The Technical Archer Austin Wargo May 14, 2010 Abstract A mathematical model of the interactions between a long bow and an arrow. The model uses the Euler-Lagrange formula, and is based off conservation of energy. It is a valid model from the time the arrow is shot to the time the arrowreachesbraceheight. Matlabwillbeusedtosolvetheequationforinitial velocities and positions as the arrow is shot. 1 Introduction Archery has exploded in popularity in the last ten to fifteen years. This is mostly due to the introductions of easier shooting compound bows. Yet, little is known by the general population of what actually happens when the string of the bow is released. This paper will deal with the simple case of a long bow being shot, and will mathematically model it. It will be a valid model from the pointwherethearrowisreleasedtothepointwherethearrowreachesthebrace heightandleavesthestring. Themodelwillbebasedonconservationofenergy, and use the Euler-Lagrange formula to find a governing differential equation. This equation will then be numerically solved to find the initial velocity of the arrow, and the position of the arrow at certain times as the arrow is being shot. The paper will be broke down into six sections. The first section is The Modelanddealswiththederivationofthemodel,andassumptionsmadeinthe process. The second section is the Set Up of the Equation, explaining how the energy equations became a differential equation. The third section is Solving of TheEquation. Statinghowtheequationwassolved, andtheresultsthatwhere found. The fourth section is Checking the Model, and does an analysis of how closethemodelistoreallife. ThefifthsectionisOtherTopicsofResearch,and will give brief account of some other topics that pertain to archery, but will not go into the solving of these topics. The sixth section is the Conclusion which will analyze what was learned from doing this project. Before the modeling process is started, some needed definitions of archery terms will be given. 1.1 Archery Terms These archery terms are terms any archer will know, but for those who are not familiar with archery will need to know before reading this paper. Full Draw When the bow is pulled all the way back. Knocking Point Where the Arrow attaches to the string. Knock The clip on the end of the arrow that attaches to the string. Draw Length Is the length between the knocking point and the handle of the bow, at full draw. Brace height The length between the handle and the string when the bow is not drawn. Riser The stiff inflexible part of the bow between the bow’s limbs. Limbs The flexible part of the bow. Draw Weight The amount of force required to pull the bow back to full draw. Thesetermswillbereferredtothroughoutthepaper,andwillneedtobeknown to fully understand the model. 2 The Model In this section of the paper assumptions will be addressed, and the energy equationswillbesetup. Thefirstpartofthissectionwilldealwithassumptions. The assumptions will be the major ones made in the paper, but some smaller oneswillbemadeastheenergyequationsarederived. Thefollowingderivation 1 of the energy equations and assumptions are not original work, but were taken from reference [1]. 