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#### Guest

·I have in my possession a partially disassembled Prius transmission. Thanks to Fred and Marion in Redmond, Wasington for donating this equipment. It is my understanding that this transmission was damaged in an accident and Fred was allowed to keep the old unit when his new transmission was installed in his Prius.

I suspect this "autopsy" may take a long time to complete as I have a couple evenings a week for this project and I wish to capture as much information as I can.

I plan to post photographs and sketches of my findings. Since this message board allows attachments I will attach documents directly to my posts. If I need to post more than one file I will zip them together and then post.

I just unloaded the transmission and it is in my back yard. My first crude measurements were of the DC resistance of MG1 and MG2 coils. I have a medium accuracy handheld Fluke digital multimeter.

MG1 measurements: I measured the resistance between U-V, U-W, and V-W terminals. In each case my meter fluctuated between 0.1 ohm and 0.2 ohm. I need a better meter. Until I get a better meter I will assume the MG1 coil resistances somewhere around 0.15 ohm.

MG2 measurements: A similar situation. In each case I measured between 0.0 ohm and 0.1 ohm. Until I get a better meter I will assume the MG2 coil resistances somewhere around 0.05 ohm.

Page 33 of the Prius New Car features book gives MG2 maximum current as 351 Amps. If we assume current flows in one wire and out a second wire (this corresponds to trapazoidal control) we can use 0.05 ohm to estimate the approximate MG2 resistor heating in this scenario. P = I^2 * R = (351)^2 * (0.05) = 6.16 kW of copper heating. Using ohms law (home slaw) we can determine the voltage drop across the 0.05 ohm resistor. V = I * R. V = 351 * 0.05 = 18 volt drop due to wire resistance when current is 351 Amps.

If we assume current flows in one wire, half out a second wire, and half out the third wire (sinusoidal control) we end up with some kind of series/parallel resistor combination that makes the total resistance look 25% or 50% smaller. We either have half the resitance in parallel with half the path or the same resistance in parallel or some combination of the two (either WYE, DELTA, or a combination of the two). This would reduce wasted power by about 25% to 50% and reduce the voltage across the resistance by about 25% to 50%. This results in copper wire heating around about 4.6 kW to 3.1 kW at 351 Amps with a voltage drop of about 13 to 9 volts due to wire resistance.

Graham, a 25% to 50% reduction in copper heating losses for the same electric current is a compelling reason to consider sinusoids over trapazoids.

Good night.

Ed