Essay Analysis

        Write 2 pages analysis about the linkages that MUST match these requirements. • These pics shows our design HOWEVER, we need to change it to what is written down. • Need a crank-slider system where the slider can reach at least 50mph and has a long enough stroke that we can load another puck between revolutions. • keep in mind that dimensions should be easily scaleable. • Make parts that can be scaled down to under 8 inches. • The motor is going to hook directly to the linkage, then gear it down to run the cam for the loader. • I want to aim for a 1/4 scale model, so no parts over 32 inches. We can only print 8in parts. • Must use Norton Software Linkage • what is the angular speed of the crank you are using in you calculations? • Must include MATLAB Codes and references. • The MATLAB codes below should work for some of the analysis clc;clear;close all;pi=3.1415926; %% Inputs (P196 in the book for diagram) a=0.1; %Crank length b=0.1; %Connecting bar between crank and slider c=0.1; %Height of the slider above the crank origin w2=0.1; %Angluar velocity of the crank t2=linspace(0,360,361); %Angle of the crank at the moment in time we are interested in, in degrees! %% Positional calculations t31=asind((a*sind(t2)-c)/b); %uncrossed angle between the slider and connecting bar t32=asind(-(a*sind(t2)-c)/b)+pi; %Crossed angle between the slider and connecting bar d1=acosd(t2)-b*cosd(t31); %X distance from the origin of the crank d2=acosd(t2)-b*cosd(t32); %% Velocity Calculations w31=(a/b)*(cosd(t2)/cosd(t31))*w2; w32=(a/b)*(cosd(t2)/cosd(t32))*w2; ddot1=-a*w2*sind(t2)+b*w31*sind(t31); ddot2=-a*w2*sind(t2)+b*w32*sind(t32); VA=a*w2*(-sind(t2)+i*cosd(t2)); VBA1=-b*w31*(-sind(t31)+i*cosd(t31)); VBA2=-b*w32*(-sind(t32)+i*cosd(t32)); VB1=VA+VBA1; VB2=VA+VBA2; %% Report max velocities Max1=max(VB1) l1=find(VB1==Max1)-1 Max2=max(VB2) l2=find(VB2==Max2)-1   Mass properties of Cranklink Configuration: Default Coordinate system: -- default -- Density = 0.06 pounds per cubic inch Mass = 17.98 pounds Volume = 279.67 cubic inches Surface area = 508.23 square inches Center of mass: ( inches ) X = 14.31 Y = 0.07 Z = 0.00 Principal axes of inertia and principal moments of inertia: ( pounds * square inches ) Taken at the center of mass. Ix = ( 1.00, -0.01, 0.00) Px = 38.17 Iy = ( 0.00, 0.00, -1.00) Py = 2290.21 Iz = ( 0.01, 1.00, 0.00) Pz = 2318.32 Moments of inertia: ( pounds * square inches ) Taken at the center of mass and aligned with the output coordinate system. Lxx = 38.31 Lxy = -18.15 Lxz = 0.00 Lyx = -18.15 Lyy = 2318.18 Lyz = 0.00 Lzx = 0.00 Lzy = 0.00 Lzz = 2290.21 Moments of inertia: ( pounds * square inches ) Taken at the output coordinate system. Ixx = 38.40 Ixy = 0.00 Ixz = 0.00 Iyx = 0.00 Iyy = 6000.24 Iyz = 0.00 Izx = 0.00 Izy = 0.00 Izz = 5972.37 Carbon fiber   Mass properties of Part2 (link3) Configuration: Default Coordinate system: -- default -- Density = 0.06 pounds per cubic inch Mass = 6.62 pounds Volume = 102.93 cubic inches Surface area = 292.74 square inches Center of mass: ( inches ) X = 0.00 Y = 0.00 Z = 0.00 Principal axes of inertia and principal moments of inertia: ( pounds * square inches ) Taken at the center of mass. Ix = ( 1.00, 0.00, 0.00) Px = 3.98 Iy = ( 0.00, 0.00, -1.00) Py = 946.06 Iz = ( 0.00, 1.00, 0.00) Pz = 948.94 Moments of inertia: ( pounds * square inches ) Taken at the center of mass and aligned with the output coordinate system. Lxx = 3.98 Lxy = 0.00 Lxz = 0.00 Lyx = 0.00 Lyy = 948.94 Lyz = 0.00 Lzx = 0.00 Lzy = 0.00 Lzz = 946.06 Moments of inertia: ( pounds * square inches ) Taken at the output coordinate system. Ixx = 3.98 Ixy = 0.00 Ixz = 0.00 Iyx = 0.00 Iyy = 948.94 Iyz = 0.00 Izx = 0.00 Izy = 0.00 Izz = 946.06 Mass properties of Piston Configuration: Default Coordinate system: -- default -- Density = 0.28 pounds per cubic inch Mass = 10.66 pounds Volume = 37.55 cubic inches Surface area = 81.73 square inches Center of mass: ( inches ) X = -0.17 Y = 0.00 Z = 0.00 Principal axes of inertia and principal moments of inertia: ( pounds * square inches ) Taken at the center of mass. Ix = ( 1.00, 0.00, 0.00) Px = 17.30 Iy = ( 0.00, 0.00, -1.00) Py = 26.40 Iz = ( 0.00, 1.00, 0.00) Pz = 36.60 Moments of inertia: ( pounds * square inches ) Taken at the center of mass and aligned with the output coordinate system. Lxx = 17.30 Lxy = 0.00 Lxz = 0.00 Lyx = 0.00 Lyy = 36.60 Lyz = 0.00 Lzx = 0.00 Lzy = 0.00 Lzz = 26.40 Moments of inertia: ( pounds * square inches ) Taken at the output coordinate system. Ixx = 17.30 Ixy = 0.00 Ixz = 0.00 Iyx = 0.00 Iyy = 36.92 Iyz = 0.00 Izx = 0.00 Izy = 0.00 Izz = 26.71 clc clear close all w=20.943951; t=linspace(0,20,1000) r=36; l=40; x=r*cosd(w*t)+l*sqrt(1-((r/l)*sind(w*t)).^2) %13.1d xdot=-r*w*(sind(w*t)+(r/(2*l))*sind(2*w*t))%13.3 %xdot=-r*w*(sind(w*t)+(r/(2*l))*(sind(2*w*t)/sqrt(1-(r*sind(w*t)/l).^2)))%13.1e max(xdot) plot(t,x)