# Proof that pi=4

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## Proof that pi=4

This paper is here: http://milesmathis.com/pi7.pdf

There is something I misunderstood maybe. Let's call the ball that goes straight A and the ball that goes around the circle B.

t0: A and B start.

t1: A and B are at the speed mark.

t2: A is at second mark and B is at the start of the circle.

now things get a bit messed up:

t3: A is at d from the second mark and B is at 1/4 of the circle. If we straighten the circle, that mark is at 0.785*d from the second mark.

t4: A is at 2*d from the second mark and B is at 1.57*d.

t5: A is at 3*d from the second mark and B is at 2.355*d.

t6: A is at 3.14*d from the second mark and B is at 2.4649*d.

t7: A is at 4*d from the second mark and B is at 3.14*d.

We can se that there is a slowing down, instantaneous and not cumulative. Is it apparent? We proposed a new experiment that calculates the final speed of the two balls.

I think that for logistic reasons we must use half a circumference.

There are 2 possible outcomes:

1) A and B have the same speed

2) A is faster than B

But that is not the case I'm afraid. If 2 is true, then in general changing curvature changes the speed as an inverse proportion. That would result in B gaining speed when getting on the straight path. So both possibilities (speed loss and physical longer distance) will produce the same final speed.

Is there anything we can do to better understand this?

PS: I love Mathis rants about mainstreams physicists

There is something I misunderstood maybe. Let's call the ball that goes straight A and the ball that goes around the circle B.

t0: A and B start.

t1: A and B are at the speed mark.

t2: A is at second mark and B is at the start of the circle.

now things get a bit messed up:

t3: A is at d from the second mark and B is at 1/4 of the circle. If we straighten the circle, that mark is at 0.785*d from the second mark.

t4: A is at 2*d from the second mark and B is at 1.57*d.

t5: A is at 3*d from the second mark and B is at 2.355*d.

t6: A is at 3.14*d from the second mark and B is at 2.4649*d.

t7: A is at 4*d from the second mark and B is at 3.14*d.

We can se that there is a slowing down, instantaneous and not cumulative. Is it apparent? We proposed a new experiment that calculates the final speed of the two balls.

I think that for logistic reasons we must use half a circumference.

There are 2 possible outcomes:

1) A and B have the same speed

2) A is faster than B

But that is not the case I'm afraid. If 2 is true, then in general changing curvature changes the speed as an inverse proportion. That would result in B gaining speed when getting on the straight path. So both possibilities (speed loss and physical longer distance) will produce the same final speed.

Is there anything we can do to better understand this?

PS: I love Mathis rants about mainstreams physicists

**Ciaolo**- Posts : 133

Join date : 2016-09-08

## Re: Proof that pi=4

The two balls:1) A on the straight track ,and; 2) B on the circular track – appear to hit their marks simultaneously. This rules out fiction. Adding a final straight section to circular track B would allow us to compare and calculate straight to straight final velocities.There is something I misunderstood maybe. Let's call the ball that goes straight A and the ball that goes around the circle B.

t0: A and B start.

t1: A and B are at the speed mark.

t2: A is at second mark and B is at the start of the circle.

now things get a bit messed up:

t3: A is at d from the second mark and B is at 1/4 of the circle. If we straighten the circle, that mark is at 0.785*d from the second mark.

t4: A is at 2*d from the second mark and B is at 1.57*d.

t5: A is at 3*d from the second mark and B is at 2.355*d.

t6: A is at 3.14*d from the second mark and B is at 2.4649*d.

t7: A is at 4*d from the second mark and B is at 3.14*d.

We can se that there is a slowing down, instantaneous and not cumulative. Is it apparent? We proposed a new experiment that calculates the final speed of the two balls.

Loop velocity is real and measurable. It also turns out to be pi/4 slower when compared to the same ‘straight’ velocity. The apparent part of loop velocity is calling it a velocity, when it is actually an acceleration.

Based on what? You need to justify your statement.I think that for logistic reasons we must use half a circumference.

Ignoring friction, I’m certain that an additional final straight would show that B’s speed would increase from the loop velocity to match A’s final straight velocity. Of course I may be wrong, but there are now plenty of people looking at this.There are 2 possible outcomes:

1) A and B have the same speed

2) A is faster than B

But that is not the case I'm afraid. If 2 is true, then in general changing curvature changes the speed as an inverse proportion. That would result in B gaining speed when getting on the straight path. So both possibilities (speed loss and physical longer distance) will produce the same final speed.

Steve was able to prove pi=4 using just rolling velocity. The classical understanding is overturned. If we could generate a constant velocity we can conduct straight and curved comparisons. We can also accept the kinematic results and race imaginary spheros around a continuous track.Is there anything we can do to better understand this?

.

**LongtimeAirman**- Admin
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## Re: Proof that pi=4

I was starting to think that if instead of the curved circular tube -- the "Pi = 3.14xx" metal ball traveled along a isosceles triangular strip only 1 atom wide at the 3 edges...that curved perfectly at the key points for perfect balance.... would the entire surface area traveled by the "Pi=3.14xxx" ball be less than the surface area traveled by the "straight path = 4" ball? The different orbital turns in the "Pi=3.14" ball path creates less ball surface area traveled?

http://mathworld.wolfram.com/SurfaceofRevolution.html

http://mathworld.wolfram.com/SurfaceofRevolution.html

**Cr6**- Admin
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## Re: Proof that pi=4

Ciaolo wrote:

We can se that there is a slowing down, instantaneous and not cumulative. Is it apparent?

I think it is important to realise that we only

**think**it slows down because we think in terms of

**pi**. The fact that it is instantaneous tells us that it is not any sort of friction or surface area relation because those would be cumulative. The fact that the velocity on exit matches the velocity of the straight path also tells us that it is the actual distance that is different because we know it can't be time changing and since v=d/t, they are the only 2 things that can change.

Ciaolo wrote:

I think that for logistic reasons we must use half a circumference.

