1/35 Modern Siege Weapons: Mechanics of the Trebuchet Shawn Rutan and Becky Wieczorek � � � � � � �
The Project 2/35 The purpose of this project is to study three simplifications of a trebuchet. We will analyze the motion over time of each of these models using differential equations and modern mechanics. � � � � � � �
Why the Trebuchet? 3/35 • The trebuchet is a historical example of mechanics and the field of engineering. • The motion of a trebuchet can used as an example of a double and a triple pendulum. • It’s an awesome example of human ingenuity! � � � � � � �
History of the Trebuchet 4/35 • The first trebuchets on record appear in Asia in the 7th century. They are rightly known as the ”heavy artillery” of the middle ages. • The chronicler Olaus Magnus writes of another trebuchet used at Kalmar. An old woman happened to sit down on the sling pouch and by mistake triggered the trebuchet with her walking stick. As a result she was hurled through the air across the streets of the town, apparently without suffering any damage. � • Another story about trebuchets is told by Froissart in connection � with the siege of Auberoche in 1334 by the French. In this case an � English messenger was captured and sent flying back to the castle � with his letters tied around his neck. � � �
Various Historical Trebuchets 5/35 � � � � � � �
Lagrangian Mechanics 6/35 • Lagrangian mechanics is a way to analyze a system that is sometimes easier to use than Newtonian mechanics. • What is the Lagrangian? The Lagrangian is defined as L = T − V , where T is the kinetic energy and V is the potential energy of the system. • Lagrangian mechanics utilizes an integral of the Lagrangian over time, called action. This path of minimum action will determine � which of an infinite choice of paths a system or particle is likely to � take. � � � � �
The Euler-Lagrange Equation 7/35 Derived from minimizing the action, the Euler-Lagrange equation gives us an simple way to solve for the equations of motion of a system. • The Euler-Lagrange equation depends on the degrees of freedom of the system (in how many ways can it move) • The general Euler-Lagrange equation (as a function of the general- ized coordinates of the system, q i ). � ∂L − d � ∂L � = 0 (1) � ∂q i dt ∂ ˙ q i � • There will be an equation for each degree of freedom, so our final � model, which will be described as a function of θ , φ , and ψ , will � require three equations. � �
The Ideal Launcher 8/35 • The counterweight falls the maximum distance allowable for any model • All of the potential energy of the counterweight goes into kinetic energy of the projectile • Projectile launched from the origin • Launch angle of 45 ◦ � � � � � � �
The Seesaw Model 9/35 m 1 ( x 1 , y 1 ) l 1 (0 , 0) θ l 2 ( x 2 , y 2 ) m 2 � � • The seesaw model is an extreme simplification of the trebuchet � • The system can be modeled as a two mass pendulum. � � • Only one degree of freedom, θ , means only one equation of motion. � �
Positions of the Masses 10/35 The coordinates of m 1 are x 1 = l 1 sin θ ( t ) and y 1 = − l 1 cos θ ( t ) . When we vectorize m 1 , this becomes � P 1 = < l 1 sin( θ ) , − l 1 cos( θ ) > . The coordinates of m 2 are � x 2 = − l 2 sin( θ ) and y 2 = l 2 cos( θ ) � � Similarly, the position vector describing the path of m 2 , P 2 , is defined � as: � � P 2 = < − l 2 sin( θ ) , l 2 cos( θ ) > . � �
Energy of the System and the Lagrangian 11/35 • From the position functions, the kinetic and potential energies of each of the masses can be derived. v is the derivative of � • Kinetic energy ( T ) = 1 v 2 , where � 2 m� P • Potential energy is given by the equation V = mg � P y . Once you have the kinetic and potential energies, the Lagrangian is simple to calculate. � L = 1 � 2 ) ˙ 2( m 1 l 2 1 + m 2 l 2 θ 2 − g cos( θ )( − m 1 l 1 + m 2 l 2 ) � = 1 θ 2 + g cos( θ )( m 1 l 1 − m 2 l 2 ) 2 ) ˙ 2( m 1 l 2 1 + m 2 l 2 � � � �
