Literary Terms and Definitions: C - Whitcraft Learning Solutions

Nancy Noell Burk

Description

• redaction - matière potentielle : the poem
• mémoire
• cours magistral - matière potentielle : on celtic literature
• expression écrite

Literary Terms and Definitions: C This list is meant to assist, not intimidate. Use it as a touchstone for important concepts and vocabulary that we will cover during the term. Vocabulary terms are listed alphabetically. [A] [B] [C] [D] [E] [F] [G] [H] [I] [J] [K] [L] [M] [N] [O] [P] [Q] [R] [S] [T] [U] [V] [W] [X] [Y] [Z] CACOPHONY (Greek

• bottom of a tiny cardboard candybox
• use of caste dialect
• flat characters with stock characters
• word canon refers to the writings
• literature
• poetry
• characters
• words
• line

Sujets

• Mémoire
• Deutsche
• Peacock
• Breastfeeding
• Rowlandson
• Manley
• Jonson
• Mistress

Informations

• Publié par: tusuf
• Nombre de lectures: 56
• Langue: English

Extrait

Spirograph Activity Sheet

The table shows the number of teeth and holes in the gears of a Spirograph:

1.

Rin Gear A

150 105 144 96

Pinion Gear B

24 30 32 40 45 48 52 56 60 63 72 80 84

Holes C

5 8 9 13 16 17 19 21 23 25 29 33 35

How many possible curves can be made, discounting different colors, with a Spirograph having the gears listed in the table?

This is a bit of a trick question, actually.

(a)

You could look at it this way: 2 curves are possible – Hypocycloids and

Epicycloids.

(b)

You could add up the number of tracing holes (253 total), multiply by

2 (two rings), and answer that there are 506 possible Epicycloids (with the

gear rolling outside rings 144/96 and 150/105) and 506 possible

Hypocycloids (with the gear rolling inside the two rings), for a total of 1012

possible curves.

Both of those answers are good ones.

Some might also have recognized that another set of Epicycloids are possible

by rolling one gear around another.

2.

(c)

With 13 gears taken 2 at a time, the natural instinct is to calculate

the number of permutations, 13 * 12 = 156. This isn’t a bad idea, but fails to

take into account the number of tracing holes. You could write a program to

loop through the 13 gears one by one, combining each in turn with the

remaining 12 gears (another loop), and keep a running total of the tracing

holes in each gear in the inner loop (see below for a VBScript program listing).

But this approach strikes me as being a bit too much work. (If programmers

hate anything, it’s doing work you don’t absolutely have to do. That’s why we

write programs; sometimes a bit of cleverness is even better.)

When taking 13 gears two at a time, you realize that one gear (the fixed one)

out of each permutation is not actually used for drawing, therefore the

number of tracing holes is irrelevant. That leaves 12 gears. All you really need

to do is multiply the total number of tracing holes for all 13 gears (253) by 12

to determine that the additional Epicycloids possible by rolling gear-on-gear

number 3036.

Add that to the number we arrived at in part (b), 1012, and you come up with

what I believe is the correct answer: 4048 possible curves.

(There is also the possibility of rolling two rings around each other, or rolling

rings on the outside of gears. Since the number of tracing holes wasn’t

provided for the wheels, I haven’t counted these possibilities toward the

total.)

How many loops will ahypocycloidhave if – A= 96 andB= 32? 3 A= 96 andB= 48? 2 A= 96 andB= 52? 24

3.

4.

5.

Write an equation showing how the number of loops depends onAandB. Does this equation LCM(A,B) work forepicycloids? Hypocycloids AND Epicycloids: B

Where LCM stands for Least Common Multiple. Why does the equation work

for both Hypocycloids and Epicycloids? (A hint may be found in the

handouts.)

Suppose that the Spirograph came with other different gears. Use your equation from question 3 to predict the number of loops that would occur if – A= 120 andB= 42? 20 A= 120 andB= 68? 30 A= 160 andB= 36? 40 A= 160 andB= 75? 32 Why did the manufacturer of the Spirograph position the holes in a spiral?

I don’t actually know the answer to this one, although I do have some ideas.

While it might be purely a matter of insuring the structural integrity of the

plastic gears (the gears might break more easily if the holes were lined up), or

perhaps for aesthetic reasons, I doubt the answer is this simple.

Does it have something to do with the angle of each hole from the center of

the gear?

