11 Oct 05
4 edits
Original argument:
Even though it's attached to a moving crank, that crank at two points in time moves perpendicular to the line of motion of the piston. At those two points in time the piston stops.
Unless you know some way to prove a single point (TDC/BDC) shows direction or is perpendicular to the pistons line, your case holds no water.
Not to mention the fact that point A travels a circular path.
Only lines, rays, segments, vectors, etc can be perpendicular to anything. Points and groups of points cannot.
Not circles either.
11 Oct 05
Originally posted by Raw760You misunderstand me. I am not arguing that the piston spent more time at the top of it's stroke. Nor did I say that point A's path was perpendicular to line P.
[b]Q:Then how can you convince me that the piston *did* spend *more* time at the top of its stroke, which is TDC(or even BDC)?
A:Why would I try to convince you of that?
Because that is your argument, along with stating that point A's circular path is perpendicular to line P at TDC and BDC. (using tangents cannot illustrate this)[/b]
I said the piston stops for an infinitesimal amount of time dt at the top of it's stroke and that point A's velocity is perpendicular to line P at TDC and BDC.
Have you taken a mechanics class? Mechanics is low level physics that deals with velocities, forces, etc.
12 Oct 05
Originally posted by Raw760You can't. So you can't prove your argument.
[b]Q: ...did the connection point spend any more time in TDC or BDC than it did in any other position while it was moving????
A: Nope.
----------------------------
Q:Then how can you convince me that the piston *did* spend *more* time at the top of its stroke, which is TDC(or even BDC)?
A:Why would I try to convince you of that?
Y ...[text shortened]... that they do in any other position. (as you said also)[/b]
This indicates continuous motion.[/b]
You are completely mistaken in your belief that my argument relies on the idea that the piston spent more time at the top of it's stroke than at any other time. I have no idea where you got this idea.
The fact that any object which reverses direction must stop in the process is well known. Are you aware that you're not just disagreeing with me, but that you're claiming that mathematics and physics as we know them are just plain wrong?
...isn't it impossible to reverse direction without stopping for just an instant?
http://www.karlscalculus.org/calc5_0.html
And take a look at the graph on this next website. Note that the green line, which represents velocity, passes through zero at TDC and BDC, indicating that the piston has zero velocity at those positions. Also note this quote:
...at TDC and again at BDC, the piston velocity is zero.
http://www.epi-eng.com/ET-PistnVelAccel.htm
Also, look at table 2 on this website and note how when the angle is 180 degrees and 360 degrees - that is, BDC and TDC - the velocity of the piston is zero, which means it's stopped.
http://www.wfu.edu/~rollins/piston/
12 Oct 05
1 edit
An explanation with cartoons!
Did you pick that out just for me?
Can't argue with Wile E. Coyote!
I find it hard to believe and harder to understand.
It seems illogical for pointA(journal) to not stop traveling its path while the piston does, even if for a miniscule period of time.
May make more sense to me when thinking of gravity(you can overcome and submit to its forces with out breaking stuff), the mechanical operation of a pistion does not equate the same to me in my mind.
But the numbers are the numbers, and they don't lie.
Still a good topic when drinking beer with your mechanic.
Though I still don't get what tangents had to do with this? Never saw any in any illustration or equation concerning a piston.
12 Oct 05
Originally posted by Raw760The velocity of point A (or any object moving in a circular path) is always tangent to it's circular path. Think of it this way: point A can be represented by a screw that holds L and C together. Now imagine that as the whole setup is working at high speed, everything but the screw at A were to disappear. What would happen to A? It would shoot out in a straight line tangent to the circular path it was previously taking, in the same way that a sling bullet shoots straight out tangent to the circular path it was on once the slinger releases it. (I'm referring to the medeival weapon sling: http://en.wikipedia.org/wiki/Sling_%28weapon%29)
An explanation with cartoons!
Did you pick that out just for me?
Can't argue with Wile E. Coyote!
I find it hard to believe and harder to understand.
It seems illogical for pointA(journal) to not stop traveling its path while the piston does, even if for a miniscule period of time.
May make more sense to me when thinking of gravity(you can overcome ...[text shortened]... tangents had to do with this? Never saw any in any illustration or equation concerning a piston.
Why does the screw move in a circular path? Because the Circle imposes a force upon it in the direction of the center of the Circle. This causes an acceleration on the screw which is perpendicular to the screw's velocity.
http://www.mcasco.com/p1cmot.html
The red arrows represent velocity. Note how they are tangent to the circular path and perpendicular to the centripetal force (green arrows) that pulls the object toward the center.
12 Oct 05
Originally posted by Raw760Maybe in a perfect world would that be true, however you are not
The point of connection is on a circular path. It will never run perpendicular or parallel to anything.
BECAUSE ITS DIRECTION IS CONSTANTLY CHANGING.
So long as the engine runs the piston will be pushed and be pulled in relation to this point. So it will always have a measurable velocity.
Even from TDC and BDC.
CAN YOU READ THIS PRINCESS???
taking into consideration the fact that the bearings, the rods, the
pistons, etc., have a bit of slop, maybe 0.002 inch or thereabouts.
Therefore when the piston gets to TDC, for instance, there in fact will
be a short but finite time in which said slop will ensure the piston
will in fact stop, then when all the slops of all the parts are taken up,
at that time and not before, the piston will start going down from TDC
and not a nanosecond before. BTW, ignition happens a few degrees
on the way down so it hasn't started firing at TDC.
13 Oct 05
Originally posted by sonhouseYour explanation may be true but it's not the only reason pistons stop. Anything which changes direction in any dimension must stop moving in that dimension in the process. It's theoretically impossible for this not to happen.
