The fence

I'm building a fence. Late is rounded up in semicircle. The tops of late in the field between the columns are to copy an imaginary circle. The tip of the first and last lath in the field is a circle whose radius is unknown. The length of the circle chord is 180cm. The height of the arc in the middle of the chord is 23cm. There are 16 lates and their axes are 12 cm apart. Please calculate the height of lates no.2-8. i.e. half of circle arc.


r =  187.6 cm
y2 =  170.6 cm
y3 =  175.6 cm
y4 =  179.6 cm
y5 =  182.8 cm
y6 =  185.2 cm
y7 =  186.7 cm
y8 =  187.5 cm


r2=(180/2)2+(r23)2 r=(8100+529)/46=187.6 cmr^2 = (180/2)^2+(r-23)^2 \ \\ r = (8100+529)/46 = 187.6 \ \text{cm}
(1290)2+(y2)2=r2 y2=170.6 cm(12-90)^2+(y_{ 2})^2 = r^2 \ \\ y_{ 2} = 170.6 \ \text{cm}
(2490)2+(y3)2=r2 y3=175.6 cm(24-90)^2+(y_{ 3})^2 = r^2 \ \\ y_{ 3} = 175.6 \ \text{cm}
(3690)2+(y4)2=r2 y4=179.6 cm(36-90)^2+(y_{ 4})^2 = r^2 \ \\ y_{ 4} = 179.6 \ \text{cm}
(4890)2+(y5)2=r2 y5=182.8 cm(48-90)^2+(y_{ 5})^2 = r^2 \ \\ y_{ 5} = 182.8 \ \text{cm}
(6090)2+(y6)2=r2 y6=185.2 cm(60-90)^2+(y_{ 6})^2 = r^2 \ \\ y_{ 6} = 185.2 \ \text{cm}
(7290)2+(y7)2=r2 y7=186.7 cm(72-90)^2+(y_{ 7})^2 = r^2 \ \\ y_{ 7} = 186.7 \ \text{cm}
(8490)2+(y8)2=r2 y8=187.5 cm(84-90)^2+(y_{ 8})^2 = r^2 \ \\ y_{ 8} = 187.5 \ \text{cm}

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