We have 2 sticks each of length a joined at one end , say A and the other ends joined by a metallic rod DE. DAE forms an isosceles triangle with DA=AE. Let angle DAE be x=xc radian. The task is to reduce the angle DAE to x1 such that DE which was earlier straight becomes a curve which is an arc of the circle with radius a. What should be the new angle x1 ? Ans:- BKC=x1a; DE= 2a sin (x/2) . Since BKC=DE => x1a=2a sin(x/2) or x1=2*sin(x/2). Suppose with new angle x1, the chord is BC. Now suppose reduce the angle x1 to x2 such that the st.line BC becomes the new arc of the circle with radius a.

Here x2=2*sin(x1/2) = 2*(sin(sin(x/2)) and so on. After n times-- xn=2*sin(n-1)(sin(x/2))

* Suppose initially we take an arbitrary value of x which is xc . The chord length is DE. Suppose we reduce the angle x to x1 from  xc such that the new arc length is DE. At this stage, the arc is still part of the circle with radius a. If the angle x1 is further reduced, the arc bulges out symmetrically about the line Ak, the peak lying on the line of extended Ak. As the angle between the sticks are reduced further, the peak shifts downward, the width of the curve reduces and height increases. Finally when the sticks coalsec i.e. the angle between them becomes zero, and if R is the peak, kR=DE/2=a*sin(x/2).Thus for  the bulging of the arc beyond angle x1, the peak hovers between Ak to AR. If h is the length of the peak from the origin, then equation can be formulated as

h(x)=A - a*sin(x/2) for the range 2sin( xc/2)  ≥  x  ≥  0  where x is in radian, A is a constant given by A=a[1+sin(sin(xc/2)] . When x=0, h(x) =A, and  h(x)=a when x=2sin( xc/2). One has to assert whether in reality, this pattern is made while the boundary conditions hold good. The equation in blue color is an equation of a st.line if sin(x/2) is taken as independent variable with slope -a and y-intercept A. Since the mapping of x to h(x) is onto and one-one within the permissible domain, an inverse function also exists.

xc in degree
any x (less than xc )in degree
xc in radian
xc/2 in radian
sin (xc/2)
x1  [ 2sin(xc/2)]
x1 as % of xc
x2 [2sin(sin (xc/2))]
x2 as % of xc
x2 as % of x1
x3 [2sin(sin(sin (xc/2)))]
x3 as % of xc
x3 as % of x1
x3 as % of x2
x4[2sin(sin(sin(sin (xc/2))))]
x4 as % of xc
x4 as % of x1
x4 as % of x2
x4 as % of x3
x5[2sin(sin(sin(sin(sin (xc/2)))))]
x5 as % of xc
x5 as % of x1
x5 as % of x2
x5 as % of x3
x5 as % of x4
x6[2sin(sin(sin(sin(sin(sin (xc/2))))))]
x6 as % of xc
x6 as % of x1
x6 as % of x2
x6 as % of x3
x6 as % of x4
x6 as % of x5
Calculation worksheet
Angle in degree
(Angle in radian)/2
Copy & paste the above angle value in radian here
xi/2=(sin of Angle in radian/2)
xi where i=1,2,3,4,........X  -1        (by clicking submit each time, you get x1,x2,x3,etc)