor of > > Prun/Pstop = 777 W / 571 W = 1.36 !?? > > This is not what we would expect in that overall the resistor turns > > orange much faster when the motor is stopped. By the time of > orange > glow, at resistor voltage and current equilibrium, we don't know > the > resistance, but the power ratio appears (assuming identical > resistance at similar temperature) to be: > > Prun/Pstop = (9.6 V)^2 / (10.8 V)^2 = 0.79 > > The equilibrium numbers make some sense in that 0.79 * (8 sec) = > 6.3 > sec, though it is off quantitatively a bit in that the orange > temperature was reached in 5 seconds. > > The initial power numbers made no sense to me in terms of the way > the > resistor acted though, and that has nothing to do with the > performance of the motor. The resistor should heat according to > the > energy applied to it. > > Finally it dawned on me. The resistor was per-heated in the second > > run. It started out with a higher resistance, but heated to orange > > faster, i.e. with less energy. I ran a quick test. Starting out > cold it took 8 seconds to heat the resistor to orange. Doing it > again, a few seconds later, it took only 3 seconds. The resistor > apparently takes a while to cool down even after it is no longer > red > or orange. Were both these second tests of heating time conducted while the motor was not running? harry
