Piston Speed Question/Quiz

[Kevin Shasteen]

I happended upon this formula while surfing the net and thought it would be interesting.

 

The formula is for determining Piston Speed. However, the answer derived is not in MPH but in G-force....as in "ft.per.sec^2".

 

So my question to you egg-heads out there is this; who can give the answer on how to convert the formula's Piston Speed in G-force to MPH?

 

Here's the formula:

 

GMax = ((N^2 x L)/2189)x(1+1/2A))

 

GMax = G-force, as in ft.per.sec^2

N = Crankshaft Speed, in RPMs

L = Stroke, in Inches

A = Connecting Rod to Stroke Ratio

 

Kevin,

(Yea,Still an Inliner)

[Dan Baldwin]

You can't convert an acceleration to a speed. Peak acceleration occurs at zero piston speed, and peak speed occurs at zero piston acceleration. In setups with different rod lengths, rpm for a given piston speed will result in differing maximum piston acceleration (greater rod length => reduced peak acceleration). Here are the equations of motion I came up with:

 

r = stroke/2

L = rod length

O = theta = crank angle (0 at TDC)

w = omega = crank rotational speed (radians/sec)

 

yp (piston position) = r*cos(O) + L*cos[(r/L)*sin(O)]

y'p (piston speed) = -r*w*sin(O) - r*w*sin[(r/L)*sin(O)]*cos(O)

y''p (piston acceleration) = r*w^2*{-cos(O) + sin[(r/L)*sin(O)]*sin(O) - (r/L)*cos^2(O)*cos[(r/L)*sin(O)]}

 

y''p simplifies to -r*w^2*[1 + (r/L)] at TDC, where peak acceleration occurs. I made a simple spreadsheet to find peak piston speed and where it occurs. Should work out to be the same as your GMax equation, but for a 3.1 liter stroker I get a peak piston acceleration of 96,000 ft/s^2 (almost 3000 gs!) at 7000 rpm, with a max piston speed of around 104 ft/s (71 mph) at around 75 deg. ATDC. GMax equation gives 131,750 ft/s^2 though.

 

Somebody lemme know if I screwed up anywhere. My brain hurts.

[Dan Baldwin]

OK, I found the problem. You list your variable "A" as connecting rod length to stroke ratio. It should the the inverse, stroke/conrod length.

[Kevin Shasteen]

Dan,

 

Thanks for trying to shed some light on the subject.

 

I also find it confusing at times. The sites and books I look at always quote the limit of piston accelerations to be in the neighborhood of 100,000-150,000 ft.per.sec^2

 

So I think the formula given was correct. My problem is that those same articles will always throw out some bit of info that pistons at peak accelerations will at times reach 80mph, which is as you put it-just before and after TDC: but then they dont say anything else on how we, the reader, can convert this acceleration to mph.

 

The only think I can think of is to run two spreadsheets side by side; one is for acceleration while the other is for speed. The spreadsheet for acceleration will generate its data based on your input of rpm, con.rod & stroke length-then transfer that data to the piston speed spreadsheet: which gives us feet per minute.

 

From the feet per minute we can calculate mph...yes/no(?): what-dya-think?

 

My brain is hurting also trying to figure this one out-someone....anyone; HELP US, we have both fallen & neither of us can get up!

 

Kevin,

(Yea,Still an Inliner)

[jeffp]

Why do you want to know the MPH of the piston. The reality of it is that the piston does not move in that fashion and really has no relivance. The feet traveled per second is the important one to know as it will give you a good idea of the

type piston you need and what type of lube you need and that information is relivant to the building of a performance engine. I would suggest you getting a book by John Lawlor, engine math book. Very informative, and will answer any engine question you may have. You can get it on Amazon.com I think it is about 20 bucks, worth every penny.

[Michael]

Dan,

 

Some time ago, I derived an alternative and somewhat convoluted formulation of the piston motion equations, which give the answers identical to yours for the speed and velocity. The acceleration answer for some reason differs by a constant multiple of exactly 6, which is almost certainly an error on my part (when multiplied by 6 and plotted vs. crank angle, our answers coincide). So my point is, your formulas make sense, and you did not make an error in typing.