2.1 Assumptions The model starts out with numerous assumptions. The first major one is the archers paradox is ignored. The archers paradox is the bending of the arrow as it is compressed by the forward action of the bow. This is an enormous assumption,butintheshorttimeframethismodelwillbevaliditisacceptable. Also,themathinvolvedwiththearchersparadoxissocomplicated,forthelittle effectitwillmake, itwillbeeasiertoleaveitoutofthemodel. Thenextmajor assumptionisthelimbsofthebowwillbetreatedastwoinflexiblebeams, with a hinge point at the ends of the riser. This does not follow the exact motion a bow’s limb will take when being shot, but due to the complexity of the math involved, different parts of the limbs changing at different rates and angles, this assumption will be made. The potential energy of the bow’s limbs will be assumed to be proportional to the square of the deflection angle, this will be explained later. The model assumes the string of the bow stays straight as the bow is being shot, and does not bend. The model will not take into account any dampening effects due to outside forces or energy losses. In other words air resistance and sound, the two main outside effects on the bow arrow system will not be taken into account. Air resistance can be neglected for the very fact of the shortness in duration the model is valid. So it’s effects should be minimal,andiftakenintoeffectitwouldmostlikelyslowthesystemdownonly slightly. Sound will effect the system by energy leaving the system. Taking it into account however, would enter too many other variables not wanted to be dealt with. The model will also, ignore stretch of the string. String stretch can be figured out, but due to time constraints this will be ignored. Also, do to modern materials string stretch is not nearly as big of a factor as what it was in years past. Finally, the arrow is assumed to be a point mass concentrated at the center of the string. This takes care of most of the major assumptions, and the model can now be set up. 2.2 Derivation of the Energy Equations ThedifferentialequationwillbesetupusingtheEuler-Lagrangeformula. Which is first set up by getting kinetic energy minus potential energy. The kinetic en- ergy of the string, arrow, and limbs need to be found, and the only potential energy is the energy stored in the limbs. The kinetic energy of the arrow is simple enough to figure out. Assuming the mass of the arrow is a point mass concentrated a the center of the string. 1 𝐾𝐸 = 𝑀𝑥˙2 (1) 𝐴 2 Where 𝑀 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑟𝑟𝑜𝑤 and 𝑥˙ is the rate of change with respect to time in the 𝑥 direction. There is no 𝑦-component because the arrow is on a rest and 2 Figure 1: Bow Diagram 1 resting on the string, perfectly horizontal. In the time frame the model is valid there is no change in the 𝑦 direction assuming the arrow is shot directly in the 𝑥 direction and not at an angle. Finding the kinetic energy of the string is a little more difficult. The mass of the string cannot be considered a point mass at the center of the string. The velocity is not constant along the whole length of the string at a specific time, so an integration is needed to find the kinetic energy. To do this integration a uniform linearly mass density is needed to be assumed as 𝜆. 