The science
What is modelled. A generic eight-wheel carrier for civil emergency crossings (flood or earthquake gaps) carries a two-half scissor bridge folded on its back. A launch boom turns the bridge about a tail pivot O, the halves are opened at the vertical, and the unfolded span is lowered over the gap. Everything is planar (side view) and quasi-static. This is a physics teaching model of a generic machine with no real model, markings or military use implied; it is not a design tool.
1. Cylinder length to boom angle (law of cosines). The lift cylinder pushes a crank through a triangle with fixed sides a = |CP| and e = |PR|; its length l fixes the angle at P:
l^2 = a^2 + e^2 - 2 a e cos(gamma) , psi = alphaC - gamma - eps
A rigid coupler k joins the crank tip Q to a bracket D on the boom (distance b, angle delta), closing a four-bar loop |D(theta) - Q(psi)| = k, solved in closed form for theta with a second law of cosines. One short stroke therefore swings the bridge through about 170 degrees, with a strongly varying mechanical advantage.
2. Cylinder force by virtual work. With U the potential energy of the hanging masses, a cylinder holding the load needs
F = dU/dl = (dU/dtheta)(dtheta/dpsi)(dpsi/dl) , dU/dtheta = g Σ mi xi
so F is the weight moment about the pivot divided by an effective lever arm, F = M g r cos(theta) / h(theta). It is checked against a free-body solution of the coupler and crank. Pressure is p = F / A with A the bore (push) or the annulus (pull); the load flips from push to pull when the bridge passes the vertical and gravity starts to help. Flow Q = n A v and hydraulic power P = p Q = F v. The pump only supplies the positive work; negative work is burnt in the counterbalance valve as heat.
3. Tipping. The span hangs beyond the tail support line xr (outriggers) or the rear axle line when they are not used:
SF = Mrestoring / Moverturning = [Mv g (xr - xv) + ...] / [ Σ mi g (xi - xr) + Fwind z ]
SF = 1 is the critical extension; with a horizontal cantilever the balance is a quadratic in the bridge length. Wind pressure q = 0.5 rho v^2 acts on the projected area, which depends on the bridge angle (largest when it stands vertical). Under the pads the soil sees p = R / Apad; settlement is estimated with the elastic half-space formula of a rigid square plate, s = p B (1 - nu^2) I / E (I about 0.82). Above the allowable bearing pressure the pads would punch in and the numbers shown are only indicative.
4. The laid bridge. A simply supported beam with equivalent stiffness EI: Mmid = wL^2/8 + PL/4, stress sigma = M c / I, deflection 5wL^4/(384 EI) + PL^3/(48 EI). The crossing load includes a dynamic factor 1.25; the rating is a generic tonne figure (an illustrative class, not a real military load class). End bearing lengths must be at least 1 m; the largest and smallest gap follow from the bridge length and the stand-off.
Simplifications. Lumped masses on the bridge axis; no inertia or hydraulic dynamics (launch speed is slow); rigid truck chassis and pads; no tyre or suspension flexibility; no wave or soil dynamics; a two-half bridge treated as a straight beam when laid; the 2:1 hinge drive is idealised. Numbers are illustrative.