Low r.p.m. urban myth

[jbcollier]

Well, Porsche was pretty insistent that you kept the revs up on their roller cranked engines when cracking the whip. Plodding about? No worries! Full steam? Revs up, please.

[machten]

I don't confess to to understand the physics of it all, but I think with experience, you can "feel" what your engine likes. Perhaps the bottom end wear isn't the dominant issue - I don't know. I think my '68 250 widecase likes the gears changed after 1st gear at no less than 5K rpm and it's smooth as silk at 6. My 450 on the other hand is quite comfortable at 3.5K and use the torque, but likes to see 4K at least.

I have a 750 Sport with the higher primary gearing compared to my 73 750GT with the lower primary gearing. Other than that, the engines (pistons, rods, etc) and their fueling are identical. They have totally different "comfort levels". The Sport does not like it wrapped on under 5000rpm, the GT is happy at 3500rpm. I too, will keep them spinning!

Kev

[Nick]

"If you (are one of those mechanically sensitive people who can) listen and feel the engine it will let you know when it's not happy."

Fixed it for you.

[Harvey]

Harvey,

I used the example of the roller to help the understanding of the forces at play. It is correct that the rod and pin always have the peak load act on the same spot if the peak pressure occurs at the same crank angle.

I disagree with your statement, -
“The slower the rpm the higher the forces acting on this shorter area.”

This seems to be going over what we discussed before. My interpretation of your statement is:

‘since the time of combustion is constant, then at slower r.p.m., the force of combustion acts on a shorter proportion of the circumference big end pin the conrod’.

Can you confirm if my interpretation is accurate, before I continue with the discussion. We might be at cross purposes otherwise.

Yes mate that is what I mean, but I can see you may have interpreted it differently.

“The slower the rpm the higher the forces acting on this shorter area.”

I don’t mean that there is a higher force acting on the area, I meant that the forces acting on this area are higher, as each of the rollers are applying this high pressure over the shorter area.

If I can explain my thoughts a bit more. Looking at the bearing it has the pressure of the piston applied through the rod, rollers, to the pin. The pin is hard steel of good quality as are the rollers. The rod is good quality steel?? with a harden track. As the rollers roll across this area, the rod track is deformed by the high load on this short area. This causes the hardening to flex, fatigue, and flake off.
At higher rpm, there is a longer rod track to take the same load, so the track is not deformed to the same extent.

[StewartD]

2014.05.05 Motoscrubs graph 2.jpg

Hi All,

Sorry for long delay in replying. It took some time to get round to redo the torque plotting on the Dynamometer graph that I reused from the thread ‘Horsepower claims, Top speed and Accuracy of Dynamometers’. I incorrectly assumed the rpm to be 8000 rpm at 80 m.p.h. for all three machines. Now I have made assumptions that are based on the machine’s peak output rpms which is given in the Clymer manual as follows.

350 Sebring 1964 - 1966: 6250 rpm
250 Mk 3 1964 - 1966: 8000 rpm

I have based my torque figures on assuming the Sebring’s peak output at 6250 rpm occurs at the dynamometer’s 75 mph, and that the 250 Mk 3’s redline of 8500 rpm occurs at the the dyno’s 71 mph.

I have used the same method to calculate torque in Newton.metre from the graph’s horsepower and rpm figures as I used in my post of the 18th of April.

The torque curves I have plotted both show that torque at low rpm, (say below 3000 rpm which seems to be a level that some on this thread suggest is a speed that is slogging the motor), is small compared to the peak torque that the motors develop.

For instance for the 350 Sebring gives 4 N.m of torque at 2500 rpm compared to 24 N.m at 4583 rpm (4/24 = 16% of peak torque), and the 250 Mk 3 gives 9 N.m at 2700 rpm compared to 23 N.m at 4167 rpm (9/23 = 39%)

The force the big end bearing experiences are directly proportional to these torque figures. This is simply a fact of geometry. Torque is a the result of a force multiplied by the ‘moment arm’ or the radius of action, or half the stroke.

Such comparatively low forces, (compared to what it is designed for), will not damage a bearing.


Bob,

I still have a Mach 1 which I bought in 1973, so I'm quite familiar with the difference in power output characteristics. As they both have a 29mm carburettor a vast difference in character is to be expected. I or anyone else, within this thread, have not made any specific comparisons between high performance models and 450's. What I wanted in this thread, was to dispel the general notion that low revs are inherently bad for the big end.

I maintain that the big end load, at just off idle revs, is very low, in any engine provided that detonation is not occurring. Of course, a Mach 1 would be more liable to detonation as it has a 10: 1 compression ratio. Kevin Cameron's article: point 4 of his 'Stopping the Show (cures), says: 'lower compression ratio. The less you squeeze it, the less it is heated.'

Graeme,

I never worried about low revs with my 450. At 2000 revs it would be doing about 30 m.p.h. and I would give it full throttle with no ill effects. This was a standard 450 and the torque was very good at this speed. As you say though; If you listen and feel the engine it will let you know when it's not happy.'

