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Drawing a Straight line. Popular Lecture Series Arif Zaman LUMS. Need for Straight Line. Sewing Machine converts rotary motion to up/down motion. Want to constrain pistons to move only in a straight line. How do you create the first straight edge in the world? (Compass is easy)
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Drawing a Straight line Popular Lecture Series Arif Zaman LUMS
Need for Straight Line • Sewing Machine converts rotary motion to up/down motion. • Want to constrain pistons to move only in a straight line. • How do you create the first straight edge in the world? (Compass is easy) • Windshield wipers, some flexible lamps made of solid pieces connected by flexible joints.
James Watt’s Steam Engine • Steam engine equipped with Watt's parallelogram. • Five hinges link the rods. • Two hinges link rods to fixed points • One hinge links a rod to the piston rod. • To force the piston rod to move in a straight line, to avoid becoming jammed in the cylinder. • The dashed line shows how this whole assembly can be simplified by using three rods and hinge.
Engineer, Mathematician, Accountant • Engineer: 19 + 20 is approximately 40 • Mathematician: 19 + 20 = 39 • Accountant: Closes the door, and in a low voice, whispers “What would you like it to be?”
A Straight Line • The linkage problem attracted designers, and pure mathematicians. • Mathematicians wanted an exact straight line. • P. L. Tschebyschev (1821-1894) tried unsuccessfully. • Some began to doubt the existence of an exact solution. • Peaucellier (1864) devised a linkage that produces straight-line motion, called the “Peaucellier's cell”. • Soon, a great many solutions of the problem were found. • Solutions were found that would produce various curves of which the straight line is only a particular case.
x→ 1/xis an inversion that maps (0,1] to [1,∞) and vice versa x → c2/x similarlymaps (0,c] to [c,∞) In two dimensions this maps the inside of a circle to the outside x y = c2 is an inversion Inversion
TO SHOW:E and be are inverses ABC and ADE are similar right triangles AC/AB = AE/AD AC · AD = AB · AE But AC · AD = c2 So AB · AE = c2 This shows that E is the inverse of B. Since B was an arbitrary point on the circle, we have shown that a circle gets mapped to a line. ASSUMPTIONS Yellow circle is circle of inversion, with center A Green circle has diameter smaller than the radius of the yellow circle, and goes through its center C and D are inverses, i.e. AC · AD = c2 B is any point on the green circle E is the point where AB intersects the perpendicular line from D Circles that are Tangent to the Center get Mapped to Straight Lines under Inversion E B A C D
AE2 + BE2 = AB2 EC2 + BE2 = BC2 AC · AD = (AE – EC)(AE + EC) = AE2 – EC2 = AB2 – BC2 is a constant! All we need to do is to fix A and move C The Peaucellier Inversion B A C E D
The Peaucellier Cell • Small red circle constrains C to move in a circle • Red vertical line is image of above circle • Large red circle is the circle of inversion • Yellow circle is the limit of movement • Green circle is just for fun
Hart’s Solution • Linkages are in middle of rods, where the green “imaginary line” intersects the blue rods. • Green circle is inversion. • Red circle mapped to red horizontal line • Centers of yellow and red circles are fixed.
Hart’s Inversion • Points O, P and Q are marked on a line parallel to FB and AD • AOP ≡ AFB, FOQ ≡ FAD • OP/FB = OA/FA is fixed • OQ/AD = FO/FA is fixed. • OP · OQ = FB · AD · cons • FB · AD = AC · AD = constant (same as Peaucellier) F B O P Q A C E D
Kempe’s Double Rhomboid • Rhomboid has two pairs of adjacent sides equal. • Here we use two similar rhomboids • This solution is not based on any inversion.
Why it works a a b C D b • By parallel lines, the two a’s and the two b’s are equal. • The two rhomboids are similar, hence a=b. • This means that CD is a horizontal line. • But C is fixed so D moves in a straight line.
What happened next… • Algebraic methods have shown how one can approximate any curve using linkages. • You can design a machine to sign your name! • This problem is not interesting any more, but other similar problems are. Seehttp://www.ams.org/new-in-math/cover/linkages1.html