Thinness/Thickness Ratio

[ArcGISUser]

http://help.arcgis.com/en/arcgisdesktop/10.0/help/index.html#/Finding_polygon_overlaps_and_gaps_that...

"The thinness ratio and polygon area can be used to define how large or small the gap or overlapping area can be in order for it to be considered a sliver."

"10. Move the Thinness Ratio slider to the position that best describes the thinness of a sliver polygon. The farther right the slider is moved, the thicker the polygon must be to be recognized as a sliver."

Where can I find more documentation or explanation on this?  What�??s the default usually?

[MichelleJohnson]

Excerpt from Microscope Image Processing by Qiang Wu, Fatima Aziz Merchant, Kenneth R. Castleman

Thinness Ratio
Thinness is typically used to define the regularity of an object. Having computed the area (A) and perimeter (P) of an object, we can define the thinness ratio as �?�. T = 4pi(A/P2) 

This measure takes a maximum value of 1 for a circle. Objects of regular shape have a higher thinness ratio than similar irregular ones.

There is not a default value that I can recommend.  That is dependent on each user's requirements.  Generally, it is less than 1.

[ColeGreene]

Excerpt from Microscope Image Processing by Qiang Wu, Fatima Aziz Merchant, Kenneth R. Castleman

Thinness Ratio
Thinness is typically used to define the regularity of an object. Having computed the area (A) and perimeter (P) of an object, we can define the thinness ratio as �?�. T = 4pi(A/P2) 

This measure takes a maximum value of 1 for a circle. Objects of regular shape have a higher thinness ratio than similar irregular ones.

There is not a default value that I can recommend.  That is dependent on each user's requirements.  Generally, it is less than 1.

I am trying to use this formula to create a "Thinness" attribute in a table. I am using VBScript in the field calculator. I am using Shape_Length as "P" and Shape area as "A" and 3.121592 as pi. The end result is way too high. Any suggestions?

[MichelleJohnson]

The formula didn't copy over correctly since the 2 in P2 isn't superscript.  It's supposed to be P squared (P x P) rather than (P * 2).

T = 4pi(A/(P*P))

Hope that helps.

[WalterSimonazzi1]

Microscope Image Processing (written by AvQiang Wu,Fatima Merchant,Kenneth Castleman). Excerpt:

Thinness is typically used to define the regularity of an object. Having computed the area (A) and perimeter (P) of an object, we can define the thinness ratio as T = 4pi(A/P²)
This measure takes a maximum value of 1 for a circle. Objects of regular shape have a higher thinness ratio than similar irregular ones.

So, you can calculate your new field in Field Calculator to be:
4 * 3.14 * [Shape_Area] / ([Shape_Length] * [Shape_Length])

Most sliver polygons would have the ratio really close to 0, but from what I've seen whatever that is lower than 0.3 might be a good candidate, too. However, consider the features areas first before running into any hasty conclusions. Large and/or multipart features with irregular shape might have a really low thinness ratio yet they are not even close to be sliver polygons. You would probably want to set a definition query on your feature class excluding most of the polygons with large area….”

Source: Fighting sliver polygons in ArcGIS: thinness ratio | Tereshenkov's Blog​