Movement of a piston

[doG]

- Piston Motion Basics - Travel, Velocity, Acceleration, Vibration

 

There are charts and everything!

 

Notice in particular the graph that includes both velocity and acceleration.

 

Did you also happen to notice the part where your link says, "Note again that at TDC and again at BDC, the piston velocity is zero, because the piston reverses direction at those points, and in order to change direction, the piston must be stopped at some point."

 

I hear they have great correspondence programs for high school coarses these days. A good many of them are now done online. You might want to do a review of the basics before entering into discussions about matters you clearly have no properly functioning knowledge of.

Maybe you could see if they offer reading. It might help you from posting links that contradict yourself.

[iNow]

[Xittenn]

Did you also happen to notice the part where your link says, "Note again that at TDC and again at BDC, the piston velocity is zero, because the piston reverses direction at those points, and in order to change direction, the piston must be stopped at some point."

 

 

Maybe you could see if they offer reading. It might help you from posting links that contradict yourself.

 

Feel free to rearrange any of my statements as you please, whatever fits your needs big guy. The two of you are still talking out your asses and seriously need a refresher in mechanics. I am not the person to teach a high school physics class and I am not going to do so.

 

If you guys want to continue to believe that objects under gravity, with no other force applied--omitting friction--, achieve an acceleration of zero through their maximum height after having been thrown in the air, well go right ahead. If you want to believe that a piston isn't under the influence of acceleration through its turning point, feel free. I had argued that the piston was stopped because v = 0 at its turning point. Swansont made a comment that made sense to me, and I agreed it could be questionable because its acceleration is not 0 and the behaviour isn't clearly defined, as far as I know.

 

As far as I'm concerned the interest here of certain individuals is purely hormonal and we all know how boys feel about their mechanics. What man isn't an expert . . . . . most of them! Please, to the lot of you, if you don't like my comments present your own f@$!@ ideas and leave me out of it. I shouldn't have to come back to dispute this. Do prove me wrong, I will only benefit from learning. But again, I ask that you let me walk away. Thanks a bunch . . .

[doG]

Feel free to rearrange any of my statements as you please, whatever fits your needs big guy. The two of you are still talking out your asses and seriously need a refresher in mechanics. I am not the person to teach a high school physics class and I am not going to do so.

I didn't rearrange anything and any claim that I did is a lie. I simply pointed out that your own link contradicts your own position and threw your childish post about remedial high school classes back in your face! It's not my fault your link contradicts you and its not my fault you didn't read it.

 

Yes, technically everything on Earth is under constant gravitational acceleration so I assume it's your position that nothing ever truly stops relative to anything else on Earth because everything is constantly accelerated downwards by gravity.

 

Now, do you agree with your link that the piston stops at BDC and TDC or do you now think your link is in error?

[Xittenn]

I would say that an instantaneous time of zero acceleration occurs at the mid point of the piston's travel and instantaneous speed of zero occurs at the top and bottom of the pistons travel. i.e a sine and cosine relationship. ( Neglecting the fact that even with a long connecting rod and small crank displacement there will be a little distortion of the sine and cosine waveforms)

 

 

I linked the synopsis to show the above graph that clearly shows the acceleration at BDC and TDC, and to have something written for people to go over.

 

 

Yes, technically everything on Earth is under constant gravitational acceleration so I assume it's your position that nothing ever truly stops relative to anything else on Earth because everything is constantly accelerated downwards by gravity.

 

Now, do you agree with your link that the piston stops at BDC and TDC or do you now think your link is in error?

 

See post #53 . . . . .

[doG]

I linked the synopsis to show the above graph that clearly shows the acceleration at BDC and TDC, and to have something written for people to go over.

I copied and pasted the statement, "Note again that at TDC and again at BDC, the piston velocity is zero, because the piston reverses direction at those points, and in order to change direction, the piston must be stopped at some point." directly from your link, from about halfway down the page. I will also point out that it says, "Note again...." because it is not the first time it is pointed out on that page.

 

Now, your link clearly states in black and white that the piston stops at BDC and TDC. Those words are directly copied from your link. Now why on Earth do you think that your link to a page that clearly says that the piston stops would be accepted as support that the piston does not stop? Do you really want the rest of us to believe that you cannot see the contradiction here between what your link says and what you say it says?

