swmn

Can anyone explain how to calculate instant piston speed, without calculus?

I failed calculus twice in college, but trig and algebra I can handle.

It seems to me piston speed, or more correctly piston velocity is going to vary continuously and will graph out as a sinusoidal wave form. At TDC and BDC piston velocity will be zero, with peak instantaneous velocity at 90 crankshaft degrees between the DCs.

So there will be a trig function in the expression, I am OK with that. Engine RPM will need to be included. It seems to me rod:stroke ratio needs to be accounted for, and I bet there is at least one exponent coming real soon.

Somehow one of my back issues of Hot Rod Mag comes up with piston speeds for a 3.48" stroke and 3.75" stroke up over 3,000 feet per second for both at 6500 rpm. I figure these have to be instantaneous velocities, as 3.75" per stroke * 2 strokes/revolution * 6500rpm * 1 foot/ 12 inches * 1 minute / 60 seconds = 67.71 feet per second average piston speed.

Maybe it was a misprint? Maybe I am a moron?

Thanks,
Scott

PS: For my project I have a red line of 5500 rpm and I am planning to use fully machined forged good name connecting rods. Can I probably ignore con rod flex when calculating piston speed, or do I need to figure that out too?

chirp_fourth

There's a few piston speed calculators out there on the web, and a few engine simulation programs that are fairly cheap. In my IC Engines class we calculated it (actually had to have a dynamic model) and it was fairly simple.

Hint #1: Piston velocity @ TDC/BDC = 0

Hint #2: Longer stroke = higher piston velocity

Hint #3: Rod flex is negligible

Hope that's what youre looking for.

white2001s10

iTrader: (4)

I've had a rough day, but I think it's like this.
The instant speed of the piston at the 90* point will equal the speed of the rod journal which can be determined by the stroke of the crank.

(((3.48" * pi) * 6500 RPM) / 60 sec) /12" = 98.6 feet/sec

Finding the other points in the stroke is too much work for me. I would say that the 45* points will be roughly half of the 90* figure and use a rough %angle against these values to find the rest and call it a day.
Of course these number will not be exactly right due to different rod lengths.
If I had a real need to get the correct numbers I'd get the simulator.


Edit: upon reflection I shouldn't have even posted... hard time focusing.
My thinking above only works if the cylinder is perpendicular to the rod journal at the 90* point, which it isn't in a real engine. My way would get you in the ballpark, but like I said is not precise due to different rod & stroke lengths.
If you were simply calculating inertia in the moving parts then my way is likely good enough, but if you're working on timing valve or exhaust events then you'd need to figure it the correct way.

Old SStroker

It's just trig, not calc, and you are not a moron.

Your calculation for "Mean (average) piston speed" was correct. It's a function of stroke and rpm only. BMW quoted mean and max. piston speeds for it's 19,200 rpm 2003 F1 engne as 25 M/sec (mean) and 40 M/sec (max). That's 82 ft/sec (mean) and 131 ft/sec(max). A 9800 rpm Nextel Cup engine has a mean PS about 88-89 ft/sec. A ProStock engine is similar.

3000 ft/second was a misprint. A 3.48 stroke has a 6500 rpm mean PS of 3770 ft/min, and a 3.75 stroke is 4063 ft/min. I've found editors/proof readers for mag articles are notoriously bad. I'm sure some authors like Jim McFarland pull their hair out when they see their writing butchered with bad numbers.

Maximum PS doesn't occur at 90 degrees unless the rod length is zero. To calculate max piston speed you need to know rpm, stroke, and rod length, but within the practical range of rods you could use, the max PS doesn't change a whole lot. Are you really looking for the crank angle for max PS? That takes some trig, but with practical rod/stroke ratios is doesn't vary much from 75 degrees ATDC.

There's almost no real good reason to know max. PS for your engine design. Max gs would be more useful, but if yours is a V8 gasoline auto engine, 5500 revs shouldn't be much of a problem, even for a long running engine. If you use 65-70 ft/sec. for mean PS, which is fairly conservative (the LS7 Vette engine has a mean PS over 77 ft/sec at max revs), your 5500 rpm redline would allow strokes from 4.25 to 4.58 in. If you go to 77 ft/sec you can go to a 5 inch stroke. Tell us more about this engine and what you are trying to achieve..

Don't get hung up on max PS or rod flex/stretch. The rod stretches the most where the piston speed is the least: TDC on the exhaust stroke. That's where the highest g loads are. F=MA applies.

