engine internals

is it better to match certain brands of engine components together (like rods & pistons, or valves, seats & and springs) ?

It's important to match the componentry to your needs, but not to a name brand. An engine is a collection of parts which must work as a cohesive whole in order to perform optimally, all you need to do is select parts that will work well together and reach your power/reliability/cost goals.

let me rephrase the question, do some companies (ie TODA) make their springs with their cams in mind so that they work better together than say TODA springs and Crower cams because Crower designed their cams to work best with their springs ? thats probably the biggest run-on-sentence youve ever seen.

Could be true, but would be a very rare occurance, and only for very high lift cams that required very strong valve springs.

You have to remember that all the aftermarket parts also have to interngrate with the stock parts, meaning they can then intergrate with each other.

I see no reason why you coudn't use TODA valve springs with Crower cams.

the crower cams and toda springs was just an example but thanks a lot for the info

Well its generaly true for most parts.

Theres no reason you cant mix rods and pistons from differnt manufactors, or even differnt cams. You just have to look at what each part does, and what other parts interact with it.
Then asses it on its own merits.

Useing the Valves springs as an example, its only important the spring will be strong enough to resist vavle bounce, and be well enough made that it won't break at high RPM. Who make's it is not important.

Originally posted by Moppie

Useing the Valves springs as an example, its only important the spring will be strong enough to resist vavle bounce, and be well enough made that it won't break at high RPM. Who make's it is not important.

I wouldn't agree that those are the ONLY criteria by which a valve spring must be judged, but they're on the list. :)

Originally posted by ivymike1031


I wouldn't agree that those are the ONLY criteria by which a valve spring must be judged, but they're on the list. :)


O.K. then I'll simplify my example


As long as the part correctly performs the required role then it dosn't matter who makes it.
:smoka:

fair enough. I was sorta hoping that you'd challenge me to come up with an exhaustive list (I'm not entirely sure I could do it, but I think I could come close). I must just like to read my own writing on the internet or something.

One obvious thing about cam selection vs valve spring selection that hasn't come forward in this thread yet would be the issue of solid clearance - if you have a valve spring design that gives a desired (small) solid clearance at max lift, and you increase max lift, there's a chance that you'll "go solid" before you get to max lift... that'd be a problem. Not that this means that you have to get the same brand cam & springs, you just have to keep in mind the intended specs of the NEW system when you're picking parts. (which is what you've been saying all along, of course)

I suppose that I might add that "valve bounce" is not as likely to be caused by improper spring selection as by excessive valvetrain vibration. I think what you were probably meaning to say is "valve float." (At least if we're giving the same meanings to the terms)

I call "valve bounce" the situation where the valve closes, then "bounces" open again (actually pushed open by vibrating valvetrain components)

I've heard of valvetrain separation (where the components at the various interfaces lose contact with each other, such as cam-follower, pushrod-lifter, etc) referred to as "valve float," but I prefer to call this behavior "separation."

OK, here would be my definitions for discussion:
Valve float: At high-RPM, the momentum of the valve is too great for the spring to control and it leaves the prescribed cam lobe profile ballistically over the nose.
[/b]Valve train seperation[/b]: Whenever any of the valvetrain units - cam, (lifters, pushrods,) rockers, valves, or springs - lose firm contact.
Valve bounce - type I (lobe bounce): After an instance of valve seperation, the spring will be forcing the components back in to contact in an uncontrolled manner. The impact of the contact will compress elastic components. As they re-expand, they will impart an opposite force to the valve, causing it to 'bounce' open again.
Valve bounce - type II (seat bounce): If the lash is too great or the closing ramp too steep, the valve face will hit the valve seat with too much velocity, elastically compressing the valve diameter and the valve seat. As they re-expand, the valve will 'bounce' off the seat and open again - also causing seperation.

I'm also under the impression that valve bounce type I is usually caused by valve float.

For example: At an excessive RPM, the opening velocity is too great for the spring to control. The momentum of the spring/valve/rocker mass continues to compress the spring (continuing to open the valve) even after the rocker has passed the nose of the cam. This is the primary cause of 'valve float'. i.e. the spring does not have enough force to overcome the momentum of the valve assembly at high speed, thus the rocker leaves the surface of the cam lobe.