[A Tripolation]

If you want to believe that a piston isn't under the influence of acceleration through its turning point, feel free. I had argued that the piston was stopped because v = 0 at its turning point. Swansont made a comment that made sense to me, and I agreed it could be questionable because its acceleration is not 0 and the behaviour isn't clearly defined, as far as I know.

 

I'm really trying to understand your argument here.

 

So everyone agrees that the velocity must equal 0 at the turning point? And then you're saying that the piston is under constant influence of acceleration due to the Earth's gravitational pull, therefore it never stops moving? Did I get it right?

[doG]

 

I linked the synopsis to show the above graph that clearly shows the acceleration at BDC and TDC, and to have something written for people to go over.

 

BTW, do note that this is a graph of the piston's motion, velocity or acceleration relative to crank rotation and it shows the pistons velocity is 0 at TDC and BDC.

 

I'm really trying to understand your argument here.

 

So everyone agrees that the velocity must equal 0 at the turning point? And then you're saying that the piston is under constant influence of acceleration due to the Earth's gravitational pull, therefore it never stops moving? Did I get it right?

No, he's saying that Swansont said:

 

I think you can make the argument that continuous motion means a and v are never simultaneously zero.

 

This in effect raises the question if a body in continuous motion can come to a stop. In the case of the piston you actually have an oscillating motion like a load bouncing on a spring or a pendulum. Both of these are also examples of continuous motion systems but can it be said that they 'stop' as they change direction?

Edited January 11, 2012 by doG

[Daedalus]

So everyone agrees that the velocity must equal 0 at the turning point? And then you're saying that the piston is under constant influence of acceleration due to the Earth's gravitational pull, therefore it never stops moving? Did I get it right?

 

I think the confusion here is force vs. acceleration. Just because the Earth's gravity will accelerate a body, does not mean that the body is accelerating.

 

In physics, acceleration is the rate of change of velocity with time.^[1] In one dimension, acceleration is the rate at which something speeds up or slows down. However, since velocity is a vector, acceleration describes the rate of change of both the magnitude and the direction of velocity.^[2][3] Acceleration has the dimensions L T ⁻². In SI units, acceleration is measured in meters per second squared (m/s²). (Negative acceleration i.e. retardation, also has the same dimensions/units.)

 

You have to remember Newton's third law of motion:

 

To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions.

 

The force of Earth's gravity is countered due to the ground pushing back. So there is no acceleration due to gravity on the piston while it is being supported by the rod, etc... When the piston is either at the top or bottom, both velocity and acceleration are zero. However, the force of gravity is always in effect. To put it more clearly... if a body was constantly being accelerated due to gravity, it would fall through the ground and eventually end up at the center of the Earth.

Edited January 11, 2012 by Daedalus

[Xittenn]

I agree with you. Imagine tossing a ball into the air. It will travel upward, slow down, reach it's highest point, stop, and then begin accelerating in the opposite direction. It travels along one vector and then has to stop in order to begin traveling along the opposite vector. It's going up, and (however momentarily) must stop before it can begin going down.

 

In my view, a piston is merely a more precise and mechanically controlled ball being tossed into the air so can be accurately described the same.

 

That's my conjecture, anyway.

 

EDIT: Cross-posted with CaptainPanic

 

 

Yes. It happened in the very first response to this thread. Post #2.

 

 

Consider 2 examples:

 

1. A ball is tossed vertically straight up into the air. At the apex of its travel it's velocity will hit zero as the acceleration changes from positive to negative. At the apex a and v with both equal 0.

 

2. A ball is launched in a parabolic arc like a cannonball fired from a cannon. At the apex of its travel its velocity will be continuously greater than zero but its acceleration will change from positive to negative. V will not equal 0 until the ball lands and comes to rest so a and v will never equal 0 simultaneously until the ball comes to rest.

 

IMO, the motion of the piston is similar to example 1 because of its axial travel in the piston bore. This becomes very evident if you place a dial indicator in the spark plug bore and bring the piston to top dead center in order to mechanically align the camshaft. You find there is a small amount of travel of the crankshaft, 1°-2°, where there is no change in the piston's position. This is caused by the change in angle of the connecting rod relative to the axis of the bore as the crankshaft moves across top dead center. During that period of travel you could consider the piston stopped.

 

 

I'm really trying to understand your argument here.

