Thread Subject: River miles

I have a file with data at lat/longs, and a file with River miles at lat/longs. How can I interpolate the River mile onto the data. I have tried using interp2, but that has been hugely unsuccessful. Please keep in mind that the river meanders and frequently doubles back on itself in both latitude and longitude.

On Nov 18, 12:01 pm, "Travis" <sinuso...@hotmail.com> wrote:
> I have a file with data at lat/longs, and a file with River miles at lat/longs. How can I interpolate the River mile onto the data. I have tried using interp2, but that has been hugely unsuccessful. Please keep in mind that the river meanders and frequently doubles back on itself in both latitude and longitude.

Lets say you have s(x,y) river miles, where x and y are the long and
lat respectively, and you want to find s for a particular xp, yp.
You need to split your problem into two parts:
1. Find the segment in x,y where xp,yp lies by firstly finding the
segments where the normal from xp,yp is inside the segment and
secondly by minimising the distance of the normal to each of those
segments.
2. Interpolate on s within the optimal segment.

NZTideMan <mulgor@gmail.com> wrote in message <4521b8d9-2021-4ff0-b430-08822a0ce2bc@s9g2000prm.googlegroups.com>...
> On Nov 18, 12:01 pm, "Travis" <sinuso...@hotmail.com> wrote:
> > I have a file with data at lat/longs, and a file with River miles at lat/longs. How can I interpolate the River mile onto the data. I have tried using interp2, but that has been hugely unsuccessful. Please keep in mind that the river meanders and frequently doubles back on itself in both latitude and longitude.
>
> Lets say you have s(x,y) river miles, where x and y are the long and
> lat respectively, and you want to find s for a particular xp, yp.
> You need to split your problem into two parts:
> 1. Find the segment in x,y where xp,yp lies by firstly finding the
> segments where the normal from xp,yp is inside the segment and
> secondly by minimising the distance of the normal to each of those
> segments.
> 2. Interpolate on s within the optimal segment.


I have tried seperating the lat and long and using interp 1, but since it double back on itself, the values are not distinct, and it doesn't work.

On Nov 18, 6:43 pm, "Travis" <sinuso...@hotmail.com> wrote:
> NZTideMan <mul...@gmail.com> wrote in message <4521b8d9-2021-4ff0-b430-08822a0ce...@s9g2000prm.googlegroups.com>...
> > On Nov 18, 12:01 pm, "Travis" <sinuso...@hotmail.com> wrote:
> > > I have a file with data at lat/longs, and a file with River miles at lat/longs. How can I interpolate the River mile onto the data. I have tried using interp2, but that has been hugely unsuccessful. Please keep in mind that the river meanders and frequently doubles back on itself in both latitude and longitude.
>
> > Lets say you have s(x,y) river miles, where x and y are the long and
> > lat respectively, and you want to find s for a particular xp, yp.
> > You need to split your problem into two parts:
> > 1. Find the segment in x,y where xp,yp lies by firstly finding the
> > segments where the normal from xp,yp is inside the segment and
> > secondly by minimising the distance of the normal to each of those
> > segments.
> > 2. Interpolate on s within the optimal segment.
>
> I have tried seperating the lat and long and using interp 1, but since it double back on itself, the values are not distinct, and it doesn't work.

It doesn't seem like you have followed my advice.
Show us the code you used for my Step 1 (which is the more important
part).

I think you need to better specify what your data looks like. I assume you have a 3xn array with x and y coordinates and river length and a 2xn array with vertex coordinates of the river course. Am I right to assume so?




"Travis" <sinusoid2@hotmail.com> wrote in message <gftkl5$6b$1@fred.mathworks.com>...
> NZTideMan <mulgor@gmail.com> wrote in message <4521b8d9-2021-4ff0-b430-08822a0ce2bc@s9g2000prm.googlegroups.com>...
> > On Nov 18, 12:01 pm, "Travis" <sinuso...@hotmail.com> wrote:
> > > I have a file with data at lat/longs, and a file with River miles at lat/longs. How can I interpolate the River mile onto the data. I have tried using interp2, but that has been hugely unsuccessful. Please keep in mind that the river meanders and frequently doubles back on itself in both latitude and longitude.
> >
> > Lets say you have s(x,y) river miles, where x and y are the long and
> > lat respectively, and you want to find s for a particular xp, yp.
> > You need to split your problem into two parts:
> > 1. Find the segment in x,y where xp,yp lies by firstly finding the
> > segments where the normal from xp,yp is inside the segment and
> > secondly by minimising the distance of the normal to each of those
> > segments.
> > 2. Interpolate on s within the optimal segment.
>
>
> I have tried seperating the lat and long and using interp 1, but since it double back on itself, the values are not distinct, and it doesn't work.

On 18 Nov, 00:01, "Travis" <sinuso...@hotmail.com> wrote:
> I have a file with data at lat/longs, and a file with River miles at lat/=
longs. =A0How can I interpolate the River mile onto the data. =A0

What do you try to do? Do you have the lat-longs and
distances for the same points? If so it ought to be
trivial to interpolate the data.

Rune

Rune Allnor <allnor@tele.ntnu.no> wrote in message <90f017a9-906d-4c7d-8ac5-7167e15e0d8c@u18g2000pro.googlegroups.com>...
> On 18 Nov, 00:01, "Travis" <sinuso...@hotmail.com> wrote:
> > I have a file with data at lat/longs, and a file with River miles at lat/=
> longs. =A0How can I interpolate the River mile onto the data. =A0
>
> What do you try to do? Do you have the lat-longs and
> distances for the same points? If so it ought to be
> trivial to interpolate the data.
>
> Rune

Consider this simple example. I have a "river" that passes
through the three points:

xy = [0 0;0 1; 1 1]

At each point, I have some other parameter or parameters
associated, some values of z at those points. Now, I have
another point, perhaps at the (x,y) coordinates (0.20, 0.75).
How should I choose to interpolate the z value onto that
new point?

I'll assume that the "river" meanders around so that it passes
through this new point, but the interpolation process is surely
not trivial in general. You must decide how the river has
meandered between the existing points, and even decide
which pair of points this new point lies between.

Yes, it is surely doable, and nztideman has suggested a
method that has serious merit. But it is not trivial.

