Analytical Solutions for Bending Of Fireworks and Similarities with the Solution of Electromagnetic Wave Diffraction
11 Analytical Solutions for Bending Of Fireworks and Similarities with the Solution of Electromagnetic Wave Diffraction
Fathan Akbar (1) and Mikrajuddin Abdullah (2,a) (1)Faculty of Mathematics and Natural Sciences Bandung Institute of Technology, (2) Department of Physics, Bandung Institute of Technology Bandung Institute of Technology Jl. Ganesa 10 Bandung 40132, Indonesia (a)
Email: [email protected]
ABSTRACT
In this paper we examine more deeply about the bending mechanism of rod-shaped fireworks which burned from the free end. We derived new analytic equations. Surprisingly, we obtained the bending patterns are similar to the cornu spiral. With a few simple steps we proved that positions of points throughout the fireworks are given by Fresnel integrals, C ( x ) and S ( x ), which are generally found in phenomena of electromagnetic wave diffraction. Although we deeply discussed bending of fireworks rods, however the proposed method is likely to explain any phenomena in nature related to an evolving length scale associated with some material that becomes progressively stiff or dry, such as the growth of resin exuded from trees. Keywords : fireworks bending, Fresnel integrals, cornu spiral
I. INTRODUCTION
Entering the
Eid celebration day, many Muslim communities, especially children, lit fireworks at the night before the
Eid day. This night is known as the
Takbeer night. There are many types of fireworks available. One famous type is a rod-shaped firework as shown in Figure 1. The fireworks are made of a metal rod (stainless steel, aluminium alloy, or other metals) coated with a flammable material. The fireworks is held at one end and burnt at the free end, usually using a matches. The firework undergoes a process of combustion from the free end to the handle end up of flammable materials burned out. There was an interesting phenomenon we have observed. If fireworks are positioned at different angles to the horizontal, the fireworks rod will bend at different patterns. Previously we have explained the bending profile numerically [1]. The simulation results can explain the measurement al results. Indeed, there are so many natural phenomena around us or everyday activities that can become interesting topics of physics researches [1-16]. For most people, such phenomena or activities could be just common things. But, for physicists such phenomena may contain a number of wonderful rules to be explained. In this paper we extend the discussion and try to develop analytic solutions. Surprisingly, we found an unexpected phenomenon. We observed the bending profile of very long fireworks resembles cornu spirals. Thus, there are similarities between bending behavior of fireworks with diffraction phenomena, which has been well known to produce cornu spiral [17]. We proved that the coordinates of bent fireworks are given by fresnel integrals, which is well known as the solution diffraction phenomena.
Figure 1.
Profiles of fireworks during combustion process. (a) fireworks before being burned, (b) fireworks at initial burning process, (c) fireworks approaches the end of the combustion process, and (d) fireworks after complete combustion. We used fireworks made of aluminum alloy rod with a diameter of 1 mm, length of 40 cm, and positioned at a tilt angle of 60 o . II. METHODS
Figure 2 (a) is the profile of the fireworks before being burned. The firework is rod shape, titled at to the horizontal .We divide the firework rod over N small segments of the same (a) (b) (c) (d) length, a = L / N , with L is the length of flammable material. Figure 2(b) is the profile of firework when the k -th segment is flaming, and the ( k +1)-th segment just finished burning. Figure 2.
