I have an unsigned 1-bit thematic forest cover raster (.img). 1 = forest and 0 = not forest. I want to calculate the distance between forest patches. It appears that the Euclidean Distance tool in ArcGIS does this but I get an output with all 0s.
Has anyone had experience with this before? Am I doing something incorrectly? Will I have to write a script to take care of it?
I can use either ArcGIS or Erdas Imagine. Whichever is easiest.
Thanks
It should work with a raster with values 0 and 1. Export your .img to .tiff with 8 bit unsigned and try it again.
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I have a list of lat/long points which form a path on the surface of the earth.
I have another point on the surface of the earth and I want to find the shortest distance from that point to a point that falls on the path.
While I could approximate the surface of the earth as a plane, accuracy is important.
The distance between points on the path could be anywhere from 1-1000m. The distance to the point not on the path is from 1-50m. The maximum acceptable error is 0.1m.
Any method of calculating this is acceptable, whether assuming a plane, sphere, or the real shape of the earth as long as the error would not exceed 0.1m for any point on land.
This question is marked language agnostic to encourage answers from people not familiar with the language used. The implementation will be in Dart.
It sounds like you need to calculate the Great Circle Distance across the surface of the earth. A Great Circle calculation permits the calculation of distance along the earth's surface between two arbitrary latitudes and longitudes.
A trivial example of a Great Circle calculation would be to migrate along the Earth's surface along the line of equator (zero degrees latitude). Each degree of longitude of migration along that line (for instance from [0 deg N, 90 deg W] to [0 deg N to 91 deg W]) equates to 1/360th of 40,070km, or 111.306km. Moving between two latitudes and two longitudes requires transformation of coordinates and is outside the scope of this quick note.
Summary and equations found here:
https://en.wikipedia.org/wiki/Great-circle_distance
The accuracy you are seeking of 0.1m requires further refinement; using a sphere to approximate the shape of the Earth will limit the accuracy to perhaps 0.5% (see paragraph from Wikipedia, below). Put another way, a 0.5% error of two points 1000 km apart would be 5000 m.
A more formal and precise calculation will use the true shape of the earth, known as a geoid. This is determined by gravitational measurements and is updated from time to time by the geodesic community.
It is possible to determine one's absolute position on Earth to +/- 0.1m (or better) with advanced GPS surveying techniques, such as RTK or satellite correction services (e.g. Omnistar), but determining the path distance to that accuracy is not the same thing. A survey-quality receiver has corrections built into it for current geoids so that it can translate the lat/long/height calculation it makes using GPS signals to the current reference geoid used by the surveying / geodesic community.
You may not actually require 0.1m accuracy for your application; very few applications require 0.1m absolute accuracy over any distance except (for example) geographic determinations of movement in the Earth's surface. Relative accuracy is more important; e.g. measuring the same point at different times. It is more important to know how much a seismic fault moved relative to its position yesterday or last week, or whether a critical point on a pipeline has moved 2 cm to the north in the past year.
I hope this helps.
Cheers
GP
Per Wikipedia:
So long as a spherical Earth is assumed, any single formula for distance on the Earth is only guaranteed correct within 0.5% (though better accuracy is possible if the formula is only intended to apply to a limited area). [7]
Warning: I know nothing about GIS. That will become very apparent in a moment, of course. My vocabulary isn't going to be spot on, either, Apologies.
I need to recreate parts of a "Strategy Map" that looks like this as "real geo-spatial" map:
Why? Because if I can manage to plot the boxes ("Maximize Shareholder Value", "Exceed Customer Expectations", etc.) on a map in correct relation to each other, I can do some very fun stuff in a data visualization tool I'm working with.
I can build the strategy map above in Visio, and then use a script to export the shapes I care about as X, Y points OR Polygons. One of the boxes above might looks like this once exported:
ShapeNo ShapeName PointNo X Y
1 Exceed Cust 2 37 155
1 Exceed Cust 4 116 155
1 Exceed Cust 6 116 234
1 Exceed Cust 8 37 234
1 Exceed Cust 10 37 155
...or it might look like this:
POLYGON ((37 155, 116 155, 116 134, 37 234, 37,155))
Regardless, I have a bunch of points, and I need to turn these into lat/lon coordinates, using lat/lon (0,0) as my point of reference. In the map above, 0,0 might be beneath the "Exceed Customer Expectations" box - more or less dead center.
Then, I suspect I can find a tool that will convert this jumble of stuff into an ESRI shapefile and I can import directly into my dataviz tool.
Are there any known (free) tools, scripts, libraries, etc.that might do some of this for me?
Your problem shouldn't be solved with a GIS but I can appreciate that you have found some cool dataviz features that require a shapefile.
The problem is that you want to take some x,y points and convert them to lat/lon. Latitude-longitude refer specifically to points on the earth's surface and the points in your problem have no relation to the earth's surface.