There are 2 possible outcomes:

1) A and B have the same speed

2) A is faster than B

But that is not the case I'm afraid. If 2 is true, then in general changing curvature changes the speed as an inverse proportion. That would result in B gaining speed when getting on the straight path. So both possibilities (speed loss and physical longer distance) will produce the same final speed.

Is there anything we can do to better understand this?

A and B have the same speed but B is traveling a distance that is different to expectation. You can only think that it slows down if you think that pi=3.14. Recalculate using pi=4 and you will find that it does not slow down.

We only have 3 variables to juggle, v, d and t. We know that t isn't changing so we can rule that out, leaving us with only v or d. You can pick whichever one you want and apply the change to it but you can't pick both and it doesn't really make any sense to pick v, since it is a derived value and all logic tells us that part of the velocity is not going to just disappear while in the curve and suddenly reappear once out of it. These are not self-propelled objects which is a huge plus for this experiment. So the only valid conclusion is that the distance is longer.

The extra distance is the sum of the changes made at each dt (delta t). To be more precise, it is the sum of the lengths of the change vectors. A change vector is the difference between the initial velocity vector, where it would go without curvature, and the resultant vector, where it did actually go. You just subtract one from the other. If you added the length of all of those change vectors you will find that it matches the difference between pi and 4. I say matches, not equals, because you have to take size differences into consideration before you can find an equality.

**Nevyn**- Admin
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## Re: Proof that pi=4

.

Sorry Cr6, I’m not seeing it. Please define ‘isosceles triangular strip’. The sphere traces a rolling contact point - a line. In what sense can the tube’s loop or straight sections’ surfaces be replaced with an ‘isosceles triangular strip’? Also, ‘surface area traveled’ doesn’t make sense to me. We are working in line distances, there are no surface areas involved that I can see.

“Juggling 3 variables v, d and t”, as in velocity, distance and time? I use v=d/t above myself, but I don’t think it’s always correct anymore. For example, d is normally a distance, a hypotenuse, the source of Pi geometry.

The velocity vector may be described by, V = a(t)X + b(t)Y. Where X and Y are the orthogonal velocity components. The functions a(t), and b(t) can describe a loop or a straight line.

I like your description of the additional 4/pi as the sum of the change vectors made at each dt. I'm still entertaining the idea that the orthogonal direction to the circular loop involves an unseen 180deg spin rotation.

I think we can nail this down.

.

Cr6 wrote:I was starting to think that if instead of the curved circular tube -- the "Pi = 3.14xx" metal ball traveled along a isosceles triangular strip only 1 atom wide at the 3 edges...that curved perfectly at the key points for perfect balance.... would the entire surface area traveled by the "Pi=3.14xxx" ball be less than the surface area traveled by the "straight path = 4" ball? The different orbital turns in the "Pi=3.14" ball path creates less ball surface area traveled?

Sorry Cr6, I’m not seeing it. Please define ‘isosceles triangular strip’. The sphere traces a rolling contact point - a line. In what sense can the tube’s loop or straight sections’ surfaces be replaced with an ‘isosceles triangular strip’? Also, ‘surface area traveled’ doesn’t make sense to me. We are working in line distances, there are no surface areas involved that I can see.

Clearly, compared side-by-side, the ball moves slower in the loop than in the straight. I’ve mistakenly said ‘speeding up’ or ‘slowing down’, the ball has only two speeds, straight or curved (the curve speed varies with curvature). Since v=d/t, we can logically expect distance along the curve be longer than the straight. This agrees with our new understanding, the distance associated with curved motion is 4/pi longer than previously thought.Nevyn wrote: we only think it (the ball) slows down because we think in terms of pi.

Nevyn, You argue that the speeds are equal. Wouldn’t that make the distances equal too? We cannot see them as equal because they are based on two different metrics. In the straight metric (which includes hypotenuses and pi), distances are equal to their geometric length. In the curved (or motion, or manhattan) metric, the hypotenuse is the c-path which is not used and pi is replaced with 4. The motion metric is new and opens up the question of where is all that additional distance coming from?Nevyn wrote:A and B have the same speed but B is traveling a distance that is different to expectation. You can only think that it slows down if you think that pi=3.14. Recalculate using pi=4 and you will find that it does not slow down.

We only have 3 variables to juggle, v, d and t. We know that t isn't changing so we can rule that out, leaving us with only v or d. You can pick whichever one you want and apply the change to it but you can't pick both and it doesn't really make any sense to pick v, since it is a derived value and all logic tells us that part of the velocity is not going to just disappear while in the curve and suddenly reappear once out of it. These are not self-propelled objects which is a huge plus for this experiment. So the only valid conclusion is that the distance is longer.

“Juggling 3 variables v, d and t”, as in velocity, distance and time? I use v=d/t above myself, but I don’t think it’s always correct anymore. For example, d is normally a distance, a hypotenuse, the source of Pi geometry.

The velocity vector may be described by, V = a(t)X + b(t)Y. Where X and Y are the orthogonal velocity components. The functions a(t), and b(t) can describe a loop or a straight line.

Nevyn wrote:The extra distance is the sum of the changes made at each dt (delta t). To be more precise, it is the sum of the lengths of the change vectors. A change vector is the difference between the initial velocity vector, where it would go without curvature, and the resultant vector, where it did actually go. You just subtract one from the other. If you added the length of all of those change vectors you will find that it matches the difference between pi and 4. I say matches, not equals, because you have to take size differences into consideration before you can find an equality.

I like your description of the additional 4/pi as the sum of the change vectors made at each dt. I'm still entertaining the idea that the orthogonal direction to the circular loop involves an unseen 180deg spin rotation.

Miles wrote:Because, to solve this with math, we have to integrate both a and b with time t. In other words, we don't integrate a with b, we integrate a with time t and b with the same time t. If you don't like the word integrate—because it is a calculus term—I can say it this way: we have to track how a changes with time and the way b changes with time. That's how we include time in the problem.

I think we can nail this down.

.