Now, the cam continues to rotate, rapidly increasing the distance between it's lobe face and the floating rocker. Eventually the spring force overcomes the momentum of the valve, and it reverses direction and begins to close, and therein lies the danger of valve bounce. The spring is now accelerating the valve closed as fast as physics allows - far faster than it would if it were following the cam lobe profile. At some point far down the closing ramp, perhaps even on the base circle, the rocker slams into the cam lobe. The elasticity in the cam, rocker, valve stem and spring absorb that impact and then reverse it right out, pushing back on the valvespring - once again causing seperation as the valve assembly 'bounces' off the cam lobe. Often times it can actually bounce several times as there is little damping in the system - keeping the valve intermittently open far past it's intended closing point.

From my personal experience, the biggest problem for street/strip engines is usually valve float - which then causes seperation and often valve bounce (type I). Although interestingly enough, in serious money racing, where valve lift and duration are restricted, some cam designers are now modifying the profile to intentionally cause valve float! A steep ramp right up to the nose will launch the valve right off. It'll open the valve further and keep it open longer, and the closing ramp is intended to 'catch' the valve smoothly to prevent bounce. But the mechanical dimensions of the cam lobe are still within spec. Tricky, eh?
:cool:

fritz, do you work in the car industry somewhere or do you just know everything?

re separation: I'd have blamed it on insufficient spring force to produce the desired valvetrain acceleration, but your description seems qualitatively adequate. ((dM/dt actual)<(dM/dt required) instead of (M)>(M controllable)) Of course, that probably counts as being anal.

re valve bounce (I): interesting. I never paid much attention to what happened after a separation event, because I've always focused on preventing the separation in the first place. The one thing that I'd have to point out, though, is that the compression and rarefaction of the valvetrain will often be the cause, rather than the result, of valve separation. There are plenty of instances that come to mind where the valve spring could provide sufficient force to produce the required kinematic acceleration profile, but the "slinky" vibration of the system launches the follower anyway (or launches the rocker off of the pushrod). For separation control, I generally just look at % contact time over the nose vs rpm, and if % contact falls below about 95% at "rated overspeed" I go back for another look at how to improve control. In a "racing" type application, greater separation is acceptable. I've often contemplated how to design a "catcher" profile, but I've never had an appropriate application for one thus far.

re valve bounce (II): it has been my experience that kinematic closing ramp velocity has little bearing on actual seating velocity of the valve when the valvetrain gets "excited" (excessive vibration, loss of control). Picture a high-frequency, medium-amplitude vibration superposed upon the kinematic lift profile - if the vibratory amplitude is as high as the closing ramp height (and it's sometimes 5 or 6 times higher), the valve can close w/o touching the ramp, then re-open partway down the ramp, then close again, then open again, etc. In fact, it has been my experience that even with "negative lash," where the kinematic cam profile doesn't allow the valve to seat at all, the valve can still bounce repeatedly off of the seat after the initial closing event, if the valvetrain vibration is severe enough. Usually when I see seat bounce it's this kind rather than the other kind, but I'm sure the other kind happens too.

Excellent and intresting points IvyMike!

dM/dt? I take it that M = momentum = mass * velocity?
So dM/dt = mass * dv/dt = mass * acceleration = Force.

The spring Force must be >= the force on the cam lobe = moving mass of valvetrain * acceleration profile of the cam lobe.

The idea of maximum 'controllable' mass is useful where the profile of the cam lobe is fixed, i.e. the acceleration is a constant. Thus we calculate the amount of mass that a given spring can control. Admittedly, this is a back of the envelope calculation, but it's really a useful parameter for us street racers that have to choose a cam out of a catalog (i.e. we pick our cam first to meet specs, then choose a combination of valvespring strength and valvetrain masses to meet our RPM goal.)

Interesting point about bounce (I) - wouldn't a good solution just be to move the natural frequency of the valvetrain significantly above (or maybe even below) the excitation frequencies of the cam profile? Or to add damping to the system? Does this still happen a lot with hydraulic lifters/rockers (which provide some damping)?

Great point about bounce (II) - I never really thought about continuous higher order vibrations in the valvetrain. What is the primary cause of that? Large-scale camshaft vibration? When are you most likely to see it? Can you actually use these vibrations to help cancel valve bounce, i.e. carefully design a system so the higher order harmonics are 180 degrees out of phase with the fundamental? Or are they usually completely uncontrollable?

3y30wnj00 - nope. I design microchips for a living. IvyMike is a real-life automotive engineer though.
:cool:

dM/dt? I take it that M = momentum = mass * velocity?
So dM/dt = mass * dv/dt = mass * acceleration = Force.

