Harry P. Posted December 30, 2011 Share Posted December 30, 2011 You answered your own question... the bug close to the center point has a shorter distance to travel, Think of it as a gear reduction... Your cam shaft and crank shaft are chained together but turn at diff speeds... Yeah, the center bug has a shorter distance to travel. But both bugs travel their respective distances in the same time! How is that possible? How can center bug, who has to travel one half mile per revolution, and outer edge bug, who has to travel one mile per revolution, do that in the same amount of time (one second)??? Quote Link to comment Share on other sites More sharing options...
moparmagiclives Posted December 30, 2011 Share Posted December 30, 2011 Yeah, the center bug has a shorter distance to travel. But both bugs travel their respective distances in the same time! How is that possible? How can center bug, who has to travel one half mile per revolution, and outer edge bug, who has to travel one mile per revolution, do that in the same amount of time (one second)??? Is this a test?? The outside bug went 1 mile in one second, the inside bug went 1/2 mile in one second AT half the speed as the out side bug. They are not both going the same speed, If your measuring the 1 mile at the outside of the disk, then the inside would be half that in distance. Quote Link to comment Share on other sites More sharing options...
Harry P. Posted December 30, 2011 Share Posted December 30, 2011 You're not following. Outside bug goes one mile in one second. Inside bug goes 1/2 mile in one second. So outside bug must be going twice as fast, because he goes twice as far in one second as inside bug does. But they're both on the same disk, spinning at the same speed! They're moving at the same speed, but outside bug goes twice as far! Quote Link to comment Share on other sites More sharing options...
Kit Basher Posted December 30, 2011 Share Posted December 30, 2011 ME BRAIN HURTS! Quote Link to comment Share on other sites More sharing options...
Darren B Posted December 30, 2011 Share Posted December 30, 2011 If you had 30 guys working double shifts on your model, yeah, it would be done in a day. Darin you hit it right on the head and include an unlimited supply of money and yeah about a day or just under a day would be about right. Quote Link to comment Share on other sites More sharing options...
David G. Posted December 30, 2011 Share Posted December 30, 2011 You're not following. Outside bug goes one mile in one second. Inside bug goes 1/2 mile in one second. So outside bug must be going twice as fast, because he goes twice as far in one second as inside bug does. But they're both on the same disk, spinning at the same speed! They're moving at the same speed, but outside bug goes twice as far! The disk and the bugs are spinning at one revolution per second relative to the disk's axis. Relative to the ground, the section of the disk under the outer bug is moving at one mile per second and the section of the disk under the middle bug is moving at 1/2 mile per second. You're mixing two points or reference- the ground and the disk's axis. Does that make sense? David G. Quote Link to comment Share on other sites More sharing options...
Harry P. Posted December 30, 2011 Share Posted December 30, 2011 You're mixing two points or reference- the ground and the disk's axis. Does that make sense? David G. I guess where I get hung up is the fact that the entire disk is rotating at the same speed, yet the two bugs (who are sitting on the disk and are both traveling at the same speed-the speed of the disk) cover different distances during each disk revolution. Imagine a stationary "finish line" above the disk, going from the axis straight out to the outer edge... and then imagine the two bugs on the disk, one sitting on the outer edge and one at a point exactly halfway between the axis and the outer bug. Both bugs will cross the finish line at the same exact time, even though during that one revolution the outer bug will have traveled one mile (the circumference of the disk) while the inner bug will have traveled only one half mile (the circumference of the disk at the halfway point between axis and outer edge). So even though each bug completes one revolution in the same time span (one second), each bug has actually traveled a different distance in that same one second of time! Maybe Christian can explain... Quote Link to comment Share on other sites More sharing options...
moparmagiclives Posted December 30, 2011 Share Posted December 30, 2011 So even though each bug completes one revolution in the same time span (one second), each bug has actually traveled a different distance in that same one second of time! Why are the start and finish lines at the track up at the high school staggered ??? Because the outside lanes are longer laps then the inside ones next to them right? I think your hung up on the disk as being one unit of time and speed. Quote Link to comment Share on other sites More sharing options...