P Mack

iTrader: (6)

I bet you they meant to print 3000 feet per minute, not feet per second. 3000 feet per second would be supersonic.

Old SStroker

Nearly triple sonic. M2.7 or so. About the muzzle velocity of a high powered rifle or a 20 mm cannon.

swmn

Actually I was looking at exhaust flow.

Briefly my plan is to drop an all motor 383 into my 2004 GTO when I get to 100k on the original motor, roughly summertime 2007. I am looking to maximize torque from 2k to 5k rpm. The application is strictly street/interstate, no track. Life expectancy shall be 100k miles. I generally cruise between 2-3k rpm, and play from 3k up to 5k. I am not planning any major changes to gearing or weight distribution, thus Al 5.7 block stroked, rather than a heavier LQ9 block, and not a Gen IV block with all those electronics updates.

Not to hijack my own thread...

I was looking at primary length for 4-2-1 headers, in pencil, at 3500 rpm, right in the middle of my desired power band. Even with ten inch primaries (including exhaust port length) at 1 5/8" OD I wasn't getting my first suction wave back to the exhaust valve seat in time to do any good at 3500 rpm. I was using 300fps for the leading edge of the exhaust gas slug and 1400 fps for the pressure and suction waves.

Pairing my primaries on 1-5, 3-7, 2-4 and 6-8 I had 270 crank degrees between exhaust pulses for each of the four secondaries. What with the firing order using up 720 degrees, the 270 separation is 450 degrees coming around the other way so that was looking like a dry hole.

What I am going to do with a handle on piston velocity is -algebraically- relate exhaust valve lift and piston speed on the exhaust stroke to the gas slug to come up with a different estimate of exhaust slug travel. And look for ways to suction cylinders with the reflected wave from some other cylinder.

At the end of the day I expect my cam shaft choice will be partly dictated by the 4-2-1s I can build to fit under my hood; but I'll need a grasp on what ideal would have been to understand the compromises I'll have to make.

Thanks,
Scott

PS: I wasn't looking for _max_, I want to know how to figure instantaneous piston velocity for any and every degree of crank rotation.

chirp_fourth

http://me.queensu.ca/courses/MECH435...erformance.ppt

Go to slide 3. Cant be any more straightforward than that.

Old SStroker

Doesn't a 6L LS2 fit your requirements? Mild head/cam work would get you there a lot cheaper than stroking an LS1, IMO.

If you are cruising at 3000 in 6th gear you cruise pretty fast! If it's not 6th, why not with a 3k-5k rpm engine?

Are you looking at part-throttle or full throttle for your header/cam/engine design?

Exhaust tuning works a lot differently from intake tuning. I believe you are overthinking it and perhaps not headed up the right road.

Placement of torque peak with exhaust is more a function of primary pipe cross section area (diameter) than length. Larger diameter moves torque peak higher, smaller moves it lower. Length can help increase the amount of torque peak, but not move it much. There are empirical formulas for this as well as some good software like "Pipemax".

If you are using the 1-8-7-3-6-5-4-2 firing order, this pairing gives 180* pulsing: 1-7, 3-5, 4-6, 2-8. I'm not sure I'd spend tons of $ getting a custom 4>2>1 for a 3k-5k engine. There's just not that much to be gained.

Have you read/studied "Scientific Design of Intake and Exhaust Systems" by Smith? This should be your first reference. Second might be a current magazing Engine Masters Racing. Jim McFarland has a few good articles which might help you.

It's always a challenge to reinvent the wheel, but others have been there, done that. My advice is to learn from them.

FWIW, if you want to go fast/quick, limiting yourself to 5000 rpm isn't going to do it.

My highly-opinionated $.02

Fireball

iTrader: (21)

technically it is calc. Trig gets you the piston position as a function of angle. knowing crank speed you get piston position as a function of time. Take the first derivative to get the instantaneous velocity.

Old SStroker

OOPS! Yeah, for velocity, acceleration, jerk, etc. you need the calc. That will probably scare swmn away.

Even though I remember how to do it, I cheat and use software to get the answers.

chirp_fourth

You don't NEED calculus to figure it out, you just need someone to solve all of the equations of motion (doing the calculus for you) and give you the simplified equations (algebraic/trig)... Sorta like my post above that gave a linky-link to the simplified equations. I think those are the only ones he needs.

