Motion of a piston be considered simple harmonic

[monty37]

can the motion of a piston be considered simple harmonic,being periodic in nature?in*case of a reciprocating engine.

 

[fleem]

It isn't sinusoidal (unless the connecting rod is infinitely long) so it isn't simple harmonic motion.

 

[Ranger Mike]

the piston has a finite dwell time ( in crankshaft degrees) at top dead center as verified with a degree wheel when properly installing a cam shaft..in other words , the piston " stops moving" for a short period of time ( number of crankshaft degrees) while at TDC

 

[xxChrisxx]

The dwell time also changes depending on conrod length/crankshaft throw ratio.

It depends on context, if it were a purely theoretical question then you could assume sinusodal motion. If you were doing analysis on a real engine i'd say the error would probably be quite large.

 

[Ranger Mike]

Excellent point...i was goping to mention it but didn't want to add another wrinkle. This was one of the old tricks used in the 1960s when we were restricted by cubic inches and " stock components"...intake, ignition etc...a longer con rod / piston set up would yeild the same cubic inches but would give way more HP than the stock con rod/piston combo

thanks

 

[monty37]

but simple harmonic motion does not necessarily mean sinusoidal,the piston for every stroke moves up and down ,this is a continuous process,why should it not be SHM? please explain the infinitely long connecting rod as well.

 

[HallsofIvy]

Uhh, yes, simple harmonic motion does necessarily mean sinusoidal. Periodic motion is not necessarily simple harmonic motion.

 

[xxChrisxx]

but simple harmonic motion does not necessarily mean sinusoidal,the piston for every stroke moves up and down ,this is a continuous process,why should it not be SHM? please explain the infinitely long connecting rod as well.

It's basically a crank-slider mechanism. The crank can rotate at a constant velocity, but a degree of rotation doesn't = a linear unit of movement of the slider.

For example, a degree of rotation at 90 aTDC will move the piston further than a degree coming up to BDC. At TDC and BDC the crank rotation is rotating the connecting rod and not moving the piston, this is the 'dwell' time.

Longer rod/throw ratios reduce this effect because they are nearly in line as you get to TDC. An infinitely long rod will produce sinusodal motion.

Also you'd probably expect the piston to be halfway through its stroke at 90 degrees, it isn't (for the same reason). I really wish my computer wouldn't have died as I had a few excel graphs showing exactly this.

 

[gmax137]

the piston has a finite dwell time ( in crankshaft degrees) at top dead center as verified with a degree wheel when properly installing a cam shaft..in other words , the piston " stops moving" for a short period of time ( number of crankshaft degrees) while at TDC

Just to clarify, the piston speed is only zero instantaneously at 0 and 180 degrees crank angle. It is pretty straightforward to develop an equation for piston position vs crank angle, and differentiating the equation gives the instantaneous piston speed. Here is the result (normalized to the mean speed (Sp-bar):

$$\frac{S_{p}}{\overline{S_{p}}}=\frac{\pi}{2} \ sin \theta\ [1 + \frac{cos \theta}{(R^2-sin^2 \theta)^{0.5}}]$$

Theta is the crank angle and R is the ratio of rod length to crank radius. If you put this into a spreadsheet and play with the R value you can see some interesting things. The 'dwell' (when the piston speed doesn't change much) really occurs at BDC (when theta is 180 degrees). This makes sense if you visualize the crank going around BDC, the small-end of the rod doesn't move much - and if you could have R=1, the small end wouldn't move at all (of course this is physically not possible, but it shows what happens in the extreme). Also, see that the effect of R at TDC is pretty minor. Finally, for large R the equation tends to

$$\frac{S_{p}}{\overline{S_{p}}}=\frac{\pi}{2} \ sin \theta$$

as someone noted above.

 

[monty37]

from where does the piston get energy to go ahead with its first stroke and
is the linear motion of piston in phase with rotatory motion of crankshaft?

 

[Ranger Mike]

Fuel/air mixture is the simple answer..the crankshaft is turned over..i.e. rotated via electric starter motor, or push start ( manual transmissions equip cars only). when the engine is spinning, it acts as a vacuum pump sucking in the fuel/air mix. the spark plug lights off the mix and you have a running engine.
the engine 4,6,8, 10, 12 cylinder is designed to ignite certain cylinders in specific sequence to make the transition as smooth as possible...12 slugs run smother than 4 since the pistons light off every 30 degrees vs. 90 degrees on the 4 banger.
the ignition is " phased" or timed to fire at a few degrees before Top Dead Center (TDC) ...i.e. the timing is advanced due to ignition fire time, flame travel, realization that the piston is revving pretty quick...don't forget the typical stock engine idles at around 750 RPM ( ignition lights 11.6 times per second) so a degree of time is required for the mechanical piston/rod assembly to hit TDC before the sparkplug fires.