John

On 18 Nov, 14:30, "John D'Errico" <woodch...@rochester.rr.com> wrote:
> Rune Allnor <all...@tele.ntnu.no> wrote in message <90f017a9-906d-4c7d-8ac5-7167e15e0...@u18g2000pro.googlegroups.com>...
> > On 18 Nov, 00:01, "Travis" <sinuso...@hotmail.com> wrote:
> > > I have a file with data at lat/longs, and a file with River miles at lat/=
> > longs. =A0How can I interpolate the River mile onto the data. =A0
>
> > What do you try to do? Do you have the lat-longs and
> > distances for the same points? If so it ought to be
> > trivial to interpolate the data.
>
> > Rune
>
> Consider this simple example. I have a "river" that passes
> through the three points:
>
> xy = [0 0;0 1; 1 1]
>
> At each point, I have some other parameter or parameters
> associated, some values of z at those points. Now, I have
> another point, perhaps at the (x,y) coordinates (0.20, 0.75).
> How should I choose to interpolate the z value onto that
> new point?

First, assign a distance d for each of the points (view
table with fixed-width fonts):

       lat long d
xy = [ 0 0 0;
       0 1 1;
       1 1 2;]

Use some measure to project the position onto the line
(shortest distance, lat-priority, long-priority) and
obtain a distance estimate for the observed point.
Go on from there.

> I'll assume that the "river" meanders around so that it passes
> through this new point, but the interpolation process is surely
> not trivial in general.

If you didn't track the river in sufficient detail you
are basically in over your head anyway.

> You must decide how the river has
> meandered between the existing points, and even decide
> which pair of points this new point lies between.

Nope. That's guesswork, as the measurements are too sparse
to be useful. Use at your own risk. If you really want
to track a meandering river, you need to sample its
positions far denser than the loops in the river.

> Yes, it is surely doable, and nztideman has suggested a
> method that has serious merit. But it is not trivial.

It's basically the same method I suggested. Don't expect
all points to map into the interior of any line segment,
though. For instance, with the track you suggested,
where do you map the point (0.9, 1.1)? And don't expect
the answets to be unique; where do you assign (1.1,0.9)?

The answers depend on what projection rule you select.

Rune

Rune Allnor <allnor@tele.ntnu.no> wrote in message <6875d2af-8eb2-443a-a831-d4c1773c1180@c36g2000prc.googlegroups.com>...
> On 18 Nov, 14:30, "John D'Errico" <woodch...@rochester.rr.com> wrote:
> > Rune Allnor <all...@tele.ntnu.no> wrote in message <90f017a9-906d-4c7d-8ac5-7167e15e0...@u18g2000pro.googlegroups.com>...
> > > On 18 Nov, 00:01, "Travis" <sinuso...@hotmail.com> wrote:
> > > > I have a file with data at lat/longs, and a file with River miles at lat/=
> > > longs. =A0How can I interpolate the River mile onto the data. =A0
> >
> > > What do you try to do? Do you have the lat-longs and
> > > distances for the same points? If so it ought to be
> > > trivial to interpolate the data.
> >
> > > Rune
> >
> > Consider this simple example. I have a "river" that passes
> > through the three points:
> >
> > xy = [0 0;0 1; 1 1]
> >
> > At each point, I have some other parameter or parameters
> > associated, some values of z at those points. Now, I have
> > another point, perhaps at the (x,y) coordinates (0.20, 0.75).
> > How should I choose to interpolate the z value onto that
> > new point?
>
> First, assign a distance d for each of the points (view
> table with fixed-width fonts):
>
> lat long d
> xy = [ 0 0 0;
> 0 1 1;
> 1 1 2;]
>
> Use some measure to project the position onto the line
> (shortest distance, lat-priority, long-priority) and
> obtain a distance estimate for the observed point.
> Go on from there.
>
> > I'll assume that the "river" meanders around so that it passes
> > through this new point, but the interpolation process is surely
> > not trivial in general.
>
> If you didn't track the river in sufficient detail you
> are basically in over your head anyway.
>
> > You must decide how the river has
> > meandered between the existing points, and even decide
> > which pair of points this new point lies between.
>
> Nope. That's guesswork, as the measurements are too sparse
> to be useful. Use at your own risk. If you really want
> to track a meandering river, you need to sample its
> positions far denser than the loops in the river.
>
> > Yes, it is surely doable, and nztideman has suggested a
> > method that has serious merit. But it is not trivial.
>
> It's basically the same method I suggested. Don't expect
> all points to map into the interior of any line segment,
> though. For instance, with the track you suggested,
> where do you map the point (0.9, 1.1)? And don't expect
> the answets to be unique; where do you assign (1.1,0.9)?
>
> The answers depend on what projection rule you select.
>
> Rune

Wolfgang, that is a correct assumption. Rune the river was mapped using a GPS that recorded time, posiion, and heading each time the heading changed. It was then interpolated so that data taken between heading chages have unique lat/long values. Are you suggesting that I break the river up into segments that do not double back on themselves and use interp1 on each segment? I have a 60 mile stretch of river to do this for.

On 18 Nov, 16:13, "Travis" <sinuso...@hotmail.com> wrote:
> Rune Allnor <all...@tele.ntnu.no> wrote in message <6875d2af-8eb2-443a-a8=
31-d4c1773c1...@c36g2000prc.googlegroups.com>...

> =A0Rune the river was mapped using a GPS that recorded time, posiion, and=
 heading each time the heading changed. =A0It was then interpolated so that=
 data taken between heading chages have unique lat/long values. =A0Are you =
suggesting that I break the river up into segments that do not double back =
on themselves

Yes. Use the segments between consecutive (lat,long) points
as basis.

> and use interp1 on each segment?

No. Interpolate the whole curve.

>=A0I have a 60 mile stretch of river to do this for. =A0

No need to do that manually. The idea is to sort
the data not according to lat or long, but according
to travelled/accumulated distance.

The distance serves as a parameter, and so the
(lat, long) path becomes an arbitrary parametric
curve in the plane. Once you see this basic idea
the rest is (almost) trivial.

