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Small challenge
06-07-2017, 11:35 AM (This post was last modified: 06-08-2017 12:54 PM by PedroLeiva.)
Post: #4
RE: Small challenge
(06-07-2017 09:53 AM)Pekis Wrote:  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 Smile
Can you provide a numerial example to test?. TYVM, Pedro
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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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