Hi, fingersfinny-ga:
Thanks for inviting me to post an "official" Answer. I'm afraid that I
have written far more below than you really wanted to know about radius
of a circle from arc length and "deflection", so let me summarize the
main points here, and you can pick through as much of what follows as
seems interesting.
I'll recap the approach outlined before, solving a pair of equations
for unknown radius and central angle from given values of arc length
and deflection (chord height).
One point not discussed before is that solutions are not unique unless
the size of the central angle is suitably restricted. That is, for
each ratio of deflection to arc length, there are at least two central
angles which correspond to that ratio. There's no ambiguity, however,
if the problem involves arcs that are no bigger than a semicircle.
At about 4.6 radians, the ratio of deflection to arc length peaks out &
begins to decrease with increasing an central angle. The maximum ratio
that can be attained is about 0.36. Therefore we will restrict our
efforts to solve the problem to positive ratios less than this, and so
to central angles between 0 and roughly 4.6 radians.
Outline of Solution
===================
A circle of unknown radius R has a chord that cuts off an arc of length
L, and the "deflection" or chord height (perpendicular distance from
midpoint of chord to midpoint of arc) is D or delta.
By introducing an unknown central angle T subtended by the chord (or by
the arc), we can formulate two equations with two unknowns:
L = T * R
D = R - R * cos(T/2)
The first equation says arc length L is the product of central angle T
(measured in radians) and radius R. Applied to the entire circle this
gives that the circumference is 2pi * R. Naturally this requires equal
units of measurement for lengths L and R.
The second equation expresses the deflection D as a difference between
a radius R through the midpoint of the chord and the included leg of a
right triangle, drawn to that midpoint from the center of the circle
(where an endpoint of the chord is the third vertex). Note T/2 is half
the central angle and that half the length of chord (the other leg of
this right triangle) is R * sin(T/2).
We propose ultimately solving by first taking the ratios of these two
equations' respective sides, which eliminates R:
D/L = (1 - cos(T/2))/T (*)
Solving this nonlinear equation for central angle T and plugging back
into our first equation would give the radius:
R = L/T
Note that radius R is proportional to L and inversely proportional to
central angle T.
It's often recommended that (*) be solved numerically, and in practice
this is easily accomplished by any of the basic root finding methods.
Cf. the "Dr. Math" links posted for you by mattstephens30-ga below.
To solving it symbolically, on the other hand, will give insight into
precisely how the solution for T varies with D/L. What we need is an
inverse function to (1 - cos(T/2))/T, ie. a function f that "undoes"
what this expression does and gives us back T:
f( (1 - cos(T/2))/T ) = T
Armed with this function we could then write from (*):
f( D/L ) = T and R = L / f(D/L)
* * * * * * * * * * * * * * * * * *
Illustration
============
Let's interrupt the narrative at this point with an example, which
almost always helps to clear up any ambiguity about the notation.
Since all the distances are relative to some unit of measurement, one
can assume without loss of generality that the arc length L = 1 where
of course the units of radius R will scale proportionately. In other
words only the ratio D/L matters, and for the sake of illustration we
take D = 0.1 and L = 1.0.
T is then f(0.1), ie. the central angle such that:
(1 - cos(T/2))/T = D/L = 0.1
The slope of the left-hand side as a function of T turns out below to
be 1/8, so a first-order approximation to T would be 8 * 0.1. In fact:
(1 - cos(0.4))/0.8 = 0.09867...
which isn't too far off. Refinement of this estimate leads to:
f(0.1) ~ 0.811055
The corresponding radius would then be R = 1/T ~ 1.232962.
As mentioned before, the solutions are only unique if we restrict our
attention to angles T in roughly (0,4.6), which corresponds to ratios
D/L between 0 and about 0.36. So with f(0.1) we have already gone a
substantial way to the outer limit of what parameters can be solved.
* * * * * * * * * * * * * * * * * *
Symbolic Solutions
==================
Let's resume analysis with the familiar power series for cosine:
+oo
cos(x) = 1 + SUM (-1)^k x^(2k) / (2k)!
k=1
= 1 - x^2/2 + x^4/12 - x^6/720 + x^8/40320 - ...
Apply this expression for cos(T/2) and simplify:
1 - cos(T/2) +oo
------------ = SUM (1/2) (-1)^(k-1) (T/2)^(2k-1) / (2k)!
T k=1
3 5 7 9
T T T T T
= - - --- + ----- - -------- + ---------- + ...
8 384 46080 10321920 3715891200
Inverting a power series (analytic) expansion about T = 0 is possible
theoretically whenever the derivative is nonzero:
[Inverse Function Theorem -- PlanetMath]
http://planetmath.org/encyclopedia/InverseFunctionTheorem.html
Note that our series derivative at T = 0 is 1/8, which is nonzero, and
conveniently the constant term is zero, so the series is zero at T = 0
and we can thus expect to define an inverse function with f(0) = 0.
In fact there's a theorem due to Lagrange which gives a sort of recipe
for the power series of this sort of inverse function:
[Lagrange Inversion Theorem -- Wikipedia]
http://en.wikipedia.org/wiki/Lagrange_inversion_theorem
[Series Reversion -- Eric Weisstein (MathWorld--A Wolfram Web Resource)]
http://mathworld.wolfram.com/SeriesReversion.html
It's not too hard to work out "by hand" the first several terms of such
a power series for an inverse function, but beyond that things get tough.