The load on the big end is directly related to the torque that a dyno test will show. The Dyno test shows exactly where the torque is high and where it is low. Where torque is low the cylinder combustion charge is not as good as where the torque is high. Having more time does not help if the air speed through the venturi is not optimal.

jbcollier,

I looked at the web and found an article about Porsche crankshafts having a problem with harmonics at low rpm. This has nothing to do with our topic though. If you can find anything that is related to big end problems at low rpm, please let us know.

Harvey,

This is not the way stress is calculated. The length of the track that a roller runs over during the course of the combustion is not taken into account. The peak of combustion pressure results in the peak of the force that the piston exerts. Depending on the geometry, this will probably be very close to the peak force on the big end bearing. This peak force results in the peak stress on some bearing rollers, peak stress at some points on the big end pin and on the con rod bearing surface. It is only this peak stress that matters.

[Harvey]

Stewart, looking back at the process that you have used, to arrive at the forces that you clam, to be very low at low rotational speeds, you have made a few assumptions. This one was the base of those calculations.

“Torque (N.m) = Power (Watt) divided by rotational speed (Radian/second)”

Now while Hp = torque x rpms, Torque does not = Hp/ rpm.
Hp is proportional to rpm. Torque is proportional to the cylinders Volumetric Efficiently

Performance Summary
-------------------
-RPM-------[Kw]-----[Nm]---------[%]-----------[bar]
-------------------------------------------------------------------------
1000----------4.45-----42.48---------79.6----------52.3
2000----------9.36-----44.69---------82.9----------53.7
3000---------14.34-----45.64---------84.5---------54.53
4000---------23.78-----56.76--------104.1---------66.38
5000---------30.33-----57.92--------110.7---------69.93
6000---------31.66-----50.38--------101.1---------63.35
7000---------27.97-----38.16----------88.2---------55.46


This table is from a simulation of a 550cc, four valve cylinder, showing the Power, Torque, VE (% of the cylinder that is filled), and the Maximum Pressure developed in the cylinder.
As you can see the torque has no relationship with the rpm. only the VE, and the maximum pressure that is developed. It is also shows that the load on the big end at low rpms is still quite high. So giving an engine full throttle at 2000 rpm has a lot more load on the bearing than your calculations show.

[StewartD]

Harvey,

You are wrong. Ask any technologist or mathematician about transforming formulas.

If Horse Power = Torque (N.m) x rpm / 7122 then (multiply both sides of equation by 7122 / r.p.m.)

Torque = Horsepower x 7122 / r.p.m.

(and further: r.p.m. = Horsepower x 7122 / Torque)

This is the same process that any electrical technologist uses with Ohms law i.e. I = V/R (current in amps = Voltage / Resistance in Ohms) can be transformed to

V = IR or R = V/I

Also horsepower is not proportional to rpm. Look at the original Dyno graphs. There is an approximately proportionate relationship between horsepower and rpm for the 350 Sebring between 41 mph and 48 mph and then another approximately proportionate relationship between 50 mph and 57 mph. At 78 mph the horsepower drops rapidly for increasing rpm. This is not a proportional relationship. Any proportional relationship will be a straight sloping line, of positive gradient, on a graph.

I do not dispute your performance figures but these figures do not show directly proportional relationship between Torque and Volumetric efficiency. Graphing these figures does not give a straight line relationship and therefore there is not a proportional relationship between VE and Torque. I tried to attach this graph but got the message 'Sorry, the board attachment quota has been reached'.

I never stated torque has a relationship with rpm. Saying that Torque = Power divided by rotational speed is correct, but this is not the same as saying torque has a relationship with rpm. The equation is simply a means of calculating an unknown value from two known values. It does not imply any direct proportional relationship between any of the variable factors.


In your figures of the 550cc four cylinder engine; the torque at 2000 rpm is 44.69 N.m which is 77% of the maximum torque (57.92 N.m) shown for this motor. This is a higher percentage than the 16% and 39% I derived from the Ducati graphs but 77% load is still not going to wreck a bearing.

Cheers,
Stewart D

[Jordan]

As the analytical thinking doesn't seem to support the popular view that WOT + Low Revs = shorter big end bearing life, the question might be, "Why is it a commonly held view?"
As empirical knowledge isn't easily dismissed, as lots of anecdotes and evidence support it, I can't help thinking there's something missing in Stewart's analysis.
But, I can't figure what it is.
Nevertheless, I'll err on the safe side and keep the engine spinning by changing down early and up late.
It just seems happier that way.

[Nick]

Look at the Ducati MotoGP bikes. The engineers did all their math, and they are convinced that they've built a bike that works — it doesn't.

Remember Hailwood and the Honda 6? He told the Japanese that their shocks were no good. They didn't believe him because their calculations showed that the shocks were fine. In exasperation Hailwood had them remove the shocks and he then threw them into an infield pond at Honda's test track. He then gave the Japanese a pair of Girlings and said: "Copy them!' (Or words to that effect)

Math is great, but it doesn't have all the answers.

[Bevel bob]

There has been very little recognition that the pump output at low rpm ( and its delivery via a leaky bush ) is poor. This is of course not a problem if the motor is used as designed. Remember it was designed not as a roadster ,but as a racer.