 

So everyone agrees that the velocity must equal 0 at the turning point? And then you're saying that the piston is under constant influence of acceleration due to the Earth's gravitational pull, therefore it never stops moving? Did I get it right?

 

 

I agree that v is 0 at the turning point. TonyMcC felt that this wasn't the case because the piston was accelerating through zero and he argued that the duration wasn't sufficient to be called an arrest in motion. I made my arguments and this led to post #19 & #29 by Swansont. He states that "you can make the argument that continuous motion means a and v are never simultaneously zero". If this is the case then we can say that the piston doesn't stop if it has acceleration.

 

Some people seem to have a misguided opinions on what exactly acceleration is, and what mechanics entails in general for that matter. And now I find me defending myself against allegations that I apparently believe the velocity doesn't achieve 0 through the turning point. Other allegations include that I am arguing that there is acceleration at the turning point for arguably incorrect reasons(vibration, solar system movements, etc.).

 

The piston is in fact designed to have acceleration through the turning point. Depending on how you want to define a body in motion this might be sufficient to say that the body does in fact remain in motion through its turning point, even despite popular hearsay. Personally I could care less either way. I did however feel the point about defining a body in motion as being any of, having v, having a, or both, was an important point. I did ask if this was properly defined as such within the field of physics in one of my posts.

 

I apologize if I got excited, I just don't know how to address argumentative situations. Usually people don't take me seriously so when I say I'm done they usually back off.

[swansont]

You have to remember Newton's third law of motion:

 

 

 

The force of Earth's gravity is countered due to the ground pushing back. So there is no acceleration due to gravity on the piston while it is being supported by the rod, etc... When the piston is either at the top or bottom, both velocity and acceleration are zero. However, the force of gravity is always in effect. To put it more clearly... if a body was constantly being accelerated due to gravity, it would fall through the ground and eventually end up at the center of the Earth.

 

That's not a third-law example. Newton's third law tells us that if the Earth exerts a force on an object, the object exerts the same magnitude force on the Earth. i.e. third-law force-pairs act on each other. What you want here is a first-law application. If the force of gravity is added to an equal but opposing force, the net force is zero and that means no acceleration. However, an acceleration from earth's gravity is always present, as long as the earth is there.

 

If the velocity and acceleration were zero, the object would not move. v=0 means it's at rest, and a=0 means the velocity is not changing — the object would just stay there. Acceleration basically tells what what the velocity is going to be an instant later ([math]v =v_0 + a dt[/math]), so if a is zero, it means the velocity is going to remain what it was.

[Daedalus]

That's not a third-law example. Newton's third law tells us that if the Earth exerts a force on an object, the object exerts the same magnitude force on the Earth. i.e. third-law force-pairs act on each other.

 

You are correct. Sometimes I mix things up.

[A Tripolation]

If the velocity and acceleration were zero, the object would not move. v=0 means it's at rest, and a=0 means the velocity is not changing — the object would just stay there. Acceleration basically tells what what the velocity is going to be an instant later ([math]v =v_0 + a dt[/math]), so if a is zero, it means the velocity is going to remain what it was.

 

So if we have v naught is equal to 0, doesn't that mean that the object is no longer moving at that exact instant? Even if it has acceleration, that doesn't mean that it is currently moving, right? Just that it will be in a moment. Or do I need to retake mechanics? Saying that a and v have to be simultaneously zero for no motion isn't entirely true, is it?

[swansont]

So if we have v naught is equal to 0, doesn't that mean that the object is no longer moving at that exact instant? Even if it has acceleration, that doesn't mean that it is currently moving, right? Just that it will be in a moment. Or do I need to retake mechanics? Saying that a and v have to be simultaneously zero for no motion isn't entirely true, is it?

 

It depends on how you define continuous motion. That's the basic problem here — the words are not well-defined. Math is much clearer.

[zapatos]

The way it was explained to me was that when a ball was tossed straight up, although velocity moved through zero, it did not spend any measurable duration of time at zero. Therefore the object was always in motion.

 

http://upload.wikimedia.org/wikipedia/commons/e/ea/Simple_Harmonic_Motion_Orbit.gif

 

In this example of simple harmonic motion, even though velocity passes through 0, it seems that point is no different than any other point. That is, why would one point be considered in motion while another would not?

