# Posts Tagged pi=4

### A reply to: “a reply to ‘The Extinction of Pi: the short version'”

This week we’ll take a look at a critique of ﻿“The Extinction of π”﻿ from a guest author at Ex Falso named Dan. At first read, Dan appears to make a slam dunk with his two proofs against Mathis’ claim that π=4, (in kinematic situations.)

Although a closer examination reveals a major flaw in Dan’s application of those proofs to Mathis’ step method which Mathis himself lays out in the long version of his paper.

First, I want make it clear that I am not really writing this paper for Dan’s benefit. In reading his recent posts on this subject it seems that his mind may already be made up. Instead, I write this for others who might read it and be encouraged to take a look at Mathis’ work.

Second, in writing this paper I am assuming that the reader has read Dan’s paper and has a basic familiarity with the subject matter. That being said, I do want to thank Dan for taking the to time to analyze Mathis’ paper because it does deserve analysis. I only wish Dan had taken extra time to more carefully read the long version of this paper and realize that Mathis unequivocally states near the beginning of the paper:

More specifically, the π that I am correcting is the constant in the orbital equation v = 2πr/t.”

Unfortunately, Dan punted on that analysis near the end of his paper,

Dan wrote,

I can only assume what he means is that in a kinematic situation he feels the constant Pi = 3.14159265… is an inappropriate constant to use and the number 4 should be used in its place. I promise you that that is not that case, but a proof of that will have to come at another time.”

I am looking forward to that paper and I hope Dan will keep his promise.

That being said, it is important to remember that just like any other author’s work in physics, Mathis’ proof is only valid under strictly defined parameters. Since Mathis begins his analysis on Newton’s Lemmae it would follow that one would be speaking about a physical context or, in other words, a kinematic reference frame. This framework was used by Newton in the same manner and was subjected to a geometric vector analysis just as Mathis does in his paper on Newton’s Lemmae.

To summarize the paper:

Mathis, while studying one of Newton’s diagrams, discovered that Newton monitored the wrong angle while trying to find the limit of his data. Newton solved the problem with a flawed presumption, that being – that at the limit; the tangent, the arc and the cord are all equal in length. Mathis found instead that a second angle would have reached its limit before the one Newton monitored, invalidating Newton’s general assumption.

Mathis gives support for these assertions in his paper. He finds that because of this re-analysis he can prove that the tangent is normally longer than the arc at the limit and that the only time the tangent is equal to the arc is when the length of the tangent is equal to the radius. This means that the analysis that Newton did could only be done on one specific orbital diagram – the one that Mathis then analyzes. Since Newton used a flawed analysis for this and subsequent lemmae this affects most of Newton assumptions after that. These assumptions are what Mathis takes issue with.

This quote sums up his discovery.

Mathis wrote,

…if we assign the tangent to the tangential velocity and the arc to the orbital velocity, as Newton did, we find they are equal not at the limit, but only when the tangent equals the radius. In fact, as I have shown, the tangent and the arc are NOT equal at the limit. At the limit, the tangent remains longer than the arc. And this means that the tangential velocity and orbital velocity are equal only when the length of the tangent is equal to the radius, or when the time is equal to 1/8th of the orbital period. An orbital velocity found by any other method will get the wrong answer. This is why 2πr/t is wrong: 2πr/8 is not equal to r.”

So this is what then brings the value of Pi, (the relationship of the orbits’ circumference to its diameter), into doubt in kinematic situations.

These are certainly bold assertions and they deal with a very specific orbital curve in Newton’s Lemmae. However, making specific assertions about specific curves does not necessarily invalidate Mathis’ proof nor does it make it some type of theoretical “cop-out” as some have implied. Mathis is very clear that Pi=4 only involves orbits and motion in the context noted above. In fact, Mathis’ proof relies on the “curve” being an orbit in kinematic terms where the length of the tangent is equal to the radius, or in other words, when the time is equal to 1/8th of the orbital period.

This discovery allows Mathis to find the length of the arc directly using the method he describes inIn theMathis explains his method of finding the length of the arc and the necessity of approaching that arc smoothy and evenly while using the method. We will come back to this issue later in this paper, but for now, let’s see how Dan sets the parameters for his proofs.

Dan made it very clear in his “Definition and Labels” that he is defining the function of a curve using two dimensions x and y. As if drawing a line on paper with a pencil. In other words, the function of the curve he is describing is using a two dimensional Cartesian grid for its values of x’s and y’s. Unfortunately, it does not adequately represent an orbit due to the fact that the curve is drawn all at once with no time factor assumed in the math. (At least it was never mentioned in the “Definitions and Labels”).

