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See ellipse as line envelope See parabola as line envelope See hyperbola as line envelope

See ellipse by drafting triangle See parabola by drafting angle See hyperbola by drafting angle

See ellipse by folding paper See parabola by folding paper See hyperbola by folding paper

For details, go to Ellipse ,Parabola and Hyperbola .

-------See cycloid_by_line envelope #0 -------See cycloid_by_line envelope #1 -------See cycloid_by_line envelope #2

For details, go to Cycloid,Epi & Hypo-cycloid as Envelopes of Lines

See Cardioid as line envelope See Nephroid as line envelope See Cremona as line envelope

For details, go to Epicycloid

See Deltoid as line envelope See Astroid as line envelope See 6-cusped hypocycloid

For details, go to
Hypocycloid

Further studies were done by Apollonius of Perga (about 262 BC - about 190 BC).

There are many excellent references on this subject, and the ones the author could purchase via internet are listed in the reference section at the bottom of this page.

*************** ellipse_manual_2.dwg** *************
****************** ellipse_auto.dwg** ****************

You can see the process in animation.

**To create this drawing and animation: **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **ellipse_auto **

To create ellipse_manual_2.dwg, type ellipse_manual_2 in command line window.

Draw a circle with radius = a. Select a point F1 with distance c from O. From a point P on the circle ,draw a line through F1 to Q on the circle. Draw a line PS perpendicualr to PQ. Make a rectangle PSRQ. It cuts x-asix at F2. Extend line F1P and make point T such taht F1P = PT Connect T-F2, This intersects PS at U. Since PQ is parallel to SR, triangles OPF1 is congruent to trianlge ORF2. Since trainlges UPF1 and UPT are congruent, F1U = UT. Therefore F2 is (c 0), and UF1 + UF2 = UF2 + UT = PR = 2a (constant) So the point U is on the ellipse with foci F1,F2. and also tangent to the curve at U.Envelope of many lines drawn like "PS" ,"QR" will make an ellipse as shown above.

Load conics_by_line.lsp

Then from command line, type

***************

You can see the process in animation.

**To create this drawing and animation: **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **ellipse_by_triangle **

************** ellipse_by_triangle.dwg** ************

When that circle is drawn , it becomes clear that line PS can also be a part of the line VW,

which is created as a crease line when a circular paper is folded so that point T falls on point F1.

**To create this drawing : **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **ellipse_model **

Answer "Yes" to "Want to continue ?" prompt.

**************** ellipse_model_2.dwg** **************

You can see the process in animation.

**To create this drawing and animation: **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **ellipse_by_folding **

*************** ellipse_by_folding.dwg** *************

*************** parabola_manual_2.dwg** *************
****************** parabola_line.dwg** ****************

You can see the process in animation.

**To create this drawing and animation: **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **parabola_auto **

To create ellipse_manual_2.dwg, type ellipse_manual_2 in command line window.

F is aFocus (c, 0), and line D_{1}D_{2}is parallel to y axis with distance c from the origin O.(calledDirectrix) B is a point on y-axis. FB intersects with directrix at C. Line AB is perpendicular to line FC at B. Draw a line from C parallel to x-axis, and this intersects AB at G. Then GC = GF-->definition of parabolaFurthermore, angle CGB = BGF = AGE.-->angle bisector of EGF is normal to AB So point G is not only a point on the parabola,but line AB touches this parabola at point G. Consequently envelopes of lines like AB will form a parabola.

Load conics_by_line.lsp

Then from command line, type

***************

Using the property that points B is on y-axis,and angle ABF is a right angle, line AB can be drawn by using a drafting triangle as follows.

You can see the process in animation.

**To create this drawing and animation: **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **parabola_by_triangle **

************** parabola_by_triangle.dwg** ************

So line AB is created by a crease on the paper when a paper is folded such that a point on directrix falls on the Focus F.

Note here that directrix D

**To create this drawing : **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **parabola_model **

The add line BK manually.

**************** parabola_model_2.dwg** **************

You can see the process in animation.

**To create this drawing and animation: **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **parabola_by_folding **

*************** parabola_by_folding.dwg** *************

************* hyperbola_manual_2.dwg** ***********
**************** hyperbola_auto.dwg** **************

You can see the process in animation.

**To create this drawing and animation: **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **hyperbola_auto **

To create hyperbola_manual_2.dwg, type hyperbola_manual_2 in command line window.

