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5.1 Segway Human Transporter in simplified Mechanics

We will focus mainly on the Segway I-Series, which is the Range and Terrain model. The only difference between the I- and E-Series (Cargo Carrying) is the potential Payload (0kg versus 34kg). Here are the parameters found on Segway's web site ( http://www.segway.com ). Note that we will be working in the metric system.

Max Speed:20 km/h
Range on a single charge:17 - 25 km
The measure for
range is as follows:
perfect conditions: 25km (15 mi)
normal conditions: 16 km (10 mi)
rough terrain: 8 km (5 mi)
Max weight of Passenger110 kg (250 lbs)
Weight of vehicle
(incl. batteries)
38 kg (83 lbs)
Weight of batteriesapprox. 9 kg
Diameter of wheels48 cm
Table 5.1
Figure 5.1
Figure 5.2

Figure 5.3 outlines a simplified model of Figure 5.1, in which the rider is standing on a Segway in a steady state. Figure 5.4 expands on the method by which the Segway is used in reality: When in motion, the rider leans slightly forward at angle θ, which places the center of mass at the point denoted by mg. The center of drag is denoted by the point D, which is usually not the same as mg. Figure 5.4 simplifies Figure 5.3 by moving the center of rotation from the center of the wheel to the ground. The distances between the center of rotation and center of mass and drag are denoted by rcm and rD, respectively. The normal force N denotes the opposing force to mg.

Figure 5.3
Figure 5.4
From the setup in the previous 4 diagrams we establish the following:

Σ Fx = 0       (Eq. 5.1)

FT Σ = D       (Eq. 5.2)
(Eq 5.1 - 5.2): In a steady state, the sum of horizontal Forces equals 0. From Figure 5.2 we see that whatever Drag exists going at velocity v, the Force FT must be of an equal opposite quanity, in order to remain in the steady state.

Σ Fy = 0       (Eq. 5.3)

Fmg = N       (Eq. 5.4)
(Eq 5.3 - 5.4): In a steady state, the sum of vertical Forces equals 0. From Figure 5.3 we see that the mass of Segway and Rider (mg) is opposed by the Normal Force N in equal magnitude.

We now try to examine the Traction Force (being equal to Drag) in more detail, as it is the force required to move the Segway and its rider. Traction Force is caused by several factors that sum up to become the total Traction Force (Drag). The most important components are: Aerodynamic drag caused by flow of air, Rolling Resistance caused by friction between the tires and the ground surface, drag caused by mass on an inclined surface, and drag caused by acceleration. Each Force is described in the following set of equations:

FAir = ½ ρ v2 A CD       (Eq. 5.5)
where ρ (rho) is the density of air at sea level = 1.225 kg/m3, A is the frontal surface area of the Segway + Rider, and CD is the coefficient of drag.

FRoll = CR m g       (Eq. 5.6)
where CR is the Rolling Resistance coefficient, and g is the acceleration due to gravity = 9.81 m/s2.

FSlope = s m g       (Eq. 5.7)
where s is the slope of the inclined surface measured by the sine of the angle 0 (0 °) <= s <= 1 (90 °).

FAccel = m a       (Eq. 5.8)

FTotal = FTraction = D = FAir + FRoll + FSlope + FAccel       (Eq. 5.9)
For Eq 5.5 - 5.8, mass m is measured in kg, Surface area A is measured in m2, velocity v is measured in m/s.

The Power required to drive the Segway at some speed v is thus:

P = FTotal * v       (Eq. 5.10)

5.1.1 Drag and the Coefficient of Drag CD

Simply and rather obviously stated, drag slows down a moving vehicle. Drag is due to many factors, ranging from the frontal area of the vehicle + rider, the shape, whether cross-sectioned from top or side, to minute details like clothing, positioning of hands and arms, etc.

