## THE REALITY, NOW AND UNDERSTANDING

### 2OBJECTS.TXT

#### 1. INTRODUCTION

This document describes the program 2OBJECTS
The document consists of three parts:
Description
Operation
Technical Description

The program performs 16 simulations and one "game"
The simulations 1 - 8 are part of CHAPTER2
The simulations 9 - 16 are part of CHAPTER3
Simulation 17 is part of chapter "PROVE"

#### 2 DESCRIPTION

The purpose of this program is to simulate the movement of two objects m0 and m1 with Newton' Law under 16 different conditions.

```                 m0        m1         v       c
test  1     >         0          0       0
test  2     =m1       =m0        0       0
test  3     >         0         up       0
test  4     =m1       =m0       up       0
test  5     >         0      right       0
test  6     =m1       =m0    right       0     ( not implemented )
test  7     >         0       left       0
test  8     =m1       =m0     left       0
test  9     >         0          0       >
test 10     =m1       =m0        0       >
test 11     >         0         up       >
test 12     =m1       =m0       up       >
test 13     >         0      right       >
test 14     =m1       =m0    right       >     ( not implemented )
test 15     >         0       left       >
test 16     =m1       =m0     left       >
test 17     mass and position calculation
```

The simulation can be subdivided into two groups:
Group 1 is from 1 to 8
Group 2 is from 9 to 16
In group 1 m0 influences m1 instantaneous and vice versa. The concept that "propagation of signals takes time" is not included. The speed of light c is 0. See chapter 2 for explanation. In group 2 the concept that "propagation of signals takes time" is included. The speed of light c is not 0. See chapter 3 for explanation.

For all the odd tests (1, 3, 5 etc.) m0 is large and m1 is small
For all the even tests (2, 4, 6 etc.) m0 is equal to m1
v = 0 means both m0 and m1 have no extra speed component
v = up means both m0 and m1 move straight upwards
v = right means both m0 and m1 move towards the right
v = left means both m0 and m1 move towards the left
test 6 is identical as test 8 (except the sign of the movement)
test 14 is identical as test 16 (except the sign of the movement)
For that reason test 6 and test 14 are not included.
For the same reason downwards movement of v is not included.

In each of the tests the position of both m0 and m1 is shown at different instances until they collide.
m0 is on the left and m1 is on the right.

As part of test 9 (in order to compare) also test 1 is shown.
The same is true for all the tests from 9 to 16.

#### 2.1 TEST 1

The first test is the most simple case.
The mass of m0 is large i.e. much greater then m1
The mass of m1 is small i.e. much smaller then m0
The initial speed of both m0 and m1 is zero.

The simulation shows that m1 moves towards m0 and that m0 moves towards m1 but the distance is very small.
The speed of m1 is much larger then the speed of m0. Newton's law describes this.

Using Newton's law the force F can be written as follows:

```            G * m0 * m1
F = -----------  = m0 * a0  = m1 * a1                   (1)
rē
```

For the acceleration a0 of m0 we get:

```             G * m1
a0 = ------                                             (2)
rē
```

For the acceleration a1 of m1 we get:

```             G * m0
a1 = ------                                             (3)
rē
```

Equation 2 describes that because m1 small (=0) is, the acceleration a0 of m0 small (=0) is. As a result is v0 of m0 small (=0).

Equation 3 describes that because m0 large is compared to m0, that the acceleration a1 of m1 is greater then a0. As a consequence v1 of m1 is greater then v0.

What the simulation also shows is that the speed of m1 increases the closer m1 comes to m0 (r becomes smaller). This is shown because the distance between the dots becomes larger.

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
A Select test 1

The time it takes before m0 and m1 collide is 1110 seconds.

#### 2.2 TEST 2

Test 2 is the opposite situation of test 1. In this test both m0 and m1 are equal. The initial speed of both m0 and m1 are zero.

The simulation shows that the two masses m0 and m1 will move towards each other, but now the speed of m0 is the same as the speed of m1. Newton's law describes this.

