tag:blogger.com,1999:blog-1606329698370612755.post325446202314594217..comments2016-10-19T08:58:58.698+02:00Comments on The Geometry of Bending: Analyzing the Drop shapeMårten Nettelbladtnoreply@blogger.comBlogger6125tag:blogger.com,1999:blog-1606329698370612755.post-53730244082353510012010-06-24T23:42:08.213+02:002010-06-24T23:42:08.213+02:00Thanks Robert, yes I remember you posted a comment...Thanks Robert, yes I remember you posted a comment earlier with <a href="http://sites.google.com/site/tetrahelix/lv-light" rel="nofollow">this link</a>.Mårtenhttps://www.blogger.com/profile/03560723401512980420noreply@blogger.comtag:blogger.com,1999:blog-1606329698370612755.post-72696637519222725122010-06-22T06:31:14.937+02:002010-06-22T06:31:14.937+02:00Marten, I thought I had replied to you before...
...Marten, I thought I had replied to you before... <br />take two rectangles of veneer (or paper) cut into the proportion 1 by 3.<br />overlay the veneers at 90 degrees to each other, glue, once set twist the open corners together and overlay again to form the cone and glue.<br /><br />the resultant cone fits perfectly over a 60 degree (equilateral cone). the bottom edge arcs have the the length 3 whilst the top arcs forming the opening have length 1. As the over lapped pieces are square the diagonal is of length root 2. the base diameter and length of the sides are therefore root 2 + root 2/2.<br /><br />general notes:<br />overlapping the veneers by 90 degrees has structural advantages as it leaves no fragile end grain exposed so the form is surprisingly resilient.<br />http://sites.google.com/site/tetrahelix/lv-light<br /><br />another form is to use a rectangle of proportion 1 by 4 and twist and overlap at 90 degrees back on itself. as I recall this fits over a 30 degree cone.robertmathiesonhttps://www.blogger.com/profile/07398992275878781323noreply@blogger.comtag:blogger.com,1999:blog-1606329698370612755.post-37420031479463416152010-06-17T22:49:43.648+02:002010-06-17T22:49:43.648+02:00Cool! That sounds very interesting! Swedish: Jag t...Cool! That sounds very interesting! Swedish: Jag tittade lite i din blogg-lista och såg att även du är intresserad av enkelkrökta ytor... Kul!Mårtenhttps://www.blogger.com/profile/03560723401512980420noreply@blogger.comtag:blogger.com,1999:blog-1606329698370612755.post-40842147102801323342010-06-17T22:06:14.404+02:002010-06-17T22:06:14.404+02:00Yes a capillary surface, an interface between two ...Yes a capillary surface, an interface between two liquids (ie air/water).<br /><br />I've been investigating these surfaces for potential uses in architecture. I figure they must have interesting statics properties.kristofferhttps://www.blogger.com/profile/01453872235668781814noreply@blogger.comtag:blogger.com,1999:blog-1606329698370612755.post-71757340000111850492010-06-17T21:31:03.410+02:002010-06-17T21:31:03.410+02:00Thanks Kristoffer! Are you talking about drops as ...Thanks Kristoffer! Are you talking about drops as in liquid drops?Mårtenhttps://www.blogger.com/profile/03560723401512980420noreply@blogger.comtag:blogger.com,1999:blog-1606329698370612755.post-6212534863305791162010-06-17T21:03:20.511+02:002010-06-17T21:03:20.511+02:00The shape of the pendant drop can be described by ...The shape of the pendant drop can be described by the equation H = -z*g + c where H is the mean curvature, g is the gravitational constant, c is constant and z is the direction of gravity.<br />In absense of gravity it reads H = 1, for which the solution is the sphere.<br />In the general case, it is not easy to solve, other than numerically. This is because curvature is an intrinsic notion, whereas the z coordinate of the solution is not, it leads to an ordinary differential equation that is non-trivial to solve.<br />It is however possible to describe the curve as a graph over the z-axis. This however, does not involve the curvature.kristofferhttps://www.blogger.com/profile/01453872235668781814noreply@blogger.com