Rune

Rune Allnor <allnor@tele.ntnu.no> wrote in message <f36d9204-1e90-4b53-b6f2-6d1b6ecb1f4b@i18g2000prf.googlegroups.com>...
> On 18 Nov, 16:13, "Travis" <sinuso...@hotmail.com> wrote:
> > Rune Allnor <all...@tele.ntnu.no> wrote in message <6875d2af-8eb2-443a-a8=
> 31-d4c1773c1...@c36g2000prc.googlegroups.com>...
>
> > =A0Rune the river was mapped using a GPS that recorded time, posiion, and=
> heading each time the heading changed. =A0It was then interpolated so that=
> data taken between heading chages have unique lat/long values. =A0Are you =
> suggesting that I break the river up into segments that do not double back =
> on themselves
>
> Yes. Use the segments between consecutive (lat,long) points
> as basis.
>
> > and use interp1 on each segment?
>
> No. Interpolate the whole curve.
>
> >=A0I have a 60 mile stretch of river to do this for. =A0
>
> No need to do that manually. The idea is to sort
> the data not according to lat or long, but according
> to travelled/accumulated distance.
>
> The distance serves as a parameter, and so the
> (lat, long) path becomes an arbitrary parametric
> curve in the plane. Once you see this basic idea
> the rest is (almost) trivial.
>
> Rune



What coding would I use in this instance?

On 19 Nov, 21:42, "Travis" <sinuso...@hotmail.com> wrote:
> Rune Allnor <all...@tele.ntnu.no> wrote in message <f36d9204-1e90-4b53-b6=
f2-6d1b6ecb1...@i18g2000prf.googlegroups.com>...
> > On 18 Nov, 16:13, "Travis" <sinuso...@hotmail.com> wrote:
> > > Rune Allnor <all...@tele.ntnu.no> wrote in message <6875d2af-8eb2-443=
a-a8=3D
> > 31-d4c1773c1...@c36g2000prc.googlegroups.com>...
>
> > > =3DA0Rune the river was mapped using a GPS that recorded time, posiio=
n, and=3D
> > =A0heading each time the heading changed. =3DA0It was then interpolated=
 so that=3D
> > =A0data taken between heading chages have unique lat/long values. =3DA0=
Are you =3D
> > suggesting that I break the river up into segments that do not double b=
ack =3D
> > on themselves
>
> > Yes. Use the segments between consecutive (lat,long) points
> > as basis.
>
> > > and use interp1 on each segment?
>
> > No. Interpolate the whole curve.
>
> > >=3DA0I have a 60 mile stretch of river to do this for. =3DA0
>
> > No need to do that manually. The idea is to sort
> > the data not according to lat or long, but according
> > to travelled/accumulated distance.
>
> > The distance serves as a parameter, and so the
> > (lat, long) path becomes an arbitrary parametric
> > curve in the plane. Once you see this basic idea
> > the rest is (almost) trivial.
>
> > Rune
>
> What coding would I use in this instance?=96 Skjul sitert tekst =96
>
> =96 Vis sitert tekst =96

Don't know what you mean by coding, but here's an example:

% lat long dist
d =3D[0.0081 0.9128 0
    0.1340 0.9210 0.9307
    0.2437 0.9485 1.9100
    0.3660 0.9609 2.9382
    0.5027 0.9581 4.0202
    0.5836 0.9375 5.1245
    0.6106 0.9114 6.2216
    0.6178 0.8592 7.2798
    0.5567 0.7837 8.2410
    0.4344 0.7658 9.1215
    0.2671 0.7782 9.9442
    0.1736 0.7919 10.7549
    0.0872 0.7617 11.5215
    0.0854 0.6985 12.2252
    0.1178 0.6738 12.9092
    0.3300 0.6078 13.6008
    0.4308 0.6147 14.3514
    0.6106 0.6490 15.2426
    0.7383 0.6834 16.2486
    0.8714 0.7012 17.3671
    0.9379 0.6463 18.5062
    0.9397 0.5968 19.6194
    0.9074 0.5391 20.6749
    0.8444 0.5364 21.6753
    0.7167 0.5474 22.5771
    0.5800 0.5776 23.3957
    0.4829 0.5254 24.1093
    0.3804 0.4512 24.6995
    0.3768 0.4045 25.2524
    0.4272 0.3674 25.8158
    0.5495 0.3633 26.4745
    0.6214 0.3743 27.2000
    0.7581 0.3565 28.0377
    0.7923 0.3207 28.8924
    0.7815 0.2507 29.7131
    0.7131 0.1655 30.4452
    0.5585 0.1669 31.0281
    0.4937 0.1875 31.5562
    0.3678 0.2026 31.9761
    0.2653 0.1806 32.2970
    0.2149 0.1133 32.5400
    0.2941 0.0680 32.8418
    0.4200 0.0694 33.2675
    0.5387 0.0556 33.8090];

% Plot the river path
subplot(2,1,1)
plot(d(:,1),d(:,2))

% Plot the lat and long vs distance
subplot(2,1,2)
plot(d(:,3),d(:,1),'b',d(:,3),d(:,2),'r')

Now you can interpolate both lats and longs with
respect to distance, and maybe be able to go on
from there.

