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Question (1):

Problems

(a) Write a Matlab M-file which generates a table of error function (erf) and its derivatives for real arguments (z) between -3 and 3 in steps of 0.25. The error function is defined by the equation below (but is rarely evaluated by performing the integration).

The values in the table should be given with 5 decimal places. The table should have headers explaining what the columns are. Explain how you designed the M-file and give an example of the output. (b) How would you change this M-file if 10 significant digits were required? Matlab M-file should also be supplied Question (2): Write an M-file that reads your name in the form <first name> <middle name> <last name> and outputs the last name first and adds a comma after the name, the first name, and initial of your middle name with a period after the middle initial. If the names start with lower case letters, then these should be capitalized. The M-file should not be specific to the lengths of your name (ie., the M-file should work with anyone’s name. programminghomeworkhelp.com

As an example. An input of thomas abram herring would generate: Herring, Thomas A. Question (3): Write a Matlab M-file that will compute the motion of a bicyclist and the energy used cycling along an oscillating, sloped straight-line path. The path followed will be expressed as where H(x) is the height of the path above the starting height, S is a slope in m/m, A and B are amplitudes of sinusoidal oscillations in the path. The wavelength of the oscillations is λ. The forces acting on the bicycle are:

where Ar is the cross-sectional area of the rider, Cd is the drag coefficient, r is the density of air and V is the velocity of the bike. For the rolling drag, Mr is the mass of the rider and bike, g is gravitation acceleration and Cr is rolling drag coefficient.

The bicyclist puts power into the bike by pedaling. The force generated by this power is given by where Fr is the force produced by the rider, Pr is power used by the rider and V is velocity that the bike is traveling (the force is assumed to act along the velocity vector of the bike). programminghomeworkhelp.com

Your M-file can assume that the power can be used at different rates along the path. The energy used will be the integrated power supplied by the rider. Assume that there is maximum value to the rider force.

Your code should allow for input of the constants above (path and force coefficients). The Mfile can assume a constant power scenario and constant force at low velocities.

As a test of your M-file use the following constants to compute: (a) Time to travel and energy used to travel 10 km along a path specified by S=0.001, A=5.0 m, B=0.0 m and λ= 2km, with constant power use of Pr =100Watts and a maximum force available of 20N. (b) The position and velocity of the bike tabulated at a 100-second interval. (c) Add graphics to your M-file which plots the velocity of the bike as a function of time and position along the path.

Assume the following values Cd = 0.9 Cr = 0.007 A r = 0.67 m 2 r = 1.226 km/m3 g = 9.8 m/s2 Mr = 80 kg

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In this case, the Matlab M-file will not be of the type used for fortran and C/C++. Look at the documentation on ODExx where xx is a pair of number for Ordinary Differential Equation solutions. Your answer to this question should include: (a) The algorithms used and the design of your M-file (b) The Matlab M-file with your code and solution (I run your M-file). (c) The results from the test case above.

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Solutions Question 1: The solution to this problem is easy in Matlab because the Erf function is built in. In the solution we use Matlab’s vector operations to form the entries for the table in one single line. By using the correct transpose on the results we are able to print the table with one print statement. We also added to the solution plots of the function and its derivative. % HW5_1: Matlab script output Erf and its derivative % arg = -3:0.25:3; % Make a table with the results in it Results = [ arg ; erf(arg) ; 2/sqrt(pi)*exp(-arg.^2) ]; fprintf('\n12.010 HW05 Q01: Table of Erf and its derivative\n'); fprintf('-------------------------------------\n'); fprintf('| Arg x | Erf(x) | d(Erf)/dx |\n'); fprintf('-------------------------------------\n'); fprintf('| %7.3f | %10.5f | %10.5f |\n',Results); fprintf('-------------------------------------\n'); %% Create Figure of results % Regenerate results at finer spacing. figure ; arg = -3:0.05:3; programminghomeworkhelp.com

Results = [ arg ; erf(arg) ; 2/sqrt(pi)*exp(-arg.^2) ]; plot(Results(1,:),Results(2,:),'k',Results(1,:),Results(3,:),'b'); xlabel('Argument'); ylabel('Erf dErf/dx'); legend('Erf','dErf/dx','Location','NorthWest') axis tight; Below is shown in the output from the program for 5 significant digits. To increase the number of significant digits, the format for the table simply needs to be changed. All of the calculations were done with double precision and therefore no change to the variables needs to be made.

