The number of turns is irrelevant.
All changing the number of turns does, assuming you also change the wire diameter so that the resulting winding fills the space available, is to change the impedance of the winding, the voltage/current ratio, not the efficiency.
Consider the space filled with two identical windings, running at the same current, voltage, field, dissipation etc. If you connect them in series, it runs at 2V and I, you have twice as many turns as each coil. If you connect them in parallel, it runs at V and 2I, you have double the copper area but the same number of turns. All of the important parameters, field, weight, torque, power lost, cost of the copper, are unaltered, even as you've doubled or halved the number of turns.
That's all to first order of course. There are many small second order things going on. The wires to the motor can be thinner if the motor impedance is higher. Thin wire tends to have a greater part of its cross sectional area occupied by insulation than thick wire, so it's not quite as efficient as medium weight wire. Very thick wire also doesn't pack as well due to its stiffness. A very high voltage motor may well dedicate more space to insulation than a lower voltage motor. There may be other considerations for voltage, for instance availability of 12v batteries, or staying below 40v so you can use 'low voltage' insulation rules.
That's enough about number of turns not affecting efficiency, so what does?
The main loss is \$I^2R\$ loss in the copper. Unfortunately there's little that can be done about the resistivity of copper. Silver is too expensive (and barely any better), cryogenics is very complicated, though superconductor is worth doing if the machine is really, really big (>> 100MW).
High field magnets produce more volts per turn, and more torque per current than low field magnets, so you should use the highest field you can afford. More volts per turn means a higher ratio of 'useful' volts (coupled to the speed of the motor) to 'useless' \$IR\$ drop volts that only generate heat, so better efficiency. A small airgap helps produce a higher field with the same magnets, but requires better machining tolerance and produces more air viscosity losses (aka windage), so it can't be reduced too much.
A high speed motor produces more power than a low speed one at the same output torque, which is another way of saying generates a higher EMF than a slow one. Again, a higher ratio of useful volts to useless volts. However, high speed requires strong materials, and careful balancing, and produces losses that increase rapidly with speed, so it can't be increased too much. The higher rotor frequency means more hysteresis loss in the magnetic materials, and the windage losses increase as the speed cubed, so will rapidly come to dominate at very high speeds.
Because of the way some power and loss terms scale, a big motor, other things being equal, will have lower losses than a small motor. But let's say you want four driven wheels. Is it better to use a small motor per wheel, or a big motor and a mechanical drive train? Almost certainly the former, for lower weight and complexity, even if it's slightly less efficient.
Motor/generator design is a compromise between efficiency and, well, cost really, once you have a motor powerful enough for your specifications. Unfortunately, even the simple things like cost, weight and volume are a multidimensional space you can't simply optimise over. When you add things you can't easily put numbers on, like time to design, maintainability, reliability, the rest of the vehicle efficiency, convenience, use of strategic materials (in strong magnets), there is no one size fits all, no one equation describes all.
Best Answer
This is a difficult one to answer with any amount of certainty. But here is a way to very roughly approximate things:
First, go to HERE and look up a radio station. I looked up KALC, which serves the Metro Denver area with a 100KW FM transmitter and is one of the bigger radio stations.
Next, I looked up Denver on Wikipedia and found that the population of the Metro Denver area is 2.6 million people. That works out to 38 mW of RF power per person.
Of course not all 2.6 million people in Denver are actually listening, all of the time. And here is where it gets tricky. We must estimate the actual number of people listening (averaged out to some time period). This turns out to be fairly difficult for someone without access to the Nielson ratings. (I should point out that Nielson does both TV and Radio ratings, but the rating system differs.)
But let's take a stab at figuring out the percentage of the population that, over a 24 hour average, is listening to that station. There are 2.6 million people times 24 hours in a day. That's 64.4 million people-hours of potential listeners. If our listener utilization were 100%, that means that all 2.6 million people are listening for 24 hours in a day. I might only listen to the radio for 1 hour/day-- 30 minutes driving to work and another 30 minutes driving home. This means that I only contribute 4.2% to the number (1/24th). If everyone in Denver listened for 1 hour/day then the average percentage for the entire population would be 4.2%.
Of course, not everyone listens to that station. It is a fairly popular station, but not that many people listen to the radio and there are a lot of radio stations in Denver. I will guess that maybe 3% of the Denver population actually listen to that station. So 3% of 4.2% is 0.125%. So of 64.4 million potential people-hours only 0.125%, or 0.078 million people-hours is actually utilized in our very loose estimate.
That station broadcasts at 100KW, and over a 24 hour period uses 2.4 million watt-hours. Divide the two and you get approximately 31 watts-hours per person-hours. Yes, this is super rough and full of errors. But it is a reasonable approximation for our purposes. I will address the potential errors in a moment.
Estimating the power required by an internet user is much harder to figure out. For starters, people probably stream Pandora or Spotify much more than they listen to FM radio. Normal radios these days is rare, while almost everyone has a computer, tablet, or phone that can stream internet radio. Further, people will tend to not stream internet radio while in the car (except for those of us that still have unlimited data plans).
Another thing that makes it difficult to estimate is the variety of devices that people use to stream the internet. On the power efficient end of the spectrum is the iPod Touch which might only consume 0.5 watts. On the other end is an old PC with a CRT monitor that could be 500+ watts. The WiFi router, Ethernet Switch, and DSL Modem (or whatever) probably contributes another 15 watts (less if those devices are shared with other users).
I am completely ignoring the power required by the Internet infrastructure that is outside of your home or office. This is fairly small because it can be efficiently shared between homes and other businesses. I would guess that the per internet subscriber power requirements is less than 10 watts and possibly as low as 1 watt.
So now let's address the errors in my numbers. For the internet solution is somewhere between 15 watts and 500 watts. Our radio solution is around 31 watts, but has a wide error margin. The actual range might be 1/10th to 10x (or worse!). So 3 watts to 300 watts! While that is lower than 15-500 watts, there is such a huge error that we can only say that they are roughly comparable-- we cannot say if one is definitively more power efficient than the other.
Indeed, we can easily come up with scenarios where either Internet or Radio is is more efficient. Further, the balance of power (hah hah, I crack myself up) is always shifting because the efficiency of radio changes greatly depending on how many listeners there are. Better efficiency during rush how, and terrible efficiency at night.