syntax252 Posted January 31, 2005 Share Posted January 31, 2005 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...
YT2095 Posted January 31, 2005 Share Posted January 31, 2005 a 180 degree inversion of motion dictates that a full STOP must occur before reversing its previous direction of travel Link to comment Share on other sites More sharing options...
syntax252 Posted January 31, 2005 Author Share Posted January 31, 2005 a 180 degree inversion of motion dictates that a full STOP must occur before reversing its previous direction of travel It seems that way don't it? But since the crank turns continually, that means that the throw on the crank is either going up, or it is going down. If the piston is attached to the rod and the rod to the crank, then why wouldn't the same thing that goes for the throw, go for the piston? Link to comment Share on other sites More sharing options...
ed84c Posted January 31, 2005 Share Posted January 31, 2005 A half elipse is 180* and yet the earth doesnt stop. Link to comment Share on other sites More sharing options...
Mart Posted January 31, 2005 Share Posted January 31, 2005 Does an object necessarily have to come to a stop before it can reverse it's direction? Is this is the same type of problem as throwing a ball vertically in the air? Or the problem of a bee hiitting the windscreen of a car? When the bee hits the car it bounces off (as bees usually do) and continues its flight in the same direction as the car. So the bee has to stop. And therefore the car has to stop. But does the bee stop the car or does the car stop the bee? Link to comment Share on other sites More sharing options...
syntax252 Posted January 31, 2005 Author Share Posted January 31, 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, tell me how long it remains motionless. Link to comment Share on other sites More sharing options...
swansont Posted January 31, 2005 Share Posted January 31, 2005 If you have an object undergoing a constant acceleration, then the v vs t curve will be continuous and pass through zero. Unless you have infinite acceleration I don't think you can get away from having a point with zero velocity. Link to comment Share on other sites More sharing options...
Mart Posted January 31, 2005 Share Posted January 31, 2005 If you have an object undergoing a constant acceleration, then the v vs t curve will be continuous and pass through zero. Constant acceleration means that the acceleration has been measured during various time intervals and found to be unchanging. But it doesn't follow that there can be a process such as an acceleration at a point in time. Drawing a continuous curve is pushing the process to a limit which it can't sustain. Newton's calculus using limits is only a workable approximation to this. Link to comment Share on other sites More sharing options...
syntax252 Posted January 31, 2005 Author Share Posted January 31, 2005 If you have an object undergoing a constant acceleration' date=' then the v vs t curve will be continuous and pass through zero. Unless you have infinite acceleration I don't think you can get away from having a point with zero velocity.[/quote'] A piston attached to a crankshaft that is turning at a constant rpm, slows down as it reaches either end of it's stroke. It accelerates from bottom to midway to the top, then decelerates the rest of the way to the top. I don't see how it could help but stop either, but since the crankshaft is moving at all times, it would also seem that the piston is moving at all times. There ain't any flat spots in the cycle. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted January 31, 2005 Share Posted January 31, 2005 In an instant. Then you have to define an instant. http://www.scienceforums.net/forums/showthread.php?t=8653 Link to comment Share on other sites More sharing options...
swansont Posted January 31, 2005 Share Posted January 31, 2005 A piston attached to a crankshaft that is turning at a constant rpm' date=' slows down as it reaches either end of it's stroke. It accelerates from bottom to midway to the top, then decelerates the rest of the way to the top. I don't see how it could help but stop either, but since the crankshaft is moving at all times, it would also seem that the piston is moving at all times. There ain't any flat spots in the cycle.[/quote'] And yet there are points where the slope is zero for x vs t. The problem is ill-defined until you say whether "comes to a stop" means for a finite time or not. Link to comment Share on other sites More sharing options...
syntax252 Posted February 1, 2005 Author Share Posted February 1, 2005 And yet there are points where the slope is zero for x vs t. The problem is ill-defined until you say whether "comes to a stop" means for a finite time or not. OK, what I am asking is, is there a time when the piston is neither rising nor lowering and if so, givin the rpm and the stroke I submitted above, what is the amount of time that it is motionless? Link to comment Share on other sites More sharing options...
