Hi ienesaglam60,

I made you an example.

X coordinate = $(Radius) *@cos(@Value(_creation_instance)* 30 * @pi() / 180)

Y coordinate = $(Radius) *@sin(@Value(_creation_instance)* 30 * @pi() / 180)

I added a userparameter for the radius.

For your case, make sure you have 12 instances per coordinate and run the rest of this workbench :) 

Hope this helps :) 

One way to do this is:

• VertexCreator to place point on desired place.
• Bufferer to replace point with circle using radius.

• GeometryCoercer to convert to line.

• ArcStroker to stroke circle to dodecagon.

Another way to do this is to calculate the vertices from the dodecagon based on insertion point and radius and put these values in an XML and feed this in the GeometryReplacer.

<?xml version="1.0" encoding="US_ASCII" standalone="no" ?>

<geometry>

    <path>

        <line>

            <coord x="5" y="0"/>

            <coord x="3.5355339059327378" y="-3.5355339059327378"/>

            <coord x="0" y="-5"/>

            <coord x="-3.5355339059327378" y="-3.5355339059327378"/>

            <coord x="-5" y="0"/>

            <coord x="-3.5355339059327378" y="3.5355339059327378"/>

            <coord x="0" y="5"/>

            <coord x="3.5355339059327378" y="3.5355339059327378"/>

            <coord x="5" y="0"/>

        </line>

    </path>

</geometry>

 I use FME 2022 version. Would you show me AttributeCreator parameters?

I have coordinate list. 

I want to draw a dodecagon with a known radius.

example

 

Hi ienesaglam60,

I made you an example.

X coordinate = $(Radius) *@cos(@Value(_creation_instance)* 30 * @pi() / 180)

Y coordinate = $(Radius) *@sin(@Value(_creation_instance)* 30 * @pi() / 180)

 

I added a userparameter for the radius.

 

For your case, make sure you have 12 instances per coordinate and run the rest of this workbench :) 

 

Hope this helps :) 

 I use FME 2022 version. Would you show me AttributeCreator parameters?

Attached an older version :) 

If you buffer a point with a round end cap style and an interpolation angle of 30 you get a dodecagon

If you buffer a point with a round end cap style and an interpolation angle of 30 you get a dodecagon

 

And we have a winner!

If you buffer a point with a round end cap style and an interpolation angle of 30 you get a dodecagon

 

“Why waste time say lot word when few word do trick”

Excellent :-)

Lol: Got me remembering this Kevin scene in The Office.   Still up in my Top 5 comedy shows.