Another way to think of this is that you are trying to take random points and say one represents the capital of Russia and the other represents a large city in Germany etc.
Another problem is that you want to have a 0,0 reference point but latitude and longitude have a datum as a reference point which is a specific geographic location.
It's hard to suggest an alternative method to solve your problem without more information on your familiarity with graphic design tools, but lat/lon with GIS are not the direction to be looking.
Many people do convert x,y points to lat/lon but this is not a direct conversion. Cartesian coordinates require a known projection and datum in order for this conversion to be accurate.
Check out this link for an in depth explanation of why arbitrary x,y cannot be converted to lat/lon.
On the other hand, +1 for an out-of-the-box original idea for strategy map design!
There are two points A, B, and distances x (miles from A), and y (miles from B). Let the distance from A to B be N. So, A is N miles away from B. How do I solve the problem: What are the points available that are (N + x + y) miles away from A? I'm not sure how to explain this any better. I really have no clue on how to attack this problem, I read Fastest Way to Find Distance Between Two Lat/Long Points and I believe the solution given calculates the distance between two points and have no idea if this solution could be used to apply to my problem, or if so, how.
If you are looking for an approximation algorithm I suggest to look for a k-means algorithm or a hierarchical cluster, especially a monster curve or a space filling curve. First off you can compute a minimal spanning tree of the graph and then remove the longest and expensivest edges. Then the tree makes many little trees and you can use the k-means to compute group of points i.e. clusters.
"The single-link k-clustering algorithm ... is precisely Kruskal's algorithm ... equivalent to finding an MST and deleting the k-1 most expensive edges." See for example here: https://stats.stackexchange.com/questions/1475/visualization-software-for-clustering.
A good example for a monster curve is the hilbert curve. The basic form of this curve is an U-shape and by copy many of it together and rotating it the curve fills the euklidian space. Surprisingly a gray code can help to find out the orientation of this U-shape. You can look up Nick's spatial index quadtree hilbert curve blog article about more details. Instead to calculate the curve's index you can put together a quadkey like in bing maps. The quadkey is unique for each coordinate and it can be used with normal string operations. Each position in the key is part of the U-shape curve and thus you can select this region of points from select partially from left to right from the quadkey.
In this image you can see the green polygon is found using a hilbert curve:
You can find my php classes here: http://www.phpclasses.org/package/6202-PHP-Generate-points-of-an-Hilbert-curve.html
I want to compute the Earth position (relative to the sun) and axis rotations for a given date and time. It's ok to assume the Sun is stationary at the 0,0,0 coordinate. Very minor deflections, due to the Moons gravitational pull for example, can also be ignored. Anything accurate within a degree or so is good enough.
Are there any libraries/source/data out there that will help me accomplish this?
The aa-56 code, which can be downloaded from here, includes a solar ephemeris that will probably meet your needs. For high-precision work you'd want something more accurate like JPL's DE421, but there are some inconveniently large tables of coefficients involved, and it's probably extreme overkill if you're happy with 1 degree accuracy.
The Earth's rotation at a given time is given by the Greenwich sidereal time.
Jean Meeus' "Astronomical Algorithms" (a good reference to have for these sorts
of calculations!) gives a formula for theta0 (cumulative rotation angle in degrees)
in terms of the Julian date JD:
T = (JD - 2451545.0 ) / 36525
theta0 = 280.46061837 + 360.98564736629*(JD-2451545.0) +
0.000387933*T*T - T*T*T/38710000.0
theta0 = 0 degrees mod 360 represents the instant when the Greenwich meridian is aligned with right ascension 0:00 in celestial coordinates.
Yes. What you need is an Ephemeris package.
The JPL has an online Ephemeris service that will do the computations for you, so if you are web-capable you can hit that up.
I found a free Ephermeris package too, but it looks like it just hits the JPL site for you. However, there's a link there to download the JPL's database and have it work off of that. So if you want to work offline and don't mind updating your database from the JPL manually every now and then, that might be an option.
In Python, using the ephem library:
>>> import ephem
>>> sun = ephem.Sun()
>>> sun.compute(ephem.now())
>>> sun.hlong, sun.hlat, sun.earth_distance
(69:41:32.6, 0:00:00.2, 0.98602390289306641)
ephem doesn't provide a convenient representation of the Earth as a body, but sun.hlong and sun.hlat give the heliocentric longitude and latitude of the Earth. This could be better documented.
For Earth rotation, maybe you can say what value you're looking for here. (Normally we use the time of day, but I think it's generally clear how to get hold of that!)
I am reverse engineering a transportation visualization app. I need to find out the latitude for the origin of their data feed. Specifically what XY 0,0 is. The only formulas I have found calculate distance between two points, or location of a bearing/distance.