Illustration of the process of bending the firework rod. The firework rod is divided over N segments of the same length. (a) firework before being burned. (b) firework after burnt up to ( k +1)-th segment and (c) fireworks after burnt up to k -th segment. The k -th segment supports the weight right part of the rod having a length L - ka . If is the mass per unit length of the burnt part, then the weight of segments that have been burnt is gkaLW k (1) However, the force responsible for bending the k -th segment is the component of the force perpendicular to that segment, namely cos kk WF (2) W k+1 F k k+1 k (a) (b) (c) We hypothesize that the magnitude of deflection angle of the k -th segment is proportional to multiplication component of force perpendicular to the segment and segment length, and inversely proportional to the cross-section of segment. The same hypothesis we have used to develop a model to explain the mechanism of palm frond bending [18]. Mathematically, this hypothesis can be written AaF k (3) with A is the cross-section of segment, and is a proportional factor. By observing Figure 2 (b) we obtain the difference in angles of the k -th and ( k +1)-th segments kk (4) Substitution of Eqs. (1) - (3) into the equation (4) we can then write AaW kkk cos (5) Furthermore, by replacement kk , sa , gsLW k )( , we can write Eq. (5) as AgsLs /cos)(/ , or ))(/cos(/ sLAgdsd , giving rise to the following solution s dssLAgs )(cos)( (6) Equation (6) applies to any fireworks rod, homogeneous or inhomogeneous, having a constant cross-section or cross-sections which are dependent on the distance from the fixed end. Especially for homogeneous fireworks, Eq. (6) becomes sLsAgs (7) Next we will discuss some properties of the Eq. (7). III. DISCUSSION
From equation (7) we obtain the angle formed by the free end of the rod, ie at s = L , satisfies LAg f (8) It appears that the angle of the free end changes with the square of the rod length and a linear function of the cosine of the fixed end angle. Rod segment forms a horizontal angle if ns )( is fulfilled, with n = 0, 1, 2, ….The negative sign informs that the rod bending directs clockwise. By using equation (7), the distance from the fixed end of the segment forming a horizontal direction, s n , satisfies nn sLsAgn or ngL A nn with Ls nn / . The solution for n is )(cos211 ngL A n (9) From equation (9) we observe that the formation of horizontal directions is formed if gLnA is fulfilled, or AgLn (10) It appears that the location of horizontal direction increases when the rod is getting longer. We can determine the maximum number of locations horizontal direction, ie
AgLroundn (11) where round ( x ) is an integer that equal to or smaller than x . Conversely, the fireworks did not produce any horizontal direction when
02 cos1
AgL or
20 0 gLA (12) Let us analyze the properties of inequality (12). If L is very small, /2 gLA is very large so that /cos becomes very large. This condition is satisfied by . But, if we can approximate such that inequality (12) can be approximated as gLA (13) or gLA giving raise to two possible solutions, ie gLAgLA (14a) or gLAgLA (14b) Solution of (14a) gives to large negative value for . However, since we have assumed that AgL (15) Conversely, if L is very large, the right side of inequality (12) approaches zero. This condition is satisfied by /2, and for this condition we can do the following approximation. )2/(2/coscos = )2/sin()2/sin()2/cos()2/cos( = )2/sin( . Since then . Therefore, in case of /2 inequality (12) can be approximated as
20 0 gLA , or
21 2/ gLA gLA (16) If the fireworks rod is very strong or has very large cross-section, the second term on the right-hand side of equation (7) approaches zero. It implies )( s for all s . This means that the rod does not undergo any deflections or remain as straight line as before. Bending Profiles . Equation (7) expresses the bending angle as a function of distance from the fixed end. To figure out the bending profiles we need to determine the positions of all segments in x and y coordinates. We assume that the fixed end is located at x = y = 0. The position of the end of the first segment is )(cos aaxx , )(sin aayy . The position of the end of the second segment is )2(cos aaxx , )2(sin aayy . In general, the position of the end to the k -th segment is )(cos kaaxx kk , (17a) )(sin kaayy kk . (17b) Figure 3 is profiles of fireworks positioned at different tilt angles: (a) 80 o , (b) 60 o , and (c) 30 o . In the simulations we uses L = 1 m and 2 g / A = 1.5 m -2 . If the tilt angle increases (closer to vertical), the rod only deflect slightly. The deflection increases if the tilt angle approaches zero. Figure 3 . The result of calculation of the rod profiles that have been positioned at different tilt angles: (a) 80 o , (b) 60 o , and (c) 30 o . In the calculations we uses L = 1 m and 2 g / A = 1.5 m -2 . Figure 4 is profiles of rod bending at various lengths: (a) L = 1 m, (b) L = 2 m, (c) L = 4 m, and (d) L = 6 m. In all the simulations we used 2 g / A = 1.5 m -2 and was fixed at 60 o . (a) (b) (c) x [m] y [ m ] L = 1 m g / A = 1.5 m -2 Figure 4 . Profilesof rod bending at various lengths: (a) L = 1 m, (b) L = 2 m, (c) L = 4 m, and (d) L = 6 m. In all the simulations we use parameter 2 g / A = 1.5 m -2 and the tilt angle is fixed at 60 o . It appears in Figure 4 that if the rod length increases, the bending profile is more likely a spiral. The bending closely resembles a cornu spiral, which are widely used in diffraction problems [17]. It is therefore interesting to examine further such bending behavior. We can further process the equation (17a) and (17b) as follows. We write )(cos kaaxx kk , (17a) )(sin kaayy kk . (17b) -1 -0.6-0.20.20.6 -0.8-0.400.40.81.2-0.5 0 0.5 1 1.5 x [m] x [m] x [m] x [m] y [ m ] y [ m ] y [ m ] y [ m ] (a) (b) (c) (d) Let us define xxx kk , yyy kk , sa , ska , and taking s so that we obtain the following equation pair )(cos sdsdx , (18a) )(sin sdsdy . (18b) Substituting ( s ) from Eq. (7) into Eqs. (18a) and (18b) resulting sLsAgdsdx cos LsL (19a) and sin
LsLdsdy (19b) with Ag (20) By using trigonometry identity then we can write sinsincoscos LsLLsLdsdx (21a) sincoscossin
LsLLsLdsdy (21b) Thus, we get the expressions for x (s) and y (s) as follows ss dsLsLdsLsLsx ''sinsin''coscos)( (22a) ss dsLsLdsLsLsy ''sincos''cossin)( (22b) where we have selected x (0) = y (0) = 0. To complete the integrals (22a) and (22b) we transforms )'( tLs so that dtds ' . Thus we can write dttdsLs )cos(')'(cos AtC )'( ALsC )()(')'(cos LCLsCdsLs s (23a) In the same process we get )()(')'(sin LSLsSdsLs s (23b) where x dttxC )cos()( and x dttxS )sin()( are Fresnel integrals. Both C ( x ) and S ( x ) are an odd functions so C (- x ) = - C ( x ) and S (- x ) = - S ( x ) and we can write )(')'(cos sLCLCdsLs s (24a) )(')'(sin sLSLSdsLs s (24b) Finally we get the following equation for rod profile )(cos)( sLCLCLsx )(sin sLSLSL (25a) )(sin)( sLCLCLsy )(cos sLSLSL (25b) From Eqs. (25a) and (25b) we get the coordinates of the rod free end as LSLLCLLx sincos)( (26a) LSLLCLLy cossin)( (26b) Let us look back at Eqs. (25a) and (25b). We inspect the condition when
0. By considering Eq. (20), this condition can be achieved if /2 or either or is very small, or A is very large. By using the properties of fresnel integral, xxxxC and xxS if x then we can approximate Eqs. (25a) and (25b) as follows LsLLLsx sL )cos( or )cos()( Lssx (27a) In the same process we obtain )sin()( Lssy (27b) As mentioned above,
0 can also be achieved by positioning the fireworks nearly vertical or /2. At this condition, L and further approximations that can be performed to Eqs. (27a) and (27b) are sinsincoscos)( LLssx Ls (28a) sincoscossin)( LLssy Ls (28b) Conversely, if L ,we can make the following approximation. We use the properties of fresnel integral at x , ie x xx xxOxsignxS (29a) x xx xxOxsignxC (29b) With sign ( x ) = -1, 0, 1 for x <0, x = 0, and x >0, respectively. Because in our case x is always positive, sign ( x ) = 1 so that x xx xxOxS x x (30a) x xxC (30b) At locations around the fixed end where s << L so that L - s >> 1,we get the following approximation )( sLCLC sL sLL L L LsL sL )sin()( ))(sin(2 1 (31a) )( sLSLS sL sLL L L LsL sL )cos()( ))(cos(2 1 (31b) By using Eqs. (31a) and (31b), Eqs. (25a) and (25b) become L LsL sLLsx )sin()( ))(sin(2 )cos()(
L LsL sLL )cos()( ))(cos(2 )sin(
L LsL sLL )sin()( ))(sin(2 )cos(
L LsL sLL )cos()( ))(cos(2 )sin( or L LLsL sLLsx )sin()cos()( ))(sin()cos()(2
L LLsL sLL )cos()sin()( ))(cos()sin(
L LsL sLL )2sin()( ))(sin( (32) Since L >> 1, L-s >>1, L , and sLL we have sx . With a similar analysis we also obtain sy . Now we inspect the location near the free end where L-s << 1 so that sL . In this condition we can approximate )())(( sLsLC , sLS , LC , and LS . Thus, Eqs. (25a) and (25b) can be rewritten as )(221)cos()( sLLsx or )cos()()cos(221)( LsLLsx (33) Using a similar process we get )sin()()sin(221)( LsLLsy (34) The analytical solutions we have derived, although have been deeply discussed for bending of fireworks rods, however it is likely to explain any phenomena in nature related to an evolving length scale associated with some material that becomes progressively stiff or dry, such as the growth of resin exuded from trees. CONCLUSION
We have derived analytical solutions to explain bending of fireworks rod or other rods that were soften sequentially from the free end. For very long rods, the bending profiles replicate the cornu spiral that commonly discussed in diffraction of electromagnetic wave, described by Fresnel integrals. Although we have been deeply discussed bending of fireworks rods, however the proposed method is likely to explain any phenomena in nature related to an evolving length scale associated with some material that becomes progressively stiff or dry, such as the growth of resin exuded from trees ACKNOWLEDGEMENT This work was supported by a research grant (No. 310y/I1.C01/PL/2015) from the Ministry of Research and Higher Education, Republic of Indonesia, 2015-2017. REFERENCES [1] Abdullah M, Khairunnisa S and Akbar F 2014 Bending of Sparklers
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