Small challenge
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06-07-2017, 09:53 AM
(This post was last modified: 06-07-2017 09:57 AM by Pekis.)
Post: #3
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RE: Small challenge
Hello,
Here is the solution: Distance AB = Distance AE + Distance EB Distance AE: d*cos(PI/6) =a*r*sqrt(3)/2 Distance EB: Arc height Formula r=h/2+c^2/(8*h), Here h=Distance EB and c=d => Solved for Distance EB=r*(1-sqrt(4-a^2)/2) Distance AB: a*r*sqrt(3)/2+r*(1-sqrt(4-a^2)/2), which must be equal to c*r (c=0.2 in the question) => Solve a*r*sqrt(3)/2+r*(1-sqrt(4-a^2)/2)=c*r => Solve a*sqrt(3)/2+1-sqrt(4-a^2)/2=c => Solved for a=(sqrt(3)*(c-1)+sqrt((c+1)*(3-c)))/2 If c=0.2 => a~=0.223694816 => d will be approx. 0.22*r while the distance AB will be 0.2*r as required If angle at B is t and (cx,cy) is the center of the outer circle with radius r: Coordinates of A: (Ax,Ay)=(cx+r*(1-c)*cos t, cy+r*(1-c)*sin t) Coordinates of C: (Cx,Cy)=(Ax+d*sin(PI/3-t),Ay+d*cos(PI/3-t)) Coordinates of D: (Dx,Dy)=(Ax+d*cos(t-PI/6),Ay+d*sin(t-PI/6)) Thanks for reading |
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Messages In This Thread |
Small challenge - Pekis - 06-06-2017, 08:05 AM
RE: Small challenge - pier4r - 06-06-2017, 01:26 PM
RE: Small challenge - Pekis - 06-07-2017 09:53 AM
RE: Small challenge - PedroLeiva - 06-07-2017, 11:35 AM
RE: Small challenge - Pekis - 06-07-2017, 03:49 PM
RE: Small challenge - PedroLeiva - 06-08-2017, 12:59 PM
RE: Small challenge - Jim Horn - 06-07-2017, 04:25 PM
RE: Small challenge - Pekis - 06-07-2017, 04:31 PM
RE: Small challenge - SlideRule - 06-07-2017, 08:42 PM
RE: Small challenge - Pekis - 06-07-2017, 09:54 PM
RE: Small challenge - SlideRule - 06-07-2017, 10:53 PM
RE: Small challenge - Pekis - 06-08-2017, 05:10 AM
RE: Small challenge - SlideRule - 06-08-2017, 12:12 PM
RE: Small challenge - Vtile - 06-09-2017, 01:14 PM
RE: Small challenge - Csaba Tizedes - 06-11-2017, 10:59 AM
RE: Small challenge - Pekis - 06-09-2017, 07:08 AM
RE: Small challenge - Pekis - 06-11-2017, 09:58 PM
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