I've resorted to an open source symbolic algebra package called Maxima:
[Maxima - a sophisticated computer algebra system]
http://maxima.sourceforge.net/
Let d = D/L be the "dimensionless" ratio on the left-hand side of (*),
so that for clarity we can write f(d) = T. Thus
3 5 7 9
32 d 1664 d 159232 d 4139008 d
T = f(d) = 8 d + ----- + ------- + --------- + ---------- + . . .
3 45 945 4725
A convergent power series expansion, carried to infinity, gives exactly
the central angle T. However convergence is only guaranteed for "small"
d = D/L. A graph of (1 - cos(T/2))/T will show that it oscillates up
and down. No single-valued inverse can be defined globally, and series
expansions around d = 0 can only converge in a symmetric interval about
the origin if they exclude the nearest singularity.
Moreover the convergence of a power series is progressively slower as
we move away from d = 0. One way to think about this slow convergence
is that we have a sequence of polynomials (power series convergents)
that "fit" the value of f(d) and successively higher derivatives at the
origin. The power series approximation is therefore optimal near d = 0
but apt to be unsuitable at modest distances from the origin.
Faster convergence is often obtained with rational approximations, and
in the context of exact expressions for the inverse, this leads to the
computation of a continued fraction formula rather than a power series.
Just as a power series can be expanded about any point interior to the
domain of the function, there is flexibility in constructing continued
fraction expansions.
Unfortunately much of the discussion of continued fractions of the Web
dwells on best rational approximations to individual real numbers. A
brief introduction to continued fraction function expansions is tacked
on to end here:
[Continued Fractions -- Numericana]
http://home.att.net/~numericana/answer/fractions.htm
As the thread in this forum mentions:
[Continued Fractions -- Ask NRICH]
http://nrich.maths.org/discus/messages/2069/6603.html?1064175671
"...[t]here is a general scrappy way of getting a continued fraction
expansion from the power series..."
which we will illustrate upon our example above.
Begin by factoring out the first term 8d, leaving an even series:
2 4 6 8
4 d 208 d 19904 d 517376 d
f(d) = 8d * ( 1 + ---- + ------ + -------- + --------- + ... )
3 45 945 4725
Now replace multiplication by this even series with division by
its reciprocal series (which exists because its constant term is
a unit):
2 4 6 8
4 d 128 d 10496 d 151552 d
f(d) = 8d / ( 1 - ---- - ------ - -------- - --------- - ... )
3 45 945 2835
Continue in this vein, factoring out -dē from each of the higher
order terms in the denominator:
2 4 6
2 4 128 d 10496 d 151552 d
f(d) = 8d/( 1 - d *( - + ------ + -------- + --------- + ... ) )
3 45 945 2835
and again rewrite the product as division by a reciprocal series:
2 4 6
2 3 8 d 496 d 12032 d
f(d) = 8d/(1 - d /( - - ---- - ------ - --------- + ... ) )
4 5 175 1125
Grind the crank a few more times in this fashion:
2 4
2 3 2 5 31 d 1955 d
f(d) = 8d/(1 - d /( - - d /( - - ----- - ------- + ... ) ) )
4 8 28 882
and:
2
2 3 2 5 2 28 15640 d
f(d) = 8d/(1 - d /( - - d /( - - d /( -- - ------- + ... ) ) )
4 8 31 8649
2 3 2 5 2 28 2 8649
= 8d/(1 - d /( - - d /( - - d /( -- - d /( ----- - ... ) ) ) )
4 8 31 15640
Until the appearance of this last "ugly" constant in the continued
fraction conversion, it certainly seemed that a simpler expression
was emerging. I suspect that with further exploration of various
alternatives, a formula with an explicit pattern of constants can
be discovered.
* * * * * * * * * * * * * * * * * *
Convergence of Power Series vs. Continued Fraction
==================================================
Let's finish by returning briefly to the numeric illustration shown
earlier, where d = D/L = 0.1, and compare how well the power series
solution converges there versus the continued fraction expansion.
First the results using n terms of the power series expansion:
n approx. value of T
--- ------------------
1 0.8
2 0.81066666666667
3 0.81103644444444
4 0.81105329439153
5 0.81105417037206
6 0.81105421959762
7 0.81105422250983
8 0.81105422268853
Because the series expansion contains only odd powers, n terms will
correspond to retaining highest power 2n-1. Thus the values shown
preserve greater accuracy in the bottom rows than we retained above
in our discussion.
Now let's present similar results extracting n continued fraction
terms from the fullest power series derived above:
n approx. value of T
--- ------------------
1 0.8
2 0.81081081081081
3 0.81104972375691
4 0.81105412922206
5 0.8110542209176
6 0.81105422266419
7 0.81105422269988
8 0.81105422272161
Even though the power series above is converging well for d = 0.1,
and despite taking the continued fraction expansion from the that
of the power series (instead of computing one independently), the
approximations from the continued fraction are better at each step.
Indeed n = 5,6,7 terms of the continued fraction are better than
n = 6,7,8 terms, respectively, in the power series.
The superiority of the continued fraction becomes more dramatic as
the value of d increases. For example at d = 0.3, just 5 terms in
the continued fraction give a more accurate solution than 8 terms
in the power series.
regards, mathtalk-ga |