

5.2 Discussion on some of the Electrical Issues
One of the most important parts of the Segway is its source of power which is derived from 2 batteries. We will need to understand a few basic concepts about how the energy in the batteries is used to power the Segway, and how much energy is needed for a given force.
Figure
5.9 is a simplified diagram of the Segway's battery B, the motor (pictured without gears), and the tire. The forces acting on a turning wheel are now measured in terms of Torque T.
An important factor in translating battery power into mechanical power are battery efficiency
η_{B} and propulsive efficiency η_{P}. While batteries function at very high efficiencies, a large quantity of power is lost in converting energy in the batteries to mechanical power.


Figure 5.9 

Starting with the diagram in Figure
5.9, we establish the following:
Σ M_{a} = 0 (Eq. 5.13)
meaning that the sum of Momenta in a steady state must equal to
0.
We further know that the force required to move the vehicle is due to torque
T on the wheel:
T = r_{w} F_{T} (Eq. 5.14)
Finally, the mechanical power required to drive the Segway at some speed
v is:
P_{mech} = T * Ω (Eq. 5.15)
where
Ω is the velocity of the wheel measured in radians per second.
Battery efficiency is typically represented by an imaginary resistor
R_{B} within the battery, as depicted in Figure 5.10. The resistance is very small, as batteries tend to operate at 90 percent efficiency. Loss of energy is caused predominantly by resistance in wires and heat dissipation of the battery.


Figure 5.10 

Battery efficiency is a function of the ideal Power supplied by the battery and the Power entering the Motor:
η_{B} = P_{Elec Out} / P_{Elec Ideal} (Eq. 5.16)
Dissipated heat from the battery and wires can thus be expressed as:
Q_{Heat in battery} = P_{Ideal out of battery}  P_{Elec to Motor} (Eq. 5.17)
Propulsive efficiency is much lower than that of the battery, as it includes a number of mechanical inefficiencies. This includes, but is not limited to heat dissipation in the electric motor, friction between gears, in bearings, and rubber bends, etc. There are a number of other factors, none of which we will describe in detail.
The total efficiency is computed as the product of all the individual efficiencies. For example, if the efficiencies for bearings, motor, and gears were 90% each, the resulting efficiency would be
(0.9)^{3} = 0.73. We could research all the individual pieces, but instead we will estimate the overall propulsive efficiency, e.g. 50%.
Propulsive efficiency is a function of Power going into the Motor and Power exerted on the wheels:
η_{P} = P_{Mech to wheel} / P_{Elec In} (Eq. 5.18)
Quite naturally, the propulsive efficiency decreases by some function
f as the force (weight) exerted on the components of the motor increases:
η_{P} = f(m) (Eq. 5.19)
However, for the upcoming exercises we will not take this factor into account, but it would be interesting to explore what this function may look like.
The overall efficiency
η_{Overall} can be computed as:
η_{Overall} = η_{P} * η_{B} (Eq. 5.20)
η_{Overall} = [ P_{Mech to wheel} / P_{Elec In} ] * [ P_{Elec Out} / P_{Elec Ideal} ] (Eq. 5.21)
η_{Overall} = [ P_{Mech to wheel} / P_{Elec Ideal} ] (Eq. 5.22)

