# Proof that pi=4

Page **3** of **3** • 1, 2, **3**

## Re: Proof that pi=4

By the way, how hard would it be to make a magnetic levitation tube to reduce friction? Or what about replacing the ball with dry ice?

**LloydK**- Posts : 448

Join date : 2014-08-10

## Re: Proof that pi=4

Proof of period, "pi" is 4 not 3.14, for circular rotation.

Defined:

x axis is the horizontal direction

y axis is the vertical direction

Radius is 1 meter.

Magnitude of velocity (v) is 1 m/s

Initial point is A at ( 1,0 )

Equation of circle: x^2 + y^2 = 1

Velocity vector is V = - v sin theta X + v cos theta Y

Where X and Y are the orthogonal velocity components

So Vx is - v sin theta

and Vy is v cos theta.

theta is an angle where:

sin theta is y/1

cos theta is x/1

The magnitude of velocity is (v^2 sin^2 theta + v^2 cos^2 theta)^(1/2) = 1

Time traveled is calculated by the distance traveled over the velocity.

The path is A -> B -> C -> D.

Focusing on AB:

Here we project the path onto the y axis and sum the accumulated time to reach B from A.

If we set the number of division in theta to be very large, say 1000.

Velocity can be calculated at a theta position.

dy/dt = cos theta dtheta/dt = cos theta

The time it takes to travel the incremental distance traveled, Vy * dtheta/dt = cos theta * 1/1000,

over velocity at the theta value , Vy = cos theta,

which is ( cos theta * 1/1000 ) / cos theta = 1/1000 sec

The total time to travel from AB is 1 sec.

Correspondingly BC, CD, and DA is also 1 sec. The total is then 4 sec.

**kug**- Posts : 4

Join date : 2018-10-23

## Re: Proof that pi=4

Hi kug. Welcome, thanks for posting. I’m no expert but I can usually work up a critical review for friends.

I must point out that the “"pi" is 4 not 3.14” comparison normally applies to the distance of the circle’s circumference and not the time it takes to travel round the circle. I would agree that Steven Oostdijk proved pi = 4 by demonstrating that the time it took for the balls to travel either of the two paths: a straight line distance of four; and a circle circumference of radius one; are both the same. I think I understand your logic.

Back to the title, the final sentence fragment is just wrong. The proof concerns circular ‘motion’ and not circular ‘rotation’. I'll assume you meant 'motion' and quote part of soostdijk‘s comment earlier in this thread - It's appropriate since you seem to be pulling a "velocity" around a circle.

Please don't take any of my criticism personally, but you must admit, it may be unreasonable of you to expect positive responses given the problems with the title alone. In my experience, Miles has been quick to reply to questions, although he might also think twice before replying to your effort here. There's too much inaccuracy to give an easy reply.

.

Sorry, the title is confusing, suspect and wrong – at least as far as I can tell. A ‘Period’ is a rate of time. A period of “pi” implies the time it takes to travel about the circle. All well and good, but the ‘rate’ interpretation is contradicted by the next sentence fragment “"pi" is 4 not 3.14”.Proof of period, "pi" is 4 not 3.14, for circular rotation.

I must point out that the “"pi" is 4 not 3.14” comparison normally applies to the distance of the circle’s circumference and not the time it takes to travel round the circle. I would agree that Steven Oostdijk proved pi = 4 by demonstrating that the time it took for the balls to travel either of the two paths: a straight line distance of four; and a circle circumference of radius one; are both the same. I think I understand your logic.

Back to the title, the final sentence fragment is just wrong. The proof concerns circular ‘motion’ and not circular ‘rotation’. I'll assume you meant 'motion' and quote part of soostdijk‘s comment earlier in this thread - It's appropriate since you seem to be pulling a "velocity" around a circle.

**Re: Proof that pi=4**. http://milesmathis.forumotion.com/t226p25-proof-that-pi4#1747.

1. Pulling a "velocity" around a circle is not the same as pulling a solid ruler around a circle. The velocity vector will need to be rotated along the path otherwise it would not follow the path. This rotation takes time. So it would not be logical that the length of a circular path measured with a velocity would be the same as measured with a ruler. For all these people shouting that Pi will always remain 3.14, sure, but only for geometric ratios.