𝑠 will be defined as the half length of string. The mass of the string can then be defined as the following 𝑚 = 2𝜆𝑠. Another variable is needed to be introduced for the point 𝑠 (𝑥 ,𝑦 )onthestringwhichis𝜉,and𝜉 =𝜖𝑠where𝜖isthefractionallengthofthe 𝑠 𝑠 string; remembering, the string is assumed to stay linear no matter what force is applied. The kinetic energy can now be found from the given information. ∫ 1 𝐾𝐸 = 𝜆𝑠(𝑥˙ 2+𝑦˙ 2)𝑑𝜖 (2) 𝑠 𝑠 𝑠 0 3 Now 𝑥 and 𝑦 can be defined by Figure 1. 𝑠 𝑠 𝑥 =𝑥+𝜉sin(𝜙) 𝑠 𝑦 =𝜉cos(𝜙) 𝑠 From this information 𝑥˙ and 𝑦˙ can be determined by taking the derivative 𝑠 𝑠 with respect to time. 𝑥˙ =𝑥˙ +𝜖𝑠𝜙˙cos(𝜙) 𝑠 𝑦˙ =−𝜖𝑠𝜙˙sin(𝜙) 𝑠 Now there is an unwanted 𝜙˙ that can be be replaced by 𝜙˙ = 𝜙 𝑥˙ were the ,𝑥 comma subscript means partial derivative. Then by making some substitutions the kinetic energy of the string becomes. ∫ 1 𝐾𝐸 = 𝜆𝑠((𝑥˙ +𝜖𝑠𝜙 𝑥˙cos𝜙)2+(𝜖𝑠𝜙 𝑥˙sin(𝜙))2)𝑑𝜖 (3) 𝑠 ,𝑥 ,𝑥 0 By carrying out this relatively simple integration, and substituting the mass of the string in the results are shown in equation (4). 1 1 𝐾𝐸 = 𝑚 (1+𝑠𝜙 cos(𝜙)+ 𝑠2𝜙 2)𝑥˙2 (4) 𝑠 2 𝑠 ,𝑥 3 ,𝑥 This sums up the kinetic energy of the bow string, and now it’s time to move onto the kinetic energy of the limbs of the bow. WorkingoffofthediagraminFigure2thekineticenergycanbedetermined. 𝑥 will be the axis of symmetry cutting the bow in half. 𝑙 is the length of the moveable part of limb. The riser or handle of the bow will be considered to be 2𝐿. The change in the position of the limb will be characterized by the angle in between limb and a vertical line that goes down from where the string meets the limbs. The limbs will be assumed to have a constant density called 𝜌. A constant thickness will also be assumed and will be called 𝜏. Finally the width can vary uniformly as it travels up the limb. 𝜔 will the width at the top of the 𝑜 limb, and 𝜔 +𝜔 is the width at the hinge point of the limb. Also, 𝑞 is the 𝑜 1 length along the limb at a given point as seen in Figure 1. The width of one limb can now be described as the following ∫𝑙(𝜔 +𝜔 −𝜔 𝑞)𝑑𝑞. The Mass of 0 𝑜 1 1𝑙 the limb can now very easily be taken from here. ∫ 𝑙 𝜔 𝜔 𝑞 𝑚 =𝜌𝜏𝜔 (1+ 1 − 1 )𝑑𝑞 (5) 𝐿 𝑜 𝜔 𝜔 𝑙 0 𝑜 𝑜 From here, the kinetic energy of the limbs can be determined, but one more factor needs to be taken into account. The limb travels in an arc as seen in Figure 1. This distance of the arc can be defined as 𝑟 =𝑞𝜃. 𝑟˙ is needed for the kinetic energy of the limbs, 𝑞 is an independent variable and 𝜃 is the dependent variableso𝑟˙ =𝑞𝜃˙. Thiscannowbesubstitutedintothekineticenergyequation, and kinetic energy can be solved for. ∫ 𝑙 𝜔 𝜔 𝑞 𝐾𝐸 =𝜌𝜏𝜔 (1+ 1 − 1 )𝑞2𝜃˙2𝑑𝑞 (6) 𝐿 𝑜 𝜔 𝜔 𝑙 0 𝑜 𝑜 4 Figure 2: Bow Diagram 2 By integrating this equation the following kinetic energy equation is obtained. ( ) 1 1𝜔 𝐾𝐸 =𝜌𝜏𝜔 𝑙3 1+ 1 𝜃˙2 (7) 𝐿 𝑜 3 4𝜔 𝑜 Thisequationneedstobeintermsof𝑥so𝜃˙ =𝜃 𝑥˙ bymakingthissubstitution, ,𝑥 and substituting in for 𝑚 the equation can be simplified down into a more 𝐿 useable form. 1 1+ 𝜔1 𝐾𝐸 =𝑚 𝑙2 ( 4𝜔𝑜)𝜃 2𝑥˙2 (8) 𝐿 𝐿 3 1+ 𝜔1 ,𝑥 2𝜔𝑜 To simplify this equation define 𝛽 as, 1(1+ 𝜔1 ) 𝛽 = 4𝜔𝑜 3 1+ 𝜔1 2𝜔𝑜 then substituted into the kinetic energy equation. 