Assuming the centerline of the cylinder intersects the centerline of the crank,...
If the stroke is 3.48, then so is the diameter of the big end circle. The circumference would be 3.48 x pi = ~ 10.933 inches. At 1000 rpm, that would be 10933 in/min ÷ 60 = 182.22 in/s, which would also be the maximum piston speed at 90° crank position. At 90°, the piston would be exactly halfway through its travel.
Thanks to all who participated.
MrMoody and Astro14 (in absentia
) had the correct answers.
This was another problem in the conversion of radial velocity to linear velocity as in,
Vlinear = Omega*radius as two hints were given.
To simplify the problem we assumed no effect of crank angle/rod/piston-pin relationships, which meant the piston is seeing only reciprocating, linear travel. Simplification also meant the connecting rod was very long so crank angles did not enter into the problem.
At 0 degrees Top Dead Center and at 180 degrees Bottom Dead Center, and with the crankshaft rotating clockwise, the piston motion is virtually zero.
At 90 degrees and at 270 degrees, the piston is halfway down or up the cylinder, respectively, and it is at these positions in which the piston velocity is maximum, or 182.22 inches per second.
For the next QOTD, we will include crank angle.