tag:blogger.com,1999:blog-1564958057737541664.post859487455551484551..comments2015-01-18T17:11:02.699-08:00Comments on On Bicycles, and.... what else is there?: Metrigear Vector Pt 6: Forces and Bending (2)djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-1564958057737541664.post-2733217032396249872014-10-16T10:09:05.579-07:002014-10-16T10:09:05.579-07:00Thanks for the note! If you catch any errors let ...Thanks for the note! If you catch any errors let me know. The reason you need two is the gauge measures the amount of bending, which is proportional to a bending moment (force times distance). I don't know either force or distance. So to solve for two unknowns, I need two constraints, which I can get by using two gauges the separation between which is already known. So by determining the bending moment relative to two reference positions I can solve for both distance and force. The complication is that real systems are more complicated than ideal cantilevers, and so the simple models I present here are imperfect.djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-47659304267547388202014-10-16T09:55:18.783-07:002014-10-16T09:55:18.783-07:00Hi DJConnel, This blog has been extremely helpful....Hi DJConnel, This blog has been extremely helpful. I'm currently working on a project for undergraduate capstone which involves building a strain gauge based pedal to measure force assymetry. <br /><br />Why are two strain gauges used to measure overall force? Could the same be accomplished with only one for each axis?Ben Russellhttp://www.blogger.com/profile/02670661936728443359noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-64611626342691115252009-10-24T03:51:40.359-07:002009-10-24T03:51:40.359-07:00Interesting! Curiously, I initially thought the s...Interesting! Curiously, I initially thought the sensors would pick up total force, not a bending moment. Jim Pampodopolous, an MIT engineer who collaborated on Bicycling Science Third Edition, said it would be a bending moment and not a force measured by the sensors. That's consistent with two parallel strain gauges, oriented longitudinally, one at the center axis and one away from the center, measuring the difference in strain.<br /><br />But if they were instead configured to measure shear strain, which is a change in vertical position along the axis of the pedal, then you're right that the driving force would be force, not a force times distance.<br /><br />So I guess the question is then whether it's relatively easier and more reliable to measure bending or to measure shear distortion. It's a hollow tapered cyclinder, so I think one would need to run some real numbers (preferably FEA) to see if there's a strong enough shear signal without second-order contamination from bending stress, and vice-versa.<br /><br />My money's would be on the bending moment. I suppose I could <i>ask</i>, but it's more fun to guess.<br /><br />It's funny I'm so interested in this topic. It sort of gives me an excuse to exercise some basic physics on a topic (bicycles) which I really enjoy.djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-7337687434119993852009-10-23T20:32:59.314-07:002009-10-23T20:32:59.314-07:00The link shows shear force and bending diagrams. ...The link shows shear force and bending diagrams. Note that unlike bending moment, shear force is not a function of displacement along the beam. I don't know how the measurements are oriented in the Metrigear design, but it is possible to arrange the sensors such that the electrical output is a function of shear force.Jim Meyerhttp://www.blogger.com/profile/08570639966758644342noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-76132030510499992072009-10-23T14:22:25.480-07:002009-10-23T14:22:25.480-07:00Thanks! The bending moment is the cross-product o...Thanks! The bending moment is the cross-product of a position and a force, which my freshman physics brain calls a torque, but I guess "bending moment" would be more clearly differentiated from the twisting moment applied to the pedal body.<br /><br />I'm trying to wrap my head around <a href="http://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.html" rel="nofollow">Jim's reference</a>, but I think what he's saying is Bernoulli's idealized cantilever fails to adequately describe a relatively short hollow tapered cylinder like a pedal spindle, at least within the 1.5% error budget, and that off-diagonal elements of the stiffness matrix are needed to really solve the problem (not just the scalar term used in the simple bending analysis). So maybe a second order calibration is needed.djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-56832333168926428152009-10-23T13:07:16.305-07:002009-10-23T13:07:16.305-07:00This is a good article, and I agree with what you ...This is a good article, and I agree with what you are saying, but it would be more clear if you referred to the bending load on the pedal spindle as a "moment" rather than a "torque".Our Crewhttp://www.blogger.com/profile/03118552935022733472noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-77753099780345939842009-10-23T08:12:57.318-07:002009-10-23T08:12:57.318-07:00http://www.roymech.co.uk/Useful_Tables/Beams/Shear...http://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.htmlJim Meyerhttp://www.blogger.com/profile/08570639966758644342noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-56635934636992772802009-10-22T19:51:24.698-07:002009-10-22T19:51:24.698-07:00Well, I suppose this shows a difference: my backgr...Well, I suppose this shows a difference: my background in mechanics is <a href="http://web.mit.edu/physics/" rel="nofollow">undergraduate physics</a> where all sensors are viewed as ideal, whereas your background is in actually making things that work :). But it seems the curvature is the only "observable" along the two axes perpendicular to the spindle (along the longitudinal axis you can measure strain directly).djconnelhttp://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-1564958057737541664.post-71300815725581671962009-10-22T15:30:19.849-07:002009-10-22T15:30:19.849-07:00Here is something to noodle on:
What if the sensor...Here is something to noodle on:<br />What if the sensors don't measure bending moment?Jim Meyerhttp://www.blogger.com/profile/08570639966758644342noreply@blogger.com