A gearing question

[BADGER.BRAD]

O.K I'm going to be shot for this but I am trying to work out an answer to a question about my other hobby/mode of transport ,motorcycles.I thought that I am more likeley to get an answer to this on cyclechat than the motorcycle forums so here goes. If my motorcycle does 35 mph at 4000rpm with a front sprocket of 14 teeth and a rear of 47 teeth what speed would it do for the same rpm with a rear sprocket of 46 teeth ? I have been trying to work this out ( after a few beers ,oh dear!) and just cannot get my head round it.Please tell me how you worked this out so that I can remember to forget how it was done.

Any help much appreciated.

 

[MajorMantra]

((14/46)/(14/47)-1)*100 = 2.2% faster

35*1.022 = 35.77 mph

Matthew

 

[GrasB]

rpm is irrelevant, the teeth on the 'box output shaft is also irrelevant. Basically you divide the the number of teeth you have now to the number of teeth you wish to move to, this tells you how much further you'll turn the rear wheel per turn of the gearbox output drive shaft. In this case 1.02174 times, you then multiply the speed by that to get 35.76mph

 

[BADGER.BRAD]

Many thanks all that now makes sense !!