Rune

Rune Allnor <allnor@tele.ntnu.no> wrote in message <01d3cdd8-c83e-47bc-9357-fb97949f0eff@r40g2000yqj.googlegroups.com>...
> On 19 Nov, 21:42, "Travis" <sinuso...@hotmail.com> wrote:
> > Rune Allnor <all...@tele.ntnu.no> wrote in message <f36d9204-1e90-4b53-b6=
> f2-6d1b6ecb1...@i18g2000prf.googlegroups.com>...
> > > On 18 Nov, 16:13, "Travis" <sinuso...@hotmail.com> wrote:
> > > > Rune Allnor <all...@tele.ntnu.no> wrote in message <6875d2af-8eb2-443=
> a-a8=3D
> > > 31-d4c1773c1...@c36g2000prc.googlegroups.com>...
> >
> > > > =3DA0Rune the river was mapped using a GPS that recorded time, posiio=
> n, and=3D
> > > =A0heading each time the heading changed. =3DA0It was then interpolated=
> so that=3D
> > > =A0data taken between heading chages have unique lat/long values. =3DA0=
> Are you =3D
> > > suggesting that I break the river up into segments that do not double b=
> ack =3D
> > > on themselves
> >
> > > Yes. Use the segments between consecutive (lat,long) points
> > > as basis.
> >
> > > > and use interp1 on each segment?
> >
> > > No. Interpolate the whole curve.
> >
> > > >=3DA0I have a 60 mile stretch of river to do this for. =3DA0
> >
> > > No need to do that manually. The idea is to sort
> > > the data not according to lat or long, but according
> > > to travelled/accumulated distance.
> >
> > > The distance serves as a parameter, and so the
> > > (lat, long) path becomes an arbitrary parametric
> > > curve in the plane. Once you see this basic idea
> > > the rest is (almost) trivial.
> >
> > > Rune
> >
> > What coding would I use in this instance?=96 Skjul sitert tekst =96
> >
> > =96 Vis sitert tekst =96
>
> Don't know what you mean by coding, but here's an example:
>
> % lat long dist
> d =3D[0.0081 0.9128 0
> 0.1340 0.9210 0.9307
> 0.2437 0.9485 1.9100
> 0.3660 0.9609 2.9382
> 0.5027 0.9581 4.0202
> 0.5836 0.9375 5.1245
> 0.6106 0.9114 6.2216
> 0.6178 0.8592 7.2798
> 0.5567 0.7837 8.2410
> 0.4344 0.7658 9.1215
> 0.2671 0.7782 9.9442
> 0.1736 0.7919 10.7549
> 0.0872 0.7617 11.5215
> 0.0854 0.6985 12.2252
> 0.1178 0.6738 12.9092
> 0.3300 0.6078 13.6008
> 0.4308 0.6147 14.3514
> 0.6106 0.6490 15.2426
> 0.7383 0.6834 16.2486
> 0.8714 0.7012 17.3671
> 0.9379 0.6463 18.5062
> 0.9397 0.5968 19.6194
> 0.9074 0.5391 20.6749
> 0.8444 0.5364 21.6753
> 0.7167 0.5474 22.5771
> 0.5800 0.5776 23.3957
> 0.4829 0.5254 24.1093
> 0.3804 0.4512 24.6995
> 0.3768 0.4045 25.2524
> 0.4272 0.3674 25.8158
> 0.5495 0.3633 26.4745
> 0.6214 0.3743 27.2000
> 0.7581 0.3565 28.0377
> 0.7923 0.3207 28.8924
> 0.7815 0.2507 29.7131
> 0.7131 0.1655 30.4452
> 0.5585 0.1669 31.0281
> 0.4937 0.1875 31.5562
> 0.3678 0.2026 31.9761
> 0.2653 0.1806 32.2970
> 0.2149 0.1133 32.5400
> 0.2941 0.0680 32.8418
> 0.4200 0.0694 33.2675
> 0.5387 0.0556 33.8090];
>
> % Plot the river path
> subplot(2,1,1)
> plot(d(:,1),d(:,2))
>
> % Plot the lat and long vs distance
> subplot(2,1,2)
> plot(d(:,3),d(:,1),'b',d(:,3),d(:,2),'r')
>
> Now you can interpolate both lats and longs with
> respect to distance, and maybe be able to go on
> from there.
>
> Rune

You are forgetting that even interpolating to either lat or long has a problem...there are mulitple river miles for numerous lats/longs, and matlab does not like that.

In article <gg22tm$k2v$1@fred.mathworks.com>, sinusoid2@hotmail.com
says...
> Rune Allnor <allnor@tele.ntnu.no> wrote in message <01d3cdd8-c83e-47bc-9357-fb97949f0eff@r40g2000yqj.googlegroups.com>...
> > On 19 Nov, 21:42, "Travis" <sinuso...@hotmail.com> wrote:
> > > Rune Allnor <all...@tele.ntnu.no> wrote in message <f36d9204-1e90-4b53-b6=
> > f2-6d1b6ecb1...@i18g2000prf.googlegroups.com>...
> > > > On 18 Nov, 16:13, "Travis" <sinuso...@hotmail.com> wrote:
> > > > > Rune Allnor <all...@tele.ntnu.no> wrote in message <6875d2af-8eb2-443=
> > a-a8=3D
> > > > 31-d4c1773c1...@c36g2000prc.googlegroups.com>...
> > >
> > > > > =3DA0Rune the river was mapped using a GPS that recorded time, posiio=
> > n, and=3D
> > > > =A0heading each time the heading changed. =3DA0It was then interpolated=
> > so that=3D
> > > > =A0data taken between heading chages have unique lat/long values. =3DA0=
> > Are you =3D
> > > > suggesting that I break the river up into segments that do not double b=
> > ack =3D
> > > > on themselves
> > >
> > > > Yes. Use the segments between consecutive (lat,long) points
> > > > as basis.
> > >
> > > > > and use interp1 on each segment?
> > >
> > > > No. Interpolate the whole curve.
> > >
> > > > >=3DA0I have a 60 mile stretch of river to do this for. =3DA0
> > >
> > > > No need to do that manually. The idea is to sort
> > > > the data not according to lat or long, but according
> > > > to travelled/accumulated distance.
> > >
> > > > The distance serves as a parameter, and so the
> > > > (lat, long) path becomes an arbitrary parametric
> > > > curve in the plane. Once you see this basic idea
> > > > the rest is (almost) trivial.
> > >
> > > > Rune
> > >
> > > What coding would I use in this instance?=96 Skjul sitert tekst =96
> > >
> > > =96 Vis sitert tekst =96
> >
> > Don't know what you mean by coding, but here's an example:
> >
> > % lat long dist
> > d =3D[0.0081 0.9128 0
> > 0.1340 0.9210 0.9307
> > 0.2437 0.9485 1.9100
> > 0.3660 0.9609 2.9382
> > 0.5027 0.9581 4.0202
> > 0.5836 0.9375 5.1245
> > 0.6106 0.9114 6.2216
> > 0.6178 0.8592 7.2798
> > 0.5567 0.7837 8.2410
> > 0.4344 0.7658 9.1215
> > 0.2671 0.7782 9.9442
> > 0.1736 0.7919 10.7549
> > 0.0872 0.7617 11.5215
> > 0.0854 0.6985 12.2252
> > 0.1178 0.6738 12.9092
> > 0.3300 0.6078 13.6008
> > 0.4308 0.6147 14.3514
> > 0.6106 0.6490 15.2426
> > 0.7383 0.6834 16.2486
> > 0.8714 0.7012 17.3671
> > 0.9379 0.6463 18.5062
> > 0.9397 0.5968 19.6194
> > 0.9074 0.5391 20.6749
> > 0.8444 0.5364 21.6753
> > 0.7167 0.5474 22.5771
> > 0.5800 0.5776 23.3957
> > 0.4829 0.5254 24.1093
> > 0.3804 0.4512 24.6995
> > 0.3768 0.4045 25.2524
> > 0.4272 0.3674 25.8158
> > 0.5495 0.3633 26.4745
> > 0.6214 0.3743 27.2000
> > 0.7581 0.3565 28.0377
> > 0.7923 0.3207 28.8924
> > 0.7815 0.2507 29.7131
> > 0.7131 0.1655 30.4452
> > 0.5585 0.1669 31.0281
> > 0.4937 0.1875 31.5562
> > 0.3678 0.2026 31.9761
> > 0.2653 0.1806 32.2970
> > 0.2149 0.1133 32.5400
> > 0.2941 0.0680 32.8418
> > 0.4200 0.0694 33.2675
> > 0.5387 0.0556 33.8090];
> >
> > % Plot the river path
> > subplot(2,1,1)
> > plot(d(:,1),d(:,2))
> >
> > % Plot the lat and long vs distance
> > subplot(2,1,2)
> > plot(d(:,3),d(:,1),'b',d(:,3),d(:,2),'r')
> >
> > Now you can interpolate both lats and longs with
> > respect to distance, and maybe be able to go on
> > from there.
> >
> > Rune
>
> You are forgetting that even interpolating to either lat or long has a problem...there are mulitple river miles for numerous lats/longs, and matlab does not like that.
>

What Rune meant, and should work, is to separately interpolate lat vs.
dist and long vs. dist, with dist the independent variable. Then you
should have new matching arrays for lat and long.

--
Loren
http://blogs.mathworks.com/loren

> What Rune meant, and should work, is to separately interpolate lat vs.
> dist and long vs. dist, with dist the independent variable. Then you
> should have new matching arrays for lat and long.
>
> --
> Loren
> http://blogs.mathworks.com/loren

How can I interpolate Lat vs. Dist and Long vs. Dist when a certain Lat or certain Long may have multiple Dist numbers associated with it? I have tried using interp1 just using lat or long and dist, and it doesn't work.

The very simplest way would be to simply find the nearest node to your
point:
[dmn,imn]=min(abs(x+i*y - (xp+i*yp)));
Then s(imn) is the river mile of the nearest node.

"Travis" <sinusoid2@hotmail.com> wrote in message <gg24kh$c5h$1@fred.mathworks.com>...
> > What Rune meant, and should work, is to separately interpolate lat vs.
> > dist and long vs. dist, with dist the independent variable. Then you
> > should have new matching arrays for lat and long.
> >
> > --
> > Loren
> > http://blogs.mathworks.com/loren
>
> How can I interpolate Lat vs. Dist and Long vs. Dist when a certain Lat or certain Long
> may have multiple Dist numbers associated with it? I have tried using interp1 just using lat
> or long and dist, and it doesn't work.
Something like this:

Lat_i = interp1(dist,lat,dist_i);

HTH,
Bjoern

On Nov 20, 9:37 pm, "Bjorn Gustavsson" <bj...@irf.se> wrote:
> "Travis" <sinuso...@hotmail.com> wrote in message <gg24kh$c5...@fred.mathworks.com>...
> > > What Rune meant, and should work, is to separately interpolate lat vs.
> > > dist and long vs. dist, with dist the independent variable. Then you
> > > should have new matching arrays for lat and long.
>
> > > --
> > > Loren
> > >http://blogs.mathworks.com/loren
>
> > How can I interpolate Lat vs. Dist and Long vs. Dist when a certain Lat or certain Long
> > may have multiple Dist numbers associated with it? I have tried using interp1 just using lat
> > or long and dist, and it doesn't work.
>
> Something like this:
>
> Lat_i = interp1(dist,lat,dist_i);
>
> HTH,
> Bjoern

But that's arse about face.
They want the distance for a given lat, long

On 20 Nov, 09:57, NZTideMan <mul...@gmail.com> wrote:
> On Nov 20, 9:37 pm, "Bjorn Gustavsson" <bj...@irf.se> wrote:
>
>
>
>
>
> > "Travis" <sinuso...@hotmail.com> wrote in message <gg24kh$c5...@fred.ma=
thworks.com>...
> > > > What Rune meant, and should work, is to separately interpolate lat =
vs.
> > > > dist and long vs. dist, with dist the independent variable. =A0Then=
 you
> > > > should have new matching arrays for lat and long.
>
> > > > --
> > > > Loren
> > > >http://blogs.mathworks.com/loren
>
> > > How can I interpolate Lat vs. Dist and Long vs. Dist when a certain L=
at or certain Long
> > > may have multiple Dist numbers associated with it? =A0I have tried us=
ing interp1 just using lat
> > > or long and dist, and it doesn't work.
>
> > Something like this:
>
> > Lat_i =3D interp1(dist,lat,dist_i);
>
> > HTH,
> > Bjoern
>
> But that's arse about face.
> They want the distance for a given lat, long

If they've sampled the positions with too poor resolution,
which it seems they have, they can't have it. The best they
can do is to get some "reasonable" distance estimate that
visits the same (dx,dy) cells and then hope that they can
deduce what leg of the river it is, by other means.

Rune

In article <gg24kh$c5h$1@fred.mathworks.com>, sinusoid2@hotmail.com
says...
> > What Rune meant, and should work, is to separately interpolate lat vs.
> > dist and long vs. dist, with dist the independent variable. Then you
> > should have new matching arrays for lat and long.
> >
> > --
> > Loren
> > http://blogs.mathworks.com/loren
>
> How can I interpolate Lat vs. Dist and Long vs. Dist when a certain Lat or certain Long may have multiple Dist numbers associated with it? I have tried using interp1 just using lat or long and dist, and it doesn't work.
>

Interpolate the other way around:

dist vs. lat
dist vs. long

then use the new lat, long as pairs

--
Loren
http://blogs.mathworks.com/loren

Loren Shure <loren@mathworks.com> wrote in message <MPG.238f225981c556229898f6@news.mathworks.com>...
> In article <gg24kh$c5h$1@fred.mathworks.com>, sinusoid2@hotmail.com
> says...
> > > What Rune meant, and should work, is to separately interpolate lat vs.
> > > dist and long vs. dist, with dist the independent variable. Then you
> > > should have new matching arrays for lat and long.
> > >
> > > --
> > > Loren
> > > http://blogs.mathworks.com/loren
> >
> > How can I interpolate Lat vs. Dist and Long vs. Dist when a certain Lat or certain Long may have multiple Dist numbers associated with it? I have tried using interp1 just using lat or long and dist, and it doesn't work.
> >
>
> Interpolate the other way around:
>
> dist vs. lat
> dist vs. long
>
> then use the new lat, long as pairs
>
> --
> Loren
> http://blogs.mathworks.com/loren

The samples are at a fine enough interval to where a river mile can be obtained. Again with the dist vs. lat/long. A single lat or single long may have multiple distances associated with it due to the meanders of the river.