Using Newton's law and the equations (2) and (3) we can see that because m0 = m1 that:

a0 = a1

This results that also v0 of m0 = v1 of m1

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 2

The time it takes before m0 and m1 collide is 786 seconds.

#### 2.3 TEST 3

Test 3 is almost identical as test 1 i.e. The mass of m0 is large i.e. much greater then m1 The mass of m1 is small i.e. much smaller then m0 The initial speed of both m0 and m1 are zero.

The only difference is that both m0 and m1 have an identical upwards speed relative to the connecting line between m0 and m1.

The simulation shows that the two masses m0 and m1 will move towards each other as in test 1 but the trajectory followed is different.

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 3

The time before collision in both test 1 and test 3 is 1110 seconds. i.e. the speed upwards has no influence.

Newton's law describes this, because the force is dependent on distance and masses which are identical in test 1 and 3. The force is independent on (the direction of) speed.

The same is true for the test 5 and 7.

#### 2.4 TEST 4

Test 4 is almost identical as test 2 i.e. In this test both m0 and m1 are equal. The initial speed of both m0 and m1 are zero.

The only difference is that both m0 and m1 have an identical upwards speed relative to the connecting line between m0 and m1.

The simulation shows that the two masses m0 and m1 will move towards each other as in test 1 but the trajectory followed is different.

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 4

The time before collision in both test 2 and test 4 is 786 seconds. i.e. the speed upwards has no influence.

Newton's law describes this, because the force is dependent on distance and masses which are identical in test 2 and 4. The force is independent on (the direction of) speed.

The same is true for the test 6 and 8.

#### 2.5 TEST 5

Test 5 is almost identical as test 3

The only difference is that both m0 and m1 have an identical speed component to the right, relative to the connecting line between m0 and m1.

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 5

The time before collision in each test 1, test 3 and test 5 is 1110 seconds. i.e. the speed to the right has no influence.

#### 2.6 TEST 6

Test 6 is identical as test 8 and is not implemented.

#### 2.7 TEST 7

Test 7 is identical as test 5, except that the speed component is to the left.

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 7

The time before collision in each test 1, test 3, test 5 and test 7 is 1110 seconds. i.e. the speed to the left has no influence.

#### 2.8 TEST 8

Test 8 is almost identical as test 4.

The only difference is that both m0 and m1 have an identical speed component to the left, relative to the connecting line between m0 and m1.

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 8

The time before collision in each test 2, test 4 and test 8 is 786 seconds. i.e. the speed to the left has no influence.

After performing test 8:

Return back to CHAPTER2.TXT

#### 2.9 TEST 9

Test 9 is identical as test 1 except that the concept "that propagation of gravitation takes time" is included.

The display shows both the results of test 1 on the top part and of test 9 on the bottom part. The bottom part shows besides a trail of dots, representing the position at each instant, also a white dot. The white dot represents the virtual position. The concept of virtual position is explained in chapter 3. The virtual position represents the position at an earlier moment. How much earlier depends on the time it takes for the propagation to travel between the two objects. The time difference between "now" and this earlier moment is equal to the distance between the two objects divided by the speed of light.

The results of test 1 and test 9 are identical. In this special case propagation time has no influence. The reason why there is no difference is because m0 is heavy (compared to m1) and as a consequence m0 does not move in test 1 and in test 9.

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 9

Because m0 does not move the force that m1 feels in test 9 is based on the true distance, which is the same as in test 1.

The time before collision in test 1 and test 9 is both 1110 seconds

#### 2.10 TEST 10

Test 10 is identical as test 2 except that the concept "that propagation of gravitation takes time" is included.

In this case both objects move compared to only one in test 9

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 10

The time between start and collision in test 2 is 786 seconds and in test 10 is 815 seconds

The reason is because the distance based on the virtual position is larger then the true distance (based on the true position). As a result the acceleration (and speed) of each object will be smaller. This in turn will give a longer time before collision.

Compare test 10 with test 12 and test 16.

