1 / 9

Loop the Loop a “Fifth Gear” stunt special

Loop the Loop a “Fifth Gear” stunt special. Hugh Hunt Cambridge University Engineering Department. At C (top of the loop) centrifugal force exactly balances weight, so mg = mV 2 2 / R or V 2 = gR. From B to C , conserve KE+PE

rupert
Download Presentation

Loop the Loop a “Fifth Gear” stunt special

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Loop the Loopa “Fifth Gear” stunt special Hugh Hunt Cambridge University Engineering Department

  2. At C (top of the loop) centrifugal force exactly balances weight, so mg= mV22 / R or V2 = gR From B to C, conserve KE+PE ½mV12= ½mV22 + 2mgR so V12 = V22 + 4gR . or V1 = 5gR . Simple physics From A to B, conserve KE+PE mgh= ½mV12 so h = V12/2g = 5R/2 G-forces are V2/R : At B : 5g + g = 6g At C : g – g = 0

  3. bad assumptions: • particle • no friction • rigid suspension • perfect tyres • short wheelbase • G-force not important • ignore transients • engine not powering • planar – no steering • rigid structure Fifth-gear Stunt Special: R = 6m  V1 = 17.2 ms-1 (39 mph) V2 = 7.7 ms-1 (17 mph) and G-force at B = 6g

  4. 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

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

  6. Transient What happens to the suspension and tyres when the weight of the car goes from 1000kg to 6000kg ? Here is a notional graph of the motion of a simple vehicle subject suddenly to a change in g to 6g . 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 30psi to …. ???

  7. Engine power and friction Toyota Aygo: 67bhp, say 40kW PE increase = 110 kJ If time to complete loop = 4.5 seconds then engine provides 90kJ on the way up Friction:  = 1 , mean “G-force” = 3g, energy lost to friction = 1000 kJ Conclude: Most important to avoid “grinding out” – friction (even as low as  = 0.1) may cause the stunt to fail.

More Related