**LongtimeAirman**- Admin
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Join date : 2014-08-10

## Re: Proof that pi=4

LongtimeAirman wrote:.Cr6 wrote:I was starting to think that if instead of the curved circular tube -- the "Pi = 3.14xx" metal ball traveled along a isosceles triangular strip only 1 atom wide at the 3 edges...that curved perfectly at the key points for perfect balance.... would the entire surface area traveled by the "Pi=3.14xxx" ball be less than the surface area traveled by the "straight path = 4" ball? The different orbital turns in the "Pi=3.14" ball path creates less ball surface area traveled?

Sorry Cr6, I’m not seeing it. Please define ‘isosceles triangular strip’. The sphere traces a rolling contact point - a line. In what sense can the tube’s loop or straight sections’ surfaces be replaced with an ‘isosceles triangular strip’? Also, ‘surface area traveled’ doesn’t make sense to me. We are working in line distances, there are no surface areas involved that I can see.

....

I think we can nail this down.

Sorry LTAM, I had more this in mind. The apex of one-side of the wire is just 1 atom thick. The ball rolls along it on both the straight and "curved" paths.

**Cr6**- Admin
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## Re: Proof that pi=4

LongtimeAirman wrote:

Nevyn, You argue that the speeds are equal. Wouldn’t that make the distances equal too?

Yes, that is the point of the experiment. Two objects with the same velocity will cover the same distance in the same time. Using velocity to measure the distance removes all need to rely on mathematical arguments. Some may argue that v=d/t does not describe angular velocity but they are missing the point. We are trying to determine the length of the circumference which means 'what is the length of this curve if it was straight'. If the velocities are equal when the ball enters the curve and they are equal when it exits the curve, then we must assume that they are equal through the curve. Anything else is absurd.

It reminds me of the old saying 'If a tree falls in the forest and no-one is around to hear it, does it make a sound?'

Well of course it does. To assume that it doesn't just because it is not being measured is absurd.

LongtimeAirman wrote:

“Juggling 3 variables v, d and t”, as in velocity, distance and time? I use v=d/t above myself, but I don’t think it’s always correct anymore. For example, d is normally a distance, a hypotenuse, the source of Pi geometry.

Yes, v=d/t means velocity = change in distance over change in time. The d is

**always**a distance. It is not a hypotenuse because distance does not require a triangle and it is not the source of pi (=3.14) because pi is determined by pure geometry which does not use distance. Miles' argument is that if you determine pi using distance, then it will equal 4 and this experiment confirms that by keeping v and t the same so that we can measure d.

We are measuring the curved path by using the straight path. You guys seem to be fixated on the curved path and looking at it in isolation but that is not the point of this experiment. The straight path tells us all of the information, v, d and t. We then use that to compare to the curved path since we can look at it at any dt and compare it to where the ball is on the straight path.

LongtimeAirman wrote:

The velocity vector may be described by, V = a(t)X + b(t)Y. Where X and Y are the orthogonal velocity components. The functions a(t), and b(t) can describe a loop or a straight line.

You have moved up to a 2D vector by splitting the X and Y components out into individual terms. But I don't agree that a(t) and b(t) can describe a curve and it still be called a velocity. A velocity is straight, never curved, if it is curved then it is an acceleration which is the addition of multiple velocity vectors. In order to maintain the equality, you have to limit the functions a and b to be linear. A linear function maintains the same relationship between any 2 consecutive output values if the input values have the same difference.

I don't think I have said that very well so I will try to explain it in math.

Let's say we have a function and we call it f(x). We calculate a value for x=1 and for x=2 and we subtract one from the other giving us the value k(1,2).

k(1,2) = f(2) - f(1)

We then do this again with x=3 and x=4 (it could be any 2 consecutive values but let's keep it simple).

k(3,4) = f(4) - f(3)

If f(x) is linear then k(1,2) = k(3,4).

Let's try it out with some real numbers.

Let f(x) = 2x.

x | f(x) |

1 | 2 |

2 | 4 |

3 | 6 |

4 | 8 |

= 4 - 2

= 2

f(4) - f(3)

= 8 - 6

= 2

So f(x) = 2x is linear.

If f(x) = x^2 then:

x | f(x) |

1 | 1 |

2 | 4 |

3 | 9 |

4 | 16 |

= 2 - 1

= 1

f(4) - f(3)

= 16 - 9

= 7

So f(x) = x^2 is not linear.

LongtimeAirman wrote:

I like your description of the additional 4/pi as the sum of the change vectors made at each dt. I'm still entertaining the idea that the orthogonal direction to the circular loop involves an unseen 180deg spin rotation.

Any spin of the ball is irrelevant to this experiment and its conclusions. We are not measuring the distance traveled by a point on the surface of the ball, we are measuring the distance traveled by the ball itself. The spin imparted to the ball may be an interesting avenue for study but it does not impact this problem.

**Nevyn**- Admin
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## Re: Proof that pi=4

Sorry Nevyn, Let me add “I agree” to my full quote here.Nevyn wrote:Yes, that is the point of the experiment.Longtime Airman wrote:

Nevyn, You argue that the speeds are equal. Wouldn’t that make the distances equal too?

Or let me add to next sentence after yours: We cannot see them as equal,Nevyn, You argue that the speeds are equal,I agree. Wouldn’t that make the distances equal too? We cannot see them as equal because they are based on two different metrics. In the straight metric (which includes hypotenuses and pi), distances are equal to their geometric length. In the curved (or motion, or manhattan) metric, the hypotenuse is the c-path which is not used and pi is replaced with 4. The motion metric is new and opens up the question of where is all that additional distance coming from?

**even though they are**, because they are based on two different metrics.

The velocities and distances may be the same in a logical or absolute number sense, but they cannot be truly equal since only one is a velocity and the other an acceleration. The equality is not complete.

Good point. Miles is also a little sloppy with d as a hypotenuse in Proof that Pi=4, http://milesmathis.com/pi7.pdf.Nevyn wrote:v=d/t means velocity = change in distance over change in time. The d is always a distance.

