I am doing a maths modelling project, and I am stuck on a part.
I have a (arc length) and L (chord length), but I want to find H, the maximum perpendicular distance between them!
Any help would be appreciated, thanks!
Note: I don't have the radius! A way of getting the radius would be brilliant, but all I have is a and L!
On
by using NEWTONS RAPHSON method we can find out the arc angle...
A Real World Example
The arc length (blue line AB) of a circle is 16 and its chord length (line AB) is 14. (Therefore arc AD = 8 and line AE = 7) How can we solve for the other parts of the circle? We can see from the graphic of the circle that:
Since both equations equal the radius AO, we then can say:
Rather than deal with two numbers, we'll let the letter 'n' equal the ratio of AE ÷ (arc AD).
Putting the equation in standard form we have:
Angle AOD can't be isolated because this is an implicit equation and requires Newton's Method to solve it.
We will need the derivative of f(x) which is:
f '(x) = cos ∠ AOD -n = 0
The value of n = (Chord AB / Arc AB) = (14 / 16) = .875
To start, let's guess that X ₀ (or angle AOD) is 1 radian. X₁ = X₀ - [ f(X₀) ÷ f'(X₀) ]
X₁ = 1 - [ {(sin(1) -(n * 1)} / (cos(1) -n) ]
X₁ = 1 - [ {0.8414709848 -(.875*1)} / (0.54030230587 -.875) ]
X₁ = 1 - [ -0.03352901519 / -0.33469769413 ]
X₁ = 0.8998229872
X₂ = X₁ - [ sin(X₁) -(n * X₁) / cos(X₁) -n ]
X₂ = 0.8998229872 - [ {sin(0.8998229872) -(.875 * 0.8998229872)} / (cos(0.8998229872) -.875) ]
X₂ = 0.8998229872 - [ {0.7832168644 -0.7873451138} / (0.6217486174 -.875) ]
X₂ = 0.8998229872 - [ -0.00412824936 / -0.25325138258 ]
X₂ = 0.8835219928
X₃ = X₂ - [ sin(X₂) -(n * X₂) / cos(X₂) -n ]
X₃ = 0.8835219928 - [ {sin(0.8835219928) -(.875 * 0.8835219928)} / (cos(0.8835219928) -.875) ]
X₃ = 0.8835219928 - [ (0.7729781357 -0.77308174371) / (0.6344326613 -.875) ]
X₃ = 0.8835219928 - [ -0.00010360801 / -0.2405673387 ]
X₃ = 0.8830913110
If we were to compute a value for X₄ it would equal 0.8830910126, which shows that X₃ is accurate to six decimal places which is adequate for most practical applications. So, now that we have determined that angle AOD equals .883091 radians ( or 50.5974 degrees), we can calculate the remaining parts of the circle.
The central angle AOB equals 2 * 50.5974 or 101.195 degrees.
The radius = line AE / sin(AOD) radius = 7 / sin (50.5974) radius = 9.0591 Now that we have this information, the rest of the circle can be easily solved.
Knowledge of $H$ and knowledge of the radius $r$ are equivalent, because $(r-H)^2+(\frac12 L)^2 = r^2$. (There is a minor technical issue about this quadratic having two solutions for $H$ once you fix $r$, which I will ignore.)
Knowledge of the radius $r$ and knowledge of the central angle $\theta$ subtended by half the arc are equivalent, because $a=2r\theta$.
Finding out $\theta$ is tricky: from \begin{align*} a &= 2r\theta \\ L &= 2r\sin\theta \end{align*} we get $$ \frac La = \frac{\sin\theta}{\theta} $$ and $\theta\in[0,\pi]$. Now, the function on the RHS (called the sinc function) is strictly decreasing on $[0,\pi]$, so there is in fact just one value of $\theta$ in that interval that satisfies this equation; but I don't know how to find it except numerically.