F cars with cam belts

That makes some sense!!

 

No mention of whether or not it being cold had an effect?

In any case, they should join FChat, they will fit right in.

 

Well, kind of. The torque is not a function of the pulley diameter, although the load on the belt is. Sort of like saying the pressure in your tires is xx lbs.

 

1) I was quoting what Goss said.

II) Torque is a function of pulley diameter, but it's actually the other way around. Torque x angular speed is constant. Smaller pulley rotates faster thus torque is lower on the smaller pulley. But as you point out, what is constant is the force applied to rotate each pulley. Assuming the smaller pulley has fewer belt teeth in contact with pulley, each tooth on the smaller pulley is stressed more. But even that is a simplistic view. What really happens is that upon start up a jerk is applied to the system to accelerate it. That is initially applied to the belt teeth on drive pulley which can shear the teeth before the force propagates through the belt to accelerate the cams. It's like crashing a car into a wall. The front end feels the impact first and deforms. By the time the compression wave propagates through the rest of the vehicle it has been so dissipated little additional damage is done.

Anyway, I found it humorous that Goss just happen to be discussing this at the time this discussion was going on.

 

Did not mean to imply otherwise.

For a given torque the force on the belt is inversely proportional to the pulley diameter. Not sure where angular speed came into the discussion. At start-up, and initial application of torque, the angular speed is zero.

 

Seems to me the force on each tooth of the belt at the crank cog would be higher not only because of the (presumed) lower number of teeth carrying the load, but also due to the fact that the crank is pulling four cams, whereas each cam cog is only carrying the load of that particular cam. So I'd think that the load applied to the belt would be *significantly* higher at the crank cog.

 

Yeah they say the oil leaks will get to you before the belt needs doing.

But if the prior eo was done well there should not be any significant oil leaks. Its either cam covers, or oil return lines. If someone knows what they are doing, and makes the effort cam covers should not leak, the rest is pretty obvious.

 

Torque is not constant at both pulleys (thinking single over head cam here). But, as I said, the torque argument is a simplistic view that applies only in steady state, as you basically point out. But to review it for others, the power transmitted through a rotating reduction system, be it gears, belts or chains, is constant (conservation of energy). Power = force x distance/time. For a rotating system this is equivalent to Power = torque x RMP. So, if you apply a torque to the smaller pulley, Ts which rotates at RMPs, the power is Ts x RPMs = Tb x RPMb. RMP is related inversely to pulley diameter, so the smaller pulley has higher RPM. Thus, Tb = Ts x RMPs / RMPb, or in terms of diameter, Tb = Ts x Db/Ds, and for a torque Ts applied to t he smaller pulley, the resulting torque applied to the bigger pulley is greater. This was only intended to point out Goss's error.

For a dual overhead cam system the power input to the lower pulley would be equal to the total of the power delivered to both cams. Assuming the work required to turn each cam was the same, then the torque generated at each cam pulley in the dual cam system would be 1/2 that applied to the cam pulley of the single cam system, assuming similar pulley sizes.

The actually issue is that the torque initially applied to the smaller pulley at start up is not instantaneously transmitted to the bigger pulley. The system has mass and inertia and elasticity. As the smaller pulley begins or attempts to turn due to the application of torque to its shaft, and tighten the belt, an impulsive force is applied to the teeth of the belt which are engaged in the smaller pulley. Of course, since the belt and the cams and the cam pulleys all have mass so they initially resist movement. Plus, due to elasticity, it takes a finite amount of time for any force applied to the belt to propagate through the length of the belt to the bigger pulley and begin to accelerate the cams. Again, since the cams have mass they will resist movement. As a result, an additional tension force is generated which reflects back through the belt to the smaller pulley in an attempt to counter the initially applied torque. This results in a force on the teeth of the smaller pulley which can exceed that generated by the application of the initial torque applied to the small pulley and shear the teeth of the belt there.

In simple words, the small pulley says to the belt, get moving and gives it a push. The belt says, I'm fine standing still. I'd rather you stop and pushed back. A fight breaks out and somebody may end up with broken teeth. Of course, the small pulley is the bully in the crowd and almost always wins the fight. Either the belt get moving or it ends up with broken teeth.