Good. I figured you'd be able to keep up, so I didn't bother breaking that one down for ya.

The spring Force must be >= the force on the cam lobe = moving mass of valvetrain * acceleration profile of the cam lobe

I'm with you for this part:
"The spring Force must be >= mass of valvetrain * acceleration profile of the cam lobe" as long as you meant "effective mass of valvetrain" since the parts are moving at different velocities depending on which side of the rocker they're on, and hence have differing contributions to "effective mass."

The part I'd disagree with:
"the force on the cam lobe = moving mass of valvetrain * acceleration profile of the cam lobe." The force on the cam lobe is 0 if the spring force is precisely equal to the force required to keep the valvetrain in contact with the cam lobe (think accelerated free fall).

wouldn't a good solution just be to move the natural frequency of the valvetrain significantly above the excitation frequencies of the cam profile?

Yep, that's a great solution, and it's the best way to make sure you don't have any vibration problems in the valvetrain. Unfortunately, it's not always possible (and there's no such thing as an easy problem). The best way to increase the natural frequency is to shorten the distance from the cam to the valve. Overhead cam valvetrains have higher natural frequencies than pushrod valvetrains, and direct-attack OHC valvetrains are better than the rocker variety. Unfortunately for cam&spring designers, many engines still use pushrods (and use them to the absolute limit, I should add).

An interesting suggestion about damping. Unfortunately, there are few places where you can practically add damping, since the motion you're trying to get rid of is not the motion with respect to "ground." Using a material with high internal damping would be an alternative, but last I checked magnesium was still pretty pricey. :alien:

re: hydraulic lash adjusters - you've found a weak point in my knowledge on the subject. I've very limited experience with HLAs, so I hesitate to answer this off of the top of my head. What I can say about them is that they're HEAVY and very flexible (both of which are bad, of course). They do have the nice feature of eliminating the need for mechanical lash. In the few systems I've designed w/ HLAs, they made things ever-so-slightly worse. I will likewise hesitate to extend a generalization about their performance in all applications, as the few situations where I've used them have been anything but typical. Tell ya what - ask me again this time next year. By then I'll probably have had a couple more chances to mess with HLAs, and I'll have a better feel for how they influence systems in general.

Here's an interesting mechanical design problem for you to mull over:
:devil: "Design the highest natural frequency pushrod possible with a length of 100mm"
* Use a generic tube of iron (or steel, if you prefer) for the pushrod. To keep this simple, don't worry about the end caps.
* Use a simple "mass on a spring" system to represent the pushrod for the purpose of calculating natural frequency. Support the entire mass of the pushrod on a spring with stiffness equal to the column stiffness of half of the pushrod. Remember, w = sqrt(k/m) for this system.
* Remember that the formula for stiffness of a column is k = E*A/Lh, where E is young's modulus (use 170 GPa for iron or 207 GPa for steel) , A is cross-sectional area, and Lh is length. Don't forget to use Lh = overall length / 2, since we're only using half of the pushrod for the stiffness calc.
* The mass of the column is d*A*Lt, where d is density (so much for greek letters, eh?), A is cross-sectional area, and Lt is total length. Use 7200 kg/m3 for iron (or 7850 kg/m3 for steel).
* Don't worry (for now) about the load on the pushrod, or buckling, or any of those other pesky things that'll slow ya down.

See if you can beat 1536.6 rad/sec for iron, or 1623.9 rad/sec for steel. I used ID = 5mm, OD = 10mm for my tubes (both iron and steel) if you'd like to know. Another thing to look at once you've found your best frequency is what crankshaft order does this correspond to? For my (iron) pushrod, here's what I get: 1536.6 rad/s = 244.6 Hz = 14763 cam cycles/min =
29347 rpm order 1
5869 rpm order 5
2935 rpm order 10
2257 rpm order 13

Since cam profiles will often have excitation orders as high as 7 or even 10, my stated pushrod design would likely vibrate quite a bit within the running range of a medium speed engine. Consider that the valvetrain as a whole will likely have a considerably lower 1-node-mode natural frequency... I think you'll see why I say it's a difficult task.

re higher order vibrations cancelling bounce: yes and no - at some operating speeds, you can get a valvetrain that's totally in control with superb seating velocities. Bump the engine speed up or down a notch, and suddenly you'd be "in-phase" again, and back in the frying pan. For a fixed-operating-speed engine, cancellation could be a useful strategy. For the more common applications of engines, it won't work (the frequency of each excitation order varies linearly with engine speed, while the components' preferred frequencies remain fixed).

