tag:blogger.com,1999:blog-7893479142930851553.post6297586881061556719..comments2023-05-29T09:56:35.339+02:00Comments on Blog of eng. Xosé Carreira (since 2007): Packaging of stonesXosé Manuel Carreirahttp://www.blogger.com/profile/11874312282479409720noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-7893479142930851553.post-38503602629329239862012-12-26T10:21:56.460+01:002012-12-26T10:21:56.460+01:00its great.
Pile Repairs
Waterproofingits great.<br /><a href="http://www.dicotech.com/pipeline-coatings/" rel="nofollow">Pile Repairs</a><br /><br /><a href="http://www.dicotech.com/water-proofing/" rel="nofollow">Waterproofing</a><br />Anonymoushttps://www.blogger.com/profile/08082535432435878588noreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-61187759424478001972008-02-12T02:17:00.000+01:002008-02-12T02:17:00.000+01:00Apreciado Xosé,Grazas polo que me dis no meu blog....Apreciado Xosé,<BR/><BR/>Grazas polo que me dis no meu blog. Pensa que eu estudei na UNED fai xa uns 10 anos ou mais, co cual me imaxino que as cousas mudarán.<BR/><BR/>Se queres irme visitando, xa sabes. Eu leo regularmente Vieiros.<BR/><BR/>Até pronto,<BR/><BR/>martíMartí Duranhttps://www.blogger.com/profile/14434569386363525709noreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-13940480803222647672008-02-06T00:31:00.000+01:002008-02-06T00:31:00.000+01:00Oh, sorry, I forgot that you only have one d=1/3 s...Oh, sorry, I forgot that you only have one d=1/3 sphere between the large ones. You are right.David Gonzalez-Rodriguezhttps://www.blogger.com/profile/00051321908354883354noreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-54748871494880449562008-02-05T23:35:00.000+01:002008-02-05T23:35:00.000+01:00Sorry, I didn't get that. A priori, the next diame...Sorry, I didn't get that. A priori, the next diameter could be greater, equal or less than 1/9, because you are not filling the holes of the d=1/3 spheres: They are not in contact to each other.<BR/><BR/>By the way, my number is optimal in some way: Given the amounts of material that I said and the diameters I chose, the maximum density we can get is 94.62% (unless I'm mistaken, of course).<BR/><BR/>But as I said before, in order to talk about an optimal solution with different diameters we would first need some constraints, otherwise the problem is not well posed and the 100% is easily attainable.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-5333340813583276282008-02-05T22:16:00.000+01:002008-02-05T22:16:00.000+01:00I see... Well, at least I can say that the next d ...I see... Well, at least I can say that the next d is grater than or equal to d=1/9, which is the diameter I can fit in the holes of the d=1/3 spheres :)David Gonzalez-Rodriguezhttps://www.blogger.com/profile/00051321908354883354noreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-60706508182886438032008-02-05T21:35:00.000+01:002008-02-05T21:35:00.000+01:00Exactly. Yes, I am not saying that it is an optima...Exactly. Yes, I am not saying that it is an optimal solution (of course, the problem is too complicated to be solved during my break!! :) )<BR/><BR/>I just wanted to get "a value". <BR/><BR/>Yes, you put the spheres, and then you see that you can place another sphere of d=1/3 in the holes. Once the holes are filled the biggest sphere you can put is of d=1/6. It is not a geometric progression because the last sphere touches 2 of the d=1/3 spheres and also 3 of the d=1/6. So you are <B>not</B> repiting the same pattern.<BR/><BR/>If you want to find the next d you can fit, you are welcome, but the calculations start to get cumbersome. Good luck!!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-36397954614610074082008-02-05T15:48:00.000+01:002008-02-05T15:48:00.000+01:00Iago, how do you calculate that? Do you:(1) place ...Iago, how do you calculate that? Do you:<BR/><B>(1)</B> place the largest spheres in close packing, <BR/><B>(2)</B> look at the geometry of the holes between those,<BR/><B>(3)</B> compute how many spheres of the next size you can put in close packing in those holes,<BR/><B>(4)</B> go to (2)<BR/>?<BR/><BR/>If you follow this procedure (which may not be optimal), I would expect the diameters to form a geometric series (1,1/3,1/9,...). How did you come up with 1/6 for the third diameter?David Gonzalez-Rodriguezhttps://www.blogger.com/profile/00051321908354883354noreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-51127917309723633272008-02-05T00:48:00.000+01:002008-02-05T00:48:00.000+01:00May I "put my little grain of sand" (or add my two...May I "put my little grain of sand" (or add my two cents -of sand-, if you prefer)?<BR/><BR/>If we have 1kg of of gravel with diameter 1 and we add 20/4*(1/3)^3 kg of gravel with d=1/3 plus 20*(1/6)^3 kg of gravel with d=1/6 we get a density of packing of 94.62% (if I'm not mistaken). Is that interesting?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-32116171488237786222008-02-04T00:36:00.000+01:002008-02-04T00:36:00.000+01:00Off-topic: One of the items showing in News at iCi...Off-topic: One of the items showing in News at iCivilEngineer on the right navigation bar of this blog, <A HREF="http://www.icivilengineer.com/News/news.php?id=8402" REL="nofollow">"River plants may play major role in health of ocean coastal waters"</A>, refers to research being done in our Fluids Group at MIT :)David Gonzalez-Rodriguezhttps://www.blogger.com/profile/00051321908354883354noreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-38594163640299139112008-02-03T22:33:00.000+01:002008-02-03T22:33:00.000+01:00Hales's Holes' Theorem?:-SHales's Holes' Theorem?<BR/><BR/>:-SDavid Gonzalez-Rodriguezhttps://www.blogger.com/profile/00051321908354883354noreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-31118548178233367012008-02-03T15:44:00.000+01:002008-02-03T15:44:00.000+01:00So the proof transfroms Kepler's conjecture into T...So the proof transfroms Kepler's conjecture into T.Hales' Theorem!!<BR/><BR/>Interesting topic. I imagine David is right and it gets even more interesting when you try to find the distribution of diameters that gives you the highest compacity. (Of course subject to a constraint in the smallest diameter you can use or something like that, otherwise a monkey could get holes=0%).<BR/><BR/>In that case I wonder how it would compare to Füller and Bolomey.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7893479142930851553.post-78389692872471851022008-02-03T04:07:00.000+01:002008-02-03T04:07:00.000+01:00Ah, the beautiful world of the packing problems :)...Ah, the beautiful world of the packing problems :) Even if that answer must be more than sufficiently accurate for your purpose, I presume the problem must become quite complicated when you start thinking in terms of the <I>distribution</I> of sizes of the oranges, I mean, stones.David Gonzalez-Rodriguezhttps://www.blogger.com/profile/00051321908354883354noreply@blogger.com