The Cardioid DESCRIPTION: The word cardioid comes from the Greek root cardi meaning heart. The Cardioid curve is a special case of the epicycloid and the limacon of Pascal. It can also be defined as the curve traced by a point of a circle that rolls around the circumference of a fixed circle of equal radius without slipping. History Studied by Ole Christensen Roemer in 1674, it was discovered during an effort to try to find the best design for gear teeth. The curve was given its name by de Castillon in the philosophical ransaction of the Royal Society of 1741. The arc length was later discovered in 1708 by La Hire. However, since this cardioid is also a special case of the limacon of Pacscal, it is believed by some to have been originated from Etiene Pascals studies. (1588-1640)
Start with two circles with centers C and C with radius R. Circle C is tangent to circle C at point T and center C is 2R from C. R T R C1 C2 Let angle TCD be the angle made by the line going through the centers of the two circles and the x-axis. Let it be known as
q1 in our explanation. So far, we know that the coordinates of C are as TC D= follows: 1 X Y C C 2 2 R 1 2 R cos( 1 )
C2 2 R sin( 1 ) R T 1 C1 D To find the coordinates of the point A we can do what we did previously, but we have two unknown angles ACE(q3) and ACB(q). Draw a line parallel to the x-axis through C and
point D (L1). AC2B=2 A B 2 3 C2 R T R 1 C1 E AC2E=3 Now draw two adjacent isosceles-right triangles with their height being the
distance from point T to the x-axis at point H, their bases are each R (along the segment C to C), and their hypotenuse be the distance from each circles center to point H. AC 2B=2 A B 2 3 C2 R T R 1 C1 H
E AC 2E=3 Using the SAS theorem (side-angle-side) we know that since these two triangles share the right angle, their base and their height, that their other angles must be equal. In turn, using another geometric proof, we now can see that q is equal to the sum of q and q, AND that q and q are equal. AC2B=2 A B
AC2E=3 2=1 1 R T R 1 C1 C2 1 H 1 3=1+2 E 3=21
By making a second triangle you can see that from C to point A is: X A Xc R cos( 1 2 ) Y A Yc R sin( 1 2 ) X A Xc R cos( 2 1 ) Y A Yc R sin( 2 1 ) 2 AC 2B=2 2 A 2 B 2
1 R C2 1 T R 1 C1 AC2E=3 2=1 H 1 3=1+2
E 3=2 1 Finally, let t be equal to the angle q. Therefore, the parametric equations (and coordinates of the point A that travels around the circle) for the Cardioid are: X 2 R cos( t ) R cos( 2 t ) Y 2 R sin( t ) R sin( 2 t ) Cardioid Using Parametric Equations
t=linspace(0,2*pi) R=1; x=2*R*cos(t)+R*cos(2*t) y=2*R*sin(t)+R*sin(2*t) plot(x,y) grid on axis square title('The Cardioid Using Parametric Equations') Cardioid Using Polar Coordinates
theta=linspace(0,2*pi); r=2*(1+cos(theta)); polar(theta,r) title('The Cardioid Using Polar Coordinates') A LOCUS! C2 T C1 R Sources: Xah: Special Place Curves http://www.best.com/~xah/SpecialPlaneCurves_dir/Cardioid_dir/cardioid.html Peter Gent: Cardioid
http://online.redwoods.cc.ca.us/instruct/darnold/CalcProj/Sp98/PeterG/Cardioid.html Sean Larson: Introducing the Cardioid http://online.redwoods.cc.ca.us/instruct/darnold/CalcProj/Sp98/seanL/cardioidf.html Limacon http://mathworld.wolfram.com/Limacon.html Cardioid http://en.wikipedia.org/wiki/Cardioid
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