Harry P. Posted December 30, 2011 Share Posted December 30, 2011 Why are the start and finish lines at the track up at the high school staggered ??? Because the outside lanes are longer laps then the inside ones next to them right? I think your hung up on the disk as being one unit of time and speed. The starting lanes on a track are staggered because the outermost lane is longer than the innermost lane. So in order for everyone to have to run the same distance per lap, the starting points are staggered. Plus, each runner is his own separate entity... one runner's speed is not connected to another runner's speed... thay all run at whatever speed they can, and the fastest runner wins the race. But that has nothing to do with what I'm talking about. My "bugs" are sitting on a rotating disk. Both bugs are sitting stlll, not moving in relation to each other. The bugs could be two spots painted on the disk instead. Both spots are moving at the exact same speed (the speed of the rotating disk), unlike runners who move at their own speed, independent of the other runners. Both dots make one revolution per second (the speed of the disk that they are painted on), both dots take the same exact amount of time to make that revolution, but the outer dot covers more distance than the inner dot in the same amount of time. How is it possible that the outer dot travels one mile of distance in one second, while the inner dot travels only one half mile in one second... yet they are both traveling at the same speed (the speed of the rotating disk)? Quote Link to comment Share on other sites More sharing options...
David G. Posted December 30, 2011 Share Posted December 30, 2011 (edited) I guess where I get hung up is the fact that the entire disk is rotating at the same speed, yet the two bugs (who are sitting on the disk and are both traveling at the same speed-the speed of the disk) cover different distances during each disk revolution. So even though each bug completes one revolution in the same time span (one second), each bug has actually traveled a different distance in that same one second of time! Maybe Christian can explain... Yes, the disk is traveling at one speed in relation to its axis, but at different speeds in relation to the ground. Since a point at the outer edge of the disk has to go a greater distance than a point closer to the center, it has to go faster in relation to the ground to cover the greater distance. You're counting RPM and MPH as the same measurement when they are two different motions (rotational vs linear) with two different reference points (axis vs ground). According to the ground the distance traveled is greater, but according to the axis of the disk it's still only 360 degrees. Motion in relation to two different points- apples vs oranges The formula is D=R*T. The Distance traveled is equal to the Rate or speed multiplied by the ammount of Time traveled. If the distance increases, but the ammount of time traveled stays the same, then the rate or speed must increase. Is that helpful? David G. edit: This is actually kind of fun... . Edited December 30, 2011 by David G. Quote Link to comment Share on other sites More sharing options...
moparmagiclives Posted December 30, 2011 Share Posted December 30, 2011 (edited) The starting lanes on a track are staggered because the outermost lane is longer than the innermost lane. So in order for everyone to have to run the same distance per lap, the starting points are staggered. Plus, each runner is his own separate entity... one runner's speed is not connected to another runner's speed... thay all run at whatever speed they can, and the fastest runner wins the race. But that has nothing to do with what I'm talking about. My "bugs" are sitting on a rotating disk. Both bugs are sitting stlll, not moving in relation to each other. The bugs could be two spots painted on the disk instead. Both spots are moving at the exact same speed (the speed of the rotating disk), unlike runners who move at their own speed, independent of the other runners. I wasnt talking about the runners them selves Harry but... So lets say they were connected all in line, all holding a pole pinned at the center of the track, wouldnt the farthest runner have to run faster then the closest runner to stay in line? Edited December 30, 2011 by moparmagiclives Quote Link to comment Share on other sites More sharing options...
