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Skill octagon on-grid

ArbLouis

Hi. I'm trying to draw an octagon on-grid using this code: 

for(i 0 7;
; Calculate coordinates for the octagonal ring (mid-width)

; Initialize point list
angle =-3*45/2+45+(i-1)*45
; Calculate x and y for each vertex
x = pog(stx + SRR_out_R * cos(angle * pi / 180)) ; 
y = pog(sty + SRR_out_R * sin(angle * pi / 180))

pointList = append(pointList list(x:y))


); end for

pog() is a custom function which sets points on-grid.

round(pt / grid_tech) * grid_tech

Unfortunately, the result I achieve is: 

As you can notice, the edges are not orthogonal and the angles are not perfectly at 45 degrees. Most probably, this is due to the fact that the pog rounds in different way the vertexes of the same edge. Do you have an easy workaround?

  • I'm surprised you hit such a rounding error if your grid is not really fine (and given that the slope is pretty visible in your picture, it seems unlikely). What is pi set to? I wonder whether you have too limited precision on your value of pi (which could certainly cause this). I would suggest:

    defMathConstants('MYCONSTS)
    pi=MYCONSTS.PI

    which you'll see is at full precision:


    sstatus(fullPrecision t)
    pi

    outputs: 3.141592653589793

    Of course, you could always ensure as you walk around the points that your delta between successive points is always either dx==dy or dx==0 or dy==0 (even then though you need to ensure that the diagonals are the same length and the horizontals/verticals are the same length - so you could compute the deltas for a diagonal and the deltas for a horizontal/vertical, and then use those as you walk around the octagon.

    If you had access to Cadence PCell Designer, there's a built-in quarter octagon command which has all the work done to ensure that everything is on grid (and in fact it's a polygon, so the outer vertices are on grid).

    Andrew

  • in reply to Andrew Beckett

    Thanks for your answer. 

    Indeed, I've defined pi as you suggest and I get the value: 3.141592653589793

    my grid is: grid_tech=0.005

  • in reply to ArbLouis

    What value of SRR_out_R are you using that shows this problem?

    Andrew

  • in reply to Andrew Beckett

    SRR_out_R=12.1

    If I reduce the grid_tech, the error is reduced. 

  • in reply to ArbLouis

    OK, I tried it, and the issue is this line:

    angle =-3*45/2+45+(i-1)*45

    The trouble is that the 3*45/2 will end up being 67 rather than 67.5 - the operands of the division are integers, so it does an integer division. If you change it to:

    angle =-3*45/2.0+45+(i-1)*45

    then that forces the division to do a floating point division as one of the operands is a float, and so that will give -67.5 degrees as the first part of the expression.

    What's happening is that your octagon is very slightly rotated...

    Andrew

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