Since we are all PhD.'s is BSology...

Originally posted by Raw760
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???

Good lord... BSology indeed.

At two points is the piston stoppped, what you call TDC and BDC.

At these points, the tangent of the point of connection to the crankshaft IS perpendicular to the movement direction of the piston.

Why you can't grasp this fundamental concept is baffling.

EDIT: And I can read just fine... it's *what* I am reading that is BS.

Sorry to make you think and answer for yourself.

No I don't quite grasp that, but I will try.

The piston had a velocity as it moved into the point of Top Dead Center(of crank revolution).
And then it moved out of that TDC position.

Did it spend any more time in that position than it did in any other position while it was moving????
I don't see how. The vertical motion follows an arc, no points would be the same consecutively.

Mechanics do graph piston velocity as passing through zero to show deceleration.
Correction they graph through zero into negative to show a continuous graph of the other half of the full revolution.

Your thought?

I suck at chess too, how bout a game?

Originally posted by Suzianne
Good lord... BSology indeed.

At two points is the piston stoppped, what you call TDC and BDC.

At these points, the tangent of the point of connection to the crankshaft IS perpendicular to the movement direction of the piston.

Why you can't grasp this fundamental concept is baffling.

EDIT: And I can read just fine... it's *what* I am reading that is BS.

Did you mean the pistons vertical line of motion is tangent to the connection point's circular line of motion at two points?
I'm not so sure that even fits the definition of a tangent, at least not for the points you are talking about.
Which represent all single points any way.

Note that in the important case of a circle, however, the tangent line will intersect the curve at only one point.
http://en.wikipedia.org/wiki/Tangent
http://en.wikipedia.org/wiki/Circle

Yes TDC and BDC are in line with the piston's line of motion, that is why mechanics call it this. They present the point of full extension of piston travel in either direction.

But you will have to try harder to convince me that the connection point travels perpendicular or parallel to the piston's line of motion, or any other straight line. Its direction is constantly changing, unlike the piston.

Can single point(tangent) be perpendicular to a line?
Can any two consecutive points on an arc be perpendicular to a vertical line?

Did the piston spend any more time in TDC or BDC than it did in any other position while it was moving????
I don't see how. The vertical motion follows an arc, no points would be the same consecutively.

I think no, no and no. So the piston did not stop.

Thoughts?

I side with logicalhippo on this.

Thank you for making more clear to me that you are wrong.

Originally posted by Raw760
Did you mean the pistons vertical line of motion is tangent to the connection point's circular line of motion at two points?
I'm not so sure that even fits the definition of a tangent, at least not for the points you are talking about.
Which represent all single points any way.

Note that in the important case of a circle, however, the d be the same consecutively.

I think no, no and no. So the piston did not stop.

Thoughts?

A tangent is not a single point. A tangent is a line that touches a circle at one point.

Let's imagine a piston P, a spinning rod which comes out of a motor R, a circle C which is attached to the spinning rod, and a linking rod that connects the circle and the piston L.

If you look at the third image on this website (under Parts)

http://en.wikipedia.org/wiki/Internal_combustion_engine

you'll see that the Rod and Circle are colored purple, the Linking Rod is brown-orange and the Piston is yellow.

(still working on this)

The Rod will spin, causing the Circle to spin, which causes the point connecting C and L (we'll call it point A) to move in a circular path. If you consider the velocity of A at any point in time, it's velocity will be a vector - which is never curved - which points along the tangent of A's path of motion.

The speed at which the the piston moves along y depends on the y component of the velocity of A. When that y component is zero, the piston stops moving for an infinitesimal amount of time. This takes place when A has its maximum y value as well as when it has its minimum y value.

Good to have a reasonable person on this debate.

What happened to the initial argument that the connecting point on the crank moves perpendicular to the pistons line of motion?

Princess was sooooo keen on this.

I am aware a tangent is not a point. I did refer to the pistons line of motion as being tangent.

Princess argued of a perpendicular tangent line to the piston's line. There is no such line, where does this line come from?

I say nothing in the engine connected to the piston moves perpendicular to the piston.

Only two points on any circle will give a slope that is a negative reciprocals of vertical line, making that line perpindicular to the piston's(and they are not consecutive).
Those are the two ponits inbetween TDC and BDC.
Not TDC and BDC in our case.

Still there is no perpendicular travel of the connecting point, to the piston's line.
Are we in agreement on this point?

(still working on this also, leave a message and I'll get back to you.)

I also ask again,

Can single point be perpendicular to a line?

Can any two consecutive points on an arc be perpendicular to a vertical line? In particular if one is TDC or BDC as in our problem?
Note: TDC means this point is the highest and central point on an arc, BDC makes it the lowest and central point in an arc.
No other point on the arc is >= TDC in Y-value. No other point is == TDC in X-value.
No other point on the arc is <= BDC in Y-value. No other point is == BDC in X-value.
Meaning you cannot use TDC or BDC to create an imaginary perpendicularl tangent to the piston's vertical line, because any other point on the arc used for this imaginary perpendicularl tangent will have an invalid Y-value.

