tag:blogger.com,1999:blog-1564958057737541664.post114915976617626675..comments2014-11-25T14:40:03.111-08:00Comments on On Bicycles, and.... what else is there?: Vector and Stages power comparisondjconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-1564958057737541664.post-26420580086393084942013-12-22T05:16:57.463-08:002013-12-22T05:16:57.463-08:00Thanks! I fixed it. I use ImageShack.us (now bec...Thanks! I fixed it. I use ImageShack.us (now becoming ImageShack.com ) for photos and sometimes there's a bit of flakiness. It's usually great, however: really easy to use.djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-72338464148972410152013-12-21T19:29:24.322-08:002013-12-21T19:29:24.322-08:00last picture do not show... last picture do not show... kamileqhttp://www.blogger.com/profile/08420974271649663998noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-9500836688564096622013-08-14T12:27:15.169-07:002013-08-14T12:27:15.169-07:00Thanks! I suspect they don't solve any equati...Thanks! I suspect they don't solve any equations, but rather use a purely heuristic approach: hang weights on the pedal and calibrate. The question is whether it can be calibrated to work no matter how far outward the weight is hung, and that it tracks the cosine of the angle relative to horizontal as the crank is rotated.djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-54228523332477136112013-08-13T08:42:19.505-07:002013-08-13T08:42:19.505-07:00If you draw a free body diagram for the stages pow...If you draw a free body diagram for the stages power meter, there are 6 degrees of freedom (Rotation XY, Rotation YZ, Rotation XY, Delta X, Delta Y, and Delta Z) that you would be solving to determine your stress, which you could convert to force. Its a bit complicated to see when you draw it on paper (2-D), but its really a 3-D problem. There are standard beam equations that are easy to solve with a computer. The equations also make certain assumptions for small deflection and that assumption is appropriate for this problem.<br /><br />The vector also has 6 degrees of freedom to solve as well. The equations are different though for a beam vs. a cylinder.<br /><br />Here is a good link to explain the beam equations a bit more:<br /><br />http://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory<br /><br />This makes the 2D problem for SRM/Quarq seem very simple.<br /><br />Eitherway, i dont think the shape and the deflection/strain relationship are difficult engineering issues to solve. I think the difficulty is implementing the solution and making it accurate and reliable. From a theoretical standpoint, its a fairly straight forward problem to solve.ian spivackhttp://www.blogger.com/profile/11242088042844554423noreply@blogger.com