𝐾𝐸 =𝑚 𝑙2𝛽𝜃 2𝑥˙2 (9) 𝐿 𝐿 ,𝑥 The kinetic energy equations of the system are now all derived, and before the Euler-Lagrange formula can be used the potential energy of the system needs to be found. 5 Before the potential energy equation can be found, a geometric relationship willneedtobemadetoget𝜃,𝜙intermsof𝑥andconstants. ReferringtoFigure 2 the following can easily be equated. 𝐷−𝑥=𝑙sin(𝜃)+𝑠sin(𝜙) (10) 𝐿+𝑙cos(𝜃)=𝑠cos(𝜙) (11) Atthispointitistimetofind𝜙 ,𝜃 ,𝜙 ,and𝜃 ,whichwillbeusedlateronin ,𝑥 ,𝑥 ,𝑥𝑥 ,𝑥𝑥 thedifferentialequationandsolvingforpotentialenergy. Thesecanbeobtained by equations (10) and (11). In taking the partial derivatives of equations (10) and (11), it is important to make sure to take the partial of 𝜙 and 𝜃 because they are both in terms of 𝑥. With the two partial derivatives expressions, and two unknowns, 𝜙 ,𝜃 , the partial derivatives equal. ,𝑥 ,𝑥 −sin(𝜃) 𝜙 = (12) ,𝑥 𝑠sin(𝜃+𝜙) −sin(𝜙) 𝜃 = (13) ,𝑥 𝑙sin(𝜃+𝜙) From the first derivatives the second derivatives can be found. −𝑠sin(𝜃+𝜙)cos(𝜃)𝜃 +𝑠sin(𝜃)cos(𝜃+𝜙)(𝜃 +𝜙 ) 𝜙 = ,𝑥 ,𝑥 ,𝑥 (14) ,𝑥𝑥 𝑠2sin2(𝜃+𝜙) −𝑙sin(𝜃+𝜙)cos(𝜙)𝜙 +𝑙sin(𝜙)cos(𝜃+𝜙)(𝜃 +𝜙 ) 𝜃 = ,𝑥 ,𝑥 ,𝑥 (15) ,𝑥𝑥 𝑙2sin2(𝜃+𝜙) Equations (14) and (15) have 𝜃 and 𝜙 in the equations, but these can be ,𝑥 ,𝑥 substituted in from equations (12) and (13). Now the potential energy of the bow can be found. There is no exact potential energy equation for a bow. So the potential energy will be assumed to be proportional to the square of the deflection angle, 𝜃. The potential energy will also be assumed to vanish at the brace height; which is not exactly accurate, because the limbs are still flexed at that point. In the context of this model though it is an accurate assumptions that should not effect the outcome, because the energy cannot be used in the system. from this information the following equation can be derived. 𝑃𝐸 =𝐶(𝜃2+𝜃2) (16) 𝐵 In this equation 𝐶 is a constant. To find 𝐶 the second derivative with respect to 𝑥 of the potential energy will be taken. By taking the second derivative the spring constant 𝑘 can be obtained by evaluating the equation at a point. The pointwillbeat𝑥 whichisthepointwherethebowisatbraceheight. Inother 𝑜 words 𝑘 = 𝑑2(𝑃𝐸) at 𝑥 = 𝑥 . By taking the second derivative the following 𝑑𝑥2 𝑜 equation is obtained. 𝑑2(𝑃𝐸) =2𝐶(𝜃𝜃 +𝜃2 ) 𝑑𝑥2 ,𝑥𝑥 ,𝑥 6 At the point 𝑥=𝑥 , 𝜃 =𝜃 , and 𝜙=0. By plugging in and solving for what is 𝑜 𝑜 known 𝐶 = 1𝑘𝑠𝑙sin(𝜃 )/𝜃 . 𝐶 can then be substituted into equation (16) the 2 𝑜 𝑜 potential energy is following equation. 1 sin(𝜃 ) 𝑃𝐸 = 𝑘𝑠𝑙 𝑜 (𝜃2−𝜃 2) (17) 2 𝜃 𝐵 𝑜 The energy equations have all been solved for, and the Lagrangian can now be set up for the system. 