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October 1, 2014

October 1, 2014

Posted by **jof** on Friday, March 30, 2012 at 12:53pm.

y = S sin(360

r t - 90

Where S is the stroke length of the cylinder, r is the rpm of the crank shaft, and t is

the time in minutes. If the stroke length is 2 inches and the crank shaft is running at

1300 rpm, find the position of the piston at t = 53 seconds. (remember that rpm

means revolutions per minute)

- Math -
**Steve**, Friday, March 30, 2012 at 2:15pmwhat's the problem?

y(t) = S sin(360rt-90)

S = 2

r = 1300

t = 53/60

y(53/60) = 2sin(360*1300*53/60-90)

= 2sin(1148.3333*360 - 90)

now, we can toss out all multiples of 360, so we are left with

y = 2sin(240-90) = 2sin(150) = 2(.5) = 1.0

- Math - PS -
**Steve**, Friday, March 30, 2012 at 2:16pmoops.

y = 2sin(120-90) = 2sin(30) = 2(.5) = 1.0

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