# Identifying the north facing edge of a polygon

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04-11-2021 07:28 PM
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New Contributor II

I'm trying to identify which edge of a land parcel is facing North.

I have found the 'front' of the parcel by finding the edge of the parcel closet to the road. Now I need to identify which edge of the parcel (labelled 'Front', 'Left', 'Back' and 'Right', clock-wise from the 'Front') is facing North.

I'm a bit stuck on where to even start with this. Any and all help is much appreciated!

1 Solution

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Occasional Contributor II

Apologies for perhaps some confusion with my explanation. There's no need to solve an orthogonal vector using the cross product, rather, we can simply solve the gradient of the line segment and solving the angle of the perpendicular gradient.

I have sketched up a basic example below with all 3 cases mentioned above. Hope this helps

4 Replies
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Occasional Contributor II

Hi OrlaHammond,

If you are looking for the a "Northern facing" line, you need to check each polygon geometry line segment's orthogonal vector in combination with the gradient of the line segment.

The orthogonal vector can be solved using the cross product. You can calculate the bearing by adding the gradient of the line segment. Also be careful of the direction of the gradient as this will affect the calculation of bearing.

I am assuming some of your parcels may not be always rectangular e.g. more than 4 lines. The condition to fine the most "Northern facing" line would be:

1. The line segment's orthogonal vector's angle +- gradient (depending on direction) is between 315 and 45 bearing,
2. The line segment's orthogonal vector angle is the minimum (closest to 0 bearing) of all line segment orthogonal vector angles.
3. The line segment's vector coordinates are max of other line segment vectors.

New Contributor II

Hi there,

I'm not quite following what you mean. Could you break it down a bit more?

by
Occasional Contributor II

Apologies for perhaps some confusion with my explanation. There's no need to solve an orthogonal vector using the cross product, rather, we can simply solve the gradient of the line segment and solving the angle of the perpendicular gradient.

I have sketched up a basic example below with all 3 cases mentioned above. Hope this helps

New Contributor II