Lumped Mass Model

For demonstration purposes I’ve assumed a four wheeled ground vehicle with all four wheels being driven. It is equipped with electric drive, a battery, a perception senor suite and some actuators to do work. Within some bounds we will often have the freedom to move these sub-systems around in order to optimise the centre of mass location of the vehicle. We can approximate our ground vehicle as a beam with masses applied to it from the separate functional sub-systems with the most significant mass contributions. Working in this way we can establish what the reactions at the wheels are, then where the vehicle’s centre of mass location is.

The table below details the sub-systems and their location in the vehicle.

A side view (\(xz\) plane) of the vehicle with locations of mass contributions is shown in figure 1. Note that some of the distance labels have been omitted for the clarity of the diagram. You will also note the reaction forces under each of the front wheels, \(R_{fw}\), and each of the rear wheels, \(R_{rw}\).

Figure 1: Longitudinal Lumped Mass Model

The total mass of our ground vehicle, \(m_{tot}\), ought to be the sum of all of the mass contributions of the sub-systems, equation (1).

\[\begin{equation} m_{tot} = m_{rw} + m_{rd} + m_b + m_e + m_c + m_a + m_{fd} + m_{fw} + m_p \tag{1} \end{equation}\]

If we call the vertical \(z\) axis and applying static equilibrium whilst summing forces yields the following.

\[\begin{equation} \sum F_z = 0 \tag{2} \end{equation}\]

\[\begin{equation} \sum F_z = (m_{rw} + m_{rd} + m_b + m_e + m_c + m_a + m_{fd} + m_{fw} + m_p)g - 2R_{rw} - 2R_{fw} \tag{3} \end{equation}\]

\[\begin{equation} R_{fw} = \frac {(m_{rw} + m_{rd} + m_b + m_e + m_c + m_a + m_{fd} + m_{fw} + m_p)g - 2R_{rw}}{2} \tag{4} \end{equation}\]

\[\begin{equation} R_{rw} = \frac {(m_{rw} + m_{rd} + m_b + m_e + m_c + m_a + m_{fd} + m_{fw} + m_p)g - 2R_{fw}}{2} \tag{5} \end{equation}\]

Now if we sum moments under equilibrium at the front wheel we can derive an expression for \(R_{rw}\). Note that clockwise rotation is positive.

\[\begin{equation} \sum M_{yy} = 0 \tag{6} \end{equation}\]

\[\begin{equation} \sum M_{yy} = (h-a)2R_{rw} - (h-a)m_{rw}g - (h-b)m_{rd}g - (h-c)m_bg - (h-d)m_eg - (h-e)m_cg - (h-f)m_ag - (h-g)m_{fd}g + (i-h)m_pg \tag{7} \end{equation}\]

\[\begin{equation} R_{rw} = \frac {g((h-a)m_{rw} + (h-b)m_{rd} + (h-c)m_b + (h-d)m_e + (h-e)m_c + (h-f)m_a + (h-g)m_{fd} - (i-h)m_p)}{2(h-a)} \tag{8} \end{equation}\]

Once we have \(R_{rw}\) we can use equation (4) to provide \(R_{fw}\). The corner weights we could expect under each wheel can be given using equation (9) and (10).

\[\begin{equation} m_f = R_{fw}g \tag{9} \end{equation}\]

\[\begin{equation} m_r = R_{rw}g \tag{10} \end{equation}\]

Now we can calculate the vehicle’s centre of mass location, \(j\), with respect to the front wheel location, \(h\), in equation (11).

\[\begin{equation} h-j = \frac {l}{m_{tot}} 2 m_r \tag{11} \end{equation}\]

Similarly the centre of mass location, \(j\), with respect to the rear wheel location, \(a\), is provided with equation (12).

\[\begin{equation} j-a = \frac {l}{m_{tot}} 2 m_f \tag{12} \end{equation}\]

A useful simple check that can be done is to check the wheelbase, \(l\), of the vehicle against the sum of the results from equations (11) and (12).

\[\begin{equation} l = h-a \tag{13} \end{equation}\]

\[\begin{equation} (h-j) + (j-a) = l \tag{14} \end{equation}\]

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