Physical π : Pi's relationship to 4

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Re: Discussing Miles W. Mathis
Well, i did a drawing, which would elaborate on the logic of pi=4.
http://sli.mg/7vxl0u
So what this drawing means: There are three trajectories an object could move. The view is topdown. In pi=3.14 logic an object 1 would get to the endpoint first, object 2 2nd and object 3 last. In pi=4 logic, all three objects would make to an endpoint at the same time. Of course, friction is not included.
I'm entertaining a thought of pi=4, which i would say it's bullshit, but since i haven't done these experiments myself, i'm not yet totally concluding this. Of course, in real life, friction would be involved as many other factors.
http://sli.mg/7vxl0u
So what this drawing means: There are three trajectories an object could move. The view is topdown. In pi=3.14 logic an object 1 would get to the endpoint first, object 2 2nd and object 3 last. In pi=4 logic, all three objects would make to an endpoint at the same time. Of course, friction is not included.
I'm entertaining a thought of pi=4, which i would say it's bullshit, but since i haven't done these experiments myself, i'm not yet totally concluding this. Of course, in real life, friction would be involved as many other factors.
Re: Discussing Miles W. Mathis
I suspect that statement is aimed mainly at me. Sorry! I did not mean to accuse you or brianv of being cointelpro shills. I was referring to dearly departed Allan Wisecracker. And yes I agree healthy skepticism is always a good thing. For me, the issue was decided 99% when I saw that Gopi vouched for Miles (they have met in person and become friends) and Simon vouched for Gopi (they have met in person and gotten to know each other).SacredCowSlayer » October 5th, 2016, 4:30 pm wrote: Until then I would simply propose that a gentleman's understanding around here is in order. That is, forum members should be able to express doubt or skepticism without hasty name calling or accusations.
Re: Discussing Miles W. Mathis
Actually there is a good reason: to make the tube go in a circle, it ends where the circle begins, so there's no where to add a straight section to then measure the speed again, unless you bent the tube up or down, in which case you're adding problematic variables to the experiment. I think this could be easily overcome by using a 1/2 or 3/4 circle and then leaving the end straight. I understand from the youtube comments that Steve is planning to do another experimental setup that measures speed at the end. You might wish to add your two cents to the comments there, or I think some of them may address the points you brought up.bongostaple » October 5th, 2016, 6:25 pm wrote: It's measured at the start, so there's no good reason not to measure it at the end as well.

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Re: Discussing Miles W. Mathis
That's a fair point  it would be difficult to measure speed at the end without having a straight section. Should be possible though. Hopefully he will do another version with measurement at the end, then there's a slightly more credible basis for discussion.
Re: Discussing Miles W. Mathis
bongostaple » October 4th, 2016, 6:40 pm wrote:
The balls are not selfpropelled, so the circular path ball will slow down  it has no other energy to give away to the opposing force that is diverting it from its preferred straight path.
...
The stills labelling the ball at 'quarter 1', 'quarter 2' etc show the ball is not actually bang on the quarter, but is progressively falling behind.
As for the first claim: that the ball must slow down as a result of the force that diverts it from its preferred straight path. That force is called centripetal force, and it should only cause the ball to change direction, not slow down. See here http://www.physicsclassroom.com/class/c ... equirement:
As for your second claim that the ball is falling progressively behind: well, I've watched the video many times, and I don't see it. But more precision from the experiment would be welcome.The centripetal force for uniform circular motion alters the direction of the object without altering its speed. The idea that an unbalanced force can change the direction of the velocity vector but not its magnitude may seem a bit strange. How could that be? There are a number of ways to approach this question. One approach involves to analyze the motion from a workenergy standpoint. Recall from Unit 5 of The Physics Classroom that work is a force acting upon an object to cause a displacement. The amount of work done upon an object is found using the equation
Work = Force * displacement * cosine (Theta)
where the Theta in the equation represents the angle between the force and the displacement. As the centripetal force acts upon an object moving in a circle at constant speed, the force always acts inward as the velocity of the object is directed tangent to the circle. This would mean that the force is always directed perpendicular to the direction that the object is being displaced. The angle Theta in the above equation is 90 degrees and the cosine of 90 degrees is 0. Thus, the work done by the centripetal force in the case of uniform circular motion is 0 Joules.
