Oiling Piston When Stopped

[EMPIRE]

Can anyone tell me where on this graph you have dx/dΘ = 0 for any dΘ !=0 (aka "no motion")
Can't see it in the graph, or even in the math on that page. Is there some dΘ where the piston motion decouples from the crank?

And if say the motion stops for 10^-20 sec, what's the oiling issue of the piston during this time? And if there is an issue, what oil property is the item to combat the wear?

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

 

[wpod]

At bottom dead center and top dead center , additives ZDPP etc is the final "layer" of protection when the oil film isn't "thick" enough
link to Bob is the oil guy ,, states of lubrication

 

[PWMDMD]

Can anyone tell me where on this graph you have dx/dΘ = 0 for any dΘ !=0 (aka "no motion")
Can't see it in the graph, or even in the math on that page. Is there some dΘ where the piston motion decouples from the crank?

And if say the motion stops for 10^-20 sec, what's the oiling issue of the piston during this time? And if there is an issue, what oil property is the item to combat the wear?

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

What do you mean? dx/dO = 0 at 0, 180, 360, 540 degrees, etc.....where the slope is zero.

 

[PimTac]

What do you mean? dx/dO = 0 at 0, 180, 360, 540 degrees, etc.....where the slope is zero.

Not satisfied that the last thread got locked, he starts a new one on the same argument.

 

[PWMDMD]

Not satisfied that the last thread got locked, he starts a new one on the same argument.

My first time on today...totally unaware of his other thread but figured I can answer a simple math question.

 

[ABN_CBT_ENGR]

Not satisfied that the last thread got locked, he starts a new one on the same argument.

The irony is that his linked article completely obliterates his original incorrect argument ( must be that anti-matter "nucular" engineering)

From the article>

As you can see, when the crank angle is at 0°, the piston is positioned at TDC and the displacement x reaches the maximum value. After half of rotation (180°), the piston is at BDC and the displacement reaches minimum value.

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. This is the main reason for which internal combustion engines have flywheels on the crankshaft, to keep the pistons moving through the dead centers.

 

[blingo]

Maybe look up the Stribeck topos to find that it ain't about the zero when not with Zeno. Or directly watch out for film thicknesses developing and dwindling along the bore.

 

[EMPIRE]

Not satisfied that the last thread got locked, he starts a new one on the same argument.

Why was the other thread locked to start with?

It was said the piston motion stops and I called BS on that. So to help get an answer I started a new thread about it.

 

[kschachn]

Why was the other thread locked to start with?

It was said the piston motion stops and I called BS on that. So to help get an answer I started a new thread about it.

Quite frankly you got boatloads of answers to that question in the other thread from multiple individuals and via multiple explanations.

Please.

 

[EMPIRE]

What do you mean? dx/dO = 0 at 0, 180, 360, 540 degrees, etc.....where the slope is zero.

dΘ is some arbitrary change in crank angle, which is just another way to represent time as the crank rotates.
dx is just some change in displacement (using those graphs) with respect to the change in crank angle. It other words, rotate the crank some degrees and measure how much the piston has moved. But instead of actual measuring some math was used to model the movement, which is in that link.

Some here on BITOG are suggesting the piston actually stops moving when it changes direction of motion. But using the math provided in that link I see no place where the piston itself decouples from the connecting rod motion, which if course is tied to the crank.

So, if motion is to halt you need to show some change in time (crank angle) where the difference in displacement (location) of the piston is zero (no change in position, aka "no motion").

I need help finding this "flat spot" in displacement while the crank is still rotating, etc.

 

[EMPIRE]

Quite frankly you got boatloads of answers to that question in the other thread from multiple individuals and via multiple explanations.

Please.

Which were all not correct. You care to help show me, or just continue to tell me the wrong answers were answers? The linked site from that thread (also posted here) has the modeled math. Can you use that to show me where the piston stops moving?

 

[Astro14]

Why was the other thread locked to start with?

It was said the piston motion stops and I called BS on that. So to help get an answer I started a new thread about it.

That thread was locked due to bickering. You “called BS” in that thread, and continued your argument, which was a large part of the bickering.

