Hi, I'm having trouble removing the distortion where the hexagons meet. Does anybody have any smart ideas as to how I could fix this issue. I've tried tracing the hexagon and leaving one side off and following the path that way but I couldn't get it to work. : (
Thanks
Solved! Go to Solution.
Hi, I'm having trouble removing the distortion where the hexagons meet. Does anybody have any smart ideas as to how I could fix this issue. I've tried tracing the hexagon and leaving one side off and following the path that way but I couldn't get it to work. : (
Thanks
Solved! Go to Solution.
Solved by chrisplyler. Go to Solution.
@TrippyLighting wrote:I though had already answered that in my post and the model provided above.
You had. I don't know why I got chosen as the solution. Perhaps my wording just made sense to richard in some way?
@TrippyLighting wrote:I though had already answered that in my post and the model provided above.
You had. I don't know why I got chosen as the solution. Perhaps my wording just made sense to richard in some way?
It's entirely possible to have regular pentagons form a circle 😁
ETFrench
It's entirely possible to have regular pentagons form a circle 😁
ETFrench
@etfrench wrote:It's entirely possible to have regular pentagons form a circle 😁
It's possible with every regular polygon of odd numbered sides. And in fact all of them that are of even numbered sides also, although using sides that aren't opposite each other. What's your point?
@etfrench wrote:It's entirely possible to have regular pentagons form a circle 😁
It's possible with every regular polygon of odd numbered sides. And in fact all of them that are of even numbered sides also, although using sides that aren't opposite each other. What's your point?
Hey, just because I can't tell the difference between a pentagon and a hexagon isn't any reason to get snippy 😀
ETFrench
Hey, just because I can't tell the difference between a pentagon and a hexagon isn't any reason to get snippy 😀
ETFrench
Ooops that's not right. It's only true for those that have any pair of sides where the angle between them is evenly divided into 360 (resulting in a whole number). But still, that's a lot!
Ooops that's not right. It's only true for those that have any pair of sides where the angle between them is evenly divided into 360 (resulting in a whole number). But still, that's a lot!
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