syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 Did you understand what I was saying syntax? The crankshaft still rotates in the given direction. It always goes in the same direction, even when the car is in reverse. It's simply a matter of inserting or removing gears as necessary to reverse the motion of the shaft that turns the wheels. Yes, I understand what you are saying, but the direction of the crankshaft is not the question posed. The question was whether or not the piston comes to a stop before it reverses direction. Link to comment Share on other sites More sharing options...
syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 this was a funny thread. syntax says "how does reverse work' date=' does the car engine---pistons valves intake exhaust and all really go into reverse?" then there is a big discussion at post #21 blike says: "Nobody answered his question yet!" [/quote'] I did not ask anything of the kind. I asked if a piston, attached to a crankshaft, necessarily had to come to a stop before it reversed direction. The following is a copy and paste of my original post: Does an object necessarily have to come to a stop before it can reverse it's direction? Take for example a piston attached to a connecting rod, which is attached to a crankshaft. Now--assuming that there is no side play of the piston in the cylinder, and assuming that the connecting rod would not either compress of stretch during the cycle, and assuming that this particular piston was not part of the fireing cycle, would it have to stop at the top before it started down? Remember, the crankshaft does not stop, so once the throw on the crank reaches the top of the stroke it immediately starts down pulling the piston behind it. If there is no clearance in the rod bearings, the rod didn't stretch or compress, how could the piston stop if the crank kept turning? Link to comment Share on other sites More sharing options...
syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 is there something about the word "instant" that people on this forum just dont like? do they not know what it means? do they not like it? what is it? How would you define "instant." Link to comment Share on other sites More sharing options...
blike Posted February 2, 2005 Share Posted February 2, 2005 Yes, I understand what you are saying, but the direction of the crankshaft is not the question posed. The question was whether or not the piston comes to a stop before it reverses direction. Aha, apologies. I misunderstood you completely. Link to comment Share on other sites More sharing options...
syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 Aha, apologies. I misunderstood you completely. No sweat. Link to comment Share on other sites More sharing options...
YT2095 Posted February 2, 2005 Share Posted February 2, 2005 So' date=' if the crank turned at high speed for many months, and all those "instants" when the piston was at rest were added up, it would still be zero, correct? Now if that is correct, the question is, can an object in motion come to a stop and then start to move again without any time passing while it was at rest? Is this what is meant by the old saying---"no time?" [/quote'] 1. correct, it would still be zero (in a perfect system). 2. yes it can, if traveling in the same direction (not in a reverse direction tho`). 3. I have no idea what that "Old saying" means beyond a simple coloquialism. Link to comment Share on other sites More sharing options...
syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 1. correct' date=' it would still be zero (in a perfect system). 2. yes it can, if traveling in the same direction (not in a reverse direction tho`). 3. I have no idea what that "Old saying" means beyond a simple coloquialism.[/quote'] It would seem that since answer #1 is correct and that answer #2 is no, at least to an object moving in an opposite direction, that the piston does not stop--correct? Link to comment Share on other sites More sharing options...
YT2095 Posted February 2, 2005 Share Posted February 2, 2005 Linear movement like that of a piston (or the wire on a rotating disc that I used as example) will indeed change direction by 180 degrees at some point. and in doing so will be required to obtain a position of "STOP" albeit only momentary. That is what I`m saying Link to comment Share on other sites More sharing options...
syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 Linear movement like that of a piston (or the wire on a rotating disc that I used as example) will indeed change direction by 180 degrees at some point. and in doing so will be required to obtain a position of "STOP" albeit only momentary. That is what I`m saying Not to be argumentative' date=' but I have a hard time understanding how..... 1)an object cannot reverse direction without stopping and 2)an object cannot stop and then move in the reverse direction without using [b']some[/b] amount of time, yet 3)the piston under discussion would not use any time to do that very thing thousands of times over. What am I missing here? Link to comment Share on other sites More sharing options...
YT2095 Posted February 2, 2005 Share Posted February 2, 2005 between the compression stage at Max and the "Bang" there is a period of "Stop". THERE will be your period of transition from 180+ and 180- anyway bud, gotta go finish the supper, we shall continue this another time if no-one`s managed to explain better than I can during Link to comment Share on other sites More sharing options...
