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The Catenary The Catenary Daniel Gent December 10, 2008 Daniel Gent The Catenary I The shape of the catenary was originally thought to be a parabola by Galileo, but this was proved false by the German mathmatitian Jungius in 1663. It was not until 1691 that the true shape of the catenary was discovered to be the hyperbolic cosine in a joint work by Leibnez, Huygens, and Bernoulli. History of the catenary. I The catenary is defined by the shape formed by a hanging chain under normal conditions. The name catenary comes from the latin word for chain catena. Daniel Gent The Catenary History of the catenary. I The catenary is defined by the shape formed by a hanging chain under normal conditions. The name catenary comes from the latin word for chain catena. I The shape of the catenary was originally thought to be a parabola by Galileo, but this was proved false by the German mathmatitian Jungius in 1663. It was not until 1691 that the true shape of the catenary was discovered to be the hyperbolic cosine in a joint work by Leibnez, Huygens, and Bernoulli. Daniel Gent The Catenary I The forces on the chain at the point (x,y) will be the tension in the chain (T) which is tangent to the chain, the weight of the chain (W), and the horizontal pull (H) at the origin. Now the magnitude of the weight is proportional to the distance (s) from the origin to the point (x,y). I Thus the sum of the forces W and H must be equal to −T , but since T is tangent to the chain the slope of H + W must equal f 0(x). Therefore jjW jj µs f 0(x) = = (where µ is the weight density) jjHjj jjW jj Derivation of the catenary. I The approach used by Leibnez, and his coworkers was as follows. Daniel Gent The Catenary I Thus the sum of the forces W and H must be equal to −T , but since T is tangent to the chain the slope of H + W must equal f 0(x). Therefore jjW jj µs f 0(x) = = (where µ is the weight density) jjHjj jjW jj Derivation of the catenary. I The approach used by Leibnez, and his coworkers was as follows. I The forces on the chain at the point (x,y) will be the tension in the chain (T) which is tangent to the chain, the weight of the chain (W), and the horizontal pull (H) at the origin. Now the magnitude of the weight is proportional to the distance (s) from the origin to the point (x,y). Daniel Gent The Catenary Derivation of the catenary. I The approach used by Leibnez, and his coworkers was as follows. I The forces on the chain at the point (x,y) will be the tension in the chain (T) which is tangent to the chain, the weight of the chain (W), and the horizontal pull (H) at the origin. Now the magnitude of the weight is proportional to the distance (s) from the origin to the point (x,y). I Thus the sum of the forces W and H must be equal to −T , but since T is tangent to the chain the slope of H + W must equal f 0(x). Therefore jjW jj µs f 0(x) = = (where µ is the weight density) jjHjj jjW jj Daniel Gent The Catenary The hanging chain. Daniel Gent The Catenary I This is the same as the second order differetial equation q µ f 00(x) = α 1 + (f 0(t))2(where α = .) jjW jj I This solves to 1 1 f (x) = − + cosh(αx): α α Derivation of the catenary. I Now this may be written. µ Z x q f 0(x) = 1 + (f 0(t))2(by the formula for arc length.) jjW jj 0 Daniel Gent The Catenary I This solves to 1 1 f (x) = − + cosh(αx): α α Derivation of the catenary. I Now this may be written. µ Z x q f 0(x) = 1 + (f 0(t))2(by the formula for arc length.) jjW jj 0 I This is the same as the second order differetial equation q µ f 00(x) = α 1 + (f 0(t))2(where α = .) jjW jj Daniel Gent The Catenary Derivation of the catenary. I Now this may be written. µ Z x q f 0(x) = 1 + (f 0(t))2(by the formula for arc length.) jjW jj 0 I This is the same as the second order differetial equation q µ f 00(x) = α 1 + (f 0(t))2(where α = .) jjW jj I This solves to 1 1 f (x) = − + cosh(αx): α α Daniel Gent The Catenary I First take