Please don't take any of my criticism personally, but you must admit, it may be unreasonable of you to expect positive responses given the problems with the title alone. In my experience, Miles has been quick to reply to questions, although he might also think twice before replying to your effort here. There's too much inaccuracy to give an easy reply.

.

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

## Re: Proof that pi=4

Well guys... the only thing about this Mathis paper that keeps me laughing is when does a real rolling ball ever actually reach "3.14xxxxxxxxxxx.....infinity" in the real world? If there was no actual t involved I suppose it could actually go on "forever".

**Cr6**- Admin
- Posts : 1085

Join date : 2014-08-09

## Re: Proof that pi=4

This is the corrected version. The important concept is the 2nd picture.

Proof of period, "pi" is 4 not 3.14, for circular rotation.

Defined:

x axis is the horizontal direction

y axis is the vertical direction

Radius is 1 meter.

Magnitude of velocity (v) is 1 m/s

Initial point is A at ( 1,0 )

Equation of circle: x^2 + y^2 = 1

Position vector is D = r cos theta X + r sin theta Y

Velocity vector is V = - v sin theta X + v cos theta Y

Where X and Y are the orthogonal velocity components

So Vx is - v sin theta

and Vy is v cos theta.

theta is an angle where:

sin theta is y/1

cos theta is x/1

The magnitude of velocity is (v^2 sin^2 theta + v^2 cos^2 theta)^(1/2) = 1

Time traveled is calculated by the distance traveled over the velocity.

The path is A -> B -> C -> D.

Focusing on AB:

Here we project the path onto the y axis and sum the accumulated time to reach B from A.

At t = 0: x = 1, y = 0

theta = 0, cos theta = 1

dx/dt = 0, dy/dt = v = 1 m/s

cos theta dtheta/dt = dy/dt

dtheta/dt = v = 1 m/s

At x = 0, y = 1

t is unknown

theta = pi/2, sin theta = 1

dx/dt = -v = -1 m/s, dy/dt = 0

-sin theta dtheta/dt = dx/dt

dtheta/dt = v = 1 m/s

dtheta/dt always proportional to the velocity

Velocity vector equation, V, does not represent the direction of the object, it represents to summation of the velocity vector before and after a point. When theta_r is very small, V is the sum of two vectors of equal magnitude.

v_real = 1/2 V

so dtheta/dt = 1/2 m/s

If we set the number of division in theta to be very large, say 1000.

Theta change from A -> B is 90 degrees, pi/2.

dy/dt = cos theta dtheta/dt = cos theta/2

If theta has 1000 divisions, then t also has 1000 divisions.

Incremental distance traveled, Vy * dt / div = cos theta * 1/1000.

Note V is used because it is the sum of all vectors at a point.

Velocity traveled: dy/dt = cos theta/2.

Time is then calculated:

Time = ( cos theta * 1/1000 ) / (cos theta/2) = 2 / 1000 sec

The total time to travel from AB is 2 sec.

The total distance traveled is sum of magnitude of vector * dt * # of divisions.

Distance traveled in y direction = 1

For a full orbit, A -> B -> C -> D > A:

Total distance is 1*4 = 4 m.

Time taken is 2*4 = 8 sec.

Period of a full circle = 2 pi

2 pi = 8

pi = 4

Proof of period, "pi" is 4 not 3.14, for circular rotation.

Defined:

x axis is the horizontal direction

y axis is the vertical direction

Radius is 1 meter.

Magnitude of velocity (v) is 1 m/s

Initial point is A at ( 1,0 )

Equation of circle: x^2 + y^2 = 1

Position vector is D = r cos theta X + r sin theta Y

Velocity vector is V = - v sin theta X + v cos theta Y

Where X and Y are the orthogonal velocity components

So Vx is - v sin theta

and Vy is v cos theta.

theta is an angle where:

sin theta is y/1

cos theta is x/1

The magnitude of velocity is (v^2 sin^2 theta + v^2 cos^2 theta)^(1/2) = 1

Time traveled is calculated by the distance traveled over the velocity.

The path is A -> B -> C -> D.

Focusing on AB:

Here we project the path onto the y axis and sum the accumulated time to reach B from A.

