[Qgis-user] Minimum bounding circle from cluster of points

SEGGIE Graeme gseggie at systra.com
Fri Jul 8 08:15:58 PDT 2016


I don’t have enough experience of coding steps in QGIS yet, but could you do the following?

For each set of points sharing a plot ID attribute value:
            Identify the distances between all points.
            Take the furthest apart pair (or one set of if more pairs are equal), which would become a diameter.
From this chosen pair, get the mid-point, which becomes the midpoint of the circle.
Buffer the identified midpoints as circles with size of radius (half max distance identified) to get circle containing all others in each set.

I think that would get all of them, or at least I can’t think of a situation where any longest pair identified could be further away from another point to be left out.

Graeme.


From: Qgis-user [mailto:qgis-user-bounces at lists.osgeo.org] On Behalf Of Tina Cormier
Sent: 08 July 2016 14:50
To: qgis-user at lists.osgeo.org
Subject: [Qgis-user] Minimum bounding circle from cluster of points

Hi all,

Looking for help aggregating some field data subplots to the plot level. In the attached screenshot, you can see clusters of 4 points (4 subplots = 1 plot). I'd like to create a circle around each cluster that is the smallest circle that would encompass all 4 points. Sort of like a convex hull, but a circle (convex hull, in this case, would give me a triangle). Is there a tool or a series of steps to accomplish this without doing it manually? In the attribute table, I have a unique ID for each plot/cluster. So for each plot ID (in this case, consisting of 4 subplots), I'd like to build a circle around all of the subplots.

I should also mention that it's not always 4 points (subplots), and they aren't always covering the same size area on the ground.

I have QGIS 2.14.2 as well as one of the nightly builds from a few weeks ago (can download the newest one if necessary).

Thanks in advance for any ideas/insights! Google has not helped me so much yet with this one!
Tina


[Inline image 2]


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