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Authorized by the cone indents can be used with a hair of margin } module shape(hsh, ird, ord, fn4, hg y0=-0.1; y1=0; y2=abs(hsh); y3=hg-abs(hsh); y4=hg; y5=hg+0.1; if ( fsh == 0 cylinder(h=chg, r=cord-cdp*smt/100, $fn=2*cfn, center=false); shape(fsh, cird, cord-cdp*smt/100, cfn*4, chg); knurled_finish(cord, cird, clf, csh, cfn, crn); else if ( fsh == 0 cylinder(h=chg, r=cord-cdp*smt/100, $fn=2*cfn, center=false); shape(fsh, cird, cord-cdp*smt/100, cfn*4, chg); module shape(hsh, ird, ord, fn4, hg x0= 0; x1 = hsh > 0 ? Ird : ord; x2 = hsh > 0 ? Ird : ord; x2 = hsh > 0 ? Ord : ird; y0=-0.1; y1=0; y2=abs(hsh); y3=hg-abs(hsh); y4=hg; y5=hg+0.1; if ( hsh >= 0 module knurled_finish(ord, ird, lf, sh, fn, rn) { for(j=[0:rn-1]) assign(h0=sh*j, h1=sh*(j+1/2), h2=sh*(j+1)) { for(i=[0:fn-1]) assign(lf0=lf*i, lf1=lf*(i+1/2), lf2=lf*(i+1)) { polyhedron( points=[ [ 0,0,h0], [ ord*cos(lf0), ord*sin(lf0), h0], [ ird*cos(lf1), ird*sin(lf1), h0], [ ird*cos(lf0), ird*sin(lf0), h1], [ ord*cos(lf1), ord*sin(lf1), h1], [ ord*cos(lf0), ord*sin(lf0), h0], [ ird*cos(lf1), ird*sin(lf1), h2], [ ird*cos(lf1), ird*sin(lf1), h2], [ ird*cos(lf1), ird*sin(lf1), h2], [ ird*cos(lf1), ird*sin(lf1), h0], [ ird*cos(lf1), ird*sin(lf1), h0], [ ord*cos(lf2), ord*sin(lf2), h0], [ ird*cos(lf1), ird*sin(lf1), h2], [ ord*cos(lf2), ord*sin(lf2), h0], [ ord*cos(lf2), ord*sin(lf2), h0], [ ird*cos(lf1), ird*sin(lf1), h0], [ ird*cos(lf1), ird*sin(lf1), h2], [ ird*cos(lf1), ird*sin(lf1), h0], [ ord*cos(lf2), ord*sin(lf2), h0], [ ird*cos(lf1), ird*sin(lf1), h2], [ ord*cos(lf2), ord*sin(lf2), h0], [ ird*cos(lf0), ird*sin(lf0), h1], [ ord*cos(lf1), ord*sin(lf1), h1], [ ord*cos(lf1), ord*sin(lf1), h1], [ 0,0,h2], Created on Tue Mar 5 20:19:51 2024 Copper Layer Stackup: ============================================================= L1 : F.Cu front L2 : B.Cu back Drill file 'precadsr-panel.drl' contains T1 3.200mm 0.1260" (4 holes) (with 4 slots T2 5.000mm 0.1969" (1 hole) T3 7.000mm 0.2756" (6 holes T4 10.000mm.

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