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Article #182Re: Geometry question.
From: Neil Green
Date: Tue, 17 Apr 2007 04:38
Date: Tue, 17 Apr 2007 04:38
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3626 bytes
"Ken Pledger" <ken.pledger@mcs.vuw.ac.nz> wrote in message news:ken.pledger-7E7134.13502417042007@bats.mcs.vuw.ac.nz... > In article > <4621e020$0$16555$afc38c87@news.optusnet.com.au>, > "Neil Green" <nrgreenNOSPAM@optusnet.com.au> wrote: > >> I need to make some curved rafters for a verandah >> roof. >> The common rafters will be 1500mm when measured in >> a >> straight line from the end points with a curve of >> 100mm, that is, the mid point of the curve will be >> 100mm from a straight line drawn from end to end, >> and >> the curve will be part of a circle. > > > O.K. That much is clear. You could use the > technical terms "arc" > (shortened from the Latin word for "bow"), "chord" > (from Greek, but > related to the English "cord" meaning a string), and > "sagitta" (the > Latin word for "arrow"). A diagram makes the > reasons for those names > pretty obvious. > > You chord is 1500 mm and your sagitta is 100 > mm. Those are enough > to calculate the radius of the circle, which I make > 2862.5 mm. (If you > want the formula for that, just ask.) > > >> I need to know the curve for the rafters at each >> corner of the verandah which are at 45deg. >> The length is obviously 1500mm x sqrt2, but is the >> curve 100mm x sqrt2? .... > > > That's more obscure. Also, I didn't follow > your words "the common > rafters" above. Will you have rafters from left to > right, from front to > back, and diagonally as well? Also, your mentioning > sqrt(2) suggests > that viewed from above your verandah roof will be > square, not just any > old rectangle. I know it's hard without a diagram, > but could you > explain in a bit more detail? > > Ken Pledger. Thanks Ken. By common rafters I mean those that run at 90deg from the building, although there will also be one common rafter at each end against the building. The verandah roof will be rectangular, 4900 long by 1500 wide approximately, which in plan view can be divided into two 1500 x 1500 squares at each corner and a rectangle of 1900 x 1500. The rafters I need to calculate run diagonally across each corner to each corner point and need to be such that the roof is in a plane, both across the front and towards the building at each corner. The arc of these rafters needs to be different to the common rafters but I'm not sure how to get to it. These rafters will span 1500(Common rafter) x sqrt2. If this isn't clear can I email you a diagram? Thanks again Neil.
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X-Google-Thread: 10d67e,3cb062b66c09d2df X-Google-Attributes: gid10d67e,public X-Google-Language: ENGLISH,ASCII-7-bit Path: arc1-ca.pugleaf.net!g2news1.google.com!news2.google.com!border1.nntp.dca.giganews.com!nntp.giganews.com!nx01.iad01.newshosting.com!newshosting.com!novia!news1.optus.net.au!optus!newsfeeder.syd.optusnet.com.au!news.optusnet.com.au!not-for-mail From: "Neil Green" <nrgreenNOSPAM@optusnet.com.au> Newsgroups: aus.mathematics References: <4621e020$0$16555$afc38c87@news.optusnet.com.au> <ken.pledger-7E7134.13502417042007@bats.mcs.vuw.ac.nz> Subject: Re: Geometry question. Date: Tue, 17 Apr 2007 14:38:06 +1000 X-Priority: 3 X-MSMail-Priority: Normal X-Newsreader: Microsoft Outlook Express 6.00.2900.3028 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 X-RFC2646: Format=Flowed; Original X-Antivirus: avast! (VPS 000734-0, 16/04/2007), Outbound message X-Antivirus-Status: Clean Lines: 76 Message-ID: <46244f1c$0$9776$afc38c87@news.optusnet.com.au> NNTP-Posting-Host: 211.28.217.98 X-Trace: 1176784669 9776 211.28.217.98