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Figure 1. Plot of the Erf function and its derivative. programminghomeworkhelp.com

Question (2): This solution is also easy in Matlab using the tolower and toupper functions. The code also detects whether two or three names have been given and changes the output accordingly. % HW5_2: Matlab script file to rearrange names. % % Ask user for name fprintf('\n----------------------------------------\n'); name = input('12.010 HW 05 Q2:\nEnter Name (first middle last) ','s'); % % Convert to lower case so that we know state name = lower(name); % % Find the blanks in the sting blanks = strfind(name,' '); if length(blanks) == 1 % Only two names passed, so skip middle initial first = name(1:blanks(1)-1); first(1) = upper(first(1)); last = name(blanks(1)+1:length(name)); last(1) = upper(last(1)); fprintf('%s, %s \n',last, first) programminghomeworkhelp.com

elseif length(blanks) == 2 % Split of the names. The function strtok could also be used to split % the string into parts. first = name(1:blanks(1)-1); middle = name(blanks(1)+1:blanks(2)-1); last = name(blanks(2)+1:length(name)); first(1) = upper(first(1)); middlei = upper(middle(1)); last(1) = upper(last(1)); fprintf('%s, %s %s.\n',last, first, middlei) else fprintf('Too many names or spaces in %s\nProgram end\n',name) End ---------------------------------------12.010 HW 05 Q2: Enter Name (first middle last) thomas abram herring Herring, Thomas A.

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Question 3: The solution to this problem is also quite easy in Matlab. We use the OED 45 1st order differential equation solver. To use this function we also have a function that computes the acceleration at the bike and another function that is the event detector that determines when the bike reaches 10 km distance. For ease of coding, the variables that control the characteristics of the bike are saved as globals. We also use an input dialog box to both set the defaults and allow the user to change the values at runtime. Also included in the solution is an animation and plot of the motion of the bike. To have the animation runs smoothly we change the output interval of the differential equation solver so that we generate more points then those that given in the standard 100 seconds separated values.

There are some interesting bugs in Matlab for plotting the velocity vectors. The quiver option is used for this plot but in the current release of Matlab the arrowheads are extremely large and are turned off in the plots. Also the automatic scaling of the vectors does not generate reasonable length vectors, and so in these plots we manually set the scale factor.

To compute the energy used during a bike ride, we simply added the derivative of the energy to the differential equations. This way the total energy was integrated automatically in the solution and could be output to the table as shown below. programminghomeworkhelp.com

The solution M-files are: % 12.010 HW 05 Q 3: Bike simulator % % This script m-file simulates the motions of a paper plane being % influenced by gravity, lift and drag % % There are many ways to solve this problem and we will use the Matlab ODE % solvers with event detection (the event in this case is the z-value % hitting zero). % global Grav Rho Mass Cd Cr Area Slope As Bs Lambda Pr Fr_Max TLength terr % Define these globals so that they will be accessible in the acceleration routine global Grav Rho Mass Cd Cr Area Slope As Bs Lambda Pr Fr_Max TLength terr % Define these globals so that they will be accessible in the acceleration routine Grav = -9.8 ; % m/sec^2 Rho = 1.226; % kg/m^3 %----- Get the inout that we need ----prompt = {'Track Grade (m/m)', 'Track Asin (m)', 'Track Bcos (m)', 'Wavelegth (km)','Track Length (km)','Mass (kg)', ...

'Rider Area (m^2)', 'Drag Coefficient','Rolling coefficient','Power programminghomeworkhelp.com

(watts)','Max Force (N)', ...