Yep, I'm an automotive engineer (of the mechanical sort). Many EE and CSE and Controls-E types would probably remind you that being a computer chip designer doesn't preclude being an automotive engineer.

Cheers.

:confused: :eek:
That is quite a post! I was following and understanding everything untill you started talking the problem at the end. That is pretty crazy stuff, but someday (hopeful) I will know how to do that stuff. I am currently studying for a degree in ME (automotive focus) at Rochester Institute of Technology. I have a couple off topic questions and a couple on topic questions.
Where did you go to school?
What is your feild of expertise? (or favorite area of auto engineering)
Any advice for a budding Mechanical Engineering?
Now back on topic:
Would the same frequency occur when using servos for valve actuation (similar the the Seimans/VDO system if I rember corectly)? :D


The part I'd disagree with:
"the force on the cam lobe = moving mass of valvetrain * acceleration profile of the cam lobe." The force on the cam lobe is 0 if the spring force is precisely equal to the force required to keep the valvetrain in contact with the cam lobe (think accelerated free fall).

As far as I know from physics is that if the spring is compressed at all it will apply a force on the cam lobe. If you meant the total force of the spring/lobe system at that given moment is zero because of a lack of acceleration it makes a little more sense.

This is definetly the weirdest path I have seen a thread take; from "should I match brands for cam/valves?" to extremly technical discussion of valvetrain dynamics. Cool

Oh, shit not more physics!!!!!!!!!! aaaagghhhhhhhhhh :( :( :( :( :( :( :(

I lost it at the fisrt mention of V=MA :rolleyes:


Anyway, my deffinition of vavle bounce is what Fritz called Lobe bounce.
At high RPM the spring is not strong enough to hold the cam follower onto the lobe, and so it simply bounces of the top of it.
Can be an extremly good way to break certain engines, while others will do it all day long. e.g. Morris A series engines will do it for ever and never fail. While a Ford Kent pushrod can suffer sever vavle train failure.




Now, Ill go read a nice simply cat in the hat book and let you guys continue the physics lesson. :cool: :smoka: :smoka:

mike, you dont fuck around when you say "putting the anal in analyst" jk ;)

Hmmm... For some reason I feel the need to post a picture of some dude holding a F1 rod/piston assembly:

http://members.aol.com/fordguy426/pics/F1Pistonb.jpg


Okay, now that's out of the way...


The best way to increase the natural frequency is to shorten the distance from the cam to the valve. Overhead cam valvetrains have higher natural frequencies than pushrod valvetrains, and direct-attack OHC valvetrains are better than the rocker variety.

But the ramp rate is extremely important isn't it?

Honda is putting that same piston into it's new 450cc 4-stroke single cylinder moto-cross race bike. It weighs close to nothing (as far as pistons go). I want those in my car.

[QUOTE]Originally posted by higgimonster
Where did you go to school?
What is your feild of expertise? (or favorite area of auto engineering)
Any advice for a budding Mechanical Engineering?

I'll address these questions in a private message.

Would the same frequency occur when using servos for valve actuation (similar the the Seimans/VDO system if I rember corectly)?

I don't think I understand the question. Which frequency are you referring to?


The part I'd disagree with:
"the force on the cam lobe = moving mass of valvetrain * acceleration profile of the cam lobe." The force on the cam lobe is 0 if the spring force is precisely equal to the force required to keep the valvetrain in contact with the cam lobe (think accelerated free fall).
As far as I know from physics is that if the spring is compressed at all it will apply a force on the cam lobe. If you meant the total force of the spring/lobe system at that given moment is zero because of a lack of acceleration it makes a little more sense.

I was perhaps less specific than I should have been. The situation I was talking about was for the nose portion of the cam profile, where the cam surface is trying to get away from the follower, and the valve spring is required to push everything back together. If the spring force is just barely enough to create the required acceleration, then the follower will stay in contact with the cam lobe but with very little force between the two. If the spring doesn't put enough force on the various components, the cam will "get away" from them, and contact force will (of course) be zero. If the spring provides more than enough force, there will be firm contact between the cam and follower. Think of this other situation - you're in your car, and a pissed-off semi truck one car behind you wants to run you over. You accelerate as rapidly as you can to get away from him. He rams into the back of the car between you and tries his best to push it into you. If he is able to accelerate that car more rapidly than you're accelerating, he'll catch you and the contact force between that car and yours will be positive. If he pushes that other car at precisely the rate at which you're accelerating, you won't get any farther away, but there won't be any force between your car and the other. If you manage to get away, there won't be any contact force (of course), and the distance between you and the other car will increase.

Originally posted by Someguy
But the ramp rate is extremely important isn't it?

The ramp rate is important, but it affects the valvetrain excitations, not the system natural frequency(s). My previous comment was in regard to increasing the system natural frequencies.

The closing ramp rate (especially for a constant velocity ramp) will establish the rate at which the valve seats, if the system is not vibrating much. The opening ramp shape (constant velocity, or constant accn, or constant jerk, etc) and rate can have a big impact on the system excitation, and thus vibratory magnitude. Flank acceleration rates and the flank periods are also critical to avoiding excessive vibration.

ivymike - you're correct about my "force on the cam lobe" statement, it was off the cuff and incorrect. More correctly, I should have said: 'the spring Force must be >= Force necessary to keep the valvetrain from seperating = effective valvetrain mass * cam profile acceleration'

(And I did mean 'effective' valvetrain mass, considering the inertia of the rocker arm, the multiplicative effect of the rocker lever on the other masses, as well as the effective mass of the spring)

I'll see if I can get some time to work on the pushrod problem. It looks like fun! :D

Another question - to determine the energy in the harmonics generated by the cam profile, do you simply take the Fourier transform of the physical shape? What does a 'typical' transform look like? Both even and odd harmonics? Does the energy die off quickly or have a significantly long tail? It would make sense why lobes have a pseudo-gaussian shape though (maximum localization of the time-frequency spectra).
:cool:

Originally posted by fritz_269
Another question - to determine the energy in the harmonics generated by the cam profile, do you simply take the Fourier transform of the physical shape? What does a 'typical' transform look like? Both even and odd harmonics? Does the energy die off quickly or have a significantly long tail? It would make sense why lobes have a pseudo-gaussian shape though (maximum localization of the time-frequency spectra).

To be completely honest, I don't have a well defined method for looking at the harmonic components of cam-related parameters, partly because the information would not be particularly useful to me even if I had it. I don't currently have a good tool for creating a cam profile that lacks certain harmonics (although I'll probably be making one as soon as I get the chance), which is where I believe this info would be most useful. The few times that I have actually tried to check harmonic content of a lift profile, I did fourier transforms of various curves - accn vs angle, vel vs angle, follower lift vs angle, etc. I also had some hearty debates over whether it was more appropriate to look just at the lift event, or to look at a whole cam cycle. Both methods have shown significant components up to, and sometimes beyond, 10th order. My current preference is to look at the lift events alone, w/o the flat parts in between. The statement I made about cam excitation containing higher orders is a "rule of thumb" design guideline - you should try to design your valve springs to have a surge frequency of at least 13X the fundamental cam excitation frequency (and some people would prefer to set the cutoff even higher. Shigley, for example, suggests 15X to 20X, in what is probably the most widely used Mechanical Engineering textbook, "Mechanical Engineering Design." Nearly every ME that I've met studied from that textbook at some point in college. http://www.amazon.com/exec/obidos/ASIN/0073659398/qid=1016142176/sr=2-1/ref=sr_2_1/103-2958492-2893425 ). The higher the better, but of course you're going to be stuck with whatever your supplier can make for a reasonable price (and whatever currently available materials will allow).

IvyMike - I don't know what I'm missing, but I can't get the same answers as you.

If k = 2 E A / L
and m = d A L
Then k/m = 2 E / (d L^2)
And w = sqrt ( 2 E / (d L^2))

The cross-sectional area cancels. Since it scales both the spring constant and the mass linearly, it is irrelevant to w!

This time I even checked the units ;)

E in Pa = N/m^2 = kg*m/s^2 / m^2 = kg/m*s^2
d in kg/m^3
and L in m

so k/m = (kg/m*s^2) / ( kg/m^3 * m^2) = 1/s^2
and sqrt(k/m) = 1/s
which is what we want.

