tag:blogger.com,1999:blog-2688138290971915830.post2437025383288289874..comments2018-01-18T20:35:54.853-08:00Comments on Interdependent Science: Bicycle Tire ShapeJimKhttp://www.blogger.com/profile/16167191806249119508noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-2688138290971915830.post-31940641212651820992018-01-18T20:35:54.853-08:002018-01-18T20:35:54.853-08:00Wow that is very cool! Please do post a link if yo...Wow that is very cool! Please do post a link if you ever put any details up on-line!JimKhttps://www.blogger.com/profile/16167191806249119508noreply@blogger.comtag:blogger.com,1999:blog-2688138290971915830.post-1173983955058181352018-01-18T20:35:07.769-08:002018-01-18T20:35:07.769-08:00yes, you are surely correct! yes, you are surely correct! JimKhttps://www.blogger.com/profile/16167191806249119508noreply@blogger.comtag:blogger.com,1999:blog-2688138290971915830.post-17889385046409002192018-01-08T18:39:22.261-08:002018-01-08T18:39:22.261-08:00It worked!
I worked out the radius of curvature, ...It worked!<br /><br />I worked out the radius of curvature, assuming a given height (and rim width and original cross section), with the constraint of the curve being tangent at the edge of the contact patch. It was a transcendental equation so I wrote a little script to numerically solve it. Then I separately calculated the net force on the rim, and it came out to exactly the same as the force on the contact patch!Charliehttps://www.blogger.com/profile/18076080703641893710noreply@blogger.comtag:blogger.com,1999:blog-2688138290971915830.post-9349126691534887182018-01-08T17:17:16.371-08:002018-01-08T17:17:16.371-08:00Just discovered this a couple of years after you p...Just discovered this a couple of years after you posted it. It seems that me that there needs to be one other condition: it seems that the arc section needs to be tangent to the flat section that is in contact with the road. Otherwise the point a the edge of the contact patch would be pulled away from the road by the tension in the arced sidewall.<br /><br />Now we have a real problem--it might not be possible to satisfy your condition of force on the rim equaling force on the road! Perhaps it will simply work out that those two conditions will give the same result, which would be very satisfying if that worked.<br /><br />So in your picture, the center of the arc should be a distance p to the right of the centerline.<br /><br />I might need to try some numerical experiments...<br /><br /> Charliehttps://www.blogger.com/profile/18076080703641893710noreply@blogger.comtag:blogger.com,1999:blog-2688138290971915830.post-70119064186118671482015-11-19T17:50:13.388-08:002015-11-19T17:50:13.388-08:00oops, that first formula should have sin theta ins...oops, that first formula should have sin theta instead of cos theta!JimKhttps://www.blogger.com/profile/16167191806249119508noreply@blogger.com