Technical Analysis

Force Analysis

Ensuring the prototype had enough torque to effectively move the MCBD was a primary factor in generating the prototype. The team determined the prototype needed to produce at least a torque of 10 lb-in or greater in order to account for the weight of the final assembly as well as the larger, heavier ball which will need more torque to move with the same desired speed of the previous team. This would ensure that the MCBD could move effectively in a factory setting and fast enough to be efficient. The torque calculations along with the parameters/variables used are in the figures to the left.

The torque calculations are below:

MatLab Code Snippets

Additionally, the provided MATLAB code calculates the forces, moments, and angles involved in the interaction of the Magnetically Coupled Ball Drive (MCBD) system with its environment, specifically when encountering an obstacle.

Key Results:

The net moment (Mf) indicates whether the forces and torques balance around the wheel. A positive moment is favorable, while a negative moment could indicate instability or issues with system design.

close all

clc

clear all

D = 9.39; %ball diameter

Ww =4; %omni wheel width (in)

Oh = 1; %Obstacle Height (in)

Hh = 5; %housing height (in) omniwheel height

v = 90; %velocity of impact (in/s)

t = 0.3; %time of impact (s)

W= .55; %Weight of ball (lbs)

Fm = 20; %magnetic force (lbf)

m = 75/3; %mass of system / 3

R = D/2; %radius

h1 = D-Ww/2;

De = 2 * sqrt(2*R*h1-h1^2) %effective diameter

Code continues from snippet on the left

Code continued from here

Ow = 2*sqrt(2*R*Oh-Oh^2); %obstacle distance from ball center

h2 = (D-Hh);

Hw = sqrt(2*R*h2-h2^2) %distance from contact point to center of the ball

theta = atan((D/2-Oh)/Ow); %angle of impact of obstacle normal to ball

thetadeg = theta*180/pi

Fo = W/32.2*v/t %force of obstacle impact

Fx = Fo*cos(theta) %x component

Fy = 2*(Fo*sin(theta)) %y component

%moment from horizontal wheels and far wheel

%convention CCW(+)

Mxo = -(Hh-Oh)*Fx %moment from x component

Myo = (Hw+Ow)*Fy %moment from y component

Mw = -De/2*W %moment from weight

Mm = De/2*Fm %moment from magnetic force

Mf = Mxo+Myo+Mw+Mm %final moment about wheel

%positve is good negative is bad

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