'Output Interval (sec)','Error (mm)'}; title = 'Track and Bike Parameters'; nlines = 1; defaults = {'0.001','5.0','0.0','2.0','10.0','80.0','0.67','0.9','0.007','100.0','20','100','1'}; % Get user inputs and save in the appropriate valuyes result = inputdlg(prompt,title,nlines, defaults,'on'); % Slope = eval(result{1}); As = eval(result{2}); Bs = eval(result{3}); Lambda = eval(result{4})*1000.0 ; % Convert to meters TLength = eval(result{5})*1000.0 ; % Convert to meters Mass = eval(result{6}) ; Area = eval(result{7}) ; Cd = eval(result{8}) ; Cr = eval(result{9}) ; Pr = eval(result{10}) ; Fr_Max = eval(result{11}) ; tstep = eval(result{12}) ; terr = eval(result{13})/1000 ; % convert height error to m programminghomeworkhelp.com

% % Set the ODE options. hit is the name of an m-file which will detect when % we hit the ground. options = odeset('AbsTol',[terr terr terr terr 100],'Events','bikehit'); % Set up the initial conditions y0 = [0.0; Bs ; 0.0 ; 0.0; 0.0]; % Set the maximum integration time to 5000 seconds. In a more general % problem we should check to see of this is enough time tmax = 5000; % Solve the ODE fprintf('\n Starting new run \n'); [t,y,te,ye,ie] = ode45(@bikeacc,[0:tstep:tmax],y0,options); % % Output the final results fprintf('---------------------------------------------------------------------\n'); fprintf('12.010 HW 05 Q 3: Bike simulation\n'); % Output table of values fprintf('--------------------------------------------------------------------------\n') fprintf(' Time X pos Z pos X Vel Z Vel Energy \n') fprintf(' (sec) (m) (m) (m/s) (m/s) (Joules)\n') fprintf(' %9.3f %12.4f %10.4f %10.4f %10.4f %12.2f\n',[t,y(:,1:5)]'); programminghomeworkhelp.com

fprintf('-------------------------------------------------------------------------------------------------\n') fprintf(‘ Time : %8.3f, Velocity %6.4f %6.4f \n',te,ye(3),ye(4)) fprintf('Energy used: %12.2f Joules, %8.2f kcals\n', ye(5), ye(5)*0.2388e-3); % Now animate (make as function) %banimate(t,y); % Re-run the solution for the animation [ta,ya,tae,yae,iae] = ode45(@bikeacc,[0:10.0:te],y0,options); fprintf('\nTo show animation: banimate(ta,ya)\n'); % function dy = bikeacc(t, y) % acc: Computes accelerations including drag (Cd), rolling (Cr) and % and gravity % y contains, [x z xdot zdot energy] and dy is the time derivative % global Grav Rho Mass Cd Cr Area Slope As Bs Lambda Pr Fr_Max TLength % Define these globals so that they will be accessible in the acceleration routine dy = zeros(5,1); % a column vector % % get the road slope for our current x position. programminghomeworkhelp.com

theta = atan(Slope + As*cos(2*pi*y(1)/Lambda)*2*pi/Lambda Bs*sin(2*pi*y(1)/Lambda)*2*pi/Lambda); % % Compute forces: Assuming bike remains on road surface % Drag: vmag = sqrt(y(3)^2+y(4)^2); thetad = atan2(y(4),y(3)); dxd = -Rho*Cd*vmag^2*Area*cos(thetad)/(2*Mass); dzd = -Rho*Cd*vmag^2*Area*sin(thetad)/(2*Mass); dxr = +Grav*Cr*cos(theta) ; dzr = +Grav*Cr*sin(theta) ; % % Get rider force Fr_mag = Fr_Max ; if ( vmag > 0 ) Fr_mag = Pr/vmag ; end if ( Fr_mag > Fr_Max ) Fr_mag = Fr_Max ; end height = Slope*y(1) + As*sin(2*pi*y(1)/Lambda) + Bs*cos(2*pi*y(1)/Lambda); %if ( y(2) > height + 0.1 ) Fr_mag = 0; end dxp = Fr_mag*cos(theta)/Mass ; dzp = Fr_mag*sin(theta)/Mass ; % % Normal force from surface (Grav is a negative value) dxn = Grav*cos(theta)*sin(theta); programminghomeworkhelp.com