Fireball

iTrader: (21)

if you do a derivative of a trig function, all you get is another trig function...so yes once the calculus is done, its just a trig function. You just have to use calc to derive it, not to necessarily use it

swmn

Thanks for slide 3 chirp forth. I'll have to chew on it a while to understand it, btu it will give me something to do.

Yes, OldSS a 6.0L LS2 will get me there, but I'll need, minimum, a new PCM that speaks CAN and a new PIM that interprets CAN to UART. If I really wanted (no offense) an 05 goat I would have waited to buy one. I am watching the over installed price of switching ls1 to ls2, but those GTO specific parts are kinda pricey.

That one aside, I generally cruise in 6th between 2000 and 2500 rpm, suitable rapid velocities on the interstate, roughly 75, 85 (@2250) and 95 axehandles per bushel if you know what I mean.

I did come up with the 1-7, 3-5 pairing on the driver's side for 180 degree exhaust pulse separation, but with primaries already quite short spanning the distance from 1-7 and clearing the steering column, and leaving space to fit the 3-5 pair inside was looking like a lot, plus from 180° (testing alt 0176) it is still 540 degrees back around the other way to the other pulse, so 270/450 it 'easier' to plumb.

I think I can derive it with trig, but it will take me a few days. I started by defining the three apices of a triangle as C for crankshaft centerline, R for centerpoint of big end of connecting rod, and P for centerpoint of little end conencting rod.

Theta xx"Θ" is degrees crankshaft rotation, but what got me addled last night was thinking about phi "φ" as the angle defined by CRP. We can go ahead and name angle RPC alpha "α".

What I will need to do is calculate the length of line CP for theta whatever, but there will be more than one correct answer for segment CP when I am working from theta, segment CR and segment RP. It is that old Angle, Side, Side proof your teacher wouldn't let you use in 8th grade.

But phi...

P Mack

iTrader: (6)

Yeah I only put supersonic because the speed of sound of the charge in the cylinder changes as it's compressed.

Old SStroker

Sorry. I took you literally. It was the piston that was allegedly going 3000 fps, which it, of course, was not. Very few mechanical parts of any automotive engine travel faster than M 1 (~ 1100 fps in air).

So what's the speed of sound of "knock" thru a cast iron block?

A. 1100 ft/sec
B. 3000 ft/sec
C. 17000 ft/sec
D. 51000 ft/sec
E. 186,000 miles/sec

P Mack

iTrader: (6)

Well i know it's not the speed of light

gun5l1ng3r

I would have to say 51000 ft/sec, Aluminum is a very dense material compared to air.

swmn

Sorry for the interuption. When the anesthesiologist says "I need you right now" the RN may, if feeling very confident, type an ellipsis and click post before bolting from the chair. The patient is fine and later I got to eat my dinner in three minutes while riding on an elevator.

So piston speed with trig. The three points are C, crankshaft centerline, R, centerpoint of the big end of the connecting rod and P, centerpoint of the little end of the connecting rod. Then theta is crankshaft rotation, phi is the angle CRP and alpha, is next so that will be the angle RPC, though I haven't needed it yet.

For my purposes for now the length of segment CR is 2.000", hereinafter called "2", the length of segment RP is 6.000", hereinafter refered to as "6".

With the piston at TDC, theta is zero degrees, alpha would be zero degrees and phi is 180 degrees. I am not going to re-invoke alpha until I need it for some reason. The length of segment CP, the length I want to solve for is clearly, in this instance, 8 inches.

At BDC theta is 180, phi is zero and the length of segment CP is 4 inches.

Since we are talking about forged steel parts torqued together and housed in a modern aluminum casting I can say within this system for each and every theta there is one and only one corresponding measurement for phi, and one and only one length for segment CP.

But look at when phi is 90 degrees. We know CR is 2, we know RP is 6, so CP has to be 6.32".

But while theta ranges from zero to 180, phi ranges from 180 to zero. How can it be that phi is already 90 degrees while CP has shrunk only from 8 to 6.32? Clearly theta can not also be 90 degrees when phi is already 90 degrees.

What I am going to do right now is go have a cold beer. then I am going to read over the slide three chirp fourth (cool handle) pointed me to this morning.

Then when I get a chance I am headed to the university book store to look for an advanced trig text book. It seems to me I ought to be able to solve for the length of CP when I know CR, RP and phi.

Thanks for helping me out.

P Mack

iTrader: (6)

I think to solve for the velocity of P involves taking the derivative of an arcsine function with the chain rule. Have fun with that.