Maybe in a perfect world would that be true, however you are not
taking into consideration the fact that the bearings, the rods, the
pistons, etc., have a bit of slop, maybe 0.002 inch or thereabouts.
Therefore when the piston gets to TDC, for instance, there in fact will
be a short but finite time in which said slop will ensure the piston
will in fact ...[text shortened]... efore. BTW, ignition happens a few degrees
on the way down so it hasn't started firing at TDC.
13 Oct 05
Originally posted by AThousandYoungThat tangent would only be a representation of point A's velocity, not any actual motion unless pointA broke free.
The velocity of point A (or any object moving in a circular path) is always tangent to it's circular path. Think of it this way: point A can be represented by a screw that holds L and C together. Now imagine that as the whole setup is working at high speed, everything but the screw at A were to disappear. What would happen to A? It would shoot out ...[text shortened]... d perpendicular to the centripetal force (green arrows) that pulls the object toward the center.
Point A only travels the circular path, and the piston only moves in relation to this path.
So I would give you a hard earned rec if I could.
And Suzianne gets a...
no wait that one goes to you too AThousandYoung, for patience and reasoning.
13 Oct 05
3 edits
Originally posted by sonhouseThis is my only problem with your post.
BTW, ignition happens a few degrees
on the way down so it hasn't started firing at TDC.
Ignition Advance
The reason for ignition advance is that the air/fuel mix doesn't burn instantaneously - it takes a little bit of time from the moment the spark is set off to the moment when peak cylinder pressure is reached. During that time, the crankshaft keeps rotating. So if you lit off the mix at TDC, the piston will be well down the bore by the time peak cylinder pressure is reached, and you'll get lousy horsepower and lots of unburned fuel out the tailpipe. The cure is to light off the air/fuel mix *before* the piston reaches TDC, so that the peak cylinder pressure is achieved at just about the time the piston is positioned to take full advantage of it. This is what we call ignition advance.
Okay, let's back up a bit. Why do we need any ignition timing advance at all? Why not light the fire off when the piston is at TDC? The answer is that it takes the flame some time to grow to fill the whole combustion chamber, and during the time the flame is growing, the piston keeps moving. The only way to get the flame to finish burning and produce lots of pressure on the piston when it's ready to be pushed down the bore, is to light it off early, while the piston is still moving up the bore. That is why ignition timing is needed.
How much advance do you need? Clearly, it depends on how fast the air/fuel mix burns, and how fast your engine is turning. Roughly speaking, if your engine is turning faster, you want more advance; this is the why distributors have mechanical advance in them, which puts out the spark earlier and earlier in the cycle as the engine rpm's climb. Once you reach a high enough rpm, the air-fuel mixture begins to whoosh into the cylinder with so much velocity that it becomes turbulent, and consequently the flame spreads very fast; increase the rpm, and the mixture becomes more turbulent in direct proportion, and the flame spreads even faster. This means that once you exceed a certain high rpm, the mixture tends to burn in about the same number of crankshaft degrees, no matter what the rpm. Now you no longer need the ignition timing to keep advancing with increasing rpm, so the distributor is designed to level off the advance above some rpm.
http://www.442.com/oldsfaq/ofign.htm
Feul ignition set off by anything other than the spark plug(called pre-ignition or detonation) is bad.
Pre-ignition/detonation could destroy an engine.
http://www.misterfixit.com/deton.htm
13 Oct 05
Originally posted by AThousandYoungI know it and you know it but I was trying to let him off easy....
Your explanation may be true but it's not the only reason pistons stop. Anything which changes direction in any dimension must stop moving in that dimension in the process. It's theoretically impossible for this not to happen.
13 Oct 05
Originally posted by Raw760Maybe part of your confusion is based on your lack of calculus experience. Even though the piston must stop in order to reverse direction, it only needs to do so for an infinitesimal amount of time. This means that the amount of time it's stopped is incredibly tiny - almost but not quite zero. It's basically zero for most intents and purposes except for theoretical discussions like this.
No need for that...
I admit I must be wrong, I do dot understand the true answer.
And I will not be satisfied with that answer till I can understand for my self!!!
but thats my problem.
Regarding how tangents apply to this problem - the velocity of A is tangent to it's circular path at all times. At TDC and BDC that tangential velocity is along the x axis. Point A pushes/pulls on L according to it's velocity. So, for that instant, A is pushing it's end of L along the x axis. In that instant L is oriented along the y axis. The force of A pushing on the end of L will tend to cause it to spin and/or translate in the x-direction. If L were to spin, the other end would move along the x axis in the opposite direction, pushing the piston only along the x axis. There would be absolutely no y movement of the piston at that instant. For that instant, A will not be pushing or pulling L in the y direction at all.
However, keep in mind we're talking about an infinitesimal amount of time. In such a short period of time, there is no "curving path". A millionth of a nanosecond later, A will begin pulling the piston down oh so slightly and the piston will be moving. A millionth of a nanosecond before TDC, A will be pushing the piston up oh so slightly and the piston will be moving. It's only for a time that is smaller than any finite amount of time you can imagine that the piston is stopped. So for most intents and purposes it's as though the piston never stopped, but in theory, it did.
14 Oct 05
Originally posted by AThousandYoungdude....
Maybe part of your confusion is based on your lack of calculus experience. Even though the piston must stop in order to reverse direction, it only needs to do so for an infinitesimal amount of time. This means that the amount of time it's stopped is incredibly tiny - almost but not quite zero. It's basically zero for most intents and purposes exc ...[text shortened]... o for most intents and purposes it's as though the piston never stopped, but in theory, it did.