Loop the Loop a Fifth Gear stunt special

Loop the Loop a “Fifth Gear” stunt special Hugh Hunt Cambridge University Engineering Department

Simple physics From A to B, conserve KE+PE mgh = ½m. V 12 so h = V 12/2 g = 5 R/2 At C (top of the loop) centrifugal force exactly balances weight, so mg = m. V 22 / R or V 2 = g. R From B to C, conserve KE+PE ½m. V 12 = ½m. V 22 + 2 mg. R 2/R : G-forces are V so V 12 = V 22 + 4 g. R. At B : 5 g + g = 6 g or V 1 = 5 g. R. At C : g – g = 0

bad assumptions: Fifth-gear Stunt Special: 1. particle 2. no friction R = 6 m V 1 = 17. 2 ms-1 (39 mph) 3. rigid suspension V 2 = 7. 7 ms-1 (17 mph) 4. perfect tyres 5. short wheelbase and G-force at B = 6 g 6. G-force not important 7. ignore transients 8. engine not powering 9. planar – no steering 10. rigid structure

Optimum shape for the loop (not a circle!) It all depends on the criteria. For these rollercoasters, height is not an issue. Fifth-gear could have constructed such a shape, but it would somehow have been less impressive… CONSTRAINT: entry and exit horizontal and coincident

Steering The line drawn in here is correct – no steering would be required The line actually painted has a kink in it – steering is needed !

Transient What happens to the suspension and tyres when the weight of the car goes from 1000 kg to 6000 kg ? Here is a notional graph of the motion of a simple vehicle subject suddenly to a change in g to 6 g. The tyre and suspension behave non-linearly. 1/6 ? The suspension bottoms out, the vehicle grinds out, the tyre blows out – all pretty undesirable. solutions: stiffen suspension increase tyre pressure from 30 psi to …. ? ? ?

Engine power and friction Toyota Aygo: 67 bhp, say 40 k. W PE increase = 110 k. J If time to complete loop = 4. 5 seconds then engine provides 90 k. J on the way up Friction: = 1 , mean “G-force” = 3 g, energy lost to friction = 1000 k. J Conclude: Most important to avoid “grinding out” – friction (even as low as = 0. 1) may cause the stunt to fail.