Mathematical model for the bell SWI: project Old Church Delft

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1 Mathematical model for the bell WI: project Old Church Delft Kees Lemmens + Hieltje Rijnks version.3 April 4, last modified July 7, Introduction This article describes a mathematical model for the swing of the Bourdon bell in the tower of the Old Church in Delft. Although a bell with clapper (or bob) is formally a coupled system of pendulums, we will use a single physical pendulum approximation for the computation of the pendulum period because of the relatively small mass of the clapper compared with the bell itself. Nevertheless, as the amplitude is relatively high (about ± 7 degrees) we do not use a linear approximation such as the harmonic oscillator but use elliptic integrals to describe this pendulum in a more accurate way. Figure () shows the double pendulum model and notations while figure () shows the single pendulum approximation. O x O x d s φ φ s d s φ φ m s m y y (a) Bourdon bell with bob (b) Generalized double pendulum Figure : Bell and bob parameters

2 O x s φ I = moment of inertia around O m = mass of pendulum = center of gravity y mg Figure : ingle pendulum approximation (schematic) Analysis for a non-linear singular pendulum Using Newton s second law for rotational motion on this pendulum yields as usual for the balance of moments : I ϕ = m g s sin(ϕ) () Here I denotes the moment of inertia, φ the amplitude, m the mass, s the distance to the centre of gravity and g the gravity constant (see also fig. ). ubstituting the commonly used radius of gyration i : yields : I = m i () ϕ + g s sin(ϕ) = (3) i With the angular speed ω defined by : g s ω = i we obtain the standard form for the non-linear pendulum : (4) We take as initial conditions (at t = ) : ϕ + ω sin(ϕ) = (5) ϕ() = ϕ max = α and ϕ() = If we multiply (5) with ϕ and integrate we find : d d dt ( ϕ) = ω (cos(ϕ + C)) dt

3 From the initial conditions we can replace the value for C : ϕ = ω {cos(ϕ) cos(α)} (6) Using cos(ϕ) = sin ( ϕ ) and cos(α) = sin ( α ) and introducing the parameter k according to : we can rewrite (6) into : k = sin( α ), k (7) ϕ = 4k ω ( sin (ϕ/) k ) (8) Introduction of a new variable y = sin(ϕ/) k transforms the problem into equation () with the help of (9) : ϕ = δϕ δy ẏ = k ( k y ) ẏ or Finally this yields as solution () : ϕ = 4k k y ẏ (9) ẏ = ω ( y )( k y ) () ω t + C = y dζ ( ζ )( k ζ ) () 3 Elliptic integrals The integral in the righthand side of () is a so-called elliptic integral of the first kind in the Legendre normal form and it s solution is available in a tabulated form ([]). It has an inverse sn(x) on y, which is an elliptic function of Jacobi and where sn(x) is a periodic function with a 4K period. K(k) is defined as follows : K(k) = dη ( η )( k η ) () o, on the interval y sn(k(k)) = k and sn(k) =. From () it follows that the solution can be written as y = sn(ωt + C). Also sn() = and with k we obtain sn(x) sin(x). Now putting all pieces together we find : 3

4 y = sin(φ/) = sn(ωt + C) (3) k With the initial conditions from above at t = (φ() = α), we obtain sin α = k sn(c), or: sn(c) =, and C = K Finally, we find the solution sin φ = ksn(ω t + K) and: φ(t) = arcsin{sn(ω t + K)} (4) We need the relation between K(k) and k on the interval <, > before we are able to see how frequency changes with the bell amplitude. We can search a few points and present the relation between k and K(k) roughly in a figure : From () we find for k = : K() = dη η = arcsin = π/ Also, because the integral diverges for k we have lim k K(k) =. For α = 9 o or π/ (remember α denotes φ max ) k = sin α/.77. We can find K(.77) numerically or by looking it up in a table such as in []. Either way we obtain K(.77).854. These values were used to sketch fig. (3). K(k) ( π ) linear case k 8 amplitude (degrees) Figure 3: Relation between k and K(k) For a more accurate approximation we have to do some extra work : 4

5 From (4) it follows, since the period of sn is 4K, that the period of the pendulum equals : T = 4K ω The influence of the nonlinearity can be estimated from : K(k) = π/ dθ k sin θ obtained by substituting η = sin θ in the integral from (). An expansion of ( k sin θ) / in the binomial series gives : ( k sin θ) / = ( / n n= Integrating term by term and using : )( k ) n sin n θ it follows that : π/ sin n (θ)dθ = n ( n n )π K(k) = π ( / n n= o for the period T we find : )( n n )( k/)n = π k { k4 +...} (5) 64 T = 4K ω = π k { + ω 4 + 9k4 +...} (6) 64 where the first term π/ω represents the period of the linearized system (the harmonic oscillator!). For k =.77 (or the amplitude α = π : ), it follows by using the first terms k /4 + 9k 4 /64.6 o for an amplitude of π to the linearized case. the nonlinear terms raise T with 6% as compared K(.77) π ( +.6).83 whereas.854 is the value from ([] we already used above. For k close to (amplitude α = π) the series converges only slowly and there we need more and more terms to find a reasonable accurate approximation, but for the Bourdon bell this is not necessary. 5