Nevyn wrote:

We are measuring the curved path by using the straight path. You guys seem to be fixated on the curved path and looking at it in isolation but that is not the point of this experiment. The straight path tells us all of the information, v, d and t. We then use that to compare to the curved path since we can look at it at any dt and compare it to where the ball is on the straight path.

They are not completely the same, one is a velocity and the other an acceleration. v=d/t is based on the geometric metric that works in the physical world - do you believe it should describe a curve, or a proportion of the curve?

Thanks for clarifying the numbers. I don't believe Miles has described how to properly integrate these curves yet. You may be right, we may not need any special transformation.

Thanks for the discussion.

.

**LongtimeAirman**- Admin
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## Re: Proof that pi=4

LongtimeAirman wrote:Sorry Nevyn, Let me add “I agree” to my full quote here.Or let me add to next sentence after yours: We cannot see them as equal,Nevyn, You argue that the speeds are equal,I agree. Wouldn’t that make the distances equal too? We cannot see them as equal because they are based on two different metrics. In the straight metric (which includes hypotenuses and pi), distances are equal to their geometric length. In the curved (or motion, or manhattan) metric, the hypotenuse is the c-path which is not used and pi is replaced with 4. The motion metric is new and opens up the question of where is all that additional distance coming from?even though they are, because they are based on two different metrics.

Yes, that is correct.

LongtimeAirman wrote:

The velocities and distances may be the same in a logical or absolute number sense, but they cannot be truly equal since only one is a velocity and the other an acceleration. The equality is not complete.

They are comparable so the equality remains. You can reduce both velocity and acceleration to distance and time.

v = d/t

a = v/t

= d/t/t

= d/t^2

But those equations are a little misleading so let me expand them into their proper forms.

Let d1, d2 and d3 be measurements of distance at t1, t2 and t3 respectively.

v = (d2 - d1)/(t2 - t1)

a = (v2 - v1)/(t2 - t1)

= ( (d3 - d2)/(t3 - t2) - (d2 - d1)/(t2 - t1) ) / (t3 - t2)

since (t3 - t2) = (t2 - t1)

a = ( (d3 - d2)/(t2 - t1) - (d2 - d1)/(t2 - t1) ) / (t2 - t1)

= ( (d3 - d2) - (d2 - d1) ) / (t2 - t1) / (t2 - t1)

= ( d3 - 2*d2 + d1 ) / (t2 - t1)^2

The experiment is looking for a distance so we can absolutely compare velocities with accelerations. My analysis using the change vectors did exactly that. The ball has a velocity as it enters the curve, at the next dt (or frame) it has deviated from that vector. We look at that deviation, in isolation, which is a velocity itself. That is the second velocity, along with the initial velocity, that forms the acceleration. Both of those velocities have a length and the sum of those lengths over the complete curve gives us the distance of the curve. I could even say that straight line motion has the same 2 vectors, but one of them will always have 0 length.

LongtimeAirman wrote:Good point. Miles is also a little sloppy with d as a hypotenuse in Proof that Pi=4, http://milesmathis.com/pi7.pdf.Nevyn wrote:v=d/t means velocity = change in distance over change in time. The d is always a distance.

Sorry, I wasn't very clear. What I meant, in a very strict sense, is that a distance is not an hypotenuse. What I didn't say is that an hypotenuse can be a distance. An hypotenuse can be many things and distance is only one of them. An hypotenuse is really a relationship and it doesn't really care what the members of that relationship are. But a distance is a real thing. It is a measurable quantity and can not be transferred to other things like a relationship can.

LongtimeAirman wrote:Nevyn wrote:

We are measuring the curved path by using the straight path. You guys seem to be fixated on the curved path and looking at it in isolation but that is not the point of this experiment. The straight path tells us all of the information, v, d and t. We then use that to compare to the curved path since we can look at it at any dt and compare it to where the ball is on the straight path.

They are not completely the same, one is a velocity and the other an acceleration. v=d/t is based on the geometric metric that works in the physical world - do you believe it should describe a curve, or a proportion of the curve?

No, v=d/t does not describe a curve, however, it can be used to measure a curve and that is because we only care about the distance part of the equation. We keep t and v equal and we measure d against a straight line path. Using it this way you can not break it down into little dt's. But if we just look at the full distance, then we can absolutely use v=d/t to compare them. We want the complete distance around the curve, not the distance between each dt.

Another way to look at it is that we are never measuring the curved path. We are measuring the straight path and comparing the curved path to it. We can definitely use v=d/t on the straight path and we want to know the length of the curved path

**if it was straight**so we can use it there too.

LongtimeAirman wrote:

Thanks for clarifying the numbers. I don't believe Miles has described how to properly integrate these curves yet. You may be right, we may not need any special transformation.

Intergration just means to add things at each step along the path. In this case we need to add the a-path with the b-path (from Miles most recent pi paper). For each dt, we start at A and the initial velocity is represented by b. The ball ends up at B so the change vector is a. The acceleration is a+b which if you do this vector addition you would get the path c, but the length of c does not represent the correct distance traveled because the ball had both a and b applied to it, not c. Simply put, reality does not use vector addition, it just applies all vectors.

**There is no a+b, there is only a and b.**

LongtimeAirman wrote:

Thanks for the discussion.

.

Always a pleasure.

**Nevyn**- Admin
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## Re: Proof that pi=4

.

Slumming again. I posted on

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448

.

Slumming again. I posted on

**A Simple Experiment Proves π = 4**http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448

.

**LongtimeAirman**- Admin
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## Re: Proof that pi=4

.

I posted this earlier today and wanted to include it here too.

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448&p=115428#p115428

Hello Aardwolf, I need to think about lost momentum. I would say that the pi calculation is used to determine distance around the circle, not time. Nevertheless I guess I need to think about that too.

Plotting distance versus time for both Pi=3.14 and Pi =4 may help everyone. I should have done this earlier.

Balls enter both tracks at the same initial speed (1.2m/s) and hit their 4 marks simultaneously:

1. The ball in the circular track is slower than the straight track. There is friction slowing the ball in the circular track pi/4 when compared to the straight track.