Recall also...that when no work is done upon an object by external forces, the total mechanical energy (potential energy plus kinetic energy) of the object remains constant. So if an object is moving in a horizontal circle at constant speed, the centripetal force does not do work and cannot alter the total mechanical energy of the object. For this reason, the kinetic energy and therefore, the speed of the object will remain constant. The force can indeed accelerate the object  by changing its direction  but it cannot change its speed. In fact, whenever the unbalanced centripetal force acts perpendicular to the direction of motion, the speed of the object will remain constant. For an unbalanced force to change the speed of the object, there would have to be a component of force in the direction of (or the opposite direction of) the motion of the object.

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Re: Discussing Miles W. Mathis
I've gone through the video again and gone frame by frame through the small apparatus run that is continuous, i.e. not the one where it stops to label where the balls are. At the point where the straight path ball has travelled to the Pi marker, the circular path ball is just a little past the threequarter point of the circular path.
Bearing in mind that the two sections of tube in question were cut at the same length, the only reasonable explanation, as far as I can see, is that the circular path ball has slowed down.
If I were to take the position proposed by Mathis/Videobloke, wouldn't it be equivalent to saying that the tube may well be pi*d long when it's straight, but if you bend it to a circular shape then it is now 4*d long?
That would leave two possible explanations for the results as they are presented:
a) The ball loses speed when it's forced into a circular path.
Or
b) A piece of plastic tubing gets longer if you shape it into a circle.
I really can't see how b) could be more likely than a) as an explanation, but I'm happy to consider both angles.
I appreciate your note of centripetal force, but in the circular situation, it is balanced with the normal force offered by the wall of the tube. This in theory should result in the ball and the tube wall, for want of a better phrase, 'pressing together harder', and I think this would increase the friction involved. The friction of gravity holding either ball down in the direction of the tabletop can be regarded as equal, but the circular path ball has some extra friction, and I believe that must slow it down during its travel.
I thought of a situation where there is a fourwheeled cart on a level track and you give it a push to a known speed, it travels a distance and friction eventually brings it to a halt.
Now let's say we add some extra weight to the cart by getting Miles Mathis to climb aboard. We then give it a push until it's going the same speed as when we let go of the unburdened cart.
Will the cart travel the same distance with Mathis on board?
I *think* the extra weight = more friction, but on the other hand, assuming it's up to the same speed, the cart+Mathis combo will have more momentum.
I'm continuing to ponder on this one  it just doesn't feel right that the circular path ball could have its course changed through a full 360 degrees without losing some energy somewhere. Nobody coasts around a Wall of Death  they are constantly putting more energy into the system to achieve the same constant speed  so why would this be different?
Look forward to your further thoughts on this, it's thorny, for sure
Bearing in mind that the two sections of tube in question were cut at the same length, the only reasonable explanation, as far as I can see, is that the circular path ball has slowed down.
If I were to take the position proposed by Mathis/Videobloke, wouldn't it be equivalent to saying that the tube may well be pi*d long when it's straight, but if you bend it to a circular shape then it is now 4*d long?
That would leave two possible explanations for the results as they are presented:
a) The ball loses speed when it's forced into a circular path.
Or
b) A piece of plastic tubing gets longer if you shape it into a circle.
I really can't see how b) could be more likely than a) as an explanation, but I'm happy to consider both angles.
I appreciate your note of centripetal force, but in the circular situation, it is balanced with the normal force offered by the wall of the tube. This in theory should result in the ball and the tube wall, for want of a better phrase, 'pressing together harder', and I think this would increase the friction involved. The friction of gravity holding either ball down in the direction of the tabletop can be regarded as equal, but the circular path ball has some extra friction, and I believe that must slow it down during its travel.
I thought of a situation where there is a fourwheeled cart on a level track and you give it a push to a known speed, it travels a distance and friction eventually brings it to a halt.
Now let's say we add some extra weight to the cart by getting Miles Mathis to climb aboard. We then give it a push until it's going the same speed as when we let go of the unburdened cart.
Will the cart travel the same distance with Mathis on board?
