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ESE 251 Presentation by Aaron Mosher A trebuchet is a  medieval siege engine that uses a massive counterweight to accurately propel a projectile great distances The trebuchet was  thought to have been invented in China, and introduced to Europe during the 13th century www.redstoneprojects.com  Constants:  M : mass of the cw counterweight  m : mass of the projectile p  h: height of the pivot  ds: sling length  dcw: CW length  dsa: length of the short arm  dla: length of the long arm  State Variables:  θ: pivot angle  φ: sling angle  ψ: pivot angle A trebuchet is a device that converts potential energy to  kinetic energy 1 Mgh → mv2 projectile cw 2 From basic physics we know that the range of a projectile  with initial velocity v and angle α is 2v2 sinαcosα R = g Thus, the maximum theoretical range of a trebuchet is given  by M R = 2 cw h max m p Given a trebuchet of fixed dimensions, you  wish to design an electronic release mechanism For a desired release angle α, you want to  find the time at which the control mechanism should release  i.e. find the time t such that sling angle φ=α . Assume that all structures are rigid, and that the  device is fixed to the ground Assume that all surfaces are smooth, and all  contacts well lubricated, so frictional effects are negligible Assume that the arm beam has negligible mass  For simplicity, suppose that the given trebuchet  has the counterweight fixed to the arm, so d =0 cw Split the model into two cases   i) The projectile slides along a smooth trough  ii) The projectile swings unconstrained through the air Use Lagrange’s equations to derive equations  of motion Use a numerical solver to solve the equations  of motion  Let T denote kinetic energy and 1  T = ml2θ2 V denote the potential energy of 2 a system. The Lagrangian of the V = mgl(1−cosθ) system is defined as 1  L = T −V L = ml2θ2 − mgl(1−cosθ) 2  For each coordinate qi, Lagrange’s equation is d  ∂L  ∂L g  0 = − 0 =θ− sinθ   dt ∂q ∂q l   i i  Ex: Consider a simple pendulum of length l and mass m  During case (i), the sling is constrained to  To model this case, adapt Lagrange’s move along a given curve equation by a Lagrange multiplier  With a little geometry, we can derive the d ∂L − ∂L −λa =0 a = ∂f following constraint dt ∂q  ∂q qi qi ∂q i i i π h−d cosθ  It can then be shown that f(θ,φ)=φ+θ− −sin−1 la  2  d  s d sin(θ) a =1− la a =1 θ d 2 −(h−d cos(θ))2 φ s la  Case (i) ends when the net force in the y direction is zero. This yields the following.  h−d cosθ  g =−x  la  pd 2 −(h−d cosθ)2 s la We are interested in the dynamics of the system in terms of  θ and ψ  We need to solve for the coordinates in terms of the two angles ( ) ( ) x = d sinθ x = d sinθ+φ −d sinθ c sa p s la x = d cosθ⋅θ x =[d cos(θ+φ)−d cos(θ)]θ+d cos(θ+φ)⋅φ c sa p s la s y = h+d cosθ ( ) ( ) c sa y = h+d cosθ+φ −d cosθ p s la  yc = −dsa sinθ⋅θ y =[−d sin(θ+φ)+d sin(θ)]θ−d sin(θ+φ)⋅φ p s la s We then plug these expressions into the Cartesian  expressions for kinetic and potential energy 1 [ ] T = M (x 2 + y 2) + m (x 2 + y 2) V = M g ⋅ y + m g ⋅ y 2 cw c c p p p cw c p p