On 20 Nov, 16:39, "Travis" <sinuso...@hotmail.com> wrote:

> The samples are at a fine enough interval to where a river mile can be ob=
tained. =A0Again with the dist vs. lat/long. =A0A single lat or single long=
 may have multiple distances associated with it due to the meanders of the =
river. =A0

So what? Go back to the example I posted, read it,
run it and then contemplate it. Everything you need
is there.

Rune

Rune Allnor <allnor@tele.ntnu.no> wrote in message <8f6d2252-4e3f-4a08-88b0-4cfd778c37bd@j38g2000yqa.googlegroups.com>...
> On 20 Nov, 16:39, "Travis" <sinuso...@hotmail.com> wrote:
>
> > The samples are at a fine enough interval to where a river mile can be ob=
> tained. =A0Again with the dist vs. lat/long. =A0A single lat or single long=
> may have multiple distances associated with it due to the meanders of the =
> river. =A0
>
> So what? Go back to the example I posted, read it,
> run it and then contemplate it. Everything you need
> is there.
>
> Rune

Please type the code you would use. From how I am reading it it won't work, but I might be reading it incorrectly.

37.3097 -120.9301
37.3097 -120.9301
37.3097 -120.9301
37.3097 -120.9301
37.3097 -120.9302
37.3097 -120.9302
37.3098 -120.9302
37.3098 -120.9302
37.3098 -120.9302
37.3098 -120.9302
37.3098 -120.9302
37.3098 -120.9302
37.3098 -120.9302
37.3098 -120.9303
37.3098 -120.9303
37.3098 -120.9303
37.3098 -120.9303
37.3098 -120.9303
37.3101 -120.9307
37.3102 -120.9309
37.3103 -120.9311
37.3106 -120.9314
37.3109 -120.9315
37.311 -120.9314
37.3113 -120.9313
37.3118 -120.9314
37.3118 -120.9317
37.3119 -120.9317
37.3121 -120.9319
37.3121 -120.9326
37.3121 -120.9333
37.312 -120.9341
37.3119 -120.9347
37.3119 -120.935
37.3118 -120.9354
37.3117 -120.9358
37.3116 -120.936
37.3115 -120.9362
37.3115 -120.9363
37.3114 -120.9365
37.3114 -120.9366
37.3113 -120.9367
37.3113 -120.9368
37.3113 -120.9369
37.3113 -120.9369
37.3113 -120.937
37.3112 -120.9371
37.311 -120.9375
37.3108 -120.9377
37.3107 -120.938
37.3105 -120.9384
37.3105 -120.9386
37.3108 -120.9393
37.3114 -120.94
37.312 -120.94
37.3125 -120.9399
37.3126 -120.9398
37.3126 -120.9398
37.3127 -120.9397
37.3127 -120.9397
37.3128 -120.9397
37.3128 -120.9397
37.3128 -120.9397
37.3195 -120.9355
37.3195 -120.9355
37.3196 -120.9355
37.3196 -120.9355
37.3196 -120.9355
37.3196 -120.9355
37.3196 -120.9355
37.3196 -120.9355
37.3196 -120.9355
37.3196 -120.9355
37.3196 -120.9355
37.3196 -120.9355
37.3197 -120.9355
37.3197 -120.9355
37.3197 -120.9355
37.3197 -120.9355
37.3197 -120.9355
37.3197 -120.9355
37.3197 -120.9355
37.3198 -120.9355
37.3199 -120.9356
37.3202 -120.9357
37.3206 -120.9358
37.3209 -120.9357
37.3212 -120.9356
37.3214 -120.9353
37.3219 -120.9351
37.3224 -120.9354
37.3229 -120.9359
37.3231 -120.9362
37.3234 -120.9365
37.3235 -120.9368
37.3237 -120.937
37.3239 -120.9375
37.3239 -120.9377
37.3314 -120.9438
37.3308 -120.9436
37.3314 -120.9431
37.3316 -120.943
37.3318 -120.9429
37.332 -120.9429
37.3323 -120.9429
37.3326 -120.9432
37.3327 -120.9434
37.3327 -120.9436
37.3326 -120.9438
37.3325 -120.9438
37.3323 -120.9441
37.3321 -120.9444
37.3319 -120.9444
37.3318 -120.9446
37.3317 -120.945
37.3316 -120.9453
37.3317 -120.9454
37.3318 -120.9456
37.3318 -120.9459
37.3318 -120.9461
37.3318 -120.9463
37.3318 -120.9464
37.3318 -120.9465