I *think* the extra weight = more friction, but on the other hand, assuming it's up to the same speed, the cart+Mathis combo will have more momentum.
I'm continuing to ponder on this one  it just doesn't feel right that the circular path ball could have its course changed through a full 360 degrees without losing some energy somewhere. Nobody coasts around a Wall of Death  they are constantly putting more energy into the system to achieve the same constant speed  so why would this be different?
Look forward to your further thoughts on this, it's thorny, for sure
Physical π : Pi's relationship to 4
In a sense, yes, it is equivalent to that, even if that is not an accurate description of what is going on. But that is why it is so hard to understand or wrap your head around. It's as if the tube is getting longer just by bending it into a circle. How is that possible? But you see it's not that the length of the tube changes. It's that in order to move around the circle, the ball has to travel a longer distance. When he says in his post that 'you'll feel like you're in a netherworld,' I think it's precisely this apparent 'warping' of spacetime that he's referring to. To be clear: there is no warping, but the explanation for what we're seeing is fairly deep and described in depth in his various papers on pi, Newton's lemmae and the virial theorem.bongostaple » October 6th, 2016, 2:49 pm wrote:
At the point where the straight path ball has traveled to the Pi marker, the circular path ball is just a little past the threequarter point of the circular path.
Bearing in mind that the two sections of tube in question were cut at the same length, the only reasonable explanation, as far as I can see, is that the circular path ball has slowed down.
If I were to take the position proposed by Mathis/Videobloke, wouldn't it be equivalent to saying that the tube may well be pi*d long when it's straight, but if you bend it to a circular shape then it is now 4*d long?
As for the two sections of tube being cut to the same length: they weren't. Unfortunately, the video doesn't show the circular tube laid out straight, but he does measure the diameter as 17.6cm, so the length of the tube is 17.6cm * 3.14 = 55.264cm. The length of the straight tube from the beginning of the circle (0) is 70.4cm.
70.4/55.264 = 1.273885.
4/3.14 = 1.273885
So the ratio of the length of the tubes are the same proportion as 4 to 3.14.
What this means is that, as the straight ball travels 17.6cm (1 diameter in length  the tick marks), the ball in the circle only travels 1/4 of the circle, which we can measure as having a length of (55.264/4=) 13.816 cm. But what Miles is trying to argue is that even though the measured length of the tube is only 3.14, the distance it travels (length/time) is 4. Another way to say this is that the ball in the circle only moves 3.14 in the time it takes the ball in the straight to travel 4, even though they are moving at the same speed.
As I tried to show above, the option is C: the ball travels a larger distance in a circular path even though the tube remains the same length. Again, the explanation for this and the distinction he draws between length and distance can be found in his papers. I encourage you to take the time to read them.That would leave two possible explanations for the results as they are presented:
a) The ball loses speed when it's forced into a circular path.
Or
b) A piece of plastic tubing gets longer if you shape it into a circle.
I really can't see how b) could be more likely than a) as an explanation, but I'm happy to consider both angles.
I appreciate your note of centripetal force, but in the circular situation, it is balanced with the normal force offered by the wall of the tube. This in theory should result in the ball and the tube wall, for want of a better phrase, 'pressing together harder', and I think this would increase the friction involved.
The friction of gravity holding either ball down in the direction of the tabletop can be regarded as equal, but the circular path ball has some extra friction, and I believe that must slow it down during its travel.
So in the scenario you are sketching, the ball is being pushed up against the outer wall of the tube (by virtue of centrifugal force), then a centripetal force acting on the ball would push it away from the outer wall of the tube. Since the centripetal and centrifugal forces are 'equal and opposite', the ball should not be pressing harder against the tube. In any event, if there were an increase of friction, we would see the ball getting progressively slower. I have watched the video, and I don't see evidence of this. I watched it again just now, and while I do think the circular ball hits the halfway mark just a smidge after the straight ball hits the 2 diameter mark, they both seem to hit their marks at the same time by the third quarter / 3rd diameter marks. So no apparent slowdown in my view. The mark at the halfway might just be a little bit off. Again, more precision in the experiment would be very welcome.