#### 2.11 TEST 11

Test 11 is identical as test 3 except that the concept "that propagation of gravitation takes time" is included.

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 11

The time between start of test and collision in test 3 is 1110.2 seconds. In test 11 this is 1110.6 seconds

The results are almost identical. The reason is because the distance between and m1 and the virtual position of m0 and the true distance is almost identical.

Compare test 11 with test 13 and test 15

#### 2.12 TEST 12

Test 12 is identical as test 4 except that the concept "that propagation of gravitation takes time" is included.

The time between start of test and collision in test 4 is 786 seconds. In test 12 this is 815 seconds

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 12

Compare test 12 with test 10 and test 16.

The results are identical. This is because the virtual positions are identical in each of those tests and the true positions are identical, resulting the same time periods. The direction of the base speed is different but this has no influence on the outcome of the experiment.

#### 2.13 TEST 13

Test 13 is identical as test 5 except that the concept "that propagation of gravitation takes time" is included.

The time between start and collision in test 5 is 1110 seconds and in test 13 is 1127 seconds

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 13

Test 13 is special because it has a speed component to the right. As a result is the virtual position of m0 to the left. This in turn means that the virtual distance is larger as the true distance, resulting in a lower acceleration and speed and in a longer time before to collide.

Compare test 13 with test 11 and test 15.

#### 2.14 TEST 14

Test 14 is identical as test 16 and is not included.

#### 2.15 TEST 15

Test 15 is identical as test 7 except that the concept "that propagation of gravitation takes time" is included.

The time between start and collision in test 7 is 1110 seconds and in test 10 is 1094 seconds

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 15

Test 15 is special because it has a speed component to the left. As a result is the virtual position of m0 to the right. This in turn means that the virtual distance is smaller as the true distance, resulting in a larger acceleration and speed and in a shorter time before to collide.

Compare test 13 with test 11 and test 15.

#### 2.16 TEST 16

Test 16 is identical as test 8 except that the concept "that propagation of gravitation takes time" is included.

The time between start and collision in test 8 is 786 seconds and in test 16 is 815 seconds

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 16

Compare test 16 with test 10 and test 12.

The results are identical. This is because the virtual positions are identical in each of those tests and the true positions are identical, resulting the same time periods. The direction of the base speed is different but this has no influence on the outcome of the experiment.

After performing test 16:

Return back to CHAPTER3.TXT

#### 2.17 MASS CALCULATION TEST

The purpose of this test is to demonstrate how we should describe the reality: doing predictions, solely based on observations.

A standard 2 body simulation starts with the following ingredients: 2 masses, 4 initial conditions (i.e. the position and velocity of both masses at t0) and Newton's law.
Newton's law services as the model to do our predictions.

In this simulation a slightly different approach is taken.
• First, solely based on observations, the positions of both m0 and m1 are established at three (equally spaced) moments: t0, t1 and t2.
The 3 positions for m0 are called: p0t0, p0t1 and p0t2
The 3 positions for m1 are called: p1t0, p1t1 and p1t2
• Next v0 of m0 at t0 and v1 of m0 at t1 are calculated using:
```          |v0| = |p0t0 - p0t1|       |v1| = |p0t1 - p0t2|
```
• Next acceleration a0 of m0 is calculated using:
```          |a0| = |v0| - |v01| = |p0t0 -2*p0t1 + p0t2|
```
• Next m1 is calculated using:
```
G*m0*m1
-------  = |a0| * m0
rē

m1 = |a0|* rē / G
```
• The same is done for acceleration a1 of m1 and m0:
```          |a1| = |p1t0 -2*p1t1 + p1t2|

m0 = |a1|* rē / G
```
• Next the speed v2 at t2 for both m0 and m1 are calculated.
• Finally the program calculates the positions of both m0 and m1, starting with t3 using Newton's Law as a model.

#### 2.17.1 OPERATION

In this two body simulation you can only change the position of m1. The position of m0 is modified equivalent.
When you start the simulation the top right side shows the position of m1 at t0, t1 and t2. The position of t0 is in red (being selected). The position of t1 and t2 are in red.