 

Resistance from the mass of the camshafts, and pulleys is negligible. The major resistance to belt motion in normal operation would be from the valve springs. Higher revving engines require stiffer springs to prevent valve float. This puts more stress on the belts.

 

The system is initially at rest, motionless. It is in static equilibrium. Nothing I rotating. That means the sum of all forces: springs, friction, what ever, which could impart rotation to the system sum to zero. Upon application of the starting torque, the mass must be accelerated. But yes, you are correct, as the system begins to rotate there are additional forces to over come as those forces from springs will move out of equilibrium and static friction must also be over come. All these simply contribute to the forces acting on the belt in the fight between the drive pulley and the belt as to what will happen, rotation or sheared teeth. I should have included them in the discussion. The physics of the starting transient doesn't change, just some force accounting. Thanks for your comment.

OK, are we having fun yet. I am.

Some real info to consider. As far as I can tell, early Ferraris like the 308 use type L timing belts. The type specifies the tooth shape and pitch. Gates provides the following specifications for their type L belts.

1) Ultimate tension strength per inch width of belt; Steel cords, 1474 lbs/in. That means a 1" wide belt will snap if the tension exceeds 1474 lbs. For Kevlar cords, 6555 lbs/in.

2) Maximum allowable belt tension. Steel cords: 186lbs/in width. Kevlar cords: 207lbs/in width.

3) Specific belt stiffness. Steel cords; ss= 92800 lbs/in, Kevlar cords: ss= 69100 lbs/in.

Specific stiffness spec is a little difference. It allows calculation of the stiffness or spring constant for a belt as follows: k = ss x b/L where b is the belt width and L the length of the belt segment in question. Knowing k, the amount of belt stretch can be found as S = T/k where T is the tension in the belt. Thus, for a 1" wide belt and a segment about 12" long, k for a steel belt would be 92800/12 = 7733 lbs/in. If the belt were tensioned to its maximum allowable tension, 186 lbs, this 12" segment would stretch by 0.024" = 0.61 mm. Assuming that the distance between the drive pulley and the cam pulley on a 308 is about 12" this amount of stretch would amount to about 0.72 degrees of rotation. A reduction of belt tension of 50 lbs reduce the stretch by 0.0065" or 0.16 mm and would change timing by 0.2 degrees. Note that this also works backwards. For example, if the center to center distance between the drive pulley and the cam pulley was reduce by only 0.0065" the tension would drop by 50 lbs. Also note that while Kevlar belts are generally stronger, they are not as stiff as steel. A Kevlar belt will stretch more that a steel belt when subjected to the same tension.


Note also that here we are talking about elastic belt stretch, not belt stretch due to wear and deformation. However, noting that a change in the length of the belt segment between the drive pulley and the cam pulley of 0.024" yields 0.72 degrees rotation we can do some addition investigation. I have read here that belt stretch due to wear can cause as much as a 3 degree shift in timing. If 0.72 degrees requires a 0.024" change in length, a 3 degree shift would require a change in length of 0.1". Now this change in length would be over just the tension side of the belt. The slack side of the belt would also have to stretch an equivalent amount since the belt continuously rotates around the system. Just assuming the slack side is of equal length would mean the belt would have to stretch over all by 0.2" in length to have a timing shift of 3 degrees. Needless to say, if a new belt were installed and tensioned to the max allowable tension, the full length of the belt would only be stretch by about 0.05". If belt wear resulted in stretching an additional 0.2 inches it should be obvious that at that point there would be no tension left in the belt at all when the motor was shut off and the belt would be loose as a goose. Additionally, that much stretch would likely prevent the belt teeth from meshing correctly with the pulleys and the belt would self destruct in a short time. One more point. I personally have measured a well used, both in miles and time, 308 belt and compared it to a new off the shelf belt. I cut the belts and placed them under 50 lbs tension. The old belt measured 915.2 mm. The new belt measured 914.5 mm under the same tension. The spec length, based on pitch and number of teeth is 914.4.mm. Belt tolerance is speced at 1.00 mm. Note that the difference between the old and new belt is only 0.7 mm, less than the tolerance for belt length when new, and much less than the estimated 5 mm (0.2") required for a 3 degree shift in timing. Even in the worst case scenario, where I assume the old belt was on the short side of the limits, 913.4 mm and then stretched do to use to 915.2 mm, the stretch is only 1.8mm (0.071") (but still enough to remove all static tension from the belt). Assigning 1/2 of that to the tension side of the belt yields a change in timing from new of only 1 degree. So, if when you do a belt job, if the belts aren't flapping in the breeze and the timing is off more than about 2 degrees, giving it was set to +/-1 degree, it was probably set incorrectly at the last belt job. And if the belts are flapping in the breeze I would suggest there might be a problem elsewhere.

