Path: news.mathworks.com!not-for-mail
From: <HIDDEN>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Downsampling with fixed intervals
Date: Wed, 24 Sep 2008 12:30:06 +0000 (UTC)
Organization: Universit&#228;t Basel
Lines: 41
Message-ID: <gbdbse$kh9$1@fred.mathworks.com>
References: <gbd9hb$9u$1@fred.mathworks.com> <gbdams$9m4$1@fred.mathworks.com>
Reply-To: <HIDDEN>
NNTP-Posting-Host: webapp-02-blr.mathworks.com
Content-Type: text/plain; charset="ISO-8859-1"
Content-Transfer-Encoding: 8bit
X-Trace: fred.mathworks.com 1222259406 21033 172.30.248.37 (24 Sep 2008 12:30:06 GMT)
X-Complaints-To: news@mathworks.com
NNTP-Posting-Date: Wed, 24 Sep 2008 12:30:06 +0000 (UTC)
X-Newsreader: MATLAB Central Newsreader 870595
Xref: news.mathworks.com comp.soft-sys.matlab:491773


Thanks for your reply, John. I think, I'll try it your way. I just wanted to be sure that something like this is not already available and I'd waste time on something others could do better.

Best regards,
Wolfgang

> 
> No, you will not find anything that does this
> explicitly, although one could write it without
> too severe mental anguish.
> 
> Assume that you have a curve defined by a
> sequence of points in the (x,y) plane. The
> curve may be piecewise linear, or it may be
> a parametric spline, defined in terms of a
> piecewise linear arclength along the curve.
> 
> Start at the beginning of the curve. Find the
> (first) intersection of your curve and a circle
> with center at the start point. Step along
> said curve, solving a sequence of root
> finding problems along the way.
> 
> You might use the intersections tool, written
> by Doug Schwarz and on the file exchange.
> It will require you to define the initial curve
> as a piecewise linear one, and then to
> approximate your circle using a piecewise
> linear, polygonal approximation to a circle.
> But doing so will then be a very fast way to
> solve your problem.
> 
> The only issue that you must deal with,
> regardless of how you choose to solve the
> problem, is if the coastline is so tightly
> convoluted that these convolutions are
> smaller than the length of your ruler. But
> in that case, it just means that your ruler
> is way too long for this problem.
> 
> HTH,
> John