Did the piston spend any more time in TDC or BDC than it did in any other position while it was moving????
Note: the piston only moves in direct relation to the connection point.

Better yet, did the connection point spend any more time in TDC or BDC than it did in any other position while it was moving????

Originally posted by Raw760
Good to have a reasonable person on this debate.

What happened to the initial argument that the connecting point on the crank moves perpendicular to the pistons line of motion?

Princess was sooooo keen on this.

I am aware a tangent is not a point. I did refer to the pistons line of motion as being tangent.

Princess argued of a perpendicular tangen ...[text shortened]... ement on this point?

(still working on this also, leave a message and I'll get back to you.)

For two infinitesimal periods of time, the connecting point A does move perpendicular to the Piston's line of motion. If the Piston is moving in the y direction, then when the Piston is fully extended or retracted (and the Linking Rod is parallel to the Piston's line of motion) then A will be moving perpendicular to the Piston's line of motion. This only occurs for an infinitesimal amount of time dt, and so A only moves the infinitesimal distance dx before it no longer is moving perpendicular to P's line of motion.

In short, Princess is right.

Of course there is a tangent line perpendicular to the piston's line of motion. There are infinite tangent lines to any circle and there are two which have any possible angle relative to the x axis.

I say nothing in the engine connected to the piston moves perpendicular to the piston.

Not true.

I don't know what TDC or BDC are but I will check.

Apparently, they mean Top Dead Center and Bottom Dead Center.

http://www.faqfarm.com/Q/What_do_BDC_and_TDC_mean_in_motorcycle_jargon

So, TDC and BDC are indeed the two points at which A moves perpendicular to P's line of motion.

Only two points on any circle will give a slope that is a negative reciprocals of vertical line, making that line perpindicular to the piston's(and they are not consecutive).
Those are the two ponits inbetween TDC and BDC.
Not TDC and BDC in our case.

I don't understand.

Can single point be perpendicular to a line? Can any two consecutive points on an arc be perpendicular to a vertical line? In particular if one is TDC or BDC as in our problem?

No, and no. Only lines, rays, segments, vectors, etc can be perpendicular to anything. Points and groups of points cannot.

No other point is == TDC in X-value...No other point is == BDC in X-value.

I don't know what the double equals signs mean, but BDC and TDC have the same x value.

Meaning you cannot use TDC or BDC to create an imaginary perpendicularl tangent to the piston's vertical line, because any other point on the arc used for this imaginary perpendicularl tangent will have an invalid Y-value.

I don't think you understand the concept of a tangent. A tangent touches a circle at only one point. Therefore your reference to using two points on the arc to define a tangent makes no sense.

Better yet, did the connection point spend any more time in TDC or BDC than it did in any other position while it was moving????

Nope.

It sounds like you've never studied calculus. Have you?

Originally posted by AThousandYoung
For two infinitesimal periods of time, the connecting point A does move perpendicular to the Piston's line of motion. If the Piston is moving in the y direction, then when the Piston is fully extended or retracted (and the Linking Rod is parallel to the Piston's line of motion) then A will be moving perpendicular to the Piston's line of motion. T ...[text shortened]... hile it was moving????

Nope.

It sounds like you've never studied calculus. Have you?[/b]

No I have hot studied calculus.
I did make some incorrect statements unclear statements, sorry.
I'll try to clear them up.

-------------------------------
I meant to say;

Only two non consecutive points on any circle will give a slope that is a negative reciprocals of vertical line's slope, making that line perpendicular to the piston's vertical line on motion.
Those are the two center points between TDC and BDC.
Not TDC and BDC in our case.

your answer:I don't understand.

Basically that means no two consecutive points on an arc or circle can be perpendicular to a vertical line. As you yourself answered later.

-------------------------------

I said:
No other point is == TDC in X-value...No other point is == BDC in X-value on an arc.
I meant to add "on an arc" because I was talking of the circle as two arcs

your answerI don't know what the double equals signs mean, but BDC and TDC have the same x value.

== means "is equal to" in program logic. != means "is not equal to."

Note: TDC means this point is the highest and central point on an arc, BDC makes it the lowest and central point in an arc.
No other point on the arc is >= TDC in Y-value. No other point on this arc == TDC in X-value.
No other point on the arc is <= BDC in Y-value. No other point on this arc == BDC in X-value.


-------------------------------

Q: Better yet, 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.

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)?
It move in relation to the connection point!

I still say it does not stop, but keeps moving at in infinitesimal velocity.

-------------------------------
I said:

Note: TDC means this point is the highest and central point on an arc, BDC makes it the lowest and central point in an arc.
No other point on the arc is >= TDC in Y-value. No other point on this arc == TDC in X-value.
No other point on the arc is <= BDC in Y-value. No other point on this arc == BDC in X-value.

Meaning you cannot use TDC or BDC to create an imaginary perpendicular tangent to the piston's vertical line, because any other point on the arc used for this imaginary perpendicular tangent will have an invalid Y-value.
my bad wording == bad logic

your answer:I don't think you understand the concept of a tangent. A tangent touches a circle at only one point. Therefore your reference to using two points on the arc to define a tangent makes no sense.