Figure 5.5 outlines a few shapes and their associated aerodynamic drag values. While we will not go into detail about what this value represents, we note that a higher value is due to more and a lower value due to less drag (resistance). From the figure it appears that a teardrop shape experiences the least amount of drag. This is due to 2 factors, both of which should be taken into consideration when designing the shape of a vehicle:
  1. The frontal area is curved, which allows the wind to transition smoothly from its straight path to around the object.
  2. The rear converges at a point.

Figure 5.5: Drag Coefficients
It is not difficult to implement a design that takes care of the first point. Cars, trucks, and motorcycles are already designed and built to counter the effect of drag in the frontal areas by making them transition between a sharp edge and the full body. However, it is uncommon for the same vehicles to have a similar tear-drop design in the rear. Truck storage trailers for example are as boxy in the front as they are in the rear. Understandably, they are designed for optimum storage capacity, which is why such trailers do not assume a teardrop shape. Passenger cars, station wagons excluded, tend to approximate the teardrop well enough to counter most of the extreme drag. Due to their size and availability of space, airplanes and most ships tend to be designed in the teardrop style to begin with.

The drag in the rear of a vehicle is the result of the presence of a wake. It is caused by displacing the vehicle in a volume of air at higher speeds, where the air is "pushed out of the way" in the front of the vehicle, and a lack of air in the rear of the vehicle causes air to be "pulled in" from the sides and top and bottom. The larger the area in the rear, e.g. for a truck trailer, the more air needs to be pulled in, which results in an increase in drag. Should the rear of the vehicle be shaped to allow the stream lines to adjust gradually, the smaller the wake. This principle is outlined in Figure 5.6.

Figure 5.6: Wake behind 2 differently shaped objects
For the Segway, we will need to estimate the coefficient of drag, because it is not known to us. We may refer to the table above for drag coefficients for some common objects. Since the rider on a Segway is completely exposed to the elements and faces to the front with her entire body, the drag coefficient for a Segway is at best as good as that for the person riding it.

Drag Coefficient CD:approx. 1.5
Table 5.2

5.1.2 Rolling Resistance Coefficient CR

Rolling Resistance is caused by the deformation of a rubber tire at the point where the tire meets the surface on which it travels. The lower the pressure and/or the higher the force exerted on the tire, the larger the coefficient. A non-deformable wheel, for example a train wheel made of steel, riding on non-deformable surfaces (rails), has the lowest Rolling Resistance, hence making it most efficient.

Figure 5.7: Rolling Resistance
Rolling Resistance is also a function of the type of rubber tire used. Rubber tires are not made entirely out of rubber, but contain some form of steel mesh, which makes the tire much more robust.

Figure 5.8: Types of Tires
Table 5.3 outlines a few types of tires along with an approximation equation for Rolling Resistance. The set of equations used is taken from Volvo Powertrain Corp.
Type of TireCR (velocity v in m/sec)
Radial-ply passenger car tire0.0136 + 0.04 * 10-6 * (v * 3.6)2
Bias-ply passenger car tire0.0169 + 0.19 * 10-6 * (v * 3.6)2
Radial-ply truck tire0.006 + 0.23 * 10-6 * (v * 3.6)2
Bias-ply truck tire0.007 + 0.45 * 10-6 * (v * 3.6)2
Table 5.3: Rolling Resistance Coefficient

5.1.3 Surface Area and Body Surface Area

For the purpose of computing FAir we need to know the total frontal area of the Segway including the rider. We will assume that the Segway itself has a frontal surface area of 0.2 m2. For the rider, we will have to compute it using some generic formula that takes into account weight and height of the person.

There exist a few methods of computing the total body surface area, i.e. not just the front. They include the Mosteller formula, the DuBois DuBois formula, the Haycock formula, the Gehan & George formula, and the Boyd formula. The most popular and most accurate of them is the Mosteller formula:

BSA = √( h * m / 3600 )       (Eq. 5.11)
where h is the person's height in cm, m is the person's weight in kg, and the result is given in m2Finally, the frontal surface area is somewhat less than half of the entire BSA:

FSA = 2/5 BSA       (Eq. 5.12)