NASCAR Technical Trivia #2

[MolaKule]

A 358 cubic inch NASCAR engine's crankshaft is turning at 9,500 rpm.

What is the linear piston speed at the center of stroke in inches/second?

Bore is 4.185 inches in diameter and has a stroke of 3.25 inches.

 

[duaneb9729]

guesses its going as fast as the rest of the car lol

 

[Ken42]

Best guess 48.7 inches per second.
or if you want to be tricky 85.66 fps and 1028 ips
about 58 mph and starting and stoping 316 times per second.
or something like that.

Ken

 

[Shannow]

What's the rod length ?

 

[MolaKule]

We have to assume the piston traverses the full stroke length.

 

[Shannow]

Yep, but the angularity of the rod (rod/stroke ratio) determines the velocity at the point you are requesting (another complicator is wrist pin offset from bore centreline), unless you have infinite rod length/scotch yoke design, where the mid stroke velocity is simply the wr of the big end.

 

[MolaKule]

Right and if we were designing a new engine or blueprinting from scratch we would concern ourselves with those variables.

Assume a perfect engine with no offset wrist pin, a conrod of sufficient length to connect piston to crankshaft etc, realizing the piston speed at TDC and at BDC is zero but maximum at midstroke, and making simple harmonic motion (SHM). We're trying to make this as uncomplicated as possible.

1. First convert Rev/min to Rev/sec

RPM/60 = RPS;

2. Realize one half crank revolution of 180 degrees (say from TDC to BDC) causes the piston to traverse 3.25". So in one complete 360 degree revolution the piston traverses a total length of 6.5" or 6.5"/Rev.

 

[MolaKule]

Quote:
or if you want to be tricky 85.66 fps and 1028 ips


KEN42 had the correct answer.

Assuming SImple Harmonic Motion (SHM) and Rounded UP:
RPM/60 = RPS;
piston speed in inches/s = [(RPM/60) X (2 X Stroke in inches)/Rev] = 1030
piston speed in feet/s = [(RPM/60) X (2 X Stroke in inches)/Rev] / 12 = 86

 

[Steve S]

Originally Posted By: MolaKule
We have to assume the piston traverses the full stroke length.
Does the piston have any choice?

 

[Steve S]

Originally Posted By: Shannow
Yep, but the angularity of the rod (rod/stroke ratio) determines the velocity at the point you are requesting (another complicator is wrist pin offset from bore centreline), unless you have infinite rod length/scotch yoke design, where the mid stroke velocity is simply the wr of the big end.
I would think the rod angle should effect the acceleration from TDC and BDC and thrust on the cyl walls over the piston speed which I am guessing would be the average?

 

[XS650]

Originally Posted By: MolaKule
. We're trying to make this as uncomplicated as possible.




Looks like you made it more uncomplicated than possible.

85.8 ft/s is average piston speed not mid stroke speed. And, you can't find mid stroke speed without knowing the rod length. Even assuming simple harmonic motion, which is still not all that close, mid stroke speed would be considerably higher. About 135 ft/s.

As Einstein said, "make everything as simple as possible, but no simpler"

 

[gtx510]

check out this piston speed formula
http://web.archive.org/web/2007020904281...erivation-1.jpg
http://web.archive.org/web/2007082302340...ratio/index.htm

http://www.devtk.com/ninja250/ninja_250_piston_speed.pdf
http://www.wfu.edu/~rollins/piston/ more complete write-up

you guys figured out the mean piston speed? which would be the piston speed at the center of the stroke, right?
http://www.ucalgary.ca/~csimpson/Tech/Calculations.html
piston speed (in ft/sec) = 0.166 x stroke (in inches) x rpm

 

[XS650]

Thats one of the best treatments of the subject I have seen online.

It's worth reading if you have any interest in such things. It would also be good idea to download it if you like it, I noticed that it's on an archive site instead of the original site, so it may go away at any time.