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I'm very new to GIS but have played around a bit with postGIS and qGIS.

I'm wondering if there's a way to generate drive time estimates like those given by google maps using open source tools.

For a small number of routes I can query google maps, and scrape results but I'm sure there's a better way.

Thanks

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Besides data containing suitable weights (driving time, or length of the road element plus allowed/estimated speed), you'll need a routing tool e.g. pgRouting for PostGIS.

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  • I'm not sure I follow. I want to get driving time I don't have it. – gisnovice Jan 11 '11 at 0:27
  • I have a set of starting and ending locations, and PostGIS (therefore pgRouting). I gather that pgRouting allows me to find the shortest path between these points, but I need to get the quickest path in terms of travel time, and return the actual travel/drive time. Similar to the drive time estimate google maps gives. – gisnovice Jan 11 '11 at 0:33
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    If your road network doesn't include information such as speed limits, stop lights/stop signs, turn or direction restrictions, etc, the calculated travel time will not be that accurate and could be wrong (for instance, if the route goes the wrong way down a one way street). – mkennedy Jan 11 '11 at 0:39
  • @gisnovice: pgRouting can find the path with the shortest length or (if you have the information) with the lowest costs (= travel time). If your graph doesn't contain the necessary info, you can't get good results. – underdark Jan 11 '11 at 9:08
  • Seems so much easier to just use Google's API even though it's not 'the right way'. Out of curiosity, is there a common source for this information for US roadways? Google has to get the data somewhere. – gisnovice Jan 12 '11 at 18:28
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GRASS' v.net.iso (http://grass.osgeo.org/grass70/manuals/html70_user/v.net.iso.html) will also do this. Again, it takes the right data, and a bit of work to massage it into shape. However, I have done this successfully in the past, so it does work! I haven't used pgRouting, but hear it may be a bit more efficient!

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