I think whether the cart will travel the same distance with added weight depends on the mechanics of the setup. If the cart is on wheels on a relatively smooth surface, then I think the added momentum would carry the day. But if the cart is, say, not on wheels but just sliding on a surface, then the added weight might add significantly to the friction so he would go a shorter distance. In short, I think it would depend on the specifics of the situation.I thought of a situation where there is a fourwheeled cart on a level track and you give it a push to a known speed, it travels a distance and friction eventually brings it to a halt.
Now let's say we add some extra weight to the cart by getting Miles Mathis to climb aboard. We then give it a push until it's going the same speed as when we let go of the unburdened cart.
Will the cart travel the same distance with Mathis on board?
I *think* the extra weight = more friction, but on the other hand, assuming it's up to the same speed, the cart+Mathis combo will have more momentum.
I'm continuing to ponder on this one  it just doesn't feel right that the circular path ball could have its course changed through a full 360 degrees without losing some energy somewhere. Nobody coasts around a Wall of Death  they are constantly putting more energy into the system to achieve the same constant speed  so why would this be different?
As for constantly putting more energy in the system of achieve a constant speed  well we do that moving in a straight as well as "around a Wall of Death." You cannot coast forever in a straight line, either. So both of the balls we see are slowing down. Don't forget also that in the Wall of Death you also have gravity pushing you down. The question is whether the curved and straight paths slow down at the same rate. Fortunately this is an empirical question: just add a straight end to see if the ball exiting the circle is traveling the same speed as the straight ball. If friction had slowed it down, then it will not magically regain speed moving into the straight path; it would remain at the slower speed.
My first thought when I saw your picture is "I need to get my eyes checked."albatrosv14 » October 6th, 2016, 8:22 am wrote:In pi=4 logic, all three objects would make to an endpoint at the same time. Of course, friction is not included.
I don't agree with your conclusion. It might hold for 1 and 3 if 1 starts the curve immediately at the point and ends the curve immediately at the other point. So in other words, if the curve is a 1/4 circle. From the way you've drawn it, it's not clear. The other condition would be that 1 is able to change direction without a loss of speed. So travelling both paths at a constant speed.
But 2 doesn't work because it's partly straight path and partly curved. If we take his running track as an analogy, lane 1 spends more time in the curve, so the distance (not length!) is greater than 2, which spends more time in the straight. So 1 and 3 should both take longer than 2.
Re: Discussing Miles W. Mathis
Not to be going in circles for forever, this is how it stands until new experiment is done. I agree here with the observations made, there is no progressive slowing down seen forcing on the ball in the experiment. Since it should be a) progressive and b) obvious to observe, it is said with certainty that nothing is affecting velocity of the ball inside the curved tube relative to the other ball. Until proven otherwise, I think, Bongostaple, you just have to go along with it with your skepticism about it.http://milesmathis.com/pi7.pdf
I predict the main response to the video will be that the ball in the curve is feeling more friction. However, it is clear at a glance this is not the case. To start with, the ball in the curve would have to be feeling over 20% more friction than the straight ball. Again, the difference between 3.14 and 4 is not marginal. It is huge. There is no way to account for a difference that large with a difference in friction. Plus, if friction were the cause, the ball in the curve should be slowing down as it progresses around the curve. Friction is of course cumulative, so we would expect a ball feeling an excess 21% of friction to be going slower in the fourth quadrant of the circle than in the first. But we see with our own eyes that isn't true. Steven marks all four quarter points in the circle, and the ball hits them all perfectly in sync with the straight ball. If the ball in the curve were feeling more friction, we would expect it to hit the ¾ mark and final mark noticeably late compared to the ¼ mark. It doesn't. This indicates very strongly that neither friction nor any other cumulative effect in the curve is causing the difference. The ball in the curve is NOT slowing relative to the straight ball.
