Will Thinner Oils Damage Your Engine?

[ZeeOSix]

dt=0 at that point where v=0, dt=0 at every point on the graph. This is how we look at a time slice, we make dt=0, which fixes the observation to some time (t).

Yes, I know. But time (or engine RPM) isn't stipulated on that graph - only piston position, velocity and acceleration vs crank angle. The engine has to be moving to produce a piston velocity graph like that. If you had super fine increments of time (like every microsecond) along with corresponding movement distance measurements of the piston, and plotted those for an engine RPM of say 1 RPM you would see a time period where the piston V=0 at TDC and BDC for some small delta-t (dt). If you can see that at 1 RPM then it's still going to be there at any RPM, only the time period the piston is at V=0 will shrink accordingly.

 

[EMPIRE]

Has nothing to do with the conversation - that graph is for one piston regardless if an engine has 1 piston or 200 pistons. They all do that same basic motion.

Fair enough. The computer model is somewhat flawed, I see no modeling for the actual combustion events. Or did I miss it?

I can say for 100% certainty, two engines, one 8cyl and one 36cyl, running at same rpm with same output power, the modeling between the two will look different whether you look at one piston or all of them.

Again, the piston never stops moving.

 

[EMPIRE]

Guess you're not very familiar with modern testing techniques. Maybe contact the SWRI and talk to them about how they measure all kinds of internal components on ICEs. And as already mentioned, it can be done with kinematics like I linked to earlier. Use the equations in that link. Write a computer program and look at the outputs as the engine RPM changes.

I not 100% familiar with their measurements, but super familiar with high speed data acquisition. I do not believe the are getting 1nsec data points in their data acquisition. I may be wrong, but...........

 

[EMPIRE]

That's where you seem to be "decoupled", so to speak. Just because the crank is moving doesn't mean the piston doesn't stop for a slice of time.

So now we are looking at a point in time and not dt? Which is it? Points in time cannot show motion, only dt's can.

If it's a slice in time then prove to me that there is no motion when t=0.2sec (the exact time v=0 when crank theta is some degrees past zero deg where v=0). Can you show/prove that (no motion) mathematically?

 

[EMPIRE]

I do not see actual author on the page, but their statement quoted is 100% false. It the statement was true then you would see a small horizontal line where V crosses zero or at zero, indicating for some degrees of crank (which is related to time via rpm) the speed remained zero. But in fact it DID NOT remain zero for any period of time, it only crossed over zero at a single point.

The piston’s speed goes through zero each time the piston is at TDC or BDC. This is because, at the dead centers, the piston changes direction, and for a fraction of the time it doesn’t move.

For any fraction of time (aka "dt") there is motion (aka dx != 0).

Kinematic analysis of the ICE piston – x-engineer.org

Tutorial on how to calculate the postion, the relative and mean speed and the accleration of an ICE piston function of the crankshaft angle

x-engineer.org

 

[Jimmy_Russells]

At top dead center (TDC) and bottom dead center (BDC), the piston is completely stationary for a split nanosecond and at boundary lubrication. Kinetic friction is at play and the additive package is doing the lubrication work. Up to approximately 20 degrees either direction from TDC and BDC, the piston is transitioning into mixed lubrication and rides in full hydrodynamic lubrication (dynamic friction) through the rest. The ring tension has the greatest effect on kinetic friction at TDC and BDC and ring thickness plays a larger role in how quickly and efficiently the rings transition into and ride in full fluid lubrication. The wrong oil for the ring thickness and tension can mean the rings don't get up in hydrodynamic lubrication like it should or increase oil transport past the rings (increasing oil consumption) or even possibly excessive lifting/loading of the rings causing them to flutter and effect ring seal. The type and style of hone on the cylinder walls can play a factor also.

Clearly you don't know what you're talking about. The piston has atoms, doesn't it? Those atoms don't stop unless the piston is at absolute zero. SHOW THE MATH DUHHHHH

 

[EMPIRE]

https://www.bobistheoilguy.com/forums/goto/post?id=5495574

At top dead center (TDC) and bottom dead center (BDC), the piston is completely stationary for a split nanosecond and at boundary lubrication. Kinetic friction is at play and the additive package is doing the lubrication work. Up to approximately 20 degrees either direction from TDC and BDC, the piston is transitioning into mixed lubrication and rides in full hydrodynamic lubrication (dynamic friction) through the rest. The ring tension has the greatest effect on kinetic friction at TDC and BDC and ring thickness plays a larger role in how quickly and efficiently the rings transition into and ride in full fluid lubrication. The wrong oil for the ring thickness and tension can mean the rings don't get up in hydrodynamic lubrication like it should or increase oil transport past the rings (increasing oil consumption) or even possibly excessive lifting/loading of the rings causing them to flutter and effect ring seal. The type and style of hone on the cylinder walls can play a factor also.

Clearly you don't know what you're talking about. The piston has atoms, doesn't it? Those atoms don't stop unless the piston is at absolute zero. SHOW THE MATH DUHHHHH

Are you are looking at a decoupling of the motion due to elasticity in the lube layer while the motion does a 180?

 

[M56959]

26 pages. So its safe to say topic has run it's course.

Majority of that was bickering and some of those folks won't be returning. .

Please refer to the Standard of Conduct and note that being civil to each other is expected.