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I'm new here and hope it is ok to post such basic questions. Sometimes things seem simple but they do not turn out that way. I'm trying to make a computer game asset of a pool triangle (15 ball)enter image description here .

All I know is that the ball in-game is exactly 1 unit. But when I create the asset it has to fit snug just like the picture. Is there a mathematical way to calculate the exact inner length of each side AND then curved corners?

Or should I just trial and error :)

Thanks for reading

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    $\begingroup$ Is the diameter or the radius of the balls $1$? In the picture the fit is not snug. There is clear space on the left between the triangle and the balls. $\endgroup$ – Ross Millikan Dec 5 '17 at 20:39
  • $\begingroup$ the diameter is 1. When I say snug, i mean approx as snug as the image so there is some margin for error. -- In reality I beleive the balls are not exactly the same size I think there is slight variations in terms of less than a mm $\endgroup$ – Big T Larrity Dec 5 '17 at 20:42
  • $\begingroup$ And I guess my question should include that I really need the length between the starts of the curves, and the lengths of the curves themselves so that I can great a mesh in Blender etc $\endgroup$ – Big T Larrity Dec 5 '17 at 20:43
  • $\begingroup$ so really i guess i just need the length of the curve, because I can calculate the length of the side-of-triangle from that multiplied by 5 balls $\endgroup$ – Big T Larrity Dec 5 '17 at 20:44
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If you draw the triangle of the centers of the corner balls, it is equilateral with a side of four diameters. The straight sides of the alignment triangle are parallel to these, offset by one radius. The rounded corners are the same radius as the balls and represent $300^\circ$ of the circle because the ball occupies $60^\circ$.

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  • $\begingroup$ thats brilliant thank you. I'm slightly confused as to why its 300d though? Wouldnt that make corners that look like pac-man facing into from each corner? $\endgroup$ – Big T Larrity Dec 5 '17 at 20:48
  • $\begingroup$ I guess I need to remove 300d of the circle and leave the remaining 60d as the corner piece?? Thanks again you've helped me a lot there $\endgroup$ – Big T Larrity Dec 5 '17 at 20:51
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    $\begingroup$ No, you remove the $60^\circ$ that is between the sides of the triangle and leave the $300^\circ$. Draw a picture. $\endgroup$ – Ross Millikan Dec 5 '17 at 21:01
  • $\begingroup$ thank you i think i get it now $\endgroup$ – Big T Larrity Dec 5 '17 at 23:26

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