Dan wrote,

(think about drawing a curve above the x-axis where you never have to lift your pencil and above each x there is only one point on the curve and it never intersects itself)”

Even though Dan uses action words like “drawing a curve” and “never have to lift your pencil” There is no assumption of time in his definitions or diagrams. Since none of Dan’s analysis was done in kinematic terms or even in the context of vector analysis this quite invalidates both proofs from the very beginning.

Mathis specifically addresses some of the issues in the long version of the paper. He refers to the diagram below in the following quote.

…I showed that the centripetal force must pull down and back in order to take any object—either a pencil tip or an orbiting spacecraft—out of its original path and into a circular path. The simplest way to think of this is to think of the original velocity as AB. Then the centripetal force creates two other velocities: a velocity of size DB, which pulls the body back from B to D; and a velocity DC, which pulls the body down to its final destination of C. This is how we break down our curve into straight velocity vectors. The motion of the body from A to C is a summation of these three vectors. All three velocities happen over the same time interval, so we sum them. It is that simple.”

Note the non-zero time interval assumed in the drawing of the orbit and the mention of the velocity vectors of the object involved. Also note, Mathis describes the summation of three different velocity vectors – meaning they “add up” or “combine” to make the amplitude and direction of the orbit itself. There is also no claim of zigzag motion like others have misinterpreted Mathis as saying.

This leaves Dan’s curve functions without enough dimensions, (only having two and assuming no time dimension), to properly describe the orbit in question and thereby invalidates these proofs beginning with the first definition.

But let’s look for further problems with Dan’s proofs.

Dan begins the written portion of his first proof with a bit of grammatical slight of hand.

Dan writes,

Assume that Mathis is correct and the length of a curve can be approximated by considering the lengths of the horizontal and vertical line segments that make up the ‘steps’ in the diagrams. Mathis is correct, no matter how few or how many of these steps we subdivide the interval [0,1] into, the total of their lengths will remain constant. These steps can be made small, medium, large, evenly sized, or unevenly sized yet their total length will remain the same.”

Mathis is very specific in the long version of this paper that his step method approaches the curve in a very specific way.

Mathis wrote,

To make this method—which I have called exhaustion but which might just as easily be called approaching a limit—work, we have to push D toward the curve in a rigorous manner. In short, all of our steps have to be approaching the curve at the same rate, or the method will not work. For instance, if we draw a different curve from A to C, one that bulges out very near to D, and then we draw our steps, we will not be able to make those steps even. Or, to put it another way, we will not be able to push D toward the curve in an even manner. Our exhaustion will not “go to infinity” at the same rate all long the curve; therefore our method will not work, mathematically or physically. But it will work with the circle arc AC, and it works for the physical reason I have shown above: both the tangential velocity and the centripetal force are constant. The arc AC is created by a constant and unvarying process, therefore that arc can be approached by the (right) orthogonal vectors in a logical and rigorous manner. In fact, the arc AC is the only curve from A to C that can be exhausted in this manner , given AD and DC. All other curves are varying curves, and cannot be approached as a limit or exhausted in this direct way.”

Mathis states that he is talking about a very specific orbit and a very specific process. In fact am not even sure how Dan’s curve, (see picture below on right), applies to Mathis paper. It is obviously not a circle, not a circular orbit, nor even the curve that Mathis is analyzing. This is the whole thesis of Mathis’ paper.

Therefore, when Dan states the following,

Dan wrote,

Now consider the curve R and B in the right side of the diagram. They can be approximated by ‘steps’ as in the Mathis proof.”

Recall that Dan claims that Mathis’ method includes unevenly sized steps. Mathis specifically excludes this approach. As one can see in the diagram of R the steps do not evenly approach the curve as Mathis method requires.

Mathis wrote,

In short, all of our steps have to be approaching the curve at the same rate, or the method will not work. For instance, if we draw a different curve from A to C, one that bulges out very near to D, and then we draw our steps, we will not be able to make those steps even. Or, to put it another way, we will not be able to push D toward the curve in an even manner. Our exhaustion will not “go to infinity” at the same rate all long the curve; therefore our method will not work,”

and this quote form Mathis’ paper,

…you can approach the circle arc AC from the line AD + DC because those vectors created the arc AC to begin with. Those vectors physically created the curved path. They are not just orthogonal vectors, chosen because they were handy. They are THE orthogonal vectors that define the path of the curve. The arc AC is approached smoothly from D because it was in some sense created from D. D is the physical balance of O, given the interval AC and motion from A to C. D was guaranteed to approach the arc AC smoothly and evenly, which is why I use the method without explanation in my gloss of this paper.”