Mart Posted February 2, 2005 Share Posted February 2, 2005 Not to be argumentative' date=' but I have a hard time understanding how..... 1)an object cannot reverse direction without stopping and 2)an object cannot stop and then move in the reverse direction without using [b']some[/b] amount of time, yet 3)the piston under discussion would not use any time to do that very thing thousands of times over. What am I missing here? You don't say what you think an object is. How big is it? Can it be any size? Or are there limits? Link to comment Share on other sites More sharing options...
syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 between the compression stage at Max and the "Bang" there is a period of "Stop". THERE will be your period of transition from 180+ and 180- anyway bud' date=' gotta go finish the supper, we shall continue this another time if no-one`s managed to explain better than I can during [/quote'] Oh, if it were in a real engin, I have no doubt that it would stop all right. The rod would compress a little on the compression stroke, there would always be a little clearance in the bearings and so forth. But I stipulated that this piston was not part of a fireing cycle. It is used only as a mechanical exercize to find out if, under the circumstances discribed, a piston would necessarily have to stop before it started to move in the opposite direction. Link to comment Share on other sites More sharing options...
syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 You don't say what you think an object is. How big is it? Can it be any[/b'] size? Or are there limits? If you will refer to post # 1 in this thread, it is laid out there. Link to comment Share on other sites More sharing options...
Mart Posted February 2, 2005 Share Posted February 2, 2005 If you will refer to post # 1 in this thread, it is laid out there. OK. Done it. Thanks The crankshaft is rotating. The crankshaft is made of smaller parts connected in some way. Focus your thoughts on one of these parts. This part is moving (it's part of the moving crankshaft). How big is that part do you think? Link to comment Share on other sites More sharing options...
Mart Posted February 2, 2005 Share Posted February 2, 2005 How would you define "instant." I don't think it can be zero. Doesn't HUP puts limits on this. Energy x time are equal to or greater than a minimum value. If time = 0 then HUP can't work. Link to comment Share on other sites More sharing options...
swansont Posted February 2, 2005 Share Posted February 2, 2005 I don't think it can be zero. Doesn't HUP puts limits on this. Energy x time are equal to or greater than a minimum[/i'] value. If time = 0 then HUP can't work. If you want to delve into the quantum regime then you can never measure any object to be at rest, because you will know the momentum. But I think this problem is classical in nature. Imagine you roll your car toward an intersection with a stop sign, going uphill, and don't touch your brakes. Your car rolls forward, and then backward, and then you step on the gas and go through the intersection. Did you obey the stop sign, or can the cop give you a ticket for running it? As I stated before - you need to decide what "stopped" means. v=0 at all, or v=0 for some finite amount of time. Having an argument incorporating ambiguous terms is just so much blather. Link to comment Share on other sites More sharing options...
Mart Posted February 2, 2005 Share Posted February 2, 2005 If you want to delve into the quantum regime then you can never measure any[/i'] object to be at rest, because you will know the momentum. I don't follow this. Do you mean "because you will know the momentum will be zero"? Link to comment Share on other sites More sharing options...
The Rebel Posted February 2, 2005 Share Posted February 2, 2005 "v=0 for some finite amount of time" Think that is the key concept here. In this scenario the time is not finite. An object under acceleration will have different velocities and different times, which could be -10ms-1, 10ms-1 and 0ms-1, etc. We can not define the period of time that an object remains at a particular velocity because it doesn't remain at all, we just know that it passes through it. Take, for example, a plot of varying velocities against time. If we take two points on this curve, a draw a straight line between them, we find the acceleration by the difference in velocity divided by the difference in time. a = (v2-v1) / (t2-t1), which in mathematical terms is just the same as a = dv/dt If we make the line shorter by making the velocities closer to each other, the time difference reduces. Until eventually when v2 = 0.00000001 and v1 = -0.00000001 etc, the time approaches zero. The difference in time never actually reaches zero though because then "a" would be undefined from the divide by zero problem. (this is the whole synopsis behind calculus methods, we approach but do not target) The words "instance" and "stopped" are subjective and can not be used to equate this scenario. Link to comment Share on other sites More sharing options...