At t = 0: x = 1, y = 0

theta = 0, cos theta = 1

dx/dt = 0, dy/dt = v = 1 m/s

cos theta dtheta/dt = dy/dt

dtheta/dt = v = 1 m/s

At x = 0, y = 1

t is unknown

theta = pi/2, sin theta = 1

dx/dt = -v = -1 m/s, dy/dt = 0

-sin theta dtheta/dt = dx/dt

dtheta/dt = v = 1 m/s

dtheta/dt always proportional to the velocity

Velocity vector equation, V, does not represent the direction of the object, it represents to summation of the velocity vector before and after a point. When theta_r is very small, V is the sum of two vectors of equal magnitude.

v_real = 1/2 V

so dtheta/dt = 1/2 m/s

If we set the number of division in theta to be very large, say 1000.

Theta change from A -> B is 90 degrees, pi/2.

dy/dt = cos theta dtheta/dt = cos theta/2

If theta has 1000 divisions, then t also has 1000 divisions.

Incremental distance traveled, Vy * dt / div = cos theta * 1/1000.

Note V is used because it is the sum of all vectors at a point.

Velocity traveled: dy/dt = cos theta/2.

Time is then calculated:

Time = ( cos theta * 1/1000 ) / (cos theta/2) = 2 / 1000 sec

The total time to travel from AB is 2 sec.

The total distance traveled is sum of magnitude of vector * dt * # of divisions.

Distance traveled in y direction = 1

For a full orbit, A -> B -> C -> D > A:

Total distance is 1*4 = 4 m.

Time taken is 2*4 = 8 sec.

Period of a full circle = 2 pi

2 pi = 8

pi = 4

Last edited by kug on Fri Oct 26, 2018 11:39 pm; edited 1 time in total

**kug**- Posts : 4

Join date : 2018-10-23

## Re: Proof that pi=4

Hi LongtimeAirman, I found about Mathis a week ago and a your forum a couple days ago. The pi = 4 problem seems simple to prove, so this is my attempt.

For example in a front wheel drive vehicle, the forward half of the vehicle is directed of the front wheels. The back half of the vehicle is directed by the rear wheel, which is fixed to the direction of the car.

If the car is driving 100 mph, the magnitude of the vector of the front and back cannot be both 100 mph. They have to be summed up together to be 100 mph in magnitude. Thus the front and back vector of the car is 50 mph.

When the car is turning, the front vector with a magnitude of 50 mph is shifted away from the center. This creates a new direction and a new magnitude that is less than 100 mph.

One interesting thing is if the front wheel is turned, there is also a vector normal to the direction of the vehicle. If the angle is high enough, there is a force limit that overcomes the static force and the vehicle begins to slide. Like in physics, where they slide a block down an angle.

Now that there are three vectors, the rear vector is always in the direction of the vehicle at the previous time. When the third vector is in play, the vehicle to us seems like it is out of control, because it is not responding as we are accustomed with two vectors.

For example in a front wheel drive vehicle, the forward half of the vehicle is directed of the front wheels. The back half of the vehicle is directed by the rear wheel, which is fixed to the direction of the car.

If the car is driving 100 mph, the magnitude of the vector of the front and back cannot be both 100 mph. They have to be summed up together to be 100 mph in magnitude. Thus the front and back vector of the car is 50 mph.

When the car is turning, the front vector with a magnitude of 50 mph is shifted away from the center. This creates a new direction and a new magnitude that is less than 100 mph.

One interesting thing is if the front wheel is turned, there is also a vector normal to the direction of the vehicle. If the angle is high enough, there is a force limit that overcomes the static force and the vehicle begins to slide. Like in physics, where they slide a block down an angle.

Now that there are three vectors, the rear vector is always in the direction of the vehicle at the previous time. When the third vector is in play, the vehicle to us seems like it is out of control, because it is not responding as we are accustomed with two vectors.

**kug**- Posts : 4

Join date : 2018-10-23

## Re: Proof that pi=4

.

Thanks for the additional background kug. In less than a week after finding out about Miles Mathis you can show us a simple proof of “pi” = 4. It's true, one need only adopt the Manhattan metric

I think it's commendable to jump into the Pi fight, although you may not have been aware of it. Pi = 4 is a point of great contention. My own posting experience began by coming out in defense of pi=4. In that light, your use of ‘circular rotation’ sounded like sarcasm. Circular rotation has a tangential velocity while circular motion is an acceleration comprised of orthogonal velocity components. Your inclusion of an additional diagram showing the tangential velocity vector as comprised of two non-orthogonal velocities makes me doubt you’ve got the metric completely figured out. Sliding vehicles on a racetrack involves curved motion, and not rotation.