Draw a circle with radius = a. Select points F_{1},F_{2}outside of this circle with distance c from O. From a point P on the circle ,draw a line through F_{1}to Q on the circle. Draw lines PB and RA perpendicualr to PQ.They intersect the circle at R & S. PSRQ is a rectangle. Extention of RS goes through F_{2}due to symmetry. Extend line F_{1}P and make point T such that F_{1}P = PT Connect T-F_{2}, This intersects PS at U. Since triangles UPF_{1}and UPT are congruent, F_{1}U = UT Therefore UF_{1}- UF_{2}= UT - UF2 = F2T = 2a (constant). So the point U is on the hyperbola with foci F_{1},F_{2}. Furthermore since angles PUF_{1}= PUT = BUV, line BP is a tangent line to this hyperbola at U.

Load conics_by_line.lsp

Then from command line, type

*************

Using the property that points P,Q,& S are on the circle,and angle SPQ is a right angle, line PS can be drawn by using a drafting triangle as follows.

You can see the process in animation.

**To create this drawing and animation: **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **hyperbola_by_triangle **

************** hyperbola_by_triangle.dwg** ************

When that circle is drawn , it becomes clear that line PS can also be a part of the line VW,

which is created as a crease line when a circular paper is folded so that point T falls on point F1.

You can see the process in animation.

**To create this drawing and animation: **

Load conics_by_line.lsp **(load "conics_by_line")**

Then from command line, type **hyperbola_by_folding **

************ hyperbola_by_folding.dwg** **********

********************* cycloid_by_line_envelope.dwg** ************************

********************* cycloid_by_line_envelope_2.dwg** *********************

You can see the process in animation.

**To create this drawing and animation: **

Load cycloid1.lsp **(load "cycloid1")**

Then from command line, type **draw_cycloid_by_line **

1-st example: no. of division = 20

2-nd example: no. of division = 200, and layer2 (yellow colored nodes) is hidden.

Point P on a circle with radius "a" is at point O initially. After this circle rolls along X-axis by θ, top, center and bottom of this circle are points Q,R & U respectively. Since R is the instantaneous center of rotation, ∠RPQ is a right angle, and line PT is tangent to the curve formed by locus of P. Additionally, it is shown that ∠PUR = θ, and ∠PRO = ∠PQR = ∠SQT = θ/2 Length of OR = a θ. This yields the following parametric equation for cycloid. x = a(θ - sin θ) y = 2a(1 - cosθ)Envelope of many lines like "PT" , will make a cycloid as shown above.This can be accomplished by either of the two ways: (1)Draw a horizontal line AB with lenght 2π, and divide in n equal lengths. Draw a line at each point with an angle incremented by 2π/n. (2)Draw a circle with radius 2a, and draw a diameter vertical initially at O. Roll this big circle along X a-xis, then the envelope of this diameter forms cycloid. The reason is that when the big circle comes to R, ∠RQP' = θ and PT is a part of the diameter.

***************

Load cycloid1.lsp

Then from command line, type

Use 20 division and stop execution at the 7-th step. Modify the drawing .

**** epicycloid_cardioid.dwg**
**** epicycloid_nephroid.dwg**
**** epicycloid_cremona.dwg**

**** draw cardioid-animation**
**** draw nephroid-animation**
**** draw cremona-animation**

T.B.A.

*************

Load cycloid1.lsp

Then from command line, type

Use 20 division and stop execution at the 7-th step. Modify the drawing .

****** string_cardioid.dwg******
****** string_nephroid.dwg******
****** string_cremona.dwg******

**** draw cardioid-animation**
**** draw nephroid-animation**
**** draw cremona-animation**

- Yates, Robert C.: A Handbook on Curves and their Properties. J.W.Edwards,1947.
- Lockwood,E.H.: A Book of Curves. Cambridge University Press, First edition published in 1961 .
- Millington, Jon: Curve Stitching. Tarquin Publications.1989.
- Downs, J.D.: Practical Conic Sections.-The geometric properties of ellipses,parabolas and hyperbolas. Dovers. First edition published in 1931.
- Kendig,Keith: Conics, Mathematical Association of America, 2005.
- Salmon, George: A Treatise on Conic Sections. Chelsea Publishing Company. Reprint. Original published in 1900.
- Hilbert,D.,Cohen-Vossen S.: Geometry and Imagination. Chelsea Publishing Company, English Translation. Original was published in 1932.

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Last Updated Aug 6-th, 2007

Copyright 2006 Takaya Iwamoto All rights reserved. .