Question

Please help, University exam in 3 days!! If someone can figure this out I will be eternally grateful and also pretty amazed! I will become a fan and give a medal!! (Topic: Gears, torque, rpm, etc...)

Answer 1

A car has an engine that develops constant torque of 150 Nm at speeds from 0 to 1800 rpm. At higher speeds, the torque decreases linearly from 150 Nm at 1800 rpm to 0 Nm at 2800 rpm. The maximum speed of the engine is 2800 rpm. The car’s roadwheels are 585 mm in diameter. You are one of a team whose task is to design a gearbox which provides gear ratios (ratio of engine speed to roadwheel speed) of 2.05 and 1.62. All gears in the gearbox should have the same circular pitch. Between the gearbox and the roadwheels there is a differential gear with a gear ratio of exactly 1.

(a) Sketch graphs showing torque and power as functions of speed for this engine. Calculate the engine’s maximum power, and the speed at which it occurs. (The graphs do not have to be accurate, but important points and values should be marked and labelled.)

(b) Explain in non-technical language how a multi-speed manual gearbox works (as in a typical car). Illustrate your answer with at least one diagram of a two-speed gearbox.

(c) Choose numbers of teeth for each gear that will provide the required gear ratios. Give an approximate solution if an exact one can’t be found.

Answer 2

@Fifciol if you can do this I would very much appreciate the help! By the way, if you don't want to bother with part (a) and (b) that's no problem. Part (c) is the priority. Thanks!

Answer 3

@Fifciol Any luck??

Answer 4

a)
\[P=\omega \tau\]\[\omega=\frac{ n2\pi }{ 60 }=\frac{ \pi n }{30 }\]
if n is less or equal to 1800 rpm power grows lineary with n :
\[P=5\pi n \]
if n is greater than 1800 torque is not constant but it decreases lineary with n
\[\tau = 420-\frac{ 15 }{ 100 }n\] (the equation of line from the first plot)
so
\[P=\pi n(14-\frac{ n }{ 200 })\]
plot: