engine internals

[Moppie]

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

I lost it at the fisrt mention of V=MA


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. :smoka: :smoka:

[3y30wnj00]

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

[Someguy]

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




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?

[higgimonster]

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.

[ivymike1031]

[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.

[ivymike1031]

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.

[fritz_269]

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!

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).

[ivymike1031]

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/AS...958492-2893425 ). 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).

[fritz_269]

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?

[fritz_269]

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.

[ivymike1031]

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.

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.

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.

[ivymike1031]

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?

[ivymike1031]

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

http://www.automotiveforums.com/vbul...threadid=35704

[fritz_269]

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.

[Moppie]

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