Martin Posted February 2, 2005 Share Posted February 2, 2005 For those of you who like to calculate things: The throw on the crankshaft is 4" and the rotation is a constant 1000 rpm. If you think the piston stops' date=' tell me how long it remains motionless. [/quote'] It sounds like by "stops" you mean "remains motionless for some length of time" for something to stop, according to this, it has to remain motionless for some duration of time, which can be brief but not zero. as you have defined the mechanical system (with no play, no looseness) it is kind of idealized and mathematical. that is how you have specified it, which is fine. so there is no stoppage. the piston never remains motionless in calculus one could study the position of the piston as a differentiable function f(t) and then one could say that if it stops for some (even very small) interval all the derivatives of f become zero. but when the shaft is turning steadily as you specify, then even tho the first derivative f'(t) equals zero not and then the piston never stops when f' equals zero one still has f'' nonzero If you have not made clear that "stops" means remaining motionless for some finite length of time then I guess I agree with swantsont: As I stated before - you need to decide what "stopped" means. v=0 at all, or v=0 for some finite amount of time. Having an argument incorporating ambiguous terms is just so much blather. Link to comment Share on other sites More sharing options...
syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 OK. Done it. Thanks The crankshaft is rotating. The crankshaft is made of smaller parts connected in some way. Focus your thoughts on one of these parts. This part is moving (it's part of the moving crankshaft). How big is that part do you think? Well, it really doesn't matter how big the parts are, since this a hypothetically perfect mechanical device. Link to comment Share on other sites More sharing options...
syntax252 Posted February 2, 2005 Author Share Posted February 2, 2005 It sounds like by "stops" you mean "remains motionless for some length of time" for something to stop' date=' according to this, it has to remain motionless for some duration of time, which can be brief but not zero. as you have defined the mechanical system (with no play, no looseness) it is kind of idealized and mathematical. that is how you have specified it, which is fine. so there is no stoppage. the piston never remains motionless in calculus one could study the position of the piston as a differentiable function f(t) and then one could say that if it stops for some (even very small) interval all the derivatives of f become zero. but when the shaft is turning steadily as you specify, then even tho the first derivative f'(t) equals zero not and then the piston never stops when f' equals zero one still has f'' nonzero If you have not made clear that "stops" means remaining motionless for some finite length of time then I guess I agree with swantsont:[/quote'] Perhaps the term we are searching for is "dwell?" I can see where an object can stop and immediately reverse direction, but if we understand that there was no dwell time between the two movements, isn't our problem solved? Link to comment Share on other sites More sharing options...
swansont Posted February 2, 2005 Share Posted February 2, 2005 I don't follow this. Do you mean "because you will know the momentum will be zero[/i']"? No, just that you will know the momentum exactly, which means the uncertainty in it would have to be zero. But, if you have any idea where the object is, the uncertainty in x isn't infinite, so the uncertainty in p can't be zero. So you can never measure anything to be precisely at rest. Link to comment Share on other sites More sharing options...
The Rebel Posted February 2, 2005 Share Posted February 2, 2005 We can't use "dwell" either. as Jakiri sais in another thread the time between it being at V=0, to "having movement" is 1/infinity. Then again technically the object is always moving if there is acceleration. Maybe we our getting confused in our minds as to how an object could instantaneously not move and then move. Momentum is overpowering our thoughts maybe. But then momentum is proportional to velocity, therefore as velocity approaches zero so does the momentum. Link to comment Share on other sites More sharing options...
Martin Posted February 3, 2005 Share Posted February 3, 2005 For those of you who like to calculate things: The throw on the crankshaft is 4" and the rotation is a constant 1000 rpm. If you think the piston stops' date=' tell me how long it remains motionless. [/quote'] Perhaps the term we are searching for is "dwell?" I can see where an object can stop and immediately reverse direction' date=' but if we understand that there was no dwell time between the two movements, isn't our problem solved?[/quote'] you posed the problem, syntax, and defined the terms, so I guess you can say when it has been solved. personally, I would agree it has been solved. you decided to define dwell as standing still for some finite (nonzero) duration----and then you say the piston velocity may pass thru zero in the course of its cycle but at no time does it dwell----which makes sense to me anyway Link to comment Share on other sites More sharing options...
john5746 Posted February 3, 2005 Share Posted February 3, 2005 http://www.rose-hulman.edu/Class/CalculusProbs/Problems/ENGINE/ENGINE.html Link to comment Share on other sites More sharing options...
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