Since you were unable to accept the math/physics in that thread, after 26 pages of posts, I don’t see why you would accept it in this thread.

Seeking an argument, for the sake of argument, is trolling.

Lock time.

 

[Astro14]

You linked the wrong photo.

Let’s look at the middle graph in the photo below...From YOUR LINK.

On the vertical axis is V, piston velocity, while theta is on the horizontal axis.

At Theta = zero degrees, 180, 360, etc, V=0. When velocity = 0 the object is stopped. It’s that simple.

 

[EMPIRE]

You linked the wrong photo.

Let’s look at the middle graph in the photo below...From YOUR LINK.

On the vertical axis is V, piston velocity, while theta is on the horizontal axis.

I cut out displacement as a function of crank angle. When does displacement ever become (stay at) zero with any change in crank angle?
You pick the change in angle, any angle, and then show me where displacement during that angle change remains zero.

The velocity graph just shows the displacement crossing over the end points of displacement, not stopping there for some time period. If it stops there then for what time period does it stay stopped there? Does it stop for 2nsec, 0.5nsec, or 0nsec ?

I can in fact also bring in the math where the linkage to piston actually decouples from the crank motion as a function of static friction, but that is not shown anywhere in that math on that linked page. Nor does the math on that page account for the constant blows of force that come from combustion events, which will in fact skew those graphs.

 

[ZeeOSix]

dΘ is some arbitrary change in crank angle, which is just another way to represent time as the crank rotates.
dx is just some change in displacement (using those graphs) with respect to the change in crank angle. It other words, rotate the crank some degrees and measure how much the piston has moved. But instead of actual measuring some math was used to model the movement, which is in that link.

Some here on BITOG are suggesting the piston actually stops moving when it changes direction of motion. But using the math provided in that link I see no place where the piston itself decouples from the connecting rod motion, which if course is tied to the crank.

I'm still waiting for proof that an object can change in linear direction exactly 180 degrees without stopping for some amount of time, regardless of how small.

So, if motion is to halt you need to show some change in time (crank angle) where the difference in displacement (location) of the piston is zero (no change in position, aka "no motion").

I need help finding this "flat spot" in displacement while the crank is still rotating, etc.

If you did a super fine analysis to spit out piston travel vs crank angle, and blew the graph up to super small angles of crank rotation you'd see the super small duration flat spot where the piston stops. The graphs posted don't have nearly that kind of resolution.

If you had an engine with the head off, setup a dial indicator to watch piston travel and watch it as you rotate a piston through TDC.

 

[Astro14]

If the velocity did not go to zero, the piston would not change direction. It has to go through zero for the direction to reverse.

Ergo, it stops. Then goes the other way.

 

[PWMDMD]

Great minds, Astro 14 - this was an IM.


"The math is not hard and pretty basic. That graph represents the displacement of the piston with respect to the crank angle. It is a FACT that at precisely 0, 180, 360, 540 degrees the derivative is zero and that the change in displacement with respect to crank angle is zero. Now the difficulty is in the idea of the limit as it approaches these crank angles. The derivative at 179.9 or 179.99 or 179.999 or 179.99999999999999999999999999999999999999999999999999999999999 is not zero but some very small number. It is only precisely at 180 degrees, to the degree that crank angle can even be measured, that the derivative is zero and the piston is not moving. Said another way you can not go from a positive displacement in the upward direction to a negative displacement in the downward direction without passing through zero - it doesn't magically skip zero."

 

[EMPIRE]

The irony is that his linked article completely obliterates his original incorrect argument ( must be that anti-matter "nucular" engineering)

From the article>

As you can see, when the crank angle is at 0°, the piston is positioned at TDC and the displacement x reaches the maximum value. After half of rotation (180°), the piston is at BDC and the displacement reaches minimum value.

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. This is the main reason for which internal combustion engines have flywheels on the crankshaft, to keep the pistons moving through the dead centers.

How much time is what I am asking.

 

[ZeeOSix]

How much time is what I am asking.

Depends on the crank throw, rod length and the engine RPM.

 

[Astro14]

Look at the equation highlighted in your link.

Answer the following question: what’s the Sine of 0? Sine of 180? Sine of 360?

Take that value, and plug it into the highlighted equation.