37.3318 -120.9465
37.3318 -120.9465
37.3318 -120.9465
37.3317 -120.9466
37.3316 -120.9467
37.3316 -120.9468
37.3315 -120.9468
37.3313 -120.9471
37.3312 -120.9474
37.3312 -120.9476
37.3313 -120.948
37.3316 -120.9481
37.3317 -120.9481
37.3318 -120.9482
37.3319 -120.9484
37.332 -120.9486
37.332 -120.9488
37.332 -120.949
37.3319 -120.9492
37.3318 -120.9496
37.3317 -120.9499
37.3316 -120.9501
37.3316 -120.9507
37.3316 -120.9509
37.3316 -120.9509
37.3316 -120.9509
37.3316 -120.9509
37.3316 -120.951
37.3316 -120.951
37.3316 -120.951
37.3316 -120.951
37.3316 -120.9511
37.3316 -120.9511
37.3316 -120.9511
37.3316 -120.9511
37.3316 -120.9513
37.3315 -120.9517
37.3313 -120.9521
37.3313 -120.9524
37.3313 -120.9524
37.3313 -120.9525
37.3313 -120.9525
37.3313 -120.9526
37.3314 -120.9526
37.3314 -120.9526
37.3314 -120.9527
37.3317 -120.9531
37.3325 -120.953
37.3332 -120.9524
37.3333 -120.9524
37.3337 -120.9522
37.3347 -120.9524
37.3353 -120.9529
37.3354 -120.9535
37.3353 -120.9544
37.3352 -120.9548
37.3351 -120.955
37.3351 -120.955
37.335 -120.955
37.3349 -120.9554
37.335 -120.9563
37.3354 -120.9569
37.3356 -120.9571
37.3359 -120.9573
37.3362 -120.9577
37.3366 -120.9581
37.3368 -120.9584
37.3369 -120.9584
37.3372 -120.9585
37.3368 -120.9593
37.3366 -120.9596
37.3361 -120.96
37.3355 -120.96
37.3354 -120.96
37.3353 -120.96
37.3352 -120.9601
37.3352 -120.9601
37.3351 -120.9601
37.3351 -120.9601
37.3351 -120.9601
37.3351 -120.9602
37.3351 -120.9602
37.335 -120.9602
37.335 -120.9602
37.335 -120.9603
37.3347 -120.9604
37.3341 -120.9607
37.3339 -120.9614
37.3343 -120.9616
37.3345 -120.962
37.3349 -120.9628
37.3352 -120.9639
37.3355 -120.9647
37.3358 -120.9652
37.3361 -120.9655
37.337 -120.966
37.3369 -120.967
37.3364 -120.9677
37.3361 -120.9684
37.3363 -120.9693
37.3366 -120.9696
37.3368 -120.9696
37.3369 -120.9697
37.3369 -120.9698
37.3372 -120.97
37.3378 -120.9703
37.3383 -120.9706
37.3388 -120.9711
37.3389 -120.9717
37.3386 -120.9721
37.3382 -120.9726
37.3378 -120.9726
37.3378 -120.9726
37.3378 -120.9727
37.3379 -120.9727
37.3379 -120.9727
37.3379 -120.9728
37.338 -120.9728
37.3393 -120.9731
37.3406 -120.9733
37.341 -120.9751
37.3405 -120.9759
37.3402 -120.976
37.3401 -120.9762
37.3401 -120.9763
37.34 -120.9765
37.3401 -120.9766
37.3402 -120.9766
37.3403 -120.9768
37.3404 -120.9768
37.3404 -120.9768
37.3404 -120.9767
37.3406 -120.9768
37.3407 -120.9768
37.3412 -120.9773
37.3413 -120.9783
37.342 -120.9787
37.3423 -120.9783
37.3424 -120.9779
37.3428 -120.9772
37.3432 -120.9767
37.3434 -120.9765
37.3435 -120.9765
37.344 -120.9762
37.3449 -120.9757
37.3455 -120.9757
37.3461 -120.9759
37.3464 -120.9765
37.3468 -120.9766
37.3471 -120.9759
37.3473 -120.9753
37.3474 -120.9753
37.3475 -120.9752
37.3475 -120.9752
37.3475 -120.9752
37.3475 -120.9752
37.3475 -120.9752
37.3475 -120.9751
37.3475 -120.9751
37.3477 -120.9747
37.348 -120.9745
37.3482 -120.9743
37.3484 -120.9743
37.3486 -120.9744
37.3487 -120.9746
37.3489 -120.9748
37.349 -120.975
37.3491 -120.9754
37.3492 -120.9758
37.3494 -120.9762
37.3495 -120.9764
37.3498 -120.9766
37.3499 -120.9767
37.3501 -120.9768
37.3504 -120.9768
37.3509 -120.9769
37.3516 -120.977
37.3519 -120.977
37.3522 -120.9771
37.3525 -120.9772
37.3528 -120.9772
37.3531 -120.9772
37.3534 -120.9773
37.3537 -120.9773
37.3538 -120.9772
37.354 -120.9772
  
