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=x_{c} 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 x_{c
}. The chord length is DE. Suppose we reduce the angle x
to x1
from x_{c} 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( x**_{c}/2)
≥ x ≥ 0 where x is in radian, A is a constant given by
A=a[1+sin(sin(x_{c}/2)] . When x=0, h(x) =A, and h(x)=a
when x=2sin( x_{c}/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. |