Dynamics of a
Trebuchet
Dr. Michael Calvisi (mcalvisi@uccs.edu)
MAE Department
UCCS
Motivation
2
Brief History
• Invented by the Chinese around 400 B.C (traction
trebuchet).
• Counterweight trebuchets used extensively during Middle
Ages in Europe and Middle East.
• Used as “Mechanical Artillery,” useful for destroying
castle walls or city defenses.
• Makes use of potential energy from counterweight rather
than human power or elastic energy (catapult).
Modern Uses
• Historical Re-enactment
• Engineering Challenges
• Recreation:
– “Pumpkin Chuckin’ ”
– Fruitcake toss
• Links:
– KKTV: http://www.youtube.com/watch?v=mLlX8spRZTk
– KOAA: http://www.youtube.com/watch?v=15MpyaDEFj4
– UCCS:http://www.youtube.com/watch?v=gDzHcifWgSE&feature=share&li
st=FLB1ioAHLX1ORfTg41mCzRyQ
Model
Assumptions
What is neglected:
• Friction from sling sliding
• Friction from the pivot
• Modeling of the release
mechanism
• Air drag on Projectile
What is assumed:
• Assumed the launch
angle
• Arm is a Rigid Body
Modeling Approach
Two Stages
Inside Trebuchet
Outside Trebuchet
Projectile motion: vimpact = 26.078 m/s
25
Height (m)
20
15
10
5
0
0
coolwoodworkplans.com
5
10
15
20
25
Distance (m) - Range = 39.009
30
35
Equations of Motion
Lagrange’s Equation
Advantages of Lagrange’s Equation:
• Use kinetic and potential energy to
solve for the motion
• No need to solve for accelerations (KE
is a velocity term)
Solved in terms of θ,Ψ,Φ
%%%phi
zdot(1)=z(2);
zdot(2)=(l2/l4)*zdot(4)-(1/(l4)^2)*((pi/2)-z(4))*cos(z(2)-((pi/2)-z(4)))...
-((l2*l3)/l4^2)*tan(z(2)-((pi/2)-z(4))).^2....
+(l3/l4)^2*((pi/2)-z(4))*tan(z(2)-((pi/2)-z(4))).^2....
-(l2/(l4*cos(z(2)-((pi/2)-z(4)))))*sin(phi-((pi/2)-theta))...
-z(2)-((pi/2)-z(4))*cos(phi-((pi/2)-theta))*sin(phi-((pi/2)-theta))...
-(l2/l4)^2*(z(2)-((pi/2)-z(4)))^2*(((sin(phi-((pi/2)-theta))*cos(phi-(pi/2)-theta)))/(cos(z(2)((pi/2)-z(4)))))...
-(2*g*l3)/(l4^2)*sin(phi-((pi/2)-psi0));
zdot(3)=z(4);
%%%theta;
zdot(4)=(1/(m4*l2*l3*cos(z(4))*sin((z(2)-(pi/2)-z(4)))))*((m3/2)*(2*l1^2+2*l1*l3*z(2)*(cos(z(4))*cos(z(6))-sin(z(4))*sin(z(6))4*l1*l3*z(4)*z(6)*cos(theta)*cos(psi0))...
-(m4/2)*(2*l2^2-2*l2*l4*zdot(2)*cos(z(2)-((pi/2)-z(4)))+pi*l2*l4*cos(z(2)-(pi/2)-z(4))...
+(pi-2)*l3^2*sin(z(2)-(pi/2)-z(4)))^2+2*l2*l3*zdot(2)*cos(z(4))*(z(2)-((pi/2)-z(4)))pi*l2*l3*sin(z(2)-((pi/2)-z(4)))...
+2*l5^2*cos(z(4))*sin(z(4))-2*l2*l4*z(4)*(z(2)-((pi/2)-z(4)))*sin(phi-((pi/2)-theta))));
%%%psi;
zdot(5)=z(6);
zdot(6)=(1/(l3^2*((cos(psi0))^2-(sin(psi0))^2)))*(2*l1*l3*(sin(z(4))*sin(z(6))+cos(z(4))*cos(z(6)))+2*l1*l3*z(4)*z(6)*(sin(z(3))*cos(z(5))cos(z(3))*sin(psi0))...