The simulation goes in two steps:
1. First select the position you want to modify
In order to select the position (one out of three) use the Tab key.
2. Second change the position.
In order to change the position use the arrow keys.

• After the Tab key is pressed the new selected position will turn in red and the old one will change in white.
• After any arrow key is pressed the program first calculates the mass of the two objects m0 and m1 and the predicted positions, starting from t3, are shown.

Now perform the program: 2OBJECTS.EXE
From the Test Selection Display:
Select test 17

It should be mentioned that the calculation in the program is not correct.
• The first reason is because the observations are arbitrary and not real ones.
• The second reason is that the calculation is based at equally spaced moments. This implies that every time at least two positions have to be changed (and not one) before the calculation is performed.
• The third reason is that no other object is included i.e. no other forces influence this simulation.

The main message of the test that the reality is based on observations only. Using those observations the future can be predicted (calculated). In this test this is only done based on Newton's Law i.e. objects have masses concentrated in one point and forces act instanteneous. It is the purpose of this book to show that this model is to simple.

Return back to PROVE.TXT

#### 3 OPERATION

The program uses the following standard feature:

When you select Esc you will terminate the program (Escape)

#### 3.1 PARAMETER SELECTION DISPLAY

From the Parameter Selection Display the following parameters can be changed:

```        0 = Select test display

1 = Set standard parameters.

2 = Screen mode. Valid values are 7,8,9 and 12. Standard value = 9
3 = Wait time in second. Physical wait time between each simulation
cycle. Standard value = 0.1
4 = Speed of light. Standard value is 10

5 = Delta time in seconds between each calculation cycle.
Standard value is 1
6 = Basic Speed. Standard value = .15

7 = Escape velocity. Standard value = 0

8 = Distance between m0 and m1. Standard value = 1000

9 = Directory name. Standard value = C:\NOW\FIG

10 = # of calculation cycles saved. Standard value = 0
0 means no calculation values saved.
```

#### 4.1 DELAY LINE

In order to calculate the position of mass m0 a time t1 ago we use a delay line. A delay line is an array of numbers, containing the position of m0 at different moments (in the past). Always we use two arrays:

one with the x co-ordinates of m0.
and one with the y co-ordinates of m0. The values in the array are stored equally spaced, for example every 5 sec. This allows us to calculate the position a time t1 ago.

The delay line with the values of the x co-ordinates is written as x(i), where the i identifies the ith element of the array x. For the y co-ordinates we use y(i), where the i identifies the ith element of the array y.

For example:
Suppose the delay line x(i) contains 6 values saved every 5 seconds then: x(0) contains the x co-ordinate now 200
x(1) contains the x co-ordinate 5 seconds ago 210
x(2) contains the x co-ordinate 10 seconds ago 219
x(3) contains the x co-ordinate 15 seconds ago 227
x(4) contains the x co-ordinate 20 seconds ago 234
x(5) contains the x co-ordinate 25 seconds ago 240

Suppose we want to know the value of the x co-ordinate 17 seconds ago. From the above list we can see this is somewhere between x(3) and x(4) i.e. some where between the 3th and 4th element of the array x(i).

Suppose we want to know the value of the x co-ordinate t1 seconds ago. This goes in three steps:

1. First you must calculate the interval for t1.
Suppose this is between i and i+1 (Above i was 3)
In order to calculate i use the following formula:
i = INTEGER (t1/deltat)
deltat is the time between intervals.

2. Second you must calculate the time dt left over in that interval
In the above example dt = 2 seconds
In order to calculate dt use the following formula:
dt = t1 - t1 * deltat

3. Third you can calculate the position using the following formula:
xc = x(i) + dt * ((x(i+1) - x(i)) / deltat
deltat is the time between intervals.
In the above example deltat = 5 seconds
In the above example:
xc = x(3) + 2 * ((x(4) - (x(3))/ 5
xc = 227 + 2 * ( 234 - 227 ) /5 = 227 + 2.8 = 229.8