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((asind(t2)-c)/b); %uncrossed angle between the slider and connecting bar t32=asind(-(asind(t2)-c)/b)+pi; %Crossed angle between the slider and connecting bar
d1=acosd(t2)-bcosd(t31); %X distance from the origin of the crank d2=acosd(t2)-bcosd(t32);
%% Velocity Calculations
w31=(a/b)(cosd(t2)/cosd(t31))w2;
w32=(a/b)(cosd(t2)/cosd(t32))w2;
ddot1=-aw2sind(t2)+bw31sind(t31);
ddot2=-aw2sind(t2)+bw32sind(t32);
VA=aw2(-sind(t2)+icosd(t2)); VBA1=-bw31(-sind(t31)+icosd(t31));
VBA2=-bw32(-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=rcosd(wt)+lsqrt(1-((r/l)sind(wt)).^2) %13.1d xdot=-rw(sind(wt)+(r/(2l))sind(2wt))%13.3
%xdot=-rw(sind(wt)+(r/(2l))(sind(2wt)/sqrt(1-(rsind(w*t)/l).^2)))%13.1e
max(xdot)
plot(t,x)

Sample Solution

ACED ESSAYS