Harry P. Posted December 30, 2011 Share Posted December 30, 2011 I wasnt talking about the runners them selves Harry but... So lets say they were connected all in line, all holding a pole pinned at the center of the track, wouldnt the farthest runner have to run faster then the closest runner to stay in line? Yes, in your example the outer runners would need to be running faster in order to keep up with the inner runners, because the outer runners have to cover a greater distance (the farther out from the center you go, the larger the circumference of the runner's lane). But in my example the "runners" (the dots painted on the disk) can't move independent of one another. The outer dot can't "run" faster than the inner dot; they're both painted onto the disk and can't change positions on the disk. Their position on the disk is fixed and immovable, therefore by definition, they both have to be spinning at the same speed as whatever speed the disk rotates. Yet the outer dot covers a greater distance per revolution. Quote Link to comment Share on other sites More sharing options...
moparmagiclives Posted December 30, 2011 Share Posted December 30, 2011 (edited) The dots both turn one revolution per second, but that does not make them turn the same mile per hour, foot per hour or inch per hour. Because they are traveling a diff distance.They are TRAVELING at diff speeds related to the diameter, but areTURNING at the same rpm Edited December 30, 2011 by moparmagiclives Quote Link to comment Share on other sites More sharing options...
heatride Posted December 30, 2011 Share Posted December 30, 2011 Whats really scary is that im reading this Quote Link to comment Share on other sites More sharing options...
Harry P. Posted December 30, 2011 Share Posted December 30, 2011 Whats really scary is that im reading this Quote Link to comment Share on other sites More sharing options...
sjordan2 Posted December 30, 2011 Share Posted December 30, 2011 The dots both turn one revolution per second, but that does not make them turn the same mile per hour, foot per hour or inch per hour. Because they are traveling a diff distance.They are TRAVELING at diff speeds related to the diameter, but areTURNING at the same rpm I don't know how it gets clearer than that. Quote Link to comment Share on other sites More sharing options...
James2 Posted December 30, 2011 Share Posted December 30, 2011 Clearer than what! Quote Link to comment Share on other sites More sharing options...
sjordan2 Posted December 31, 2011 Share Posted December 31, 2011 I asked my daughter, who is an engineer, if she could think of a simpler way of putting this. She has quite a sense of humor... "The best way I can explain it right now would be this: Here's the equation for the circumference of a circle: C = 2*pi*r, where r = radius of the circle So let's say say the inner point is ri, and the outer point is ro. And assume ri = 1 in, and ro = 2 in (just to make it easy). So Ci = 2*pi*ri = 2*pi*1 = 2*pi ~ 6in And Co = 2*pi*ro = 2*pi*2 = 4*pi ~ 12in Now let's say it takes 1 minute for the dots to make an entire revolution. The two rates of rotation would then be 6 in/min and 12 in/min." Quote Link to comment Share on other sites More sharing options...
bill_rules Posted December 31, 2011 Share Posted December 31, 2011 (edited) That makes it as easy as pi. Edited December 31, 2011 by bill_rules Quote Link to comment Share on other sites More sharing options...
59 Impala Posted December 31, 2011 Share Posted December 31, 2011 Question. Does the bug setting on the hub get dizzy before the bug setting on the outer rim when they stop spinning? Quote Link to comment Share on other sites More sharing options...
sjordan2 Posted December 31, 2011 Share Posted December 31, 2011 (edited) I think that the point of the equation (and what has been said by many people here) is that while the revolutions per minute are the same, the bug on the outer rim is still traveling at twice the speed of the bug on the hub based on the example. Edited December 31, 2011 by sjordan2 Quote Link to comment Share on other sites More sharing options...
James2 Posted December 31, 2011 Share Posted December 31, 2011 I'm not sure if ri or ro bugs me more, but either way I am now dizzy. Quote Link to comment Share on other sites More sharing options...
sjordan2 Posted December 31, 2011 Share Posted December 31, 2011 (edited) I'm not sure if ri or ro bugs me more, but either way I am now dizzy. Then you must be sitting on the hub. (Nice bug pun, by the way.) Edited December 31, 2011 by sjordan2 Quote Link to comment Share on other sites More sharing options...
David G. Posted December 31, 2011 Share Posted December 31, 2011 That makes it as easy as pi. Too Funny!!! ...wish I' have thought of that one.. David G Quote Link to comment Share on other sites More sharing options...
heatride Posted December 31, 2011 Share Posted December 31, 2011 Ok Cajun understan now yah ? Boudreaux ro da boat when we get dare I ri up out da boat. Quote Link to comment Share on other sites More sharing options...
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