But you telling me the single points of TDC and BDC means perpendicular travel makes no sense to me either!

You need two points to make a line. We are talking about an arc, so no points on this arc will be perpendicular or parallel to any line.

You cannot draw a line that is perpendicular to the P's line using TDC or BDC. So you cannot argue that the connection point moves perpendicular to the P's line using TDC or BDC as a reference, or at all.

Tangent lines are outside the connection points line of travel and only touch any single point. including TDC or BDC, so you can't use Tangents to show perpendicular travel of the connection Point.
They have no meaning or bearing on the mechanical operation.


-------------------------------
you said:
In short, Princess is right.

Of course there is a tangent line perpendicular to the piston's line of motion. There are infinite tangent lines to any circle and there are two which have any possible angle relative to the x axis.

my answer:
Infinite tangent lines or not, these lines which are outside the connection points line of travel and only touch any single point have no meaning or bearing on the mechanical operation.

Where do these lines come from, nothing moves perpindicular to the P's line.

I still say the piston does not stop, but keeps moving at in infinitesimal velocity near at TDC and BDC.

Remember?:
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.

ponit A is conecting point on crank, Line P is piston's line of travel.

You cannot use a tangent line touching TDC or BDC to show perpendicular travel of point A to Line P.
All other points on the tangent line are not on point A's circular path, or even with in it.

A single point cannot show perpendicular travel of point A to line P, not even TDC or BDC.

You cannot show perpendicular travel of point A to line P.
Any two consecutive points (including TDC or BDC as one of the points) on a circle or arc will not be perpendicular to a vertical line because one of the two points will always have an invalid Y-value. The slope of any two consecutive points on a circle/arc will not be a negative reciprocals of a vertical line's slope. I'm not so sure vertical or horizontal line has a slope.

Only lines, rays, segments, vectors, etc can be perpendicular to anything

So long as point A does not stop its circular motion, the piston travels its vertical line.
No matter how slowly or if at the pace of 10k+ rpms.

Originally posted by Raw760
No I have hot studied calculus.
I did make some incorrect statements unclear statements, sorry.
I'll try to clear them up.

-------------------------------
I meant to say;

Only two [b]non consecutive points on any circle will give a slope that is a negative reciprocals of vertical line's slope, making that line perpendicular to the piston ...[text shortened]... y more time in TDC or BDC than it did in any other position while it was moving????

A: Nope.[/b]

Only two non consecutive points on any circle will give a slope that is a negative reciprocals of vertical line's slope...

A vertical line has an infinite slope, so I don't know if you can take a reciprocal of it. But if we call the reciprocal of infinity zero it seems to work out. I don't think that's relevant to our disagreement anyway.

Now, if you're trying to find two 'nonconsecutive' points for which, if a line is drawn through them, the slope of that line will be zero, there are an infinite number of such pairs of points. Any two points with the same y value will do. However that is totally irrelevant to this problem. What are 'consecutive' points anyway?

Basically that means no two consecutive points on an [b]arc or circle can be perpendicular to a vertical line.[/b]

Points cannot be perpendicular to lines at all. Lines, segments, vectors etc can be perpendicular to lines; not points or pairs of points.

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)?

Why would I try to convince you of that?

I still say it does not stop, but keeps moving at in infinitesimal velocity.

Would that infinitesimal velocity be upwards or downwards?

Take a look at this webpage:

http://www.mcasco.com/p1cmot.html

It describes circular motion, such as that of the connection point A. Notice the red arrows. They describe the velocity of the particle (A in our case) at any particular moment in time. That velocity is straight - it's a vector, so it has to be straight. It's also tangent to the circular path. At TDC, that velocity will be along the x axis. Does that help?

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)

Points cannot be perpendicular to lines at all. Lines, segments, vectors etc can be perpendicular to lines; not points or pairs of points.

Only lines, rays, segments, vectors, etc can be perpendicular to anything. Points and groups of points cannot.



But not arcs, curves or circles, as you say yourself.

The *circular* path of point A is never perpendicular to line P. So ponit A *never travels * perpendicular to line P.

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?

You can't. So you can't prove your argument.
The piston travels its vertical line in relation to point A, which never stops on it circular path.

Therefore point A never travels perpendicular to line P and the piston never stops.

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RAW: I still say it(piston) does not stop, but keeps moving at in infinitesimal velocity.

ATY: Would that infinitesimal velocity be upwards or downwards?

Depends on the position and direction of point A to TDC or BDC.
The closer to the point of TDC/BDC the slower the piston moves.

It reaches that point, but spend no great amount of time there than any other ponit before it moves on.

The piston travels straight through these points without stopping because the piston moves in relation to points A's circular path (which never stops as you say also) and cannot ever travel perpindicular to line P.

So neither point A or the Piston stay in the same position for any great amount of time that they do in any other position. (as you said also)

This indicates continuous motion.