If it is so obvious to you, that experiment is showing otherwise, I'd encourage you to repeat it with measurements of speed on multiple points within a new setup. Only then can you refute with arguments, that is if they show up in your new experiment, and afterwards I will be the first one to admit Mathis was wrong when he said that the velocity loss/ additional friction is not involved. I predict though that there will be no velocity loss measured. And I predict as well Oostdijk will do it before you. Perhaps just to work on his lousy engineering skills
As in chess game, let's play with the outcomes of a new experiment. Let's presume firstly that you don't measure any statistically significant loss of velocity inside the curved path, then all is as it was before. Now let's presume you'd actually confirm some statistically significant velocity loss, would it refute the point being made? Since in the original experiment setup one cannot see any velocity loss that would be really obvious, but you've anyway found some, I presume it would be shown in less than 1% change. Would that be suffice to account for 21% ? It's a rather rhetorical question I believe. As it seems, the only thing that you could achieve if your new experiment's setup shows some velocity loss, is to prove Mathis made an observational error and concluded wrongly on part of it. I'm always for some more evidence on any subject, so please do bring it in this "ring".

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Re: Discussing Miles W. Mathis
*
QUESTION to Miles Mathis (and his Pi=4 followers)
Let's say that the blue and green locomotives are electricallypowered  and that they both travel at say, 1km/h.
Let's say that, in order to be perfectly sure that they always travel at exactly 1km/h, we equip them with some highprecision airspeed speedometer  so that we may eliminate any concern about whatever friction / centrifugal issues (or whatnot) that may be involved with the green train (travelling around the green loop).
(To be more precise, the loop's diameter would be 63.6620310097753)  as measured from centertocenter of rail.
So the question is :
A ) Will the green locomotive complete one loop in the same time that the blue locomotive travels from A to B? (I would personally expect so).
or...
B ) Will the green locomotive be still 21% short from completing one loop  when the blue locomotive has covered the A to B distance? (due to 63.662 X 4 adding up to 254,648cm  or due to any other reason that currently escapes me?)
or...
C ) ?
It may well be that I'm missing something with regards to the Pi=4 thesis proposed by MM  and will gladly be lectured as to what exactly I'm missing.
QUESTION to Miles Mathis (and his Pi=4 followers)
Let's say that the blue and green locomotives are electricallypowered  and that they both travel at say, 1km/h.
Let's say that, in order to be perfectly sure that they always travel at exactly 1km/h, we equip them with some highprecision airspeed speedometer  so that we may eliminate any concern about whatever friction / centrifugal issues (or whatnot) that may be involved with the green train (travelling around the green loop).
(To be more precise, the loop's diameter would be 63.6620310097753)  as measured from centertocenter of rail.
So the question is :
A ) Will the green locomotive complete one loop in the same time that the blue locomotive travels from A to B? (I would personally expect so).
or...
B ) Will the green locomotive be still 21% short from completing one loop  when the blue locomotive has covered the A to B distance? (due to 63.662 X 4 adding up to 254,648cm  or due to any other reason that currently escapes me?)
or...
C ) ?
It may well be that I'm missing something with regards to the Pi=4 thesis proposed by MM  and will gladly be lectured as to what exactly I'm missing.
Re: Discussing Miles W. Mathis
Simon, that's an ingenious question! And just goes to show once again what a very useful engine that Thomas is. I think the answer has to be B according to the theory, but I can't say for certain.

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Re: Discussing Miles W. Mathis
I'm hoping his next version with the speed measured at the end will also be properly circular, and shot at a high frame rate from directly above. Then whatever speed the circular or straight ball is doing at any point can be worked out, so long as the frame rate is constant.
Tell you what, I'm not poring over individual frames and taking relative measurements twice
I am tempted to build one myself, but I am prone to underestimating the time I would need for such things.
In the meantime, Simon  I would plump for answer a), assuming that the selfpropelled engines maintain constant speed.
Let's wait and see!
Tell you what, I'm not poring over individual frames and taking relative measurements twice
I am tempted to build one myself, but I am prone to underestimating the time I would need for such things.
In the meantime, Simon  I would plump for answer a), assuming that the selfpropelled engines maintain constant speed.
Let's wait and see!
Re: Physical π : Pi's relationship to 4
I choose B as correct answer.

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Re: Discussing Miles W. Mathis
daddie_o wrote: I think the answer has to be B according to the theory, but I can't say for certain.
Ok, Vexman and daddie_o.VexMan wrote:I choose B as correct answer.
Now, would it be too much to ask of you two to elaborate in a  ehrm  scientific manner as to why B would be the correct answer? Thanks!
Also, would daddie_o please contact MM by email and ask him to comment on my above post? Thanks.