the first an second derivitives. f 0(x) = sinh(αx) f 00(x) = α cosh(αx) I Now input these to our differential equation. q (α cosh(αx)) = α 1 + (sinh(αx))2 q α cosh(αx) = α cosh2(αx) α cosh(αx) = α cosh(αx) 1 1 I Thus f (x) = − α + α cosh(αx) is a solution. Derivation of the catenary. I Now, although it is difficult to arrive at this solution check to ensure that it holds is an easy matter. Daniel Gent The Catenary I Now input these to our differential equation. q (α cosh(αx)) = α 1 + (sinh(αx))2 q α cosh(αx) = α cosh2(αx) α cosh(αx) = α cosh(αx) 1 1 I Thus f (x) = − α + α cosh(αx) is a solution. Derivation of the catenary. I Now, although it is difficult to arrive at this solution check to ensure that it holds is an easy matter. I First take the first an second derivitives. f 0(x) = sinh(αx) f 00(x) = α cosh(αx) Daniel Gent The Catenary 1 1 I Thus f (x) = − α + α cosh(αx) is a solution. Derivation of the catenary. I Now, although it is difficult to arrive at this solution check to ensure that it holds is an easy matter. I First take the first an second derivitives. f 0(x) = sinh(αx) f 00(x) = α cosh(αx) I Now input these to our differential equation. q (α cosh(αx)) = α 1 + (sinh(αx))2 q α cosh(αx) = α cosh2(αx) α cosh(αx) = α cosh(αx) Daniel Gent The Catenary Derivation of the catenary. I Now, although it is difficult to arrive at this solution check to ensure that it holds is an easy matter. I First take the first an second derivitives. f 0(x) = sinh(αx) f 00(x) = α cosh(αx) I Now input these to our differential equation. q (α cosh(αx)) = α 1 + (sinh(αx))2 q α cosh(αx) = α cosh2(αx) α cosh(αx) = α cosh(αx) 1 1 I Thus f (x) = − α + α cosh(αx) is a solution. Daniel Gent The Catenary I In fact any normal polygon can roll on a series of catenaries. Though a triangle will not work in a real life system even though it should in theory. I A second way to derive the catenary is to trace the vertex of a parabola as it rolls along the x-axis. Properties of the catenary. I A square can roll smoothly without slippage on a series of inverted catenaries. Daniel Gent The Catenary I A second way to derive the catenary is to trace the vertex of a parabola as it rolls along the x-axis. Properties of the catenary. I A square can roll smoothly without slippage on a series of inverted catenaries. I In fact any normal polygon can roll on a series of catenaries. Though a triangle will not work in a real life system even though it should in theory. Daniel Gent The Catenary Properties of the catenary. I A square can roll smoothly without slippage on a series of inverted catenaries. I In fact any normal polygon can roll on a series of catenaries. Though a triangle will not work in a real life system even though it should in theory. I A second way to derive the catenary is to trace the vertex of a parabola as it rolls along the x-axis. Daniel Gent The Catenary ex +e−x I Recall that cosh(x) = 2 . Also recall Euler's equation eiθ = cos θ + i sin θ I (cos θ + i sin θ) + (cos θ − i sin θ) Thus cosh(iθ) = 2 2 cos θ = 2 = cos θ I A similar process will show that sinh iθ = i sin θ. The catenary with complex numbers. I One of the more odd proberties of the catenary is how it behaves when numbers with only an imaginary part are the inputs. Ex: cosh ix Daniel Gent The Catenary I (cos θ + i sin θ) + (cos θ − i sin θ) Thus cosh(iθ) = 2 2 cos θ = 2 = cos θ I A similar process will show that sinh iθ = i sin θ. The catenary with complex numbers. I One of the more odd proberties of the catenary is how it behaves when numbers with only an imaginary part are the inputs. Ex: cosh ix ex +e−x I Recall that cosh(x) = 2 . Also recall Euler's equation eiθ = cos θ + i sin θ Daniel Gent The Catenary I A similar process will show that sinh iθ = i sin θ. The catenary with complex numbers. I One of the more odd proberties of the catenary is how it behaves when numbers with only an imaginary part are the inputs.
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