-4*l3^2*cos(psi0)*sin(psi0)+2*g*l3*cos(psi0));
Attempt to Solve ODE’s
Used MATLAB to solve the ODE’s in terms of θ,Ψ,Φ
Put ODE’s into state variable form
Problems:
Double derivatives on left hand side of state variables for theta and phi
Solution:
Solve through substitution for the double derivatives
Complications:
It’s a very long process
Unable to use MATLAB for this process
Projectile Motion
Equations:
Initial Velocity of Projectile:
𝑣0 =
2𝑔 ℎ + 𝑙2 sin
𝜋𝜃
180
− 𝑙4 cos
𝜋 𝜃−90 −𝜋𝜃
180
[1]
Note: 𝑣3 = 𝑙1 𝜃 sin 𝜃 + 𝑙3 𝜑 sin 𝜑 𝑖 + −𝑙1 𝜃 cos 𝜃 − 𝑙3 𝜑 cos 𝜑 𝑗 ; we assumed psi is 0˚ for this calculation
For time of flight:
2𝜋𝜃 2
2𝜋𝜃
2 𝑣0 sin 180 ± 𝑣 2 sin 180
𝑡𝑓𝑙𝑖𝑔ℎ𝑡 =
𝑔
Range of Projectile:
𝑅 = 𝑣0 cos
𝜋𝜃
𝑡
180 𝑓𝑙𝑖𝑔ℎ𝑡
2
+ ℎ0
 with repect to time
200
4
d1/dt
3.5
d/dt (rad/s)
100
50
30
d2/dt
3
2.5
2
1.5
Total Projectile Trajectory
1
0
0.5
-50
0
0.2
0.4
0.6
Time (s)
0.8
1
0
Velocity of Projectile
2
150
40
Projectile Plots
1
Degree (degree)
Veloity of Projectile with repect to 1
Change in  with repect to time
20
10
0
Vtotal
Vx
-10
Vy
0
0.2
0.4
0.6
Time (s)
0.8
1
-20
-50
0
50
1 (rad)
Postion of Projectile
50
40
In sling
Free flight
Heigth (m)
30
20
10
0
-10
-20
0
20
40
60
Distance(m)
80
100
120
140
Projectile Plots
Change in Launch Angle
30°
50°˚
Projectile motion: vimpact = 26.078 m/s
5°
Projectile motion: vimpact = 21.925 m/s
Projectile motion: vimpact = 15.405 m/s
20
18
9
25
16
8
14
20
7
12
10
10
Height (m)
Height (m)
Height (m)
6
15
8
6
5
5
10
15
20
25
Distance (m) - Range = 39.009
30
35
3
2
2
0
1
0
5
10
15
20
Distance (m) - Range = 28.000
25
0
20°
45°
Projectile motion: vimpact = 25.160 m/s
Projectile motion: vimpact = 19.444 m/s
15
25
20
Height (m)
Height (m)
10
15
10
5
5
0
0
0
5
10
15
20
25
Distance (m) - Range = 37.543
30
4
4
0
0
5
35
0
2
4
6
8
10
12
14
Distance (m) - Range = 20.196
16
18
20
2
4
6
8
Distance (m) - Range = 10.384
10
Simplified Analysis
Conservation of Energy + Σ𝐹 = 𝑚𝑎
• Initial velocity determined by conservation of energy:
Δ𝐾𝐸 = Δ𝑃𝐸
• Range determined by integrating Newton’s 2nd Law with/without air drag:
Σ𝐹 = 𝑚𝑎
case No
M(lbs)
1
2
3
4
5
6
7
8
200
400
m(lbs)l(ft) V0(ft/s)
Range (ft)
10
41.86621 61.03395595
25 1.5 8.838783 5.620556001
10
87.8189 246.6925646
25 4
49.6173 83.25171665
10
71.06975 163.9409537
25 1.5 37.3815 49.84010442
10
128.4802
519.94132
25 4
77.3297 192.8371451
Simplified Analysis
Projectile Results
Acknowledgment
The modeling and simulation work presented here is credited to
the following students:
• Brandon Fortune, MAE 4402, Fall 2012
• Hannah Tackett, MAE 4402, Fall 2012
• Akihiko Ohnaka, Historical Engineering Society, 2011-2012
Questions?
Cited Works
[1] Mosher, Aaron. 2009. ‘Mathematical Model for a Trebuchet’.
http://classes.engineering.wustl.edu/2009/fall/ese251/presentations/(AAM_13)Tre
buchet.pdf