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9 AUTONOMOUS VEHICLE MISSION DESIGN, WITH A SIMPLE BATTERY MODEL139 Autonomous Vehicle Mission Design, with a SimpleBattery ModelAn autonomous land robot carries its own energy in the form of a E(t = 0) = 700W h(Watt-hour) battery. The robot can stay stationary at a cost of 18W to run its processorand instrumentation, or it can move with the additional locomotion cost of 35U3Watts,where U is the speed in m/s. The battery has a nonlinear discharge curve, that penalizeshigh loads: −dE/dt = P + 0.005P2, where P is the total power load (taken in Watts), anddE/dt is also in Watts.1. What is the longest duration mission that we can run, if the mission ends when E = 0?The longest duration mission is one where we don’t move. We simply have the hotelload eating away at the battery, drawing 19.6W ; the mission time is T = 35.7h.Formulas are in the attached MATLAB code.2. What is the longest distance mission that we can run? Give the distance, duration,and vehicle speed for this mission.The battery energy loss rate is a static function of load, so we have to just pick the bestspeed and run until the battery is dead. My code takes a numerical approach, sweepingthrough a number of possible speeds and picking out the case with greatest distance.You could also do it directly, by writing one equation for the distance as a function ofthe speed, and then finding the maximum using l’Hopital’s rule (zero slope). Eitherway, we find that the greatest distance is about 52.5km, over about 24.3h at a speedof 0.60m/s.3. What is the longest distance mission, if we alternate resting periods and moving ata specific speed, with about half of the time spent in each state? Give the distance,duration, and the speed when moving.The key here is that half of the time, we have only the hotel load to satisfy, andotherwise we have hotel plus locomotion. In all, the battery energy discharges at theaverage of the two rates. But the distance traveled is only the moving speed times halfof the mission duration. I get a distance of 32.7km, over about 24.6h at a speed of0.74m/s.4. Comment on how you would use your findings in the design of effective mission plans.The longer the mission, the more important it is to understand the hotel load as distinctfrom locomotion. This is especially true for a vehicle that has a lot of power-hungrysensors or communication gear on board, but may not have much locomotion cost. Inthis example, the hotel load costs us 20km of range when we operate in the stop-gomode.9 AUTONOMOUS VEHICLE MISSION DESIGN, WITH A SIMPLE BATTERY MODEL14 0 10 20 30 40 50 60 70 80 90 100 0 50 100 150 Load, Watts Battery Energy Loss Rate, Watts Constant Speed Case Stop/Go Case 0 0.2 0.4 0.6 0.8 1 1.2 0 20 40 60 80 100 120 U, m/s Distance, km Duration, hours Load, W 0 0.2 0.4 0.6 0.8 1 1.2 U when moving, m/s 0 10 20 30 40 50 60 70 Distance, km Duration, hours Average Load, W9 AUTONOMOUS VEHICLE MISSION DESIGN, WITH A SIMPLE BATTERY MODEL15 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Autonomous Vehicle Energy with Battery Model % MIT 2.017 FSH Sept 2009 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear all; E0 = 700*3600 ; % battery capacity, Joules hotelLoad = 18 ; % Watts Uvec = 0:.01:1.3 ; % a vector of speeds we’ll look at squareCoefficient = 0.005 ; propLoadCoefficient = 35 ; % Show the discharge curve of the battery %%%%%%%%%%%%%%%%%%%%% P = 0:1:100 ; Edot = P + squareCoefficient*P.^2 ; figure(1);clf;hold off;subplot(211); plot(P,Edot,’LineWidth’,2); hold on; plot(P,P,’--k’); xlabel(’Load, Watts’); legend(’Battery Energy Loss Rate, Watts’,2); grid; % Problem 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Edot = hotelLoad + squareCoefficient*hotelLoad^2 ; duration1 = E0 / Edot ; disp(sprintf(’Max duration mission is %g hours, 0 km.’, duration1/3600)); % Problems 2 and 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% i=1; for U = Uvec, propulsionLoad = propLoadCoefficient*U^3 ; totalLoad = hotelLoad + propulsionLoad ; % Problem 2 Edot2(i) = totalLoad + squareCoefficient*totalLoad^2 ; duration2(i) = E0 / Edot2(i) ; distance2(i) = duration2(i) * U ; % Problem 3 Edot3(i) = (hotelLoad + squareCoefficient*hotelLoad^2) / 2 + ... (totalLoad + squareCoefficient*totalLoad^2) / 2 ; % time-averaged value of Edot duration3(i) = E0 / Edot3(i) ;9 AUTONOMOUS VEHICLE MISSION DESIGN, WITH A SIMPLE BATTERY MODEL16 distance3(i) = duration3(i) * U / 2 ; % note division by two i=i+1; end; figure(2);clf;hold off; subplot(211); plot(Uvec,distance2/1000,Uvec,duration2/3600,’--’,Uvec,Edot2,’:’,... ’LineWidth’,2) ; axis(’tight’); xlabel(’U, m/s’); title(’Constant Speed Case’); legend(’Distance, km’, ’Duration, hours’, ’Load, W’,2); [junk,ind] = sort(distance2) ; ix = ind(end); hold on; plot(Uvec(ix),distance2(ix)/1000,’ro’); disp(sprintf(’Max constant-speed distance mission is %g hours, %g km’, ... duration2(ix)/3600, distance2(ix)/1000)); subplot(212); plot(Uvec,distance3/1000,Uvec,duration3/3600,’--’,Uvec,Edot3,’:’,... ’LineWidth’,2) ; axis(’tight’); xlabel(’U when moving, m/s’); title(’Stop/Go Case’); legend(’Distance, km’, ’Duration, hours’,’Average Load, W’,2); [junk,ind] = sort(distance3) ; ix = ind(end); hold on; plot(Uvec(ix),distance3(ix)/1000,’ro’); disp(sprintf(’Max stop/go distance mission is %g hours, %g km’, ... duration3(ix)/3600, distance3(ix)/1000)); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%MIT OpenCourseWarehttp://ocw.mit.edu 2.017J Design of Electromechanical Robotic Systems Fall 2009 For information about citing these materials or our Terms of Use, visit:


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