# What is the general formula for a trebuchet?

What I'm really looking for is a formula for a trebuchet that I can input the desired initial velocity after launch and mass of the object, and from that figure out how long the long arm, short arm, sling and pivot need to be, and how much mass I need to have for a counterweight. Does this exist?

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Because of the sling trebuchet mechanics are complicated. –  dmckee Sep 12 '12 at 14:38
More on trebuchets: physics.stackexchange.com/q/2279/2451 –  Qmechanic Mar 10 '13 at 22:17

The math is quite abstruse but if you want you can go to this calculator and punch in numbers to find what you need.

http://virtualtrebuchet.com/Trebuchet.aspx

It's the best simple trebuchet calculator I've seen.

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I have been playing with it, and asked for the formula from them, however it does not give me the initial velocity that I'm looking for. –  Ehryk Sep 12 '12 at 6:15

To my knowledge such a thing does not exist. A trebuchet is a rather complex instrument. Just describing the path the payload makes as the trebuchet fires is complicated.

Some information can be gathered by using Google. One interesting page is

http://www.algobeautytreb.com/

and the accompanying mechanics page. Be warned that the analysis presented there is simplified.

---- Paul J. Gans

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If you assume essentially all of the potential energy goes into the projectile, then by setting the kinetic energy of the projectile $\frac{1}{2} (\mathrm{mass of projectile}) *(\mathrm{initial velocity of projectile})^2$ equal to the potential energy before launch $g*(\mathrm{height of weight})*(\mathrm{mass of weight})$, you get the sort of equation you desire.

A good estimate of the efficiency of such a mechanism would be quite difficult, but my impression is that they are rather good at what they do.

Another possible modification would be to include the mass and moment of the arm throwing the projectile, but this is also simple to do.

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Assuming a perfect efficiency then: $1/2*m_{projectile}*v^2 = g*h*m_{weight}$, so $h*m_{weight} = \frac{m_{projectile}*v^2}{2g}$, so to get 10kg to 6km/s I would need a 200,000kg counterweight dropping 91.8m? –  Ehryk Sep 12 '12 at 6:28
Sounds reasonable to me. 6km/s is mighty fast. –  Ryan Thorngren Sep 12 '12 at 7:49
What's the most efficient machine to transfer gravitational potential energy to kinetic energy? –  Ehryk Sep 12 '12 at 9:27
But you have to take into account drag. I did a similar calculation for a Napoleon 12-pounder, and without drag you get an incorrect solution by an order of magnitude(!). –  Alex Nelson Sep 24 '13 at 13:56

You could find the moment of inertia of the apparatus around the pivot as a function of three arguments (angle between sling and vertical, angle between arm and vertical, sling tension) and use x=cos(angle) and y=sin(angle) to get three equations and unknowns. Then evaluate the differential equation numerically.

A word of caution: this took me several hours. Approximation with conservation of energy is the best way to go if you're not fond of differential equations.

Hope this helps.

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If it took you several hours, there maybe is a bit more to share in this post? –  Bernhard Mar 10 '13 at 22:03
It mostly took hours because I made so many mistakes. An easier way to do it might be to find the Lagrangian of the system, KE-U, and find the equations of motion from that. I chose the most obvious coordinate system, using three angles, but perhaps a Cartesian system is easier. Good luck. –  user21033 Mar 24 '13 at 16:03