If you haven't done so already, please read a few of Miles’ other papers on pi before jumping to any more conclusions. Here's Miles' pi list.

//////\\\\\///////////\\\\\/////

http://milesmathis.com/

85.

350.

351.

http://milesmathis.com/pi2.html

352.

http://milesmathis.com/pi4.html

353a.

353b.

353c.

.

Thanks for the additional background kug. In less than a week after finding out about Miles Mathis you can show us a simple proof of “pi” = 4. It's true, one need only adopt the Manhattan metric

*****in order to show that pi=4. The metric can be used to explain the cycloid******. The Manhattan metric is used when the tangent line to the circle is replaced with the orthogonal components making up that tangent line. Please see*****353a and******353b.I think it's commendable to jump into the Pi fight, although you may not have been aware of it. Pi = 4 is a point of great contention. My own posting experience began by coming out in defense of pi=4. In that light, your use of ‘circular rotation’ sounded like sarcasm. Circular rotation has a tangential velocity while circular motion is an acceleration comprised of orthogonal velocity components. Your inclusion of an additional diagram showing the tangential velocity vector as comprised of two non-orthogonal velocities makes me doubt you’ve got the metric completely figured out. Sliding vehicles on a racetrack involves curved motion, and not rotation.

If you haven't done so already, please read a few of Miles’ other papers on pi before jumping to any more conclusions. Here's Miles' pi list.

//////\\\\\///////////\\\\\/////

**THE GREATEST STANDING ERRORS IN PHYSICS AND MATHEMATICS**http://milesmathis.com/

85.

**A Clarification of the Equation a=v2/r**. I import new information from my π papers to extend my findings regarding orbits. I also show that the centripetal acceleration is not instantaneous. 6pp. http://milesmathis.com/avr2.html350.

**What is π?**Here I show that π is a centripetal acceleration and that the circumference of any circle has complex dimensions. 6pp. http://milesmathis.com/pi.html351.

**The Extinction of π**. Here I show that the true value of π, defined as the ratio of circumference to diameter, is 4. 15pp.http://milesmathis.com/pi2.html

352.

**Proof from NASA that π is 4**. Simple proof from Explorer that overturns π. 5pp.http://milesmathis.com/pi4.html

353a.

**The Manhattan Metric**. Where I show the specific links to Hilbert's math, further supporting my contention that π=4 in kinematic situations. 11pp. http://milesmathis.com/manh.pdf353b.

**The Cycloid and the Kinematic Circumference**. I use the cycloid again to explain very briefly why pi=4 when you have motion around a circle. 12pp. http://milesmathis.com/cycloid.pdf353c.

**More on the Running Track**. This is to clarify my recent addendum to my paper on pi=4. I show that both the distances and velocities are being miscalculated in the curves on normal running tracks. 6pp. http://milesmathis.com/track.pdf.

**LongtimeAirman**- Admin
- Posts : 1044

Join date : 2014-08-10

## Re: Proof that pi=4

Hi Kug, and welcome to the forum!

I too think it's awesome and commendable that you're jumping into things feet-first. So again, as LongtimeAirman stated, don't take it too personally if you're very critical. It's the best and even only way to keep ourselves honest and accurate, at the same time!

Good stuff here though. I like your second presentation.

I too think it's awesome and commendable that you're jumping into things feet-first. So again, as LongtimeAirman stated, don't take it too personally if you're very critical. It's the best and even only way to keep ourselves honest and accurate, at the same time!

Good stuff here though. I like your second presentation.

**Jared Magneson**- Posts : 442

Join date : 2016-10-11

## Re: Proof that pi=4

Hi Jared, thanks for the welcome.

I won't take the criticism too personally, I'm figuring out along the way to to illustrate the problem and then solve it mathematically with a realistic example.

I was trying to show how the velocity vector equation is calculating the direction and magnitude at a point. The correct velocity vector equation to use for summing vectors is actually 1/2 of the magnitude velocity vector equation. Inadvertently, I found out acceleration is equal to v^2/2R.

Fig 1. Velocity vector equation as is.

