� 16 ROAD VEHICLE ON RANDOM TERRAIN 40 16 Road Vehicle on Random Terrain An autonomous, four-wheeled road vehicle travels over an uncertain terrain. You are asked to characterize the vertical motion of the center of gravity, and the pitching (nose-up vs. nose-down) response of the vehicle, as it crosses the terrain, using a linear model.We know that the vehicle chassis has a mass of m =42kg and a pitching moment of inertia of J =25kg m2 . Consider the tires and suspension to have zero mass, but the front and back suspension systems are each modeled as having a spring of stiffness k = 1000N/m,in parallel with a linear damper having coefficient b = 100N/(m/s). The distance between the front and rear wheels is 2l =0.9m. The vehicle travels at a speed U, which we will vary below. Becauseweconsider U to be constant, you can use it directly to map between the horizontal terrain coordinate x and time t in this problem. The terrain roughness has been described only statistically (e.g., using an orbiting cam-era or some other remote sensing system). Its content is given by the following harmonic components: n 1 2 3 4 5 6 7 8 wave number, 1/m (kn =2π/λn) 0.250 0.275 0.300 0.325 0.350 0.375 0.400 0.425 0.450 0.475 0.500 amplitude, m (an) 0.01 0.02 0.02 0.04 0.03 0.02 0.01 0.03 0.05 0.01 0.02 9 10 11 Here, kn is the n’th wave number (spatial frequency) and λn is the n’th wavelength (spatial period). We model the ground elevation as a function of the horizontal coordinate: h(x)= an sin(knx + φn). n Notice that 2l<< min(λ); the vehicle is much shorter than the smallest wavelength, so you can approximate h(x) ≈ [h(x + l)+ h(x − l)]/2, h (x)= dh(x)/dx ≈ [h(x + l) − h(x − l)]/2l, h (x)= d 2h(x)/dx2 ≈ [h (x + l) − h (x − l)]/2l, and so on.� � 16 ROAD VEHICLE ON RANDOM TERRAIN 41 1. Develop a model that describes the response of the vehicle to terrain variations h(x). It should comprise two uncoupled second-order differential equations, one for vertical motion and the other for pitch. Make an annotated figure to go with the equations. You may like to use the fact that the vertical velocity of the ground seen from a reference frame moving horizontally at velocity U,is Uh (x). Newton’s laws and some geometry give us mz¨ = −k[(z + lθ − h(x + l)) + (z − lθ − h(x −l))] −b[( ˙z + lθ˙− Uh (x + l)) + ( ˙z − lθ˙− Uh (x − l))] ≈−k[2z − 2h(x)] − b[2 ˙z − 2Uh (x)] ¨ Jθ = −kl[(z + lθ − h(x + l)) − (z − lθ − h(x −l))] −bl[( ˙z + lθ˙− Uh (x + l)) − (˙z − lθ˙− Uh (x − l))] ≈−kl[2lθ − 2lh (x)] − bl[2lθ˙ − 2lUh (x)]. As advertised, these equations are uncoupled, except through the input h(x) and its derivatives. 2. What are the damped and undamped natural frequencies of the two systems? What are the damping ratios? List all six numbers, with units. For the vertical motion, the undamped natural frequency is ωn = 2k/m ≈ 6.9rad/s, and the damping ratio is ζ =2b/2mωn ≈ 0.35, giving a damped natural frequency of √ ωd = ωn 1 − ζ2 ≈ 6.5rad/s. For the pitch motion, the undamped natural frequency is ωn = 2l2k/J ≈ 4.0rad/s, and the damping ratio is ζ =2l2b/2Jωn ≈ 0.20, giving √ a damped natural frequency of ωd = ωn 1 − ζ2 ≈ 3.9rad/s. Both of these modes are rather lightly damped and so may exhibit resonance behavior. 3. Create a set of elevation data h(x) based on the above (spatial) frequency content, and plot it. Consider the range of x from zero to twenty times the longest wavelength; your spacing in x should be no more than one-tenth of the smallest wavelength, to get a good picture. Use a random phase angle φn for each of your components. See the attached figures and code. Note that I used fifty cycles of the longest wavelength (not twenty). 4. What is the maximum amplitude of elevation h(x) in the terrain data you created? What is the maximum amplitude of the slope? The maximum amplitude shown is about 0.21m, with a maximum slope of about 4.8◦ . These values come out differently each time the program is run because the phases are determined randomly. 5. Now run your vehicle model over this terrain at U =5m/s. What is the maximum amplitude of the vertical motion? What is the maximum amplitude of the pitch mo-tion?16 ROAD VEHICLE ON RANDOM TERRAIN 42 The maximum vertical motion amplitude is about 0.23m and the pitch is about 6.4◦ -both of these are larger than the terrain itself, indicating that some dynamic effect is occurring. 6. Repeat this calculation for speeds of 10, 15, 20, and 25m/s. Make a tabular listing for the vehicle’s maximum amplitudes of vertical motion and pitch, as a function of U. Are the motions consistent with the systems’ resonant frequencies and damping properties? The vertical motion amplitudes for U =[5, 10, 15, 20, 25]m/s are approximately [0.23, 0.29, 0.34, 0.29, 0.23]m, respectively. The peak amplification is 0.34/0.23 or al-most fifty percent, consistent with the damping ratio we found. The driving frequencies from the terrain are given by ωn = Ukn,so that 15m/s × 0.45rad/m =6.75rad/s -a high-amplitude harmonic input (0.05m) is occurring very close to the damped reso-nant frequency. The pitch motion amplitudes are approximately [6.4, 9.9, 4.9, 2.2, 1.3]◦ . The peak amplification is over two, consistent with the lower damping ratio in pitch. The slope of any particular terrain harmonic goes as knan, so again the input at k =0.45rad/m is the dominant one. 10m/s × 0.45rad/m =4.5rad/s, again pretty close to our damped natural frequency. 7. What different effects could we expect if the wheelbase became long compared to some of the terrain harmonic components? (No calculations.) We would start to see spatial aliasing. This would grotesquely deform our neat expla-nations above, and invalidate the approximations we used for the slopes and average values of h(x). This latter point could, however, be easily fixed in the code simply by carrying out the full expressions, that is, using h(x + l), h(x − l),and so on.16 ROAD VEHICLE ON RANDOM TERRAIN 43 0.08 0.06 0.04 0.02 0 Bretschneider Spectrum, H1/3 = 0.90m 0 0.5 1 1.5 2 2.5 3 3.5 4 rad/s 1 0.5 0 −0.5 −1 Simulation +/− H1/3/2 0 100 200 300 400 500 600 Time, seconds 0.2 0.18 0.16 0.14 0.12 0.1 0.08