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?
[Physics] the general formula for a trebuchet
newtonian-gravitynewtonian-mechanicsprojectile
Related Solutions
This types of problems are solved by observing projectile movements in $x$ and $y$ direction separately. In $x$ direction you have constant velocity movement
$$v_x = v_{x0} = v_0 \cos(\theta), \; (1)$$
$$x = v_{x0} t +x_0 = v_0 \cos(\theta) \; t +x_0, \; (2)$$
and in $y$ direction you have constant acceleration movement with negative acceleration $-g$
$$v_y = - g t + v_{y0} = - g t + v_0 \sin(\theta), \; (3)$$
$$y = - \frac{1}{2} g t^2 + v_{y0} t + y_0 = - \frac{1}{2} g t^2 + v_0 \sin(\theta) \; t + y_0. \; (4)$$
Your initial conditions are
$$x_0 = 0, \; y_0 \ne 0,$$
and final conditions (at moment $t=T$ projectile falls back on the ground) are
$$t = T, \; x = d, \; y = 0.$$
If you put initial and final conditions into equations (2) and (4) you end up with two equations and two unknowns $v_0, T$. By eliminating $T$ you get expression for $v_0$.
My calculations show that
$$v_0 = \frac{1}{\cos(\theta)}\sqrt{\frac{\frac{1}{2} g d^2}{d \tan(\theta)+y_0}}$$
which is I believe equal to your equation. Maybe your problem is that $d$ means displacement in direction $x$, while the total displacement is $\sqrt{d^2+y_0^2}$?
This question was asked four months ago, but none of the existing answers mentions trebuchets.
To my knowledge the trebuchet design is the only design that is purely mechanical. Other projectile throwing devices store elastic energy and on release transfer that to the projectile.
So a trebuchet it is.
(In effect 'lever' and 'trebuchet' are synonymous. A trebuchet gets its leverage by being a lever (pun intended)).
It may be worthwhile to research the noble art of pumpkin chunkin'. There's a town in the US where there is a yearly pumpkin chunkin' contest that also features a trebuchet category.
Can a trebuchet deliver close to 100% efficency?
For one thing, frictional losses can be kept quite low, that's good.
The problem is this: to throw the projectile the lever arm must be accelerated to a large angular velocity. After the projectile has been released the lever arm is swinging violently. That kinetic energy of the lever arm is energy from the counterweight that was not transferred to the projectile.
The trebuchet design involves trade-offs. A longer lever arm gives the potential for faster throws, but a longer arm also has a bigger moment of inertia.
An ideal trebuchet would transfer all of its kinetic energy to the projectile, so that right after releasing the projectile it would just sit there, nearly motionless. I'm not aware of any trebuchet design that achieves that.
To reduce the inefficiency the mass of the lever arm must be as low as possible. That is what the pumpkin chunkers do: they make the lever arm as flimsy as they dare. For every throw they stand well back, as their machines tend to self-destruct when the trigger is pulled.
Best Answer
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.