----- Original Message -----
From: Serge Bedard <[EMAIL PROTECTED]>
To: Quantitative Decisions <[EMAIL PROTECTED]>;
<[EMAIL PROTECTED]>
Sent: Thursday, April 12, 2001 2:54 PM
Subject: RE: MI-L Re: Calculating distance from a point to bodies of water
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>
> You're perfectly right Bill, a Thiessen/Vorono=EF approach would =
> probably be
> much better. Sorry about that Adam.
>
> -----Original Message-----
> From: Quantitative Decisions [mailto:[EMAIL PROTECTED]]
> Sent: Wednesday, April 11, 2001 5:28 PM
> To: [EMAIL PROTECTED]
> Subject: MI-L Re: Calculating distance from a point
> to bodies of water
>
> At 03:43 PM 4/11/01 -0400, a respondent wrote:
> >At first sight, if your bodies of water are closed
> polygons, you could
> >determine which body of water is the closest of a
> particular point. To
> >narrow the search, you could calculate the distance between
> your point and
> >the centroid of each body of water and keep, say, the 10
> closest ones.
>
> This clever idea will fail in some exceptional cases, such
> as when the=20
> bodies of water are long rivers: New Orleans, for instance,
> is right on the=20
> Mississippi River, but it is closer to the centroids of
> hundreds of=20
> tributaries (and hundreds of other rivers, lakes, and bayous
> unconnected to=20
> the Mississippi) than it is to the centroid of the river
> itself.
>
> The computational geometry literature contains many
> efficient and correct=20
> solutions to the problem of finding the distances from each
> point to its=20
> nearest polygon. One of them, a vector approach,
> generalizes the Thiessen=20
> polygon construction; references to this are in Preparata &
> Shamos'=20
> book. Another, a raster approach, constructs the distance
> grid for the=20
> collection of water body polygons. After this preliminary
> computation,=20
> finding the required distances is very rapid: you just
> interpolate within=20
> the distance grid. Other approaches use quadtrees or other
> spatial data=20
> structures precomputed from the polygons. Most GIS software
> has one (or=20
> more) of these solutions built in to its basic
> functionality.
>
>
> >Since the distance between a point and a
> >line is proportional to the area of the triangle formed by
> the point and the
> >two nodes of the line,
>
> This statement assumes the perpendicular from the point to
> the line falls=20
> between the two nodes, which will usually *not* be the case.
>
> The general formula is easily expressed in vector notation.
> Let the=20
> endpoints of the line segment be a and b and let the point
> be p. Let v =3D=20
> b-a. Each of a, b, p, and v is a vector. Compute t =3D inner
> product of v=20
> with (p-a) and v2 =3D inner product of v with itself; t and v2
> are=20
> numbers. If t <=3D 0, the closest point to p is a and the
> distance is=20
> |p-a|. If t >=3D v2, the closest point to p is b and the
> distance is=20
> |p-b|. Otherwise, v2 <> 0, the point c =3D a + t*v/v2 is the
> closest point=20
> to p along the line segment between a and b, and the
> distance is |p-c|.
>
>
> >I would use the determinant method to calculate it.
> >This method works as follow: calculate the determinant of
> this matrix:
> >xp yp 1
> >x1 y1 1
> >x2 y2 1
>
> None of these possibilities can be expressed by a
> determinant based on the=20
> point and node coordinates alone: a square root operation is
> unavoidable=20
> (except for some special configurations). This is clear
> when you consider=20
> the effect of changing the units of measurement. If all
> coordinates are=20
> multiplied by z, to be consistent the answer had better be
> multiplied by=20
> |z|; however, the determinant above (or any similar looking
> expression)=20
> will be multiplied by z^2.
>
> --Bill Huber
> Quantitative Decisions
>
>
>
> =09
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