.

or

2. Both tracks are the same length when the curve is measured with Pi=4.

.

I posted this earlier today and wanted to include it here too.

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448&p=115428#p115428

Hello Aardwolf, I need to think about lost momentum. I would say that the pi calculation is used to determine distance around the circle, not time. Nevertheless I guess I need to think about that too.

Plotting distance versus time for both Pi=3.14 and Pi =4 may help everyone. I should have done this earlier.

Balls enter both tracks at the same initial speed (1.2m/s) and hit their 4 marks simultaneously:

1. The ball in the circular track is slower than the straight track. There is friction slowing the ball in the circular track pi/4 when compared to the straight track.

.

or

2. Both tracks are the same length when the curve is measured with Pi=4.

.

Last edited by LongtimeAirman on Fri Oct 14, 2016 9:54 pm; edited 1 time in total (Reason for editing : Dang typos. Changed)

**LongtimeAirman**- Admin
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## Re: Proof that pi=4

.

Double posting - My post from TB a little earlier today.

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448&p=115637#p115637

...........................

A Simple Experiment Proves π = 4

http://milesmathis.com/pi7.pdf

...........................

Straight paths allow the single dimensional equation, (rate)*(time) = distance.

Unfortunately, the formula doesn’t “fit” curves, since curves are two dimensional.

Pi = Circumference/diameter, was invented in order to solve that. Pi=3.14 is the curve to linear metric that allows us to measure any curved path as a linear distance.

Steven’s experiment includes that. The circular track length is based on diameter 17.6cm.

Pi * diameter = 3.14 * 17.6cm = 55.3cm.

A steel ball is launched at 120cm/s into the circular track and completes it at a constant speed 0.5867 seconds later. The circular track velocity is thus: distance/time = 55.3cm/0.5867s = 94.3cm/s

Another steel ball is launched at the exact same time and initial velocity into a separate straight track, four diameters long (4*17.6cm), 70.4cm. Both balls travel for the same amount of time. The straight track velocity was a constant 120cm/s.

Two apparently different velocities: circular - 94.3cm/s; and straight - 120cm/s.

You might say there’s no problem here. The velocities cannot be equal because the distances 55.3cm and 70.4cm are different. If you watch the experiment (hypothetical you) you might even believe, as most would agree, the ball appears to travel slower through the circular track. The time may be the same but the two distances are different.

There’s a big problem. Steven was kind enough to include the 55.3cm distance (the Pi mark) on the straight track. When the ball in the circular track completes its circle, the straight track ball should be at the Pi mark, and not at the 4 mark, as is clearly shown in the video. A dilemma. This violates the known geometrical distance calculation (pi = 3.14 = 55.3cm/17.6cm = circumference / diameter). Repeat, geometry is broken here.

The two balls: 1) started with the same initial velocity; 2) experienced no significant friction; and 3) traveled the same amount of time. It’s not unreasonable to suspect that the two distances and velocities are actually the same.

A new motion metric is needed. The same experiment that demonstrates a failure in pi=3.14 geometry also shows us the correction, replace 3.14 with 4. We then see that the two distances and velocities are equal (70.4cm and 120cm/sec).

For objects in curved motion. Pi=4.

We should observe that we’ve always been underestimating certain distances and velocities. The ball is traveling ((4-3.14)/4) or 21 percent further and faster through the circular track than we thought.

...........................

.

Double posting - My post from TB a little earlier today.

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448&p=115637#p115637

...........................

A Simple Experiment Proves π = 4

http://milesmathis.com/pi7.pdf

...........................

Airman said. That’s what the experiment shows. We have two apparently different constant velocities: circular - 94.3cm/s; and straight - 120cm/s; that are both equal to 120cm/s when pi=4.

lw1990. Are you serious right now

They 'appear' to be different velocities because they are.

Your false idols of Mathis and his student have made you a foolish believer in the superstition of pi=4

By convincing you that the very real phenomenon of speed of the ball does not matter, but whatever bull they feed you, like 'distance warping magic' affects the experiment instead.

How do we know that a ball in any given experiment is traveling at any given speed, how do we know it is not just 'appearing to do so'? This is the kind of problem you get when you invent things with no basis in reality.

**Airman**. To calculate velocity, one needs to know distance and time.Straight paths allow the single dimensional equation, (rate)*(time) = distance.

Unfortunately, the formula doesn’t “fit” curves, since curves are two dimensional.

Pi = Circumference/diameter, was invented in order to solve that. Pi=3.14 is the curve to linear metric that allows us to measure any curved path as a linear distance.

Steven’s experiment includes that. The circular track length is based on diameter 17.6cm.

Pi * diameter = 3.14 * 17.6cm = 55.3cm.

A steel ball is launched at 120cm/s into the circular track and completes it at a constant speed 0.5867 seconds later. The circular track velocity is thus: distance/time = 55.3cm/0.5867s = 94.3cm/s

Another steel ball is launched at the exact same time and initial velocity into a separate straight track, four diameters long (4*17.6cm), 70.4cm. Both balls travel for the same amount of time. The straight track velocity was a constant 120cm/s.

Two apparently different velocities: circular - 94.3cm/s; and straight - 120cm/s.

You might say there’s no problem here. The velocities cannot be equal because the distances 55.3cm and 70.4cm are different. If you watch the experiment (hypothetical you) you might even believe, as most would agree, the ball appears to travel slower through the circular track. The time may be the same but the two distances are different.

There’s a big problem. Steven was kind enough to include the 55.3cm distance (the Pi mark) on the straight track. When the ball in the circular track completes its circle, the straight track ball should be at the Pi mark, and not at the 4 mark, as is clearly shown in the video. A dilemma. This violates the known geometrical distance calculation (pi = 3.14 = 55.3cm/17.6cm = circumference / diameter). Repeat, geometry is broken here.