 

For typical rod/stroke ratios, piston peak velocity indeed occurs at around 75 deg ATDC (and again at 285 deg); for example, for my engine (stroke of 4.00", rod of 6.135"). The MEAN piston velocity is 4000 ft/min, but the peak is around 6600 ft/m. Another typical result. For that combo, the piston peak acceleration comes out to ~87,000 ft/s^2. Peak acceleration at TDC is of course maximum. By the way, local maxima of piston acceleration occur at 135 deg and 225 deg ATDC (~47,400 ft/s^2). Another local extremum is at BDC. The acceleration curve has inflection points at around 60, 155, 205 and 300 deg ATDC.

 

Playing with these curves illustrates an interesting point about the relevance (or rather, irrelevance) of rod/stroke ratio. Doubling the Chevy 454's stock 6.135" connecting rod length does reduce the piston peak acceleration, but only by about 12%!

 

Also, for the fans of dynamic compression ratio - exactly the same formulas are used to compute DCR; get the intake valve closing angle, go to the corresponding point in the piston position curve, and compute the remaining swept volume.

[Michael]

oops, first paragraph above - I meant displacement and velocity. late at night...

[Pop N Wood]

Speed = acceleration multiplied by time.

 

Thus if you were accelerating at say 32 feet/sec/sec, then after 10 seconds you would be going 320 feet per second (or 218 MPH).

 

But, in the case of a piston, the linear acceleration is constantly changing, thus you need to integrate the acceleration as a function of time.

 

If would be a straight forward trigonometric calculation to produces a closed form solution of the piston speed and acceleration as a function of RPM., rod length, stroke and crankshaft degrees BTDC.

 

Or simply give Grumpy a call and let him post a link (or several) with all the details.

[Pop N Wood]

BTW, to convert acceleration into G's simply divide by 1 G (32 Ft/sec/sec)!!

[Dan Baldwin]

Kevin, it's not your formula for acceleration that's wrong, it's your definition of variable "A". Should be stroke/rodlength. Peak piston acceleration is AT TDC. 2nd time, acceleration and speed are different, you can't convert one to the other. Peak SPEED occurs at ZERO ACCELERATION, and vice versa. mph, ft/s, are in units of SPEED (distance/time). ft/sec^2 and g are in units of ACCELERATION (distance/time^2). There is and can be no conversion.

 

Jim, check post #2 for the equations of motion for the piston.

 

Michael, I'll go through my calcs again, but I think they're alright. Good points on the local extrema. Most people would think BDC should be the location of maximum upward piston acceleration, but it's not, that would be at 145 ATDC and 215 for my engine (apparently 135 and 225 for yours). Acceleration at BDC is actually a local minimum, which may come as no surprise to anyone who's observed piston motion as an engine is turned over. The conrod "standing up" increases piston acceleration at TDC, but reduces it at BDC, to the extent that for my engine the maximum upward acceleration near BDC is only 55% of the maximum downward acceleration at TDC.

 

Interesting stuff, no?

[Dan Baldwin]

Originally posted by Dan Baldwin:

mph, ft/s, are in units of SPEED (distance/time). ft/sec^2 and g are in units of ACCELERATION (distance/time^2). There is and can be no conversion.

Maybe I shouldn't have worded it exactly like that. What I meant is that you can't take a value in units of length/time (speed) and "convert" it into a length/time^2 (acceleration) value, just like you can't convert linear feet into acres (the question "How many feet of land is your house on" has no meaning). You CAN, however, derive equations for speed and acceleration if you have an equation for position, which isn't too hard to come up with for piston motion. You do have to have to perform derivative calculus, though. Take the derivative of the equation for position with respect to time, and you get the equation for speed. Then take the derivative of the equation for speed to get the equation for acceleration.

[Kevin Shasteen]

Originally posted by Dan Baldwin:

Interesting stuff, no?

Thanks everyone for your input; and Yes Dan I find it very interesting.

 

In fact I have been searching, when time allows, for some info that allows me to create a spreadsheet that measures "Piston Motion"....my new term-thanks for introducing the new vernacular as it appears I've been dancing all around it.

 

Last year I taught myself basic Trig in order to perform the Crank Angle Charts; I have a spreadsheet for that & can do the Dynamic Comp.Ratio in my sleep. Are you telling me that I can not perform the "Piston Motion" charts w/out leaning [CALCULUS]? I have seen a few charts mapping the Piston Motion and have always desired to create my own Piston Motion Charts as it was the natural follow up to the Crank Angle Charts.

 

Dan,

 

In the formulas you gave in your 2nd Post you mention "Omega" as Crank Angle Speed and that is referenced as Radians/sec...(Radians always confuse me); I have a few questions:

 

1) Is this Radians/sec synonimous for "RPM"?

2) What unit of measurement is the Piston Motion calculated in: a) ft.per.sec, ft.per.sec^2, or c)ft.per.minute?

 

Thanks for the assistance.

 

Kevin,

(Yea,Still an Inliner)

[Pop N Wood]

There are 2 pi radians in 360 degrees.

 

It is really as simple as that.

 

Just like feet and meters. They measure the same thing, just different units.

 

One "revolution" is 360 degrees (one full turn).

 

so 1 "Revolution per minute" = 360 degrees per 60 seconds = 6 deg per second = 0.1047 radians per second

[Dan Baldwin]

What Jim said.

 

The equations work with whatever length units you want to use, you just have to be CONSISTENT. If you want feet/second, input stroke and rod length in FEET.

 

If anybody wants my spreadsheet, I could email it to ya.

 

Dan

[Kevin Shasteen]

another duplicate entry-please stop the madness here

[Kevin Shasteen]

Originally posted by Kevin Shasteen:

quote:

---

Originally posted by Dan Baldwin:

.., If anybody wants my spreadsheet, I could email it to ya.., Dan

Dan,

 

I'ld be interested in seeing it-so email away please.

 

What I would like to emulate is a line graph that maps the piston motion.

 

If you have Forbes Aird's book entitled "Automtive Math Handbook" he has this line graph on page#51. I realize this is just another form of visulalizing Piston Motion; but this is the graph I'ld like to duplicate.

 

He also has a formula for piston acceleration. It appears to be similar to one of your previous formulas, here it is-what'dya think?:

 

ACCp = {(RPM^2*S/2189)*[(cos(O) + r/l*cos2(O)]}

 

ACCp = Piston Acceleration in ft/sec^2

S = Stroke in inches

r = Crank Offset

l = Rod length in inches; rod center to center

cos(O) = cosine of the crank angle

cos2(O) =cosine of twice the crank angle

 

His formula seems to be somewhat simpler to me I'm not a real egg-head....only an egg-head wanna'be

an egg-head imposter if you will tripping my way thru the math!

 

Thanks Jim/Dan for enlightening me-any other suggestions that may help-feel free to keep'em coming.

 

Kevin,

(Yea,Still an Inliner)

---

[Kevin Shasteen]

oops; accidently duplicated the post-ignore this but not the previous post (confused? )

[Dan Baldwin]

That equation gives the correct piston accelerations at TDC and BDC, but it is incorrect elsewhere. It misses the maximum upward piston acceleration that occurs before and after BDC. I'd bet money (if not my Z) on my equations. I'll send a plot of that equation along with mine in the spreadsheet I'm sending you.

[Dan Baldwin]

Whoops, I blew it. His equation DOES match mine. I thought the cos2(O) in his eqn was cosine of theta SQUARED instead of cosine of twice theta. My bad. It certainly is a much tidier equation, guess I've got some cleaning up to do.

[Kevin Shasteen]

Okay, found the solutioin to convert ft.per.sec & ft.per.minute to MPH.

 

1) Ft.per.sec. x .6818182 = MPH

2) Ft.per.minute / 88 = MPH

 

This doesnt address acceleration as someone said earlier in the thread: acceleration is measured in g-force; which is written in Ft/sec^2.

 

JeffP,

 

FWIW: I should've looked in the back of my J.Lawler book to begin with-as that is where I found the conversions...oh well-the thread was thought provoking none the less.

 

Kevin,

(Yea,Still an Inliner)