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of the cam-follower mechanism Since we are concerned with the maximum pressure angle tan a m = tan t m = E M (148)

Summary

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EXAMPLE An eccentric circle cam drives a translating roller follower This circle has an eccentricity of 0500 in and a diameter of 2500 in and rotates at 180 rpm The roller follower has a diameter of 0750 in Find (1) the characteristics of the follower motion after the cam has rotated 30 from its lowest position and (2) the maximum pressure angle Solution The equivalent mechanism has a crank length E equal to 0500 in The connecting rod length equals the eccentric circle radius plus the roller radius, giving

M = 1250 + 0375 = 1625 in The cam angular velocity w = 180 From Eq (144) sin t = E sin e M 0500 t = sin -1 sin 30 = 885 1625 2p = 1885 rad sec 60

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In re National Institute of Standards and Technology Use of Electronic Data Interchange to Create Valid Obligations, file B-245714 (Comptroller Gen l, 1991) Mesenbourg, Thomas L Measuring Electronic Business: Definitions, Underlying Concepts, and Measurement Plans US Census Bureau http://www censusgov/epcd/www/ebusineshtm Mitchell, Mary General Services Administration The State of Federal eGovernment and eBusiness January 23, 2001 http://wwwegovgov/presentations/DigGovMI12001/sld001htm Office of Management and Budget Guidance on Implementing the Electronic Signatures in Global and National Commerce Act (E-SIGN Act) http:// wwwwhitehousegov/omb/memoranda/esign-guidancepdf Smedinghoff, Thomas J, and Ruth Hill Bro Electronic Signature Legislation http://profslpfindlawcom/signatures/signature_4html US Department of Commerce Economic and Statistics Administration National Telecommunications and Information Administration A Nation Online: How Americans Are Expanding Their Use of the Internet February 2002 http:// wwwntiadocgov/ntiahome/dn/ US Department of Commerce E-stats March 18, 2002 http://wwwcensus gov/eos/www/papers/estatstextpdf

The displacement from Eq (142) y = E - E cos e + M cos t - M = 0500 - 1625 - 0500 cos 30 + 1625 cos 885 = 00475 in The velocity from Eq (146) y = Ew (sin e - cos e tan t ) = ( 0500)(1885)(sin 30 - cos 30 tan 885) = 344 lb in The acceleration, Eq (147) E cos 2 e y = Ew 2 cos e + sin e tan t M cos3 t 0500 cos 2 30 2 + sin 30 tan 885 = ( 0500)(1885) cos 30 1625 cos3 885 = 1252 in sec The maximum pressure angle from Eq (148)

Testable designs have virtual methods, nonsealed classes, interfaces, and a clear separation of concerns. They have fewer static classes and methods, and many more instances of logic classes. In fact, testable designs are what SOLID design principles have stood for. Perhaps it s time that the end goal should not be testability, but good design instead.

t m = tan -1

The eccentric is a type of cam It is a circular member having its center of rotation offset from its geometric center The follower may be at-faced or roller type A translating atfaced follower will be discussed The eccentric represents the simplest cam mechanism possible Its zero pressure angle eliminates the problem of the follower jamming In Fig 148a we see an eccentric rotating about point A with its geometric center at a distance E In Fig 148b we see the equivalent mechanism which is the Scotch yoke mechanism in which the follower movement is a simple harmonic function Let y = follower displacement, in ye = equivalent mechanism follower displacement, in q = cam angle of rotation for displacement y, rad e = crank angle of rotation (equivalent mechanism) for displacement ye = rad E = distance from cam center to circular-arc center of curvature, in h = 2E = maximum displacement of follower, in w = cam and equivalent mechanism angular velocity, rad/sec In this example, y = ye and q = e From Fig 148b we see that the follower displacement

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