Re: Discussing Miles W. Mathis
I will contact him, but no guarantee that he'll answer.simonshack » October 7th, 2016, 1:10 am wrote:
Now, would it be too much to ask of you two to elaborate in a  ehrm  scientific manner as to why B would be the correct answer? Thanks!
Also, would daddie_o please contact MM by email and ask him to comment on my above post? Thanks.
As for an elaboration of why B would be the correct answer, well the best I can do is refer you to Miles's papers on Pi. He explains things much more clearly than I can, and with a much firmer grasp on things. Here are some links, with the first one being the most thorough:
http://milesmathis.com/pi2.html
One of his previous papers that he links to there is his paper on Newton's principia (A Disproof of Newton's Lemmae), which is dense but worth reading if you really want to delve into this: http://milesmathis.com/lemma.html
http://milesmathis.com/pi.html
http://milesmathis.com/pi3.html
http://milesmathis.com/manh.pdf
http://milesmathis.com/cycloid.pdf
http://milesmathis.com/angle.html
Here are the two most recent papers that everyone here seems to be reading (if at all), but these are in some ways the most superficial papers about the topic. You really need to read the papers above to get a deeper sense of what his argument is.
http://milesmathis.com/track.pdf
http://milesmathis.com/pi7.pdf
The second best I can try to do is to gloss the key argument(s):
We are used to thinking of curved motion as being the same as straightline motion. But they're not. When motion is involved, so is time, and geometry doesn't take that into consideration. What is more, our understanding of orbits (moving around a circle) has developed under a mistake in Newton's Principia.
That mistake has led us to believe that real bodies traveling curves travel at the limit of the hypotenuses of the little triangles in the diagram below (which can be found on page 2 of http://milesmathis.com/manh.pdf). But no: real bodies travel at the limit of the orthogonal legs.
Here is more from that Manhattan metric paper:
And here from the pi2.html paper:This means that the curve does not approach the hypotenuses of these steps, no matter
how many there are. The hypotenuses are the chords, and they cannot be approached by
the arc or tangent.
To put it still another way, the limit we are approaching when we increase the number of steps is not the
sum of the hypotenuses. The hypotenuses are not the limit. The hypotenuses don't enter the correct
math at all, neither the calculus nor the algebra nor the geometry. This is because—as I prove in an
earlier paper—the tangent never approaches the chord as we go to the limit. The hypotenuse is the
chord here, and the chord is always shorter than the tangent, even at the limit. Due to this, the chord is
always shorter than the arc as well, even at the limit.
Modern physicists and other readers can't comprehend this, because they simply accept what Newton
told them, and Newton told them that the tangent and the chord and the arc all went to equality at the
limit. They don't, and I prove that in very simple fashion, with the simplest possible math and
explanation. But rather than show where my proof or explanation fails, they shout that I haven't proved
it to their satisfaction, in the terms they demand. As if I have to create a new proof for every person in
the world, in their own preferred symbolisms, and tutor them on it personally until it penetrates their
skulls.
...
Again, the proof that the tangent and chord don't approach equality is in my paper titled A Disproof of
Newton's Fundamental Lemmae. You have to go there to get the proof, since I can't include everything
I know in every paper I write. Given that proof, we find that the circle is NOT composed at the limit of
chords or hypotenuses, as in Archimedes, Newton, or anyone after. Given motion and the time
variable, the circle is composed at the limit of the orthogonal vectors. In other words, it is composed of
the two shorter sides of the right triangle, not the long side. Which means that real objects in orbit
travel a path that is represented not by the limit of the Euclidean metric, but by the limit of the
Manhattan metric. And this means that in the kinematic circle, π=4 and C = 8r.
So there you have it: comparing the distance traveled on a curve (an acceleration) to the distance traveled in a straight line (a velocity) is like comparing apples to oranges. So I'm quite sure his answer will be B, but I will ask anyway.But let us start at the beginning. By definition, a velocity vector cannot curve. A velocity takes place in one dimension or direction only. In a velocity, there is only one distance in the numerator and one time in the denominator. These times and distances are also vectors, and may not curve. But to create a curve, either mathematically or physically, requires at least two velocities happening over the same interval. Or, to put it another way, it requires two distances measured over the same time interval. If we sum these velocities over the same interval, we achieve an acceleration, and thereby—assuming the two velocities are at an angle—a curve.