[doG]

If the velocity and acceleration were zero, the object would not move. v=0 means it's at rest, and a=0 means the velocity is not changing — the object would just stay there. Acceleration basically tells what what the velocity is going to be an instant later ([math]v =v_0 + a dt[/math]), so if a is zero, it means the velocity is going to remain what it was.

The question here though is not what the velocity will be an instant later but what is the velocity this very instant, i.e. is the object 'stopped' when v=0?

[Xittenn]

If we start by saying that a body is in motion when a body has a v greater than zero then we can simply say that a body in motion has momentum. There are two clear arguments against a body in motion simply having acceleration under this condition. The first and most obvious is that it has no v. The second is a mathematical argument.

 

[math] \mathbf I = \mathbf p [/math]

 

[math] \mathbf F \cdot t = m \cdot \mathbf v [/math]

 

Being that we are examining the value of momentum imparted by an instantaneous value in impulse we will always get a value of zero because an instance occurs in zero time. Being that instantaneous I imparts zero momentum, one could say that this is sufficient argument to say that acceleration is not, in of itself, an argument for a body in motion.

 

Conversely, we could examine the absolute displacement through a v of zero. Being that a body will have a value of absolute displacement, or distance, through a turning point in velocity, one could argue that the body remains in motion through a velocity of zero if there is a consistent force of greater than zero and the object has acceleration.

 

Neither argument is sufficient to contradict the other and we are left with normative definitions. Unless we can conclusively test either argument as being definitively correct, a normative definition has to be established and adhered to. Judging by the outcome of this thread this will never happen and we are left without a positive definition for a physical process. This to me seems unusual because I would think most physical processes involve positive definitions as ambiguity can create problems in detailing solutions. That said I don't feel by defining motion as solely being related to velocity covers the true breadth of the topic. This, is quite unlike the definition that includes acceleration, which covers all case presented by mechanics. This of course is simply my opinion.

Edited January 11, 2012 by Xittenn

[swansont]

The way it was explained to me was that when a ball was tossed straight up, although velocity moved through zero, it did not spend any measurable duration

 

Which is why you have to either define what you mean, or ask if v=0.

[Phi for All]

The way it was explained to me was that when a ball was tossed straight up, although velocity moved through zero, it did not spend any measurable duration of time at zero. Therefore the object was always in motion.

The time a piston spends at TDC and BDC at zero has to be measurable. Even if it's a nanosecond it's still measurable.

[Daedalus]

That said I don't feel by defining motion as solely being related to velocity covers the true breadth of the topic. This, is quite unlike the definition that includes acceleration, which covers all case presented by mechanics. This of course is simply my opinion.

Then why define motion with just v and a? Wouldn't we have to state that jerk, the rate of change of acceleration, would also have to be zero? And if we state that the rate of change of acceleration would also have to be zero, wouldn't we have to state that all higher order derivatives of a body's position must also equal zero?

 

If x(t) represents the position of an object at time t, then the higher-order derivatives of x have physical interpretations. The second derivative of x is the derivative of x′(t), the velocity, and by definition this is the object's acceleration. The third derivative of x is defined to be the jerk, and the fourth derivative is defined to be the jounce.

I'm just throwing this out there because it would make sense that we would have to do so.

Edited January 12, 2012 by Daedalus

[Xittenn]

Then why define motion with just v and a? Wouldn't we have to state that jerk, the rate of change of acceleration, would also have to be zero? And if we state that the rate of change of acceleration would also have to be zero, wouldn't we have to state that all higher order derivatives of a body's position must also equal zero?

 

 

I'm just throwing this out there because it would make sense that we would have to do so.

 

I think it would suffice to say that the body would have to be acted upon by an unbalanced force, which I believe would imply all the rest. Individuals could then omit details as necessary!

 

**as a replacement for the a term in a possible redefinition

[TonyMcC]

The time a piston spends at TDC and BDC at zero has to be measurable. Even if it's a nanosecond it's still measurable.

 

Assuming the crank pivot is on the centreline of the cylinder the height of the piston depends on the joining of two lines. A line from the crank pivot to the crank point and a line the length of the connecting rod. At top dead centre these will be a straight line. The slightest movement of the crank will destroy the straightness of the line and lower the height of the piston. That is why I think the time at top dead centre is an infinitely small instant of time leading me to the conclusion that the piston doesn't actually stop. (Not even for a picosecond ).

Edited January 13, 2012 by TonyMcC