I don't know if anyone still follows this topic; but I'm writing here just to make it a reference to those who question this behaviour.
Now, if you chose to connect your circuit element (say, it's a motor but it doesn't matter) to your emitter output, you are actually causing that element to be "shared" between your base-emitter circuit, and collector-emitter circuit. Considering a transistor, in a simplistic manner, two diodes together; the circuits happen like attached image.
Now, left side of that image is the Base-Emitter side. Notice the load there? Yeah, it's there, because you decided to place that to the "common emitter" line. Ergo, you caused the load to affect the base current. Now, if you do read about transistors, there is an equation as follows:
Ic = B * Ib
"B" in this equation (also knows as Hfe) is the "current gain" of the transistor. Simply put, it's the ratio of the current that goes from collector to the emitter to the current that comes from the base to the transistor - that means, a transistor can only allow to pass "B times Ib" amount of current from it's collecter-emitter circuit (Ic). Why so? It's because of how diodes and semiconductors do work there.
There is also another very trivial but important equation:
Ie = Ic + Ib
which means that the emitter current is dependent of the collector and base current. Putting the first eq. into the second, we get:
Ie = B * Ib + Ib = B (1 + Ib)
Now, this is very important in our case! By putting the load to the common emitter line; you increased the resistance (or should I say, impedance?) of the base-emitter circuit; ergo decreased the total current that circulates at the base-emitter loop. Because of this, emitter current automatically decreases, even though there is a current coming from the collector base and transistor working (not efficiently, though). Due to Ohm's law; if current decreases; voltage decreases (if the resistance is constant - we assume it to be so); thus you measure much less voltage on the pins of the engine if you connect it to emitter line.
Trying to saturate a transistor (i.e. making it work efficiently) in this condition is virtually impossible without increasing the collector voltage, because as you try to saturate the transistor, the B value changes, so the transistor behaviour changes... It's like a dog, trying to catch it's tail
On the other aspect; due to the Kirchoff's Law; you cannot measure any voltage higher than the Vb in the emitter line - even if there is no load - anyway, because the base-emitter circuit is a loop and there is nothing to increase the voltage there!
Sorry, this has been a kind of long one; but I hope it makes it simpler to understand.