dzn = -Grav*cos(theta)^2; % Get the curvature dy2dx2 = -As*sin(2*pi*y(1)/Lambda)*(2*pi/Lambda)^2 Bs*cos(2*pi*y(1)/Lambda)*(2*pi/Lambda)^2; IRc = (dy2dx2)/(1+theta^2)^(3/2); dxc = vmag^2*sin(theta)*IRc; dzc = vmag^2*cos(theta)*IRc; %fprintf(' %5.2f %9.5f %9.5f %5.2f %6.4f %10.1f \n',vmag, dxc, dzc, dxd, dzd, Rc) % % Update the dy velocity acceleration vector dy(1) = y(3); dy(2) = y(4); dy(3) = dxd + dxr + dxp + dxn + dxc ; dy(4) = dzd + dzr + dzp + dzn + dzc + Grav ; dy(5) = Fr_mag * vmag; function [value,isterminal,direction] = bikehit(t,y) % Locate the time when height passes through zero in a decreasing direction % and stop integration. global Grav Rho Mass Cd Cr Area Slope As Bs Lambda Pr Fr_Max TLength value = y(1)-TLength; isterminal = 1; direction = 1;

% detect when end is 0. % stop the integration % negative direction

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function banimate(t,y) % Function to animate the motion qs = 50; num = length(t); hf1 = figure(1); set(hf1, 'Position',[10 100 750 400]); hold off; pc = plot(y(1,1),y(1,2),'o','MarkerFaceColor','r','EraseMode','xor'); hold on pq = quiver(y(1,1),y(1,2),y(1,3),y(1,4),0); xm = y(num,1)*1.01; ym = max(y(:,2))*1.01; xs = min(y(:,1))*1.01; ys = min(y(:,2))*1.01; xlim([xs xm]); ylim([ys ym]); xlabel('Distance (m)'); ylabel('Height (m)'); pt = text(xm/100,ym*0.96,'Time 0.00 s'); %daspect([1 1 1]); drawnow; % Release 2011a of Matlab has some strange scaling in quiver and so we % manually set properities that should be automatically set. Also arrow % heads are very wide so we turn them off here. for i = 1: num % Replace the data point set(pc,'XData',y(i,1),'YData',y(i,2)); delete(pq); pq = quiver(y(i,1),y(i,2),y(i,3),y(i,4),qs); set(pq,'ShowArrowHead','off','AutoScaleFactor',qs,'AutoScale','on'); otext = sprintf('Time %6.1f s',t(i)); programminghomeworkhelp.com

set(pt,'String',otext); drawnow; end plot(y(:,1),y(:,2),'g'); %% hf2 = figure(2); set(hf2, 'Position',[760 100 350 200]); hold off; pc = plot(y(:,1),y(:,2),'o','MarkerFaceColor','r','EraseMode','normal'); hold on pq = quiver(y(:,1),y(:,2),y(:,3),y(:,4)); set(pq,'ShowArrowHead','off'); xlabel('Distance (m)'); ylabel('Height (m)'); axis tight; ylim([-10 15]); %% hf3 = figure(3); set(hf3, 'Position',[760 380 350 200]); plot(t,(sqrt(y(:,3).^2+y(:,4).^2))); xlabel('Time (sec)'); ylabel('Total Velocity (m/s)'); The table and figures below show the results for the default solution.

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Figure 2: Trajectory of the bike along with vectors showing its velocity (green lines). Due to poor scaling in Matlab, no arrowheads are shown.

Figure 3: Velocity as a function of time for the bike. programminghomeworkhelp.com