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I start a new project with base on OpenLayers plugin but if I check the distance on QGIS with distance tool (I choose a straight road) I view 5.466m on QGIS and 4.300m on google (by walking).

How is it possible?

On QGIS 2.6.1 I do not set any CRS

Joseph
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Ale
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2 Answers2

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I am no expert in this but from my understanding:

  • The OpenLayers plugin in QGIS uses the EPSG:3857 CRS which is a projected CRS on a flat surface (here's a very good post describing it). Therefore, it calculates a straight-line distance as you would on a paper map.
  • I can't find how Google Maps calculates its distances but a common method would be to use the Haversine formula or "as the crow flies". This calculates distances from two points on a sphere. There is a forum here suggesting what Google Maps uses to calculate distances.

So in terms of calculating distances, Google Maps provides a more accurate measurement assuming your CRS is set to EPSG:3857 in QGIS. You should always use the CRS of the local country to give you the best distance measurement.

Hope this helps and if anyone spots a mistake, please correct me :)

Community
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Joseph
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    In my situation the difference is over 20%, I think there is an error in the project but I am not able to find it – Ale Jul 15 '15 at 08:56
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    You should never do measurements that you hope to do accurately in any sort of projected coordinate system. It is best to use a geographic coordinate system (such as 4326) and use the Haversine formula to do any sort of mapping. If you are concerned with having a higher accuracy over smaller distances than use a local projection system that preserves distances. – onakua Dec 11 '17 at 16:43
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Joseph's answer assuming it has something to do with straight-line distance is not correct.

Although the question is different, the answer is the same as this and this.

The QGIS ruler calculates the distance in the viewers projection which I assume is Web Mercator (EPSG:3857) in your case.

WGS 84 / Pseudo-Mercator (EPSG:3857) projection is heavily distorted when moving away from the equator. Thus, it could be discussed if the units should be called "Pseudo-meters". One meter in reality is approximately 1/cos(lat) pseudo-meters.

4300 m / cos(38.1°) = 5466 pseudo-meters

enter image description here

CC BY-SA 3.0, Author: Stefan Kühn

pLumo
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  • Just found this old question when working on a similar question and thought I should answer it too for other people having the same problem. – pLumo Dec 11 '17 at 15:41