Let me summarize as before. If when you do a belt job the old belts still have significant tension in them they haven't stretched enough to significantly alter the timing from when they were installed. A rough estimate is 0.2 degree shift for every 50lbs change in belt tension.

 

On any given cog I think you would find that the first tooth, on the 'tension' side of the belt, has substantially higher load than the others.

 

Obviously, and older belt will experience reduced strength due to fatigue of the element in the cord. On the other hand, shearing of the teeth has little to do with the tensile strength of the belt.

As for what each tooth on the pulley sees at a load you have to look at it as each tooth is connected to the next by a spring. So if the load on the first tooth is greater than the load on the next, the spring connecting them will see a differential force that will elongate the spring to even out the load. That is, the difference in force will cause the pitch to change a little so that the second tooth fall back against the cog on the pulley. And so on and so fourth for each additional tooth. Of course, this will be a dynamic process as different teeth rotate on and off the pulley and the forces won't be exactly equally distributed.

Anyone interested, here is a link to a pdf by gates on belt failure. It's a non technical discussion of failure types and causes.

http://www.bbman.com/assets/pdf/Maintenance/Belt_failure_analysis_Final.pdf

I currently have a request on an engineer at gates for a paper on "Probabilistic modeling for timing belt fatigue life predictions using accelerated testing" . Should be interesting.
I also downloaded some papers from Gates on belt design and belt sizing, but a little to deep for this type of discussion.

I know what you are thinking, NERD ALERT. Sorry, as a retired researcher I get into stuff like this from time to time, particularly when it cold and with snow cover and I can't take my cars out.

Of course, don't know what this will tell us about belt life on a Ferrari. Tomorrow someone will post, "I just picked my car up from a belt service and the belt broke on the way home."

 

I learned one thing from thgis thread already. Despite their maintanace requirements, belts are a great way to go for all sorts of reasons.

 

And that is likely one major reason why Ferrari amended their initial 7-year 52,000 miles to 3-year 30,000 miles

 

That certainly seems logical.

 

This has turned into a long winded thread.

 

Please don't tell me you expected any different?

You've been a member long enough to know the score (when it comes to cambelt threads)

 

You appear to be attempting to prove the earlier statement that belts fail more often on startup, which I put down to other factors and was rubbished. But you are on the wrong track.
The inertia at startup is negligible. The polar moment of the camshaft is tiny, a relatively light narrow tube. The load on the belt is from the valve springs which impart a number of superimposed waveforms on the belt. These are no different at startup. The load increases only with RPM as the camshaft has to accelerate the valves against the spring at a higher rate.

Belts failing more often at startup is owing to other factors, tension being lower possibly causing jumped teeth, belt stiffer and more brittle so any cracks can turn to breaks, maybe grease in the tensioner bearing stiffer when cold and other factors. None of these factors affecting a serviceable belt or other parts of the system but exacerbating an existant issue. Also this startup factor is unlikely to be heard of by someone in a climate which never gets very cold.

You were right about the BMW issue I mentioned though, the issue is with chains. I suppose I had assumed belts as breaking timing chains really is a serious design failure!