The two balls: 1) started with the same initial velocity; 2) experienced no significant friction; and 3) traveled the same amount of time. It’s not unreasonable to suspect that the two distances and velocities are actually the same.

A new motion metric is needed. The same experiment that demonstrates a failure in pi=3.14 geometry also shows us the correction, replace 3.14 with 4. We then see that the two distances and velocities are equal (70.4cm and 120cm/sec).

For objects in curved motion. Pi=4.

We should observe that we’ve always been underestimating certain distances and velocities. The ball is traveling ((4-3.14)/4) or 21 percent further and faster through the circular track than we thought.

...........................

**Maol**. Well, this certainly solves the paradoxical question of "What is the radius of a square?"**Airman**. Pi are square.**Maol**. Well, and I'm sure you easily see, Pi are square solves for the radii of a rectangle, which is sort of an ellipse with corners, like the difference between the tracks at Indy and Talladega, but only to an approximation, or several approximations, as required. Pi are cubed then solves for the multi-inclusive radii of an ellipsoid and requires more approximations, many more.**Airman**. It’s not easy, but I think I see. Pi are hypercubed diagonalizes the universe, allowing us to solve for the radius in any corner of space and time. Thanks..

**LongtimeAirman**- Admin
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## Re: Proof that pi=4

Nice presentation LTAM. Steven's experiment gives even the old-timers something to chew on. I bet they all have their own "spin" on the ball for a complete explanation.

If the ball traveled along a steel ribbon instead of a "tube" what would be the shape of the "curved" path? Would it twist?

If the ball traveled along a steel ribbon instead of a "tube" what would be the shape of the "curved" path? Would it twist?

**Cr6**- Admin
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## Re: Proof that pi=4

Cr6. Nice presentation LTAM. Steven's experiment gives even the old-timers something to chew on. I bet they all have their own "spin" on the ball for a complete explanation.

If the ball traveled along a steel ribbon instead of a "tube" what would be the shape of the "curved" path? Would it twist?

Hi Cr6. It's fair to say our imaginations were spinning up too.

I believe that Steven said the ball in the curve "banks a few tens of degrees". I currently interpret that "fact" to mean that we can replace the circular track with large radius gently concave surface, like a shallow serving bowl. If we were accurate enough, a gently banked steel ribbon along a plate's rim would work just as well. Simply replacing the steel balls with patterned glass marbles would show that there weren't any additional rotations present; the experimental velocities may vary, but the outcome would remain the same. It's still fun.

.

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## Re: Proof that pi=4

And again, today's post:

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448&p=115660#p115660

.

The circular track length of 55.3cm is marked into quarters (0,1,2,3,4) (5 marks inclusive). The straight track has 6 marks, (-1,0,1,2,3,4) with unit separations of 17.6cm; 17.6 the straight track also has a “PI mark” at 3.14 showing where the circular track’s original 55.3cm distance would end. The -1 (-17.6cm) and 0 marks allow one to calculate and compare each ball’s observed initial velocity of 120cm/s. A time vs distance plot of the balls’ path through the track markers are shown.

The horizontal values 0.0000s, 0.1467s, 0.2933s, 0.4400s, 0.5867s, are the observed times the steel balls crossed their marks (0,1,2,3,4) in both tracks. The corresponding distances are indicated on the plot lines.

Five measured points for each track lets us calculate 4 interval (quarter track) velocities. The velocity calculation for each interval is given by:

velocity = change of distance over the change in time, or

velocity(current) = (currentPosition – previousPosition)/(currentTime-previuosTime)

Circular track:

(0,1). 1st interval. (13.8-0)/(0.1467-0) = 94.3cm/s

(1,2). 2nd interval. (27.7-13./(0.2933-0.1467) = 94.3cm/s

(2,3). 3rd interval. (41.5-27.7)/(0.4400-0.2933) = 94.3cm/s

(3,4). 4th interval. (55.3-41.5)/0.5867-0.4400) = 94.3cm/s

Straight track:

(0,1). 1st interval. (17.6-0)/(0.1467-0) = 120cm/s

(1,2). 2nd interval. (35.2-17.6)/(0.2933-0.1467) = 120cm/s

(2,3). 3rd interval. (52.8-35.2)/(0.4400-0.2933) = 120cm/s

(3,4). 4th interval. (70.4-52./0.5867-0.4400) = 120cm/s

As you can see, our two apparently different velocities of 94.3cm/s and 120cm/s occurred over each of our quarter intervals.

All velocities, including our initial velocities are 120cm/s when Pi=4.

We also have a camera with frame rate. Measurement marks may be placed along the entire length of transparent tubing. Make that a future refinement. It isn’t necessary, the demonstration is clear.

My previous post referred to constant velocities. That’s what these plots show. There are no decelerations observed. Any friction present is acting over both tracks equally and is at least an order of magnitude less than the observed curve-to-linear (pi=3.14) metric error.

Pi=4 for bodies in curved motion.

.

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448&p=115660#p115660

.

Airman wrote. There’s a big problem. Steven was kind enough to include the 55.3cm distance (the Pi mark) on the straight track. When the ball in the circular track completes its circle, the straight track ball should be at the Pi mark, and not at the 4 mark, as is clearly shown in the video. A dilemma. This violates the known geometrical distance calculation (pi = 3.14 = 55.3cm/17.6cm = circumference / diameter). Repeat, geometry is broken here.

lw1990 wrote.It is not a dilemma if the speed of one ball is slower than another ball, even if the speed slowly decelerates through the track, this still means normal physics is happening and nothing is broken or wrong. The experiment does not even come close to showing a shred of evidence that the speeds of the two balls are identical throughout the track. If it did, you might have a point, but it absolutely does not.

You're calculating 'velocity' based on a start distance and end distance of marks on the track, rather than calculating if there's actually a change in velocity while going through the track (like deceleration or even acceleration difference). ...There is a whole lot of horrible science here, but let's start with the fact that you don't measure velocity through the track except the overall start point-end mark. ...... The point here is that you have to measure the speed of the balls throughout the track to find out why, ...