If we bring time back into the problem of the circle, we find that every line or distance becomes a velocity and every curve becomes an acceleration. So the diameter becomes a velocity and the circumference becomes an acceleration. All we have to do is imagine the lines being drawn. The pencil must have some velocity or acceleration as it moves along the line or curve. Likewise with a planet drawing out an orbit, or any other possible creation of a circle in the real world.
Once we do this, we see that in comparing the diameter to the circumference in any real circle we are comparing a velocity to an acceleration. But you cannot directly compare two numbers, when one is a velocity and one is an acceleration. Or, you can if you like, but the new number you get from the ratio is not going to be a number that carries any real meaning in it. It is certainly not the same as comparing one distance to another. For example, if you compare one distance to another by putting them into a fraction and achieving a new number, this new number will contain useful information. It will tell you how long one line is compared to the other, obviously. But if you compare a velocity and an acceleration, what information do you get? Say you achieve the number 5, which tells you the acceleration is five times the velocity. Does that tell you anything about distances? Yes, it might, if you develop a transform. But without a transform and a bit of thinking, the number 5 isn’t telling you anything. It certainly isn’t telling you that some distance is 5 times some other distance.
To give a specific example, what if my acceleration is 3 and your velocity is 1. Can we compare those two numbers directly? No, we cannot put them into a fraction or any other equation without doing some more work upon them. We cannot claim that I have done anything three times as much as you. With an acceleration of 3, my velocity could be anything at a given interval, and my distance traveled likewise. What if my acceleration is π and your velocity is 1? Is π the value of any real relationship between us? No. You can’t compare an acceleration to a velocity. You need more information.
This is important because this is precisely what we think π is telling us. We think it is telling us that the circumference is 3.14 times the diameter. But it isn’t. About real circles, π is telling us nothing. About abstract circles, π is telling us that if the circumference were a straight line, it would be π times the diameter. But since the circumference is not a straight line, π is telling us nothing useful there either. In reality, π is precisely as useful as some numerical relationship between apples and oranges, one that began with the postulate “if oranges were apples” and finds “then oranges would be π times redder than they are.” All very edifying I am sure, but since oranges are not apples, any number found is just a ghost.
Last edited by daddie_o on Fri Oct 07, 2016 7:51 am, edited 1 time in total.
Re: Discussing Miles W. Mathis
"Sometimes the perceptions of reality may be the mere shadows of a real thing". When an object is in circular motion, it travels along a path, that is longer that what we believed is it's actual traveled path/distance. It turns as such that while in circular motion, a point on circle's boundary covers distance equal to 8R instead of suggested by 2R*Pi as per circle's circumference equation. The mechanics of this is cycloid:simonshack » October 7th, 2016, 12:10 am wrote:
Why B would be the correct answer (?)
http://www.f.waseda.jp/takezawa/matheng ... fcyclo.htm .
As one can clearly see in gif from the link, we were taught to observe the "wrong" path of actual distance traveled by such point on the circle's boundary. We were observing shadows of the real motion. The chosen B answer suggests that time/distance is involved, to understand what above gif implies : time needed for a point on the boundary taken to travel it's distance (8R) while in 1 full rotation of the circle = time needed for the same point on the boundary to finish 1 full rotation around its center (2R*Pi) while static (EDIT: static is referring to the circle being static, not the point itself)
Both trains starting at the same time, the time needed for blue locomotive to finish its AB path is NOT EQUAL to time needed for green locomotive to finish its circular path. As blue one would be crossing its finish line, green locomotive would be missing 21% distance needed to finish its circle. In other words, in time needed for green locomotive to finish its circular path, blue locomotive would be at 8R traveled distance (R = same as with green's locomotive circle/path).
This only confirms the discussed case as with the running track and Oostdijk's experiment, they all show that observing circle's circumference while in motion doesn't follow real path traveled by the point on circle's boundary.
https://www.youtube.com/watch?v=CwZBX2XcA , published yesterday by MM, from 11:15 .... that's where I took the intro from .