6 Bell : Height 8 mm Royal Eijsbouts Mass (without crown) 77 kg Royal Eijsbouts Centre of Gravity (from bottom bell) 78 mm Royal Eijsbouts Moment of inertia in CG () 5 kg.m Royal Eijsbouts Rotation axis (from bottom bell) 46 mm TNO Delft Crown : Mass 55 kg Royal Eijsbouts Distance from rotation axis 3 mm estimated Counterweight : Mass kg estimated Distance from rotation axis mm TNO Delft / estimated Table : ome properties of the Bourdon bell 4 Computing the bell period To be able to use the formulas above we need detailed values for many properties of the Bourdon bell. Unfortunately some of the values below are at this moment only rough estimations. The Royal Eijsbouts Company and TNO Delft (both in the Netherlands) provided us with some numbers. Other were estimated by ourselves using photographs. The most important ones are listed in table (). If we assume that the maximum amplitude of the Bourdon bell α = 7 o (= 7/8π) we have k = sin( α ) =.57. Using the formula from (5) or the graph from figure (3) we find : K(.57) π.57 { } =.73 The Bourdon bell has a counterweight on the opposite side of the rotation axis O (see also figure (4) ). This implies that the new center of gravity tot for the total combined system must be shifted towards O. This new center can be easily computed : M counter.( counter + s tot ) = M bell.( bell s tot ).( + s tot ) = 77.(73 s tot ) s tot = 45364/87 = 5 mm (.5 m) M tot bell + crown + counterbalance : 95 kg I bell around O (using teiners rule) : (73) = 935 kg.m 6

7 counter weight 87 O tot bell s tot swing axis Figure 4: Bourdon bell dimensions Total moment of inertia : I = I bell + I counter + I crown = (.) + 55.(.3) 5 kg.m Now we can gather all other numbers and compute the period : i = ω = I M = g stot i T = 4 K ω = m =. rad/sec sec As there are about chimes per period (the clapper hits the bell twice per period) this implies n =. 6 T.6 = chimes/min. Linear approximation : If we would have used the linear approach (simple harmonic oscillator) the result would be T lin = Π ω =.98 and 4 chimes/min. o the difference is K() K(α) = =.9 or about 9 %. 7

8 O H V mg x y ϕ a r sin ϕ s a t a r a t cos ϕ Figure 5: Contribution of accelerations to force (here only shown for H) 5 The forces on the axis of rotation The acceleration of the centre of gravity consists of components (see also figure 5) : in radial direction : a r = s ϕ in tangential direction : a t = s ϕ o using the contributions of both in the x-direction we find : ẍ s = a r sin ϕ + a t cos ϕ = s ϕ sin ϕ + s ϕ cos ϕ = s ω sin ϕ{4(k sin ( ϕ )) + cos ϕ} The last line was obtained by using equations (5) and (8)). With Fx = m ẍ s the horizontal force H on the pendulum in the reference point O (to the rightside) is given by : H = m s ω {4(k sin ( ϕ )) sin ϕ + sin ϕcosϕ} (7) and with Fy = m ÿ s the vertical force V in the reference point (in upwards direction) is given in a similar way by : ÿ s = a r cos ϕ + a t sin ϕ = s ω {4(k sin ( ϕ )) cos ϕ sin ϕ} This yields by using the same substitutions as for the horizontal force : V = m s ω {4(k sin ( ϕ )) cos ϕ sin ϕ} + m g (8) 8

9 Here of course ϕ is defined by equation (5). By taking the derivative for H and V and see where these are equal to we obtain the maximum amplitudes for H and V. Doing this in Maple yields : V max occurs for φ = rad. H max occurs for φ.7 rad. Using the numerical values we found in the previous section this yields for the forces : H 35 N V 5 N (or 33 N without the m g component) Note that the signs of the forces are relative to the directions as shown in figure (5). 6 Conclusion Given the fact that for several parameters we only have a very rough estimation (especially for the counterweight), the error caused by a linear approximation of the system is after all not such a big issue and may be even less than the error caused by incorrect parameters... The bell is only tolled at very special occasions of which the most important one is a funeral of a member of the Royal Dutch family, which is of course not a very common occasion. However, just one day after the workshop ended the former Queen of the Netherlands Juliana died unexpectedly and so the bell was tolled at her funeral only a couple of days later. We were able to obtain a short video recording from the swinging bell at the funeral, so we could measure the period straight from this video. This turned out to be 3. sec or about 37 chimes/min, so notwithstanding the estimated and therefore possibly incorrect parameters the results from above resemble reality quite well! References [] Ferdinand Verhulst, Nonlinear Differential Equations and Dynamical ystems, (th ed.), (Berlin: pringer Verlag, 996). 9

10 [] Jahnke-Ende, Tafeln Höhern Funktionen, (5th ed.), (Leipzig: B.G. Teubner, 96). [3] Maria L. Beconcini, tefano Bennati, Walter alvatore, tructural characterisation of a medieval bell tower: First historical, experimental and numerical investigations, (University of Pisa, Dept. of structural engineering, Pisa, Italy, ).