 

Andy:

Actually, I think the only place we are in disagreement is with the initiation of rotation. With the system at rest all the forces acting on it sum to zero and there is no acceleration. The only thing resisting the initiation of motion when a torque is applied is the total mass that is to be put in motion: cams, belts, valves, tappets, pulleys, effective mass of the springs.... Doesn't matter whether it is a large or small mass. All that effects is the magnitude of the initial acceleration. So initially the torque attempts to accelerate the mass. Fist thing that happens is that it is resisted by friction in the valve train, so belt tension increase, but still no motion. Does the belt fail then? By teeth shearing or any other mechanism? Maybe, maybe not. Once the threshold of friction is surpassed the system will begin to undergo angular displacement and the changing spring forces will come into play. But I would not be surprised if the belt actually failed before the cams move, or just after they begin to move, due to the impulsive loading, i.e. the jerk, J = da/dt =(dF/dt ) /m. It's like asking when you hit a nail with a hammer does it bend before the nail starts to penetrate further into the wood or a micro second after it begins to. Either way, it's the impulsive force that is the cause. If the nail gets moving w/o bending, it's unlikely to bend on that strike.

Mike, Jcurry:

With regard to tooth loading, what I have found is the forces are greatest on the tooth closest to the tension side of the pulley, either the driving or driven pulley, and smallest on the slack side. From this attach picture from a Gates paper, tension varies more or less smoothly across the pulley's engaged teeth. The forces on each tooth is such that the sum of the tension forces and the tooth force is zero, as shown at the right. A couple to things to help understand the figure. This is for what they refer to as a constant slack side tension set up. I.e. it has a movable tensioner that maintains constant tension of the tensioner bearing, like a 355.

The dotted line around the system at a distance Ti from the belt represents the initial, static tension or how you would tension the belt. T1 is the tension on the tension side of the belt when running. Note that it is constant between the pulleys on the tension side. T2 is the tension on the slack side and is kept equal to Ti by the tensioner. So looking at the force triangle you can see that for the firs tooth Tk = T1 and Tk+1 would be something lower. On the last tooth, Tk+1 = T2 and Tk would be something greater. Anyway, it is apparent that for each successive tooth around the pulley both values of T decrease and, therefore the force on the tooth decreases. This doesn't suggest a substantially high force on the first tooth on the tension side, but a gradual reduction around the pulley.
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The total mass is not directly relevant (other than the mass of the belt which moves in a linear direction). The relevant figure is the total polar moment of inertia. Which is tiny in relation to the valve spring resistance. Also friction is negligible otherwise the bearings would fail.

Polar moment is related to mass but it also depends on how far the mass is situated relative to the axis of rotation. The pulleys would have a relatively high polar moment related to their mass owing to their shape but they are made of material so light that its of no consequence. Camshafts would have a tiny polar moment owing to being narrow despite being made from heavy material.

 

True enough!!

 

Total mass is the only thing that is relevant. It's all F = ma. Nothing else. The torque applied at the drive pulley generates a tension in the belt. That tension generates a torque in the driven pulley. That torque is either sufficient to over come the forces resisting rotation and accelerate the effective moving mass of the valve train or not. The varying forces in the valve train do not alter the tension in the belt, they alter the acceleration of the cams, valves, etc.

It may be easier to think about just cranking of the engine. The starter motor applies a fixed torque to the fly wheel. When the engine cranks we all hear, chug, chug, chug... as consecutive cylinders come up on compression because the cranking speed slows down, then increases as they go off compression. That is, the rotational speed of the engine accelerates and decelerates under the influence of the constant applied torque by the starter because the forces resisting acceleration due to compression change.

Now, once the car is running, belt tension will be dependent on RMP. Why? The forces in compressing the springs don't change. So what changes? Two things. The valves and tappets have to open faster. Part of the springs have to move faster even though they have the same stiffness. That means they all have to accelerate faster. That means greater force must be applied to the tappets by the cams (F = ma). Also the greater force on the cam lobes means greater frictional forces. The net result is more torque is required to turn the cams at higher RPM which translates to greater tension in the belts. Since the engine is running at constant RPM, the torque requirement will vary periodically as different valves open and close which means the belt tension with vary periodically as well, and the torque load on the engine to drive the cams will also vary periodically.

So this is the difference between running at constant RPM, where torque and tension loads vary periodically, and cranking under a constant applied torque where RMP varies periodically but bet tension is constant.

Technically there can be some change sin belt tension when cranking but that comes down to characteristics of the starter motor and also, again, acceleration of other pars of the engine, pistons, rods, crank....