德博士的 Notes About Circles, ज्य, & कोज्य

What in the world is a hacovercosine?

Historically, a multitude of functions that can be derived from sine and cosine were provided in tables along with their logarithms for early global navigators. Today with the help of electronic computing, most of these are derived directly from sine and cosine (ज्य "jyā" and कोज्य "kotijyā") when needed, or as complex exponentials e^jθ.

sin

sine

cos

cosine

tan

tangent

cot

cotangent

sec

secant

csc

cosecant

exs

exsecant

exc

excosecant

ver

versine

cvs

coversine

vcs

vercosine

cvc

covercosine

hav

haversine

hcv

hacoversine

hvc

havercosine

hcc

hacovercosine

E^͡A

arc

angle

crd

chord

sag

sagitta

apo

apothem

A hacked over cosine is a sine wave that oscillates between zero and one. Here it is on the unit circle and unit square with hcc and 19 of its trigonometric relatives. (Actually hacovercosine is short for halved-co-versed-co-sine, but hacked-over cosine seems to describe it more colorfully )

Sines and cosines marking the coordinate of a point on a unit circle corresponding to a given angle measured in radians (i.e. arclength on a unit circle) can be expressed through Taylor series expansion. When the imaginary number i is used to represent an imaginary unit orthogonal to one, (which when squared is equal to -1), the complex exponential emerges from Taylor expansion.

e^{i θ} = ∑_{n = 0}^{∞} \frac{{(i θ)}^{n}}{n!} = 1 + i \frac{θ^{1}}{1!} - \frac{θ^{2}}{2!} - i \frac{θ^{3}}{3!} + \frac{θ^{4}}{4!} + i \frac{θ^{5}}{5!} - \frac{θ^{6}}{6!} - i \frac{θ^{7}}{7!} + ... = cos θ + i sin θ

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where

i^{2} = -1

with a repeating pattern that repeats in circles like:

e^{i 0} = i^{0} = 1

e^{i π/2} = i^{1} = i

e^{i π2/2} = i^{2} = -1

e^{i π3/2} = i^{3} = -i

e^{i π4/2} = i^{4} = 1

e^{i π5/2} = i^{5} = i

e^{i π6/2} = i^{6} = -1

e^{i π7/2} = i^{7} = -i

e^{i π8/2} = i^{8} = 1

e^{i π9/2} = i^{9} = i

e^{i π10/2} = i^{10} = -1

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or if you're an engineer, avoid $i$ by using $j$ , (since the letter i is used for intensité de courant in engineering)

e^{j θ} = ∑_{n = 0}^{∞} \frac{{(j θ)}^{n}}{n!} = 1 + j \frac{θ^{1}}{1!} - \frac{θ^{2}}{2!} - j \frac{θ^{3}}{3!} + \frac{θ^{4}}{4!} + j \frac{θ^{5}}{5!} - \frac{θ^{6}}{6!} - j \frac{θ^{7}}{7!} + ... = cos θ + j sin θ

where

j^{2} = -1

Just because both $i^{2} = -1$ and $j^{2} = -1$ does not mean $i$ and $j$ are the same imaginary number. In fact, we can multiply $i$ and $j$ together to get a different value than multiplying them together in reverse order.

i j = - j i

So what is $i j$ ? It is yet another complex unit distinct from both $i$ and $j$ . When you work with these complex numbers $i$ , $j$ , and $ij$ together, they are hypercomplex numbers called quaternions.

i j i j = - j i i j = - j (i^{2}) j = - j (-1) j = j^{2} = -1

For convenience, k is defined as ij, and symetrically, it behaves exactly like i and j, mutatus mutandus.

Using imaginary values i,j, and k individually with real numbers constructs complex numbers. Using them together constructs hypercomplex numbers called quaternions. Quaternions are nice for representating rotations in 3 dimensional space for robotics, or virtual reality and other 3D rendering applications. They can also be used for representing composite color. In either application, the "complex" parts are more related to real-life quantities than the so-called "real" parts... so maybe they should have been named the other way around.

I used quaternions for color object recognition that I use for feedback for a crane robot in simulation. It's fairly easy to calculate inverse kinematics to determine the length that you want the ropes to be to control the robotic crane, but the calculations are more involved when you want to determine the position of the platform from the lengths of the ropes. Fortunately that doesn't need to be calculated to control the robot, but I looked into how to do it anyway.

A horizontal beam runs s₀ units between two points P₁ on the left, and P₂ on the right. String S₁ of length s₁ is attached at its upper end to P₁, and string S₂ of length s₂ is attached at its upper end to P₂. The two strings are connected together at their lower ends to a heavy object at point to Z to keep them taut, and creating a swing. The swing has radius r along a circle about point Q on the horizontal beam.

Brahmagupta's Formula calculates the area of a cyclic quadrilateral, given the side lengths:

K_{4} = \frac{\sqrt{{(\sum s^{2})}^{2} + 8 \prod s - 2 \sum s^{4}}}{4}

or Heron's a for cyclic triangle

K_{3} = \frac{\sqrt{{(\sum s^{2})}^{2} - 2 \sum s^{4}}}{4}

The radius of the swing then is the height of the triangle

r = \frac{2 K_{3}}{s_{0}} = \frac{2 \sqrt{{(\sum s^{2})}^{2} - 2 \sum s^{4}}}{4 s_{0}}

The distance P₁Q and the distance P₂Q are

d_{k} = P_{k} Q = s_{0} \sqrt{{s_{k}}^{2} - 4 {K_{3}}^{2}}

Three such swings, joined at the endpoints of their beams to form a horizontal equilateral triangle P₁P₂P₃ mutatus mutandus to form swing traingles P₁Z₁P₂, P₂Z₂P₃, and P₃Z₃P₁. The beams have a common size, s₀s, but the string lengths s₁ and s₂ are independently variable, having s₁₁ and s₂₁ for the first swing, s₁₂ and s₂₂ for the second swing, and s₁₃ and s₂₃ for the third swing. The swinging points Z₁,Z₂,Z₃ are constrained to be equadistant from one another, connecting them with straight line beams of length one results in another equilateral triangle similar in configuration to an inverted Stewart platform, or a NIST RoboCrane®. Here is a diagram showing circles of swing of one such a robotic crane.

The forward kinematic problem is reduced to finding the set of three points (x₁,y₁,z₁), (x₂,y₂,z₂), (x₃,y₃,z₃) lower than the support such that the distance separating each pair of points one. Then since $x = T x_{0}$ , and $x_{0}$ is known at the zero-reference pose of the platform, then then

X_{0} = (\begin{matrix} x_{10} & x_{20} & x_{30} \\ y_{10} & y_{20} & y_{30} \\ z_{10} & z_{20} & z_{30} \end{matrix})

X = (\begin{matrix} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{matrix})

X = T X_{0}

T = X {X_{0}}^{-1}

The 8th order direct kinematic solution was published by Nanua, Waldron, and Murthy (1990). Yes, after deriving all of that, I found it was already done, for it is written: ט מַה-שֶּׁהָיָה, הוּא שֶׁיִּהְיֶה, וּמַה-שֶּׁנַּעֲשָׂה, הוּא שֶׁיֵּעָשֶׂה; וְאֵין כָּל-חָדָשׁ, תַּחַת הַשָּׁמֶשׁ. (What has been will be again, what has been done will be done again; there is nothing new under the sun.)