Therefore, Dan’s statements and diagram are obvious red herrings and simply do not represent Mathis’ method. This then further invalidates proof #1.

Also, note the curve that Dan chooses to use in his diagram across from Mathis’ diagram. I studied the diagram and I noticed something familiar about it – it is precisely the type of curve that Mathis gives us in his text as an example of a curve that will not work with his method!

It might be difficult to see at first because Dan has the diagram of his curve turned 90° counter-clockwise from Mathis’. I have edited Dan’s side of the diagram on the right and added the “D” to show you where it corresponds to Maths’ “D” in the diagram on the left.

Here’s the quote again,

Mathis wrote,

For instance, if we draw a different curve from A to C, one that bulges out very near to D, and then we draw our steps, we will not be able to make those steps even. Or, to put it another way, we will not be able to push D toward the curve in an even manner. Our exhaustion will not “go to infinity” at the same rate all long the curve; therefore our method will not work,”

So what Dan has really proven is that Mathis was correct in this analysis above and nothing more. Also note the blue line B in the diagram above. I will be referring to this line again when we examine Dan’s second proof.

Dan considers this right triangle and attempts to use the step method to measure the hypotenuse. Dan claims that AC is a curve in his definitions and in a general way he is correct. As it turns out, it happens to be a very specific curve. Namely a straight line. The 90° angle at B guarantees this. In fact, the line segment AC in Dan’s diagram directly corresponds to Mathis’ AC in his diagram. Dan’s hypotenuse is Mathis’ cord.

I have edited Dan’s diagram again, added an impression of a circle like in Mathis’ diagram, and added Mathis’ back on the left for comparison. The “D” right above Dan’s 90° was also added to again correspond to Mathis’ “D” in his diagram. I had to rotate the triangle 180° to make it correspond to the right triangle in Mathis’ side of the picture. Mathis’ cord is shown in red. It is clear to see that Dan’s hypotenuse acts exactly like Mathis’ cord AC in this case.

Mathis also makes it very clear in the long version of his paper how futile it is to measure the length of the cord or any straight line from D with his method. This includes Dan’s hypotenuse above and also includes line B in Dan’s other diagram. B also acts like a cord.

Mathis wrote,

To show why the chord AC cannot be approached like this, we use much the same analysis. At first look, it appears that you could draw steps along the chord AC in the same way, “exhaust” them in the same way by increasing the number of steps higher and higher, and find by this magic that the straight line AC was the same length as AD + DC. The reason you cannot do this is because once again you cannot approach the chord AC in an even manner from the point D. Therefore all your steps won’t go to the limit at the same rate, and your “method” won’t work.”

and this quote from the same paper,

The straight line is actually the most difficult thing to approach, and the impossibility of this approach is actually the easiest to discover. For instance, draw four equal steps along AC, then look at them from the point D. There isn’t any way you could have approached those four equal steps from D. In the method, you aren’t just drawing any steps you like. You are supposed to be drawing steps that would occur if you pushed the line AD + DC toward AC. Exhausting a process or going to a limit is not a willy-nilly process, it is a defined and rigorous process.”

and this quote,

…the distance of a line cannot be approached from off the line, because the line is already the distance itself. It is “even to start with”; therefore, it cannot be approached evenly (except by a parallel line of the same length).

It appears from this that Dan has actually proven Mathis’ analysis correct again!

In summary, at the very best, Dan is having trouble choosing the proofs that are relevant to Mathis’ theory. It is possible this was just a lack of reading comprehension or perhaps a lack of understanding of the material in question.

In the worst case scenario, Dan – reading these two examples of the “limitations” of Mathis’ method, (the only two that Mathis cites in the whole paper by the way), has knowingly attempted use them to “disprove” Mathis’ method. I hope this is not true because it would be very disingenuous to Dan’s readers.

I want to make it clear that I don’t dispute Dan’s proofs in isolation. I maintain that their application to Mathis’ method is faulty in this case. Therefore, Dan’s proofs fail to be relevant as to the veracity of Mathis’ procedure one way or the other.