**Airman**. Steven uses 13 marks – a baker’s dozen – in order to calculate, compare and monitor for a change of speed in the two tracks. Not, you might notice, just the “overall start point-end mark”.The circular track length of 55.3cm is marked into quarters (0,1,2,3,4) (5 marks inclusive). The straight track has 6 marks, (-1,0,1,2,3,4) with unit separations of 17.6cm; 17.6 the straight track also has a “PI mark” at 3.14 showing where the circular track’s original 55.3cm distance would end. The -1 (-17.6cm) and 0 marks allow one to calculate and compare each ball’s observed initial velocity of 120cm/s. A time vs distance plot of the balls’ path through the track markers are shown.

The horizontal values 0.0000s, 0.1467s, 0.2933s, 0.4400s, 0.5867s, are the observed times the steel balls crossed their marks (0,1,2,3,4) in both tracks. The corresponding distances are indicated on the plot lines.

Five measured points for each track lets us calculate 4 interval (quarter track) velocities. The velocity calculation for each interval is given by:

velocity = change of distance over the change in time, or

velocity(current) = (currentPosition – previousPosition)/(currentTime-previuosTime)

Circular track:

(0,1). 1st interval. (13.8-0)/(0.1467-0) = 94.3cm/s

(1,2). 2nd interval. (27.7-13./(0.2933-0.1467) = 94.3cm/s

(2,3). 3rd interval. (41.5-27.7)/(0.4400-0.2933) = 94.3cm/s

(3,4). 4th interval. (55.3-41.5)/0.5867-0.4400) = 94.3cm/s

Straight track:

(0,1). 1st interval. (17.6-0)/(0.1467-0) = 120cm/s

(1,2). 2nd interval. (35.2-17.6)/(0.2933-0.1467) = 120cm/s

(2,3). 3rd interval. (52.8-35.2)/(0.4400-0.2933) = 120cm/s

(3,4). 4th interval. (70.4-52./0.5867-0.4400) = 120cm/s

As you can see, our two apparently different velocities of 94.3cm/s and 120cm/s occurred over each of our quarter intervals.

All velocities, including our initial velocities are 120cm/s when Pi=4.

We also have a camera with frame rate. Measurement marks may be placed along the entire length of transparent tubing. Make that a future refinement. It isn’t necessary, the demonstration is clear.

My previous post referred to constant velocities. That’s what these plots show. There are no decelerations observed. Any friction present is acting over both tracks equally and is at least an order of magnitude less than the observed curve-to-linear (pi=3.14) metric error.

Pi=4 for bodies in curved motion.

.

**LongtimeAirman**- Admin
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## Re: Proof that pi=4

Here is a video showing how the motions of the circle correspond with the motion on the two axes. Hosted on Vimeo, the link should take you there directly.

https://vimeo.com/189106809

https://vimeo.com/189106809

**Jared Magneson**- Posts : 372

Join date : 2016-10-11

## Re: Proof that pi=4

Thanks Jared.

Unfortunately, it's a bit too unwieldy. 8 seconds is too short, each time it stops we have to view the vimeo site application in order to start it up again. Too many graph lines and numbers are unnecessary.

The only image app I have is windows superpaint.

Can we turn this into a gif?

.

Unfortunately, it's a bit too unwieldy. 8 seconds is too short, each time it stops we have to view the vimeo site application in order to start it up again. Too many graph lines and numbers are unnecessary.

The only image app I have is windows superpaint.

Can we turn this into a gif?

.

**LongtimeAirman**- Admin
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## Re: Proof that pi=4

Sure, but I'll have to get to it tonight. I have all the big dog programs for my CAD and arch/viz career, I just never had much chance to do animations in Maya so mine are still pretty rough. I'll dress them up and make them pretty later on, once things are "right" for any particular topic. I'll do an animated .gif tonight.

**Jared Magneson**- Posts : 372

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## Re: Proof that pi=4

Don't rush. It's just something good to have. Thanks

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## Re: Proof that pi=4

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448&p=115713#p115713

.

Thanks too, in a cosmic sense. I see TB member StevenO was also part of the discussion.

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=8&t=3183&start=15#p34628

StevenO wrote: Re:

StevenO went and created an actual demonstration of the dynamic circle vs. geometric circle problem. I daresay you played a part in StevenO’s motivation (hence cosmic).

Steven Oostdijk,

http://milesmathis.com/pi7.pdf

StevenO‘s experiment “shows that, while Pi as a distance is 3.14, Pi as a distance/time is 4”

1) 55.3cm=3.14 * 17.6cm and 2) 70.4cm=4 * 17.6cm; and marked into quarters (0,1,2,3,4).

For bodies in curved motion (accelerations comprised of orthogonal velocities) Pi=4

.

.

Siggy_G. We actually discussed this topic a few years back here: Mathis and pi

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=8&t=3183

**Airman**. Thanks for the reading material, what an eye-opener. I wasn’t a member yet and hadn’t seen it. Mathis and pi is a sensitive issue here and I think I understand that and a few of the earlier responses in this discussion much better now.Thanks too, in a cosmic sense. I see TB member StevenO was also part of the discussion.

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=8&t=3183&start=15#p34628

StevenO wrote: Re:

**Mathis and pi**Hi Siggy,

As I stated before, to get a good grasp of this dynamic circle vs. geometric circle problem, you should really study Miles' papers on the calculus. The mistakes in the calculus spill over to a lot of area's in physics and lead to many misconceptions. A math description is always less complete than the actual physics.

**Airman**. After reading that whole discussion I don’t know if you did or not. You may have won the discussion at the time, unfortunately the whole Mathis and pi thing has developed into a schism. This board has suffered because of it.**That discussion is history. Everyone needs to move on**.StevenO went and created an actual demonstration of the dynamic circle vs. geometric circle problem. I daresay you played a part in StevenO’s motivation (hence cosmic).

Siggy_G. As for the video (maybe it already has been mentioned), I believe the bent tube will add a slight amount of friction to the motion due to the radial redirectioning.