Looking at the printed instructions that come with Spirograph, you realize

that the angle is irrelevant. If one wishes to create successively smaller-scale

patterns, one rotates the gear so the new hole aligns with the mark on the

wheel used by the prior pattern. If one wishes to rotate the pattern, the

directions usually indicate an offset of so many teeth, NOT by aligning tracing

holes.

Do we need to know the equation of the spiral in order to answer this

question?

Purely by visual inspection, the spiral of the holes looks most like either a

Hyperbolic or Equiangular (Logarithmic) spiral. That means the equation

(  cot b) probably looks something like r = a/or r = a . The forms of these

equations suggest a geometric progression as opposed to an arithmetical

one, and indeed, reading entries on MathWorld

<http://mathworld.wolfram.com/ > for the relevant spirals tell us that “Any

radius from the origin meets the spiral at distances which are in geometric

progression.”

In other words, the tracing holes, numbered 1 ton, begin with 1 being closest

to the circumference of the gear and the holes increasingly close to the

center. As the holes get closer to the center, the pattern looks more and

more like a circle. (If it were possible in the physical version of the Spirograph

for hole #1 to be on the circumference, the curves traced would be

Hypotrochoids and Epitrochoids. Also, in the physical version, no tracing hole

is exactly at the center of the gear. If it were, the pattern produced would be

a precise circle.)

What happens when the gears change size?

As B increases, or the gears become larger, the number of tracing holes also

increases, but never equals B. (This is probably, at least in part, a physical

constraint, so that smaller gears are not riddled with holes.)

These are the ideas I have come up with so far; I don’t know that I have fully

answered the question, however.

6.

7.

8.

I’d love to hear your thoughts. Please e-mail your answers to: [e-mail]

Does a relationship exist betweenC, or the number of holes, andB, or the number of teeth, in the pinion gear? If so, what is it?

Given B, you can predict C using this equation:

B 7 2 Round the answer up to the nearest whole number to get the number of

tracing holes. (Try it!)

Use the definition of ahypocycloidto derive the set of parametric equations  BA   BA  x(A B) cost Ccost,y(A) sinsint. 1      1 B t C    B   B 

See: http://www.iw.net/~nnburk/Handouts.pdf

If the manufacturer could make a Spirograph with A = B = C, what kind of shape would it

generate?

For an Epicycloid, a Spirograph where A=B=C would form a Cardioid. Simplify

the equations as shown in the illustration below, and graph.

See: http://www-groups.dcs.st-and.ac.uk/~history/Curves/Cardioid.html

For the Hypocycloid, simplifying the equations shows us that a device having

A=B=C would produce a single point: (A,0). (Why?)

* * * * * * *

VBScript to brute-force calculate the number of gear-on-gear Epicycloids:

dim GearsAndHoles(12,2) dim patterns, i, j

 Dim fso  Dim filOutput

 Set fso = CreateObject("Scripting.FileSystemObject")

 Set filOutput = _  fso.CreateTextFile("patterns.txt", True)

(j,1)

GearsAndHoles(0,0)=24:GearsAndHoles(0,1)=5 GearsAndHoles(1,0)=30:GearsAndHoles(1,1)=8 GearsAndHoles(2,0)=32:GearsAndHoles(2,1)=9 GearsAndHoles(3,0)=40:GearsAndHoles(3,1)=13 GearsAndHoles(4,0)=45:GearsAndHoles(4,1)=16 GearsAndHoles(5,0)=48:GearsAndHoles(5,1)=17 GearsAndHoles(6,0)=52:GearsAndHoles(6,1)=19 GearsAndHoles(7,0)=56:GearsAndHoles(7,1)=21 GearsAndHoles(8,0)=60:GearsAndHoles(8,1)=23 GearsAndHoles(9,0)=63:GearsAndHoles(9,1)=25 GearsAndHoles(10,0)=72:GearsAndHoles(10,1)=29 GearsAndHoles(11,0)=80:GearsAndHoles(11,1)=33 GearsAndHoles(12,0)=84:GearsAndHoles(12,1)=35

patterns = 0

for i = 0 to 12 for j = 0 to 12 if not(i=j) then patterns = patterns + GearsAndHoles

next

next

filOutput.Write GearsAndHoles(i,0) filOutput.Write chr(9) filOutput.Write GearsAndHoles(j,0) filOutput.Write chr(9) filOutput.WriteLine patterns end if

 filOutput.Close

 Set filOutput = nothing  Set fso = nothing