YT2095 Posted February 1, 2005 Share Posted February 1, 2005 it`s the design of the crank shaft. it`s a "Converter" if you like, from linear to angular momentum. just like the reverse, you have a pin in a slow rotating disc. you hook a wire over that pin, as the disc rotates the wire pulls away from you for 180 deg of the rotation and them pushes back at you for the rest of the 180 deg. at the point of +180 and -180 there is a Zero. as for how long it remains there is like trying to divide infinity. IF we COULD have a unit of of infinately small time units (and we can`t) there would be a clear and distinct time presented. this is NOT possible though. if we say one revolution takes 1000 milliseconds, then you can bet that the Zero occurs somewhere around the 500 millisecond mark though (as a purely hypothetical example) Link to comment Share on other sites More sharing options...
syntax252 Posted February 1, 2005 Author Share Posted February 1, 2005 So, your answer is that it does indeed stop, but for an infnately short period of time? Link to comment Share on other sites More sharing options...
YT2095 Posted February 1, 2005 Share Posted February 1, 2005 I guess you could word it like that yeah Link to comment Share on other sites More sharing options...
Callipygous Posted February 1, 2005 Share Posted February 1, 2005 It seems that way don't it? But since the crank turns continually' date=' that means that the throw on the crank is either going up, or it is going down. [/quote'] except for that one instant at the very top where it is moving sideways and therefor neither pushing nor pulling the piston. i swear one of these days im gonna read an entire thread before replying... : P Link to comment Share on other sites More sharing options...
syntax252 Posted February 1, 2005 Author Share Posted February 1, 2005 except for that one instant at the very top where it is moving sideways and therefor neither pushing nor pulling the piston. i swear one of these days im gonna read an entire thread before replying... : P At the very top of it's arc, if it moves sideways, it also moves down. Link to comment Share on other sites More sharing options...
syntax252 Posted February 1, 2005 Author Share Posted February 1, 2005 I guess you could word it like that yeah So, 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?" Link to comment Share on other sites More sharing options...
Mart Posted February 1, 2005 Share Posted February 1, 2005 Or even "no time like the present" Link to comment Share on other sites More sharing options...
ecoli Posted February 1, 2005 Share Posted February 1, 2005 An object that changes direction must stop first...or the whole rectilinear motion section of Calculus would have to be thrown out the window... Link to comment Share on other sites More sharing options...
blike Posted February 1, 2005 Share Posted February 1, 2005 No one has answered his question. The crankshaft does not stop. Reversing does not stop the direction of the crankshaft motion, it just changes the gears around so that the direction of gear rotation is reversed. Here's an excellent animation to illustrate what I mean: http://auto.howstuffworks.com/transmission5.htm Link to comment Share on other sites More sharing options...
syntax252 Posted February 1, 2005 Author Share Posted February 1, 2005 An object that changes direction must stop first...or the whole rectilinear motion section of Calculus would have to be thrown out the window... Well OK, I am willing to admit that I do not see how it (the piston) could reverse directions without stopping. However, since the time it is stopped (or the rime we think it is stopped) is incalculable, how can one really say for sure that it is stopped, and not constantly in motion, to and fro? Link to comment Share on other sites More sharing options...
blike Posted February 1, 2005 Share Posted February 1, 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. Link to comment Share on other sites More sharing options...
Martin Posted February 2, 2005 Share Posted February 2, 2005 No one has answered his question.... ... Here's an excellent animation to illustrate what I mean: http://auto.howstuffworks.com/transmission5.htm the animation is really clear and instructive. I agree. you can see the idler gear slide in when you shift into reverse and the bottom shaft changes direction but the top shaft, which comes from the engine, never changes direction the order of the gears (except for Reverse) is from left to right, in the animation-----first second third fourth. this was a funny thread. syntax says "how does reverse work, 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!" Syntax, a standard fourstroke piston internal comb. engine cannot run backwards. the timing of the valves would be wrong. they would open and close at the wrong time. the spark would come at the wrong time. you could not get it to run in reverse direction without totally rebuilding the engine. however transmissions are beautiful and one may benefit from understanding how they work Blike, I am pining for LaTex to be restored. I hope hope hope it will be sooner rather than later Link to comment Share on other sites More sharing options...
Callipygous Posted February 2, 2005 Share Posted February 2, 2005 At the very top of it's arc, if it moves sideways, it also moves down. 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? Link to comment Share on other sites More sharing options...
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