Fig 2. Shift left of velocity vector equation by 1/2 unit, where 1 unit is 7.5 degrees.

Velocity vector equation:

V = - v sin theta X + v cos theta Y

The angle from A to C is 15 degrees and the angle from A to B is 7.5 degrees. The orange vector is first velocity vector and the green is the second velocity vector. Orange and green vectors are the same size, just rotated due to acceleration vector (small vector colored in brown). The red vector is the orange or green vector added to the acceleration vector.

The fundamental problem is that we assume the starting position A is at rest, ie: v_0 = 0, before we even set t = 0. You need to have momentum to arrive a point A. You don't just appear with a velocity, and then start moving. However, that is exactly what the velocity vector equation does.

Halving the velocity vector equation solves this problem by splitting the vector into two and also doubles the number of divisions. This split separates the preceding and succeeding vectors, and allows the sum of the vectors at any point be equal to the magnitude of V.

If Fig 1. at point B, there are two vectors, AB and BC. The sum of both would be AC which is nearly double of both the orange and green.

If Fig 2. If the velocity vector equation did not put the first vector at point A, but instead put the middle of the first vector at point A, I can show how a = v^2/2R.

The orange and green vectors are the same as in Fig 1, only the initial configuration is different. Notice how much of an acceleration vector is needed to keep the path onto the circular path. In Fig 1, the acceleration vector need to correct a full orange vector distance. But in Fig 2, the amount of acceleration vector needed is only half the orange vector distance.

Derivation of acceleration value in Cartesian coordinates:

similar to http://milesmathis.com/avr3.pdf where it is derived in r and theta.

R is the radius

x^2 + y^2 = R^2

y = sqrt(R-x^2)

dy/dt = 1/2 (R-x^2)^(-.5) *(-2x) dx/dt

= -1/2y * 2x ds/dt

= x/y dx/dt

dx/dt = -y/x dy/dt

Velocity magnitude is constant, C

(dy/dt)^2 + (dx/dt)^2 = C^2

(dy/dt)^2 + (y/x)^2 (dy/dt)^2 = C^2

(1+ y^2/x^2) (dy/dt)^2 = C^2

(x^2 + y^2)/x^2 (dy/dt)^2 = C^2

R^2/x^2 (dy/dt)^2 = C^2

dy/dt = C/R x

dx/dt = - C/R y

Acceleration magnitude is constant ,C1

(d2y/dt2)^2 + (d2x/dt2)^2 = C1^2

d2y/dt2 = (C/R) dx/dt = -(C/R)^2 y

d2x/dt2 = -(C/R) dy/dt = -(C/R)^2 x

(C/R)^4 x^2 + (C/R)^4 y^2 = C1^2

(C/R)^4 (x^2 + y^2) = C1^2

(C/R)^4 R^2 = C1^2

C1 = C^2/R

With the acceleration vector cut by 2:

a = v^2/(2 R)

I won't take the criticism too personally, I'm figuring out along the way to to illustrate the problem and then solve it mathematically with a realistic example.

I was trying to show how the velocity vector equation is calculating the direction and magnitude at a point. The correct velocity vector equation to use for summing vectors is actually 1/2 of the magnitude velocity vector equation. Inadvertently, I found out acceleration is equal to v^2/2R.

Fig 1. Velocity vector equation as is.

Fig 2. Shift left of velocity vector equation by 1/2 unit, where 1 unit is 7.5 degrees.

Velocity vector equation:

V = - v sin theta X + v cos theta Y

The angle from A to C is 15 degrees and the angle from A to B is 7.5 degrees. The orange vector is first velocity vector and the green is the second velocity vector. Orange and green vectors are the same size, just rotated due to acceleration vector (small vector colored in brown). The red vector is the orange or green vector added to the acceleration vector.

The fundamental problem is that we assume the starting position A is at rest, ie: v_0 = 0, before we even set t = 0. You need to have momentum to arrive a point A. You don't just appear with a velocity, and then start moving. However, that is exactly what the velocity vector equation does.

Halving the velocity vector equation solves this problem by splitting the vector into two and also doubles the number of divisions. This split separates the preceding and succeeding vectors, and allows the sum of the vectors at any point be equal to the magnitude of V.

If Fig 1. at point B, there are two vectors, AB and BC. The sum of both would be AC which is nearly double of both the orange and green.