and

37.2943 -120.9174 127.3714
37.2943 -120.9174 127.3714
37.2943 -120.9174 127.3714
37.2943 -120.9174 127.3714
37.2943 -120.9174 127.3714
37.2943 -120.9174 127.3714
37.2943 -120.9174 127.3703
37.2943 -120.9174 127.3703
37.2943 -120.9174 127.3703
37.2943 -120.9174 127.3703
37.2943 -120.9174 127.3703
37.2943 -120.9174 127.3703
37.2943 -120.9174 127.3691
37.2943 -120.9174 127.3691
37.2943 -120.9174 127.3691
37.2943 -120.9174 127.3691
37.2943 -120.9174 127.3691
37.2943 -120.9175 127.3646
37.2944 -120.9177 127.3513
37.2944 -120.9178 127.3477
37.2945 -120.918 127.3328
37.2947 -120.9183 127.314
37.2948 -120.9185 127.2983
37.295 -120.9187 127.2805
37.2951 -120.9188 127.2716
37.2953 -120.9188 127.2551
37.2956 -120.9189 127.2348
37.2958 -120.9191 127.2152
37.296 -120.9194 127.1947
37.2959 -120.9197 127.179
37.2959 -120.92 127.1636
37.2958 -120.9203 127.1472
37.2957 -120.9206 127.1294
37.2955 -120.9209 127.108
37.2954 -120.9212 127.0881
37.2953 -120.9215 127.0678
37.2953 -120.9219 127.0466
37.2954 -120.9223 127.0259
37.2954 -120.9224 127.0182
37.2955 -120.9226 127.0095
37.2956 -120.9228 126.9928
37.2958 -120.9231 126.9742
37.296 -120.9233 126.9545
37.2962 -120.9236 126.9343
37.2964 -120.9238 126.9172
37.2967 -120.9239 126.897
37.297 -120.924 126.8771
37.2972 -120.924 126.8574
37.2976 -120.924 126.8366
37.2977 -120.924 126.8292
37.2977 -120.924 126.8259
37.2979 -120.9239 126.8117
37.2981 -120.9238 126.7979
37.2982 -120.9238 126.7915
37.2982 -120.9238 126.7915
37.2982 -120.9238 126.7915
37.2982 -120.9238 126.7915
37.2982 -120.9237 126.7896
37.2982 -120.9237 126.7896
37.2982 -120.9237 126.7881
37.2982 -120.9237 126.7869
37.2982 -120.9237 126.7858
37.2982 -120.9237 126.7847
37.2982 -120.9236 126.7828
37.2982 -120.9236 126.7799
37.2982 -120.9235 126.7761
37.2982 -120.9235 126.7725
37.2982 -120.9234 126.7678
37.2983 -120.9233 126.7629
37.2983 -120.9231 126.7545
37.2982 -120.923 126.7485
37.2982 -120.9229 126.7402
37.2982 -120.9227 126.7284
37.2983 -120.9224 126.7157
37.2983 -120.9224 126.7097
37.2984 -120.9223 126.7078
37.2984 -120.9222 126.6983
37.2985 -120.9221 126.6896
37.2987 -120.922 126.6754
37.2989 -120.922 126.6617
37.2992 -120.922 126.6439
37.2994 -120.922 126.6277
37.2997 -120.9219 126.6097
37.2999 -120.9219 126.5934
37.3002 -120.9218 126.5737
37.3004 -120.9218 126.5544
37.3007 -120.9217 126.5356
37.301 -120.9217 126.5148
37.3012 -120.9217 126.5
37.3013 -120.9217 126.497
37.3013 -120.9217 126.4939
37.3013 -120.9217 126.4924
37.3013 -120.9217 126.4913
37.3013 -120.9217 126.4913
37.3014 -120.9218 126.4894
37.3014 -120.9218 126.4879
37.3014 -120.9218 126.4879
37.3014 -120.9218 126.486
37.3014 -120.9218 126.486
37.3014 -120.9218 126.486
37.3014 -120.9218 126.486
37.3014 -120.9218 126.4845
37.3014 -120.9218 126.4845
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37.3668 -120.981 116.164
37.3671 -120.9804 116.1233
37.3674 -120.9795 116.0712
37.3676 -120.9787 116.0241
37.3678 -120.978 115.9841
37.3678 -120.9775 115.9568
37.3676 -120.9771 115.9265
37.3672 -120.9764 115.8831
37.3669 -120.9756 115.8341
37.3672 -120.975 115.7938
37.3676 -120.9749 115.7655
37.368 -120.9744 115.7229
37.3684 -120.9739 115.6866
37.3687 -120.9736 115.6608
37.3691 -120.9733 115.6299
37.3693 -120.9732 115.6148
37.3693 -120.9732 115.6129
37.3693 -120.9731 115.6117
37.3693 -120.9731 115.6102
37.3693 -120.9731 115.6091
37.3694 -120.9731 115.6059
37.3695 -120.9731 115.5939
37.3699 -120.9732 115.5652
37.3704 -120.9737 115.5216
37.3706 -120.9743 115.4847
37.3706 -120.975 115.4504
37.3706 -120.9756 115.4148
37.3702 -120.9761 115.3788
37.37 -120.9768 115.3419
37.3699 -120.9773 115.3108
37.37 -120.9777 115.2866
37.3702 -120.9783 115.2545
37.3705 -120.9787 115.2235
37.371 -120.9788 115.1913
37.3715 -120.9789 115.154
37.3721 -120.9788 115.1152
37.3725 -120.9786 115.0822
37.373 -120.9784 115.0466
37.3735 -120.9782 115.0108
37.374 -120.9782 114.9737
37.3746 -120.978 114.9328
37.3751 -120.9777 114.8922
37.3754 -120.9774 114.865
37.3757 -120.9769 114.8297
37.3761 -120.9761 114.7805

Here is the data I am using for this. If you plot the data you will see that linear interpolation won't work due to the river meanders. All I am trying to do is to interpolate the rivermile (3rd column) onto the first set of data so that the lat/longs can be referenced with a rivermile of their own.

...
> Here is the data I am using for this. If you plot the data you will see that linear interpolation won't work due to the river meanders. All I am trying to do is to interpolate the rivermile (3rd column) onto the first set of data so that the lat/longs can be referenced with a rivermile of their own.

Looks like your the dataset with the river miles is higher resolution than the GPS points you took...This probably is not the best idea, but may be a workaround if you can't get interpolation to work. You could just find the closest points in your second data set to the points in your first dataset.

suppose your first dataset is stored in variable "data" and the second one (with the river miles) is stored in variable "xyr'.

xd=cellfun(@(x)(abs(x-xyr(:,2))),...
    num2cell(data(:,2)),'uni',0);
yd=cellfun(@(x)(abs(x-xyr(:,1))),...
    num2cell(data(:,1)),'uni',0);
[distr,ind]=cellfun(@(x,y)(min(sqrt(x.^2+y.^2))),xd,yd);

riverMile=xym(ind,3);

%check the results
figure
scatter(data(:,2),data(:,1),40,riverMile);
hold on
scatter(xyr(:,2),xyr(:,1),20,xyr(:,3),'filled')

P.S. Dont try this with a large dataset...

Andrew

Why can't you just griddata? I copied the data you posted here, stored it in rivermiles.txt and rivermiles2.txt, then ran this:



fid = fopen('rivermiles.txt'); d1=fscanf(fid,'%f %f'); fclose(fid);
fid = fopen('rivermiles2.txt'); d2=fscanf(fid,'%f %f %f'); fclose(fid);
d1 = reshape(d1,2,length(d1)/2);
d2 = reshape(d2,3,length(d2)/3);
t1 = 1:length(d1);
t2 = 1:length(d2);

d3 = griddata(d2(1,:),d2(2,:),d2(3,:),d1(1,:),d1(2,:));

figure(10), clf
subplot(231), plot(t1,d1(1,:),t2,d2(1,:)), title('y')
subplot(234), plot(t1,d1(2,:),t2,d2(2,:)), title('x')
subplot(232), plot(t2,d2(3,:),t1,d3), title('dist')
subplot(233),plot(d2(2,:),d2(1,:),'y.',d1(2,:),d1(1,:),'ro'), title('spatial')

figure(11), clf
plot3(d2(2,:),d2(1,:),d2(3,:),'y.')
plot3(d1(2,:),d1(1,:),d3,'ro')


and the 3d plot looked correct to me, as seen here: http://img128.imageshack.us/img128/3489/rivermilesfp3.png (hopefully that is self explanatory)

But I must have missed something...