3 Set Up of Equation Withthekineticenergyandthepotentialenergyfoundthedifferentialequation can now be set up using the Euler-Lagrange formula. The Euler-Lagrange is a way to take the energy equations of a system and turn them into a system of differential equations. In our case however, there is just one differential equation, but the process is the same. By taking some partial derivatives of the lagrangian, which is shown in equation (19), The differential equation can be found. First, a simplification will be made to make the equation more user friendly. 𝐾 will be defined as follows. 1 𝐾 =𝑚 (1+𝑠𝜙 cos(𝜙)+ 𝑠2𝜙 2)+2𝑚 𝛽𝑙2𝜃 2 (18) 𝑠 ,𝑥 3 ,𝑥 𝐿 ,𝑥 The Lagrangian can now be written as, 1 𝐿= (𝑀 +𝐾)𝑥˙2−𝑃𝐸 (19) 2 The Euler-Lagrange formula is now used. First, the partial derivative with respect to 𝑥 is taken so the equation becomes. ∂𝐿 1 = 𝐾 𝑥˙2−𝑃𝐸 ∂𝑥 2 ,𝑥 ,𝑥 Where 𝑀 is a constant and does not depend on 𝑥, but 𝐾 depends on 𝑥, and so does 𝑃𝐸. The next step is to take the partial derivative with respect to 𝑥˙ and differ- entiate with respect to time, which gives the following. 𝑑 ∂𝐿 =(𝑀 +𝐾)𝑥¨ 𝑑𝑡∂𝑥˙ Where only 𝑥˙2 depends on 𝑥˙. Bysettingthesetwoequationsequaltoeachotherthegoverningdifferential equation is as follows, according to the Euler-Lagrange formula. 1 (𝑀 +𝐾)𝑥¨= 𝐾 𝑥˙2−𝑃𝐸 (20) 2 ,𝑥 ,𝑥 7 Substitutions can be made to make the differential equation into useable form, butduetothelengthoftheequation,Iwillleaveitseparate. However,equation (18) can be substituted in for 𝐾, and 𝐾 can be substituted in from equation ,𝑥 (21). Which is a matter of taking the partial derivative of 𝐾 with respect to 𝑥. In taking the partial derivative with respect to 𝑥 one has to be careful because 𝜙 and 𝜃 are defined in terms of 𝑥 so the partial of 𝜙 and 𝜃 needed to be taken, again. 2 𝐾 =𝑚 (𝑠𝜙 cos(𝜙)−𝑠𝜙2 sin(𝜙)+ 𝑠2𝜙 𝜙 )+4𝑚 𝛽𝑙2𝜃 𝜃 (21) ,𝑥 𝑠 ,𝑥𝑥 ,𝑥 3 ,𝑥 ,𝑥𝑥 𝐿 ,𝑥 ,𝑥𝑥 Next the partial derivative of potential energy can be taken. Again, taking into account 𝜙 and 𝜃 are defined in terms of 𝑥. sin(𝜃 ) 𝑃𝐸 =𝑘𝑠𝑙 𝑜 𝜃𝜃 (22) ,𝑥 𝜃 ,𝑥 𝑜 The differential equation is now set up, but it can be seen solving the equation will be rather difficult with the equation in terms of three variables. This prob- lem will be resolved in the next section with some geometry already displayed. 4 Solving of Equation The differential equation being in terms of three variables creates problems for any way of trying to solve this equation. Also, there are numerous partial derivatives in the equation, but they have already been found from earlier. By referringtoequations(12),(13),(14),and(15)theycanbesubstitutedintothe differential equation. Thistakescareofallpartialderivatives,buttheproblemofhavingtheequa- tioninthreedifferentvariablesisstillthere. Ichosetosolveforsin(𝜃),sin(𝜙),cos(𝜃), andcos(𝜙). Tosolveforthesefourtermsfromequations(11)and(12)whichare 𝐷−𝑥=𝑙sin(𝜃)+𝑠sin(𝜙),𝐿+𝑙cos(𝜃)=𝑠cos(𝜙),andknowingsin2(𝜃)+cos2(𝜃)= 1 and sin2(𝜙)+cos2(𝜙)=1, it takes a not quite so simple algebra step. Results however, of solving for these terms are as follows. To fit it on the paper, first, a term has to be defined. 𝑠2−𝑥2+2𝐷𝑥−𝐷2−𝐿2−𝑙2 𝐴= (23) 2𝐿 √ 2𝐿𝐴+ 4𝐿2𝐴2−4(𝐴2−(𝑥−𝐷)2)(𝐿2+(𝑥−𝐷)2) cos(𝜃)= (24) 2(𝐿2+(𝑥−𝐷)2) √ 2𝐴(𝑥−𝐷)+ 4𝐴2(𝑥−𝐷)2−4(𝐴2−𝐿2)(𝐿2+(𝑥−𝐷)2) sin(𝜃)= (25) 2(𝐿2+(𝑥−𝐷)2) √ 2𝐴(𝑥−𝐷)+ 4𝐴2(𝑥−𝐷)2−4(𝐴2−𝐿2)(𝐿2+(𝑥−𝐷)2) 𝐷−𝑥−𝑙( ) 2(𝐿2+(𝑥−𝐷)2) sin(𝜙)= (26) 𝑠 8