If Fig 2. If the velocity vector equation did not put the first vector at point A, but instead put the middle of the first vector at point A, I can show how a = v^2/2R.

The orange and green vectors are the same as in Fig 1, only the initial configuration is different. Notice how much of an acceleration vector is needed to keep the path onto the circular path. In Fig 1, the acceleration vector need to correct a full orange vector distance. But in Fig 2, the amount of acceleration vector needed is only half the orange vector distance.

Derivation of acceleration value in Cartesian coordinates:

similar to http://milesmathis.com/avr3.pdf where it is derived in r and theta.

R is the radius

x^2 + y^2 = R^2

y = sqrt(R-x^2)

dy/dt = 1/2 (R-x^2)^(-.5) *(-2x) dx/dt

= -1/2y * 2x ds/dt

= x/y dx/dt

dx/dt = -y/x dy/dt

Velocity magnitude is constant, C

(dy/dt)^2 + (dx/dt)^2 = C^2

(dy/dt)^2 + (y/x)^2 (dy/dt)^2 = C^2

(1+ y^2/x^2) (dy/dt)^2 = C^2

(x^2 + y^2)/x^2 (dy/dt)^2 = C^2

R^2/x^2 (dy/dt)^2 = C^2

dy/dt = C/R x

dx/dt = - C/R y

Acceleration magnitude is constant ,C1

(d2y/dt2)^2 + (d2x/dt2)^2 = C1^2

d2y/dt2 = (C/R) dx/dt = -(C/R)^2 y

d2x/dt2 = -(C/R) dy/dt = -(C/R)^2 x

(C/R)^4 x^2 + (C/R)^4 y^2 = C1^2

(C/R)^4 (x^2 + y^2) = C1^2

(C/R)^4 R^2 = C1^2

C1 = C^2/R

With the acceleration vector cut by 2:

a = v^2/(2 R)

**kug**- Posts : 4

Join date : 2018-10-23

## Re: Proof that pi=4

Unfortunately, Fig 2 is completely invalid. You can not just slide the velocity vector along like that. It is illegal and not physical because it does not adhere to the definition of a velocity.

A velocity is not a thing in its own right. It does not exist, as say a table exists, but is a property of something that does exist. You can say 'That car has a velocity', but you can't say 'That velocity has a car'. Further more, a velocity is a relative property. It is always relative to the current position of that which it belongs to. A velocity is also a future event. The velocity you have

In the case of circular motion, we are looking at a

That is why you can't just

A velocity is not a thing in its own right. It does not exist, as say a table exists, but is a property of something that does exist. You can say 'That car has a velocity', but you can't say 'That velocity has a car'. Further more, a velocity is a relative property. It is always relative to the current position of that which it belongs to. A velocity is also a future event. The velocity you have

*now*dictates where you*will*end up.In the case of circular motion, we are looking at a

*point*on the circumference of a circle. You can choose any point you like, it is completely arbitrary and will not change the results. Whatever point you do choose, though, is where your velocity vector is*based*. That is the point that represents*now*and the velocity points from that position to show where it*will*cause that point to go.That is why you can't just

*base*that vector somewhere else. It*belongs*to that point because it is*relative*to that point. A velocity is the motion, over some set time period, of some*thing*from one position to another.**Nevyn**- Admin
- Posts : 1360

Join date : 2014-09-11

## Re: Proof that pi=4

And this is why you're so critical here, Nevyn. I hate to put that stress on your shoulders, but there it is. Bear it however you must. I didn't take the time to dig through the math here, since I was already convinced of the premise from all the other directions. Good lookin' out, Nevyn.

Don't be discouraged, Kug. Just keep pluggin' away, and take these criticisms for what they are. We NEED to be accurate on these topics, and plug in any holes or perceived weaknesses by beating each other up a bit. Nevyn has helped me personally on many projects, especially stacked spins, and please take his analysis under heavy consideration.

Don't be discouraged, Kug. Just keep pluggin' away, and take these criticisms for what they are. We NEED to be accurate on these topics, and plug in any holes or perceived weaknesses by beating each other up a bit. Nevyn has helped me personally on many projects, especially stacked spins, and please take his analysis under heavy consideration.

**Jared Magneson**- Posts : 442

Join date : 2016-10-11

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