Equations of Motion 12/35 Applying the Euler-Lagrange equation to the Lagrangian gives us the equation ∂L ∂θ − d � ∂L � = 0 ∂ ˙ dt θ Solving this, we have our equation of motion to be 2 )¨ − ( m 1 l 1 − m 2 l 2 ) g sin( θ ) − ( m 1 l 2 1 + m 2 l 2 θ = 0 � � � � � � �
13/35 Figure 1: A strobe-like image of the seesaw model in action � � � � � � �
Range and Efficiency of the Seesaw Model 14/35 The range is easily calculated from the x and y velocity of the pro- jectile upon moment of release. Using the following parameters θ (0) = 135 ◦ , ˙ θ (0) = 0 g = 9 . 8 m/s 2 , l 1 = 10 m, l 2 = 100 m, m 1 = 1000 kg , m 2 = 1 kg • The range is calculated to be 2570 meters. � � • Optimal release angle of 38 . 5 ◦ . � • Maximum range for a projectile shot by the ideal launcher is 34 , 142 � meters. � • This means the seesaw model is only 8% efficient. � �
2000 Range(ft) 1000 15/35 0 –80 –60 –40 –20 20 40 60 80 Angle(degrees) –1000 –2000 Figure 2: A graph of the release angle versus the range of the projectile � � � � � � �
Trebuchet With a Hinged Counterweight 16/35 l 4 l 1 φ (0 , 0) m 1 ( x 4 , y 4 ) θ l 2 ( x 2 , y 2 ) m 2 � • An example of a double pendulum � • More of the potential energy of the counterweight is translated into � kinetic energy of the projectile � � • This system is dependent on both θ and φ , so there will be two � equations of motion. �
Positions of the Masses 17/35 • The only change is the position of the counterweight • This can be seen as a geometric addition of the original position vector for m 1 and a new position vector to the mass from that point. � Q 1 = < l 1 sin( θ ) − l 4 sin( θ + φ ) , − l 1 cos( θ ) + l 4 cos( θ + φ ) > � � Q 2 = < − l 2 sin( θ ) , l 2 cos( θ ) > � � � � � �
The Lagrangian 18/35 The Lagrangian of this system is dependent on both θ and φ . L = T − V = 1 1 ˙ θ 2 − 2 l 1 l 4 ˙ θ ( ˙ θ + ˙ 4 ( ˙ θ + ˙ 2 m 1 [ l 2 φ ) cos ( θ ) + l 2 φ ) 2 ] + 1 θ 2 + m 1 gl 1 cos( θ ) − m 1 gl 4 cos( θ + φ ) − m 2 l 2 g cos( θ ) 2 ˙ 2 m 2 l 2 � � � � � � �
Equations of Motion 19/35 • We solve the ELE with respect to θ ∂L ∂θ − d � ∂L � = 0 ∂ ˙ dt θ From this we see that the first equation of motion is Eq 1 = − m 1 g sin( θ ) l 1 + m 2 gl 2 sin( θ ) + gm 1 l 4 cos( θ ) sin( φ ) 1 ¨ θ + 2¨ + gm 1 l 4 sin( θ ) cos( φ ) − m 1 l 2 θ cos( φ ) m 1 l 4 l 1 (2) � − 2 ˙ θ sin( φ ) ˙ φm 1 l 4 l 1 − m 1 l 4 l 1 sin( φ )( ˙ φ ) 2 � + m 1 l 4 l 1 cos( φ )¨ 2 ¨ 4 ¨ 4 ¨ φ − m 2 l 2 θ − m 1 l 2 θ − m 1 l 2 φ � • Solving the ELE with respect to φ � � ∂φ − d ∂L dt ( ∂L ) = 0 ∂ ˙ � φ �
Thus, the second equation of motion is θ ) 2 sin( φ ) − sin( φ ) cos( θ ) g − g sin( θ ) cos( φ ) Eq 2 = − m 1 l 4 ( − l 1 ( ˙ − l 1 ¨ θ cos( φ ) + l 4 ¨ θ + l 4 ¨ φ ) 20/35 (3) � � � � � � �
21/35 Figure 3: A strobe-like image of the hinged counterweight trebuchet in action � � � � � � �
Range and Efficiency 22/35 We used the following parameters in our calculations. θ (0) = 135 ◦ , ˙ θ (0) = 0 φ (0) = 45 ◦ , ˙ φ (0) = 0 g = 9 . 8 , l 1 = 10 , l 2 = 100 , l 4 = 21 , m 1 = 1000 m 2 = 1 � � � � � � �
16000 14000 12000 10000 Range(ft) 23/35 8000 6000 4000 2000 –40 –20 20 40 60 –2000 Angle(degrees) Figure 4: A graph of the release angle versus the range of the projectile � � � � � � �
Properties of the Hinged Counterweight Model 24/35 • Angular speed of this model is greater • Optimal angle of release is 19 ◦ . • Maximum range of 16 , 050 meters (approximately six times the range of the seesaw model) • Range efficiency is 47% � � � � � � �
Trebuchet with a Hinged Counterweight and Sling 25/35 l 4 l 1 φ (0 , 0) m 1 ( x 4 , y 4 ) θ l 2 � ψ ( x 3 , y 3 ) l 3 � m 2 � � � � �
Benefits of the Sling 26/35 • Increase in the length of the throwing arm • Rotational energy of the system is greater • Angular velocity of the projectile is greater • The overall range is increased � � � � � � �
Positions of the Masses 27/35 The position of m 1 is � R 1 = < l 1 sin( θ ) − l 4 sin( θ + φ ) , − l 1 cos( θ ) + l 4 cos( θ + φ ) > . However, the position of m 2 has changed. Now, the new position of m 2 is � � R 2 = < − l 2 sin( θ ) − l 3 sin( − θ + ψ ) , l 2 cos( θ ) − l 3 cos( − θ + ψ ) > . � � � � � �
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