They use the XY to display a map in a very legacy application. The XY is in FEET.
I have these coordinates:
47.70446615506108, -122.34469839507263: x=1268314, y=260622
47.774182540800616,-122.3412994737105: x=1269649, y=286031
47.60024792289405, -122.32767331735774: x=1271767, y=222532
47.57012494413499, -122.29129609983679: x=1280532, y=211374
I need to find out what the latitude and longitude of x=0, y=0 is and what the formula would be to find this out.
They have two data feeds, one is more current than the other. The feed with the most current data does NOT include latitude, longitude, but only XY. I am trying to extrapolate based on their less current, yet more informative (includes lat, lon) data feed what 0,0 is so I can simply convert their (more current) data feed's XY coordinates to latitude and longitude.
If you look at the first 2 lines of data, and subtract the latitude
47.7044 - 47.7741 = -0.06972 degrees
There are 60 nautical miles per degree of latitude, and 6076 feet per nautical mile.
-.06972 * 60 * 6076 = 25,415 ft
Subtracting the two 'Y' values:
260662 - 286031 = 25,409 ft
So indeed that seems to prove the X and Y values are in feet.
If you take any of the Y values, and convert back to degrees, for example
260622 ft / ( 6076 ft/nm ) / ( 60 nm/degree ) = .71
286031 ft / 6076 / 60 = .78
So subtracting those values from the latitudes of (47.70 and 47.77) gives you very close to exactly 47 degrees, which should be your y=0 point.
For longitude, a degree is 60 nautical miles at the equator and 0 miles at the poles. So the number of miles per degree has to be multiplied by the cosine of the latitude, so approx cos(47 degrees), or .68. So instead of 6076 nm per degree, it's about 4145 nm.
So for the X values,
1268314 ft / ( 4145 ft/nm ) / ( 60 nm/degree ) = 5.10 degrees
1269649 ft / 4145 / 60 = 5.10 degrees
These X numbers increase as the latitude increases (less negative), so I believe you should add 5.1 degrees, which means the X base point is about
-122.3 + 5.1 = 117.2 West longitude for your x=0 point.
This is roughly the position of Spokane WA.
So given X=1280532, Y=211374
Lat = 47 + ( 211374 / 6096 / 60 ) = 47.58
Lon = -117.2 - ( 1280532 / ( 6096 * cos(47.58)) / 60 ) = -122.35
Which is roughly equivalent to the given data 47.57 and -122.29
The variance may be due to different projections - the X,Y system may be a "flattened" projection as opposed to lat/long which apply to a spherical projection? So to be accurate you may yet need more advanced math or that open source library :)
This question may also be helpful, it contains code for calculating great circle distances:
Calculate distance between two latitude-longitude points? (Haversine formula)
There are many different coordinate systems. You need to find out the what the coordinate systems are for both the lat/lon's (e.g. WGS84 etc) and x/y's first (e.g. some sort of projected system probably).
Once you have that information there are several tools you can use to do conversions and manipulations. One example (of a free open source coding library) is proj4.
Ask them what coordinate system they're using! (or if you got the dataset from some database, look at the metadata for the dataset and it should tell you. Otherwise I'd be skeptical of its value)
Most likely this is one of the state plane coordinate systems. They're for localized areas of the earth (kind of like UTM), and are frequently used for surveying.
You can use CORPSCON (or other GIS programs; ExpertGPS will do this if you have the GIS Option Pack but it's not free. I forget whether GPSBabel does conversion) to convert between lat/long and any of the state plane coordinate systems. You'll also need to know which datum the coordinates are in. WGS84 and NAD83 are very close but NAD27 is different.
You've got good advice on coordinate systems already, so I'll just chime in with the library I've used with great success in the past.
Geotrans is approved for use by the US Department of Defence, so you can be sure that it is well tested. You can grab it from here:
http://earth-info.nga.mil/GandG/geotrans/index.html
That might not be the right link as that page talks about the application, not the library. I expect the library is in the Developers package. Licensing terms were very liberal from memory, but make sure you review the terms before using it commercially.
Edit:
An interesting discussion on Geotrans licensing can be found here:
http://www.mail-archive.com/debian-legal#lists.debian.org/msg39263.html
Over here, I said this:
In Java, I would use the OpenMap converter from a point's expression in UTM to one using Latitude and Longitude (assuming a WGS-84 ellipsoid which is most commonly used in GPS).
OpenMap is open source and I would post a link to their download page but they have a short license script in the way. So, to avoid being rude, I won't deep link. Instead, head to their homepage and click Downloads.
That should either solve your problem directly or at least point you towards a useful algorithm.
I've used Brenor Brophey's gPoint PHP class to do this on a couple of occasions. Solid results, GPL code, and easily deployed. Recommended.