The original rolling motion of the ball will meet some resistance from the tube walls, at some points counter-roll the motion. If the ball had color patterns, this would show up. If the tube was very elastic, the ball would clearly lose energy to tube deformation.

**Airman**. "(maybe it already has been mentioned)" – yes, I’ll review:Steven Oostdijk,

**A Simple Experiment Proves π = 4**http://milesmathis.com/pi7.pdf

StevenO‘s experiment “shows that, while Pi as a distance is 3.14, Pi as a distance/time is 4”

**Two transparent pvc tube tracks, side-by-side**;**Length**. 1) circular and 2) straight; cut to two different lengths:1) 55.3cm=3.14 * 17.6cm and 2) 70.4cm=4 * 17.6cm; and marked into quarters (0,1,2,3,4).

**Motion**. Two balls are released into the tracks; same initial velocities – 120cm/s; same start(0.0000), middle (0.1467s, 0.2933s, 0.4400s,) and stop(0.5867s) times; Speeds are constant throughout the experiment; 1) 94.3cm/s and 2) 120cm/s; indicating no apparent friction in either track.**Airman**. About the apparent speed difference (94.3cm/s and 120cm/s) between the two tracks I wrote:I agree the experiment could be improved. I’ve remarked that colored marbles can be used to see whether any strange rotations are present. Nevertheless, for those who will see, it is clear. The fact of the matter is that this wonderful, world changing demonstration was created by a TB member as a result of discussion here.More than 20% different? With a ratio involving 3.14 and 4? They started at the same speed, and their track speeds remained constant. Friction’s been eliminated. Motion thru both tracks are synchronized. But why call it a slowdown? A constant velocity difference ratio of 20% without a change in time isn’t a slowdown, it’s a distance scaling change.

And so the solution is simple. The experiment shows us a direct comparison of curved and linear motion that happens to conflict with our current understanding of geometry. A new motion metric where we are shown that both distances are equal (70.4cm) and both velocities are equal (120cm/s). For bodies in motion pi=4.

For bodies in curved motion (accelerations comprised of orthogonal velocities) Pi=4

.

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## Re: Proof that pi=4

http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=16448&p=115745#p115745

Thanks Jared! It worked perfectly.

by LongtimeAirman » Sun Oct 30, 2016 11:35 am

.

Cycloids are not clear examples of curved motion. Calculating their paths, however, do require two variables and time (total of 3 variables).

.................

A circle is more than a circle when it is being created by an object moving through space. Curved motion means that an object is subject to two orthogonal velocities*.

"Here is a video showing how the motions of the circle correspond with the motion on the two axes. Hosted on Vimeo, the link should take you there directly."

https://vimeo.com/189106809

* Oops, no, an object is subject to forces resulting in orthogonal velocities.

I’m still learning.

.

Thanks Jared! It worked perfectly.

by LongtimeAirman » Sun Oct 30, 2016 11:35 am

.

Cycloids are not clear examples of curved motion. Calculating their paths, however, do require two variables and time (total of 3 variables).

.................

A circle is more than a circle when it is being created by an object moving through space. Curved motion means that an object is subject to two orthogonal velocities*.

"Here is a video showing how the motions of the circle correspond with the motion on the two axes. Hosted on Vimeo, the link should take you there directly."

https://vimeo.com/189106809

* Oops, no, an object is subject to forces resulting in orthogonal velocities.

I’m still learning.

.

**LongtimeAirman**- Admin
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## Re: Proof that pi=4

Here is a newer, cleaner version of that animation that should loop, although I don't think Vimeo gives you the choice. Download it if you like.

https://vimeo.com/189647953

https://vimeo.com/189647953

**Jared Magneson**- Posts : 372

Join date : 2016-10-11

## Re: Proof that pi=4

Great. The bars give a whole 'nother view. I've posted too many times today so I've gotta quit soon.

Question - not a tasking - I recall the video that was there and gone again, a scene showing a silhouette where two cyclically moving orthogonally overlapping bars presented a moving circular image. I'm not sure the memory is real or not right now.

I would cite you but you seemed to indicate otherwise. Dragon Face is due for a lot more hits.

Thanks for sharing.

Question - not a tasking - I recall the video that was there and gone again, a scene showing a silhouette where two cyclically moving orthogonally overlapping bars presented a moving circular image. I'm not sure the memory is real or not right now.

I would cite you but you seemed to indicate otherwise. Dragon Face is due for a lot more hits.

Thanks for sharing.

**LongtimeAirman**- Admin
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## Re: Proof that pi=4

Here is a short video of the cycloid, showing Pi=3.14 "racing" it down below.

Pi Cycloid Comparison

Hmm, I did just remove my previous Pi video (on my Vimeo page), but I don't know if that's the one you were talking about. I think these two are better.

Pi Cycloid Comparison

Hmm, I did just remove my previous Pi video (on my Vimeo page), but I don't know if that's the one you were talking about. I think these two are better.

**Jared Magneson**- Posts : 372

Join date : 2016-10-11

## Re: Proof that pi=4

Jared,

I'm confused by your cycloid example. Why is there a velocity difference between the yellow and gray balls? I see that the gray is moving 4/pi faster than the yellow ball, but why? The cycloid has nothing to do with the yellow ball.

I see that your original circular motion vimeo that I posted at TB ( https://vimeo.com/189106809 ) is dead. I can't go back to edit it. I'll post your new circular motion with bars ( https://vimeo.com/189647953 ).

I'm still breaking this little app in, hope you don't mind; I'll just leave this here.

.

I'm confused by your cycloid example. Why is there a velocity difference between the yellow and gray balls? I see that the gray is moving 4/pi faster than the yellow ball, but why? The cycloid has nothing to do with the yellow ball.

I see that your original circular motion vimeo that I posted at TB ( https://vimeo.com/189106809 ) is dead. I can't go back to edit it. I'll post your new circular motion with bars ( https://vimeo.com/189647953 ).

I'm still breaking this little app in, hope you don't mind; I'll just leave this here.

.

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