If I choose Steel (of any cross section), I get:
k/m = 2 * 207E9 / (7200 * 0.1^2) = 5.75E9 / s^2
and w = sqrt (k/m) = 75,828 radians / second = 12,068 Hz

Following your analysis then,
12,068 Hz = 724,000 cam cycles / minute
and 1,448,000 crank RPM = O(1)
144,800 crank RPM = O(10)
14,800 crank RPM = O(100)
7,400 crank RPM = O(150)

So the 150th order (!) vibration will be right near redline of most high performace pushrod engines. That doesn't seem like it would be a factor at all.

Did I miss something here? :confused:

Originally posted by ivymike1031
I also had some hearty debates over whether it was more appropriate to look just at the lift event, or to look at a whole cam cycle. Both methods have shown significant components up to, and sometimes beyond, 10th order. My current preference is to look at the lift events alone, w/o the flat parts in between.
If you don't mind, I'd like you to revisit that debate. It seems to me that, a priori, you should consider the entire circular profile. The Fourier transform is based on the idea of an infintely periodic signal. If the signal isn't infinite, you have to use a circular approximation and usually windowing functions to make it appear infinitely periodic (or you could just use finite time-base wavelet transforms).

I was just thinking about how cool it was that you actually wouldn't have to make any of those approximations by forcing periodicity into your calculation, becase it was just natural since the cam is rotating.

Can you give me any intuition as to why you wouldn't want include the base circle in your transform? It's completely non-obvious to me.
:cool:

well, it wouldn't surprise me any to find that I'd botched something... I was doing the calcs on a hand calculator after a couple of beers. :bloated:

You did find the "rub" that I was trying to point out - that the natural frequency of the pushrod by itself is pretty much dictated by the material properties and the length, no matter what you do to the section. When I simplify the equations, I come up with w = sqrt(2E/d) / L

Today I can't seem to get my numbers to agree with yours OR my previous ones. I'm pretty confident in the accuracy of the following figures, though (who knows what I did last night?):

sqrt(2*207*10^9/7850)/.1 = 72622 rad/sec = 11558Hz
sqrt(2*170*10^9/7200)/.1 = 68718Hz

I thought that the frequency looked a bit too low last night. (who ever heard of "pushrod surge" anyway?) The overall system natural frequency is usually much lower than what you get for the pushrod, of course. That's what I get for trying to drink & calculate. :rolleyes:

The only thing that I was trying to get across was that it can be very difficult to affect the system vibrational frequencies in a useful manner simply by changing the system geometry. Moving the cam closer to the valve is a great way to do it, if you're allowed to. If not, you can be stuck with a very hard to control system.

I recently had the "pleasure" of doing some valvetrain work on an engine with 0.5m long pushrods and valves that each weighed ~3kg. (with 2 intake & 2 exhaust valves per cylinder) 2400rpm was asking a heck of a lot from this engine...

Anyway, good catch.

Originally posted by fritz_269
If you don't mind, I'd like you to revisit that debate. It seems to me that, a priori, you should consider the entire circular profile. The Fourier transform is based on the idea of an infintely periodic signal. If the signal isn't infinite, you have to use a circular approximation and usually windowing functions to make it appear infinitely periodic (or you could just use finite time-base wavelet transforms).

Sure, we can do that. Perhaps it would be more appropriate to start a new thread?

I started a new thread over here (engineering / technical forum):

http://www.automotiveforums.com/vbulletin/showthread.php?s=&threadid=35704

Originally posted by ivymike1031
Today I can't seem to get my numbers to agree with yours OR my previous ones. I'm pretty confident in the accuracy of the following figures, though (who knows what I did last night?):

sqrt(2*207*10^9/7850)/.1 = 72622 rad/sec = 11558Hz
sqrt(2*170*10^9/7200)/.1 = 68718Hz

I thought that the frequency looked a bit too low last night. (who ever heard of "pushrod surge" anyway?)
Bah - I just accidentaly used the Iron density with the Steel's modulus. I agree with your figures above.

Yes. 215 Hz seemed awful low for a 10mmx100mm tube of steel.

Point taken though - I never did that calc, so I never noticed that the cross-sectional areas would cancel. Thanks for pointing it out! I definitely learned something today.
:D

Originally posted by fritz_269

a priori,


Yay a word I understand!!!! :flash:

Now the cat in the hat sat with a piston in his lap.

He played with cam, which when hit with valve, made an awful loud wham.

Then he remembered, to fit a push rod, and low and behold it all just followed.

(Ok, so I dont have a picture of a guy holding a F1 piston and rod assembly.) :ylsuper