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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 .
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
************* 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.To create this drawing :
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 ************
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 a Focus (c, 0), and line D1 D2 is parallel to y axis with distance c from the origin O.(called Directrix) 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 parabola Furthermore, 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.
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 ************
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 F1,F2 outside of this circle with distance c from O. From a point P on the circle ,draw a line through F1 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 F2 due to symmetry. Extend line F1P and make point T such that F1P = PT Connect T-F2, This intersects PS at U. Since triangles UPF1 and UPT are congruent, F1U = UT Therefore UF1 - UF2 = UT - UF2 = F2T = 2a (constant). So the point U is on the hyperbola with foci F1,F2. Furthermore since angles PUF1 = PUT = BUV, line BP is a tangent line to this hyperbola at U.
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 ************
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.
** epicycloid_cardioid.dwg
** epicycloid_nephroid.dwg
** epicycloid_cremona.dwg
** draw cardioid-animation
** draw nephroid-animation
** draw cremona-animation
T.B.A.
**** string_cardioid.dwg****
**** string_nephroid.dwg****
**** string_cremona.dwg****
** draw cardioid-animation
** draw nephroid-animation
** draw cremona-animation
All questions/complaints/suggestions should be sent to Takaya Iwamoto
Last Updated Aug 6-th, 2007
Copyright 2006 Takaya Iwamoto All rights reserved. .