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1
Flat Earth General / Re: Cavendish experiment
« on: June 27, 2019, 06:27:29 PM »
Quote
Electricity doesn't kill people.  Current kills people.
it is the voltage which means push the greater the push/voltage the greater will be its killing effect

2
Flat Earth General / Re: Cavendish experiment
« on: June 26, 2019, 12:49:00 PM »
Quote
If 'gravity' is not real, why does the Cavendish experiment work then, showing 'gravity' between masses?

we all know things on fall earth but don't why yet. Human try to represent it in mathematical models but haven't succeeded yet.

Quote
why does the Cavendish experiment work then, showing 'gravity' between masses

one word hocus-pocus

3
Flat Earth General / Re: Flat earth and physics
« on: June 25, 2019, 05:07:54 PM »
Oh no, guys don't be confused with sideral and solar day

Let's start from square one

Number of solar middays (noon) = Number of solar midnights;- Always - Right

Earth traced it's orbit (either circular or elliptical) in between outer circle and inner circle around the sun as shown in Fig 2  (not to the scale) - Right 
Any point on the outer circle represents solar midnight while on inner circle solar noon - Right
The length of an outer circle is greater than the length of the inner circle - Right

This means the # of Midnights > # of Middays or Occurrence of midnights > Occurrence of middays when the earth revolves around the sun in its orbit – Right

Although its clearly visible either from the outer and inner circles or we can straight them (outer, middle and inner circles) as shown i Fig 3

Detail analysis of why the # of Midnights > # of Middays or Occurrence of midnights > Occurrence of middays

Fig #1: Noon and Midnight occur at fixed point N and AN respectively when the earth rotates on its axis but has zero orbital motion - Right

Now as the earth has orbital motion too besides its axial rotation therefore 

Fig #4 represents the occurrence of one (1) or single Noon and Midnight when the earth rotates on its axis and moves in its orbit around the sun

We know, Relative to the sun

The orbital velocity of the earth at midnight  > The orbital velocity of the earth at Noon

As soon as Midnight occurs at B, Noon occurs at D - Right (Fig 4)

Midnight has to travel more in space as compared to Noon because of its greater orbital velocity - Right

Thus Midnight travels from B to A while Noon travels from D to C

Why - Because

The earth spins on its axis at the speed of 30 m/sec. The traveling speed of the earth is 30,000 m/sec in its orbit. (30,000 m/sec >>>> 30 m/sec). As

Circumference of outer circle > Circumference of an inner circle

Therefore there are two possibilities

EITHER # of Midnights > # of Middays as shown in Fig of the orbit of the earth around the sun (which is impossible)

OR there must be the happening of traveling of Noon and Midnight without the axial rotation (relative to the sun) in the orbit of the earth as explained above

Thus

Relative to the sun,
The earth moves in its orbit but has zero axial rotation (no spinning) due to its high orbital speed as compared to its spinning
Noon moves from D to C (there is constant noon from D to C) - 
Midnight moves from B to A (there is constant Midnight from B to A)
I haven't calculated the time-span (DURATION) of the above drag w/o spin

Since, Relative to sun, the earth doesn't spin from D to C and there is a constant Noon from D to C, therefore, the growth of shadow stops on earth when the earth moves from D to C in its orbit or during the occurrence of single Noon

4
Flat Earth General / Re: Flat earth and physics
« on: June 25, 2019, 01:53:21 PM »
Quote
Nothing is "is dragged from B to A" nor from "D to C". The earth has rotated smoothly half a turn relative to the sun in this time.

During the time from "B to A" the earth has rotated half a turn relative to the sun and half of the earth has experienced midnight.
And during the time from "D to C" the earth has rotated the same half a turn and the other half of the earth has experienced midday.

PS I put in the "relative to the sun" each time because in 12 hours the earth rotates, relative to the stars, 0.49° more than half a turn.

i dont understand  "half a turn" relative to the sun - But
Quote
Nothing is "is dragged from B to A" nor from "D to C"
- By dragging i mean
Relative to the sun, the earth moves in its orbit but has zero axial rotation of (i.e no spinning) dut to its high speed as compared to its spinning when the Midnight moves from B to A and Noon moves from D to A - do you agree

Quote
The earth has rotated smoothly half a turn relative to the sun in this time.
No, because otherwise, # of midnights will be > # of noons as explained earlier.

5
Flat Earth General / Re: Flat earth and physics
« on: June 24, 2019, 11:50:25 PM »
Quote
gathered that but why is it a problem?
For a 149,000,000 km orbital radius for the centre of the earth the outer orbital velocity is 29.66958 km/s and the inner is 29.667044 km/s.
The difference is only 2.53 m/s. The earth's radius is tiny compared to the orbital radius.

The surface velocity of the earth, relative to it centre is 463.31 m/s and this adds to the orbital velocity at midnight but subtracts at midday.
And this surface velocity dominates the difference in velocities but the change occurs gradually from midday to midnight so there is no sudden change.

A more extreme case, closer to home is a tyre on a car travelling at 25 m/s (90 km/hr or 56 mph).
Relative to the road the bottom of the tyre is stationary but the top is travelling forward at twice the speed of the car or 50 m/s.
Again this change does not happen instantaneously but in the time it takes the tyre to complete half a revolution.
If you look more closely the magnitude of the acceleration is constant but is always directed to the centre of rotation - it is just the centripetal acceleration.

Again, I hope my "sums" are right.

I didn't do the calculation either but it means the growth of shadow stops on earth for a very short period of time e.g. at NOON when it is dragged from D to C and would that drag affect SEASONS or CALENDAR due to the change in the position of the earth in its orbit?

6
Flat Earth General / Re: Flat earth and physics
« on: June 24, 2019, 10:33:06 PM »
Quote
Just why do you claim that the "Circumference of outer circle must be equal to the Circumference of inner circle". I see no reason for that at all.
no idea at this time why i wrote that but I struck it off

here is the detail of occurring of one NOON and MIDNIGHT in the orbit



MIDNIGHT is dragged from B to A
Noon is dragged from D to C
length of BA > length of the DC because the orbital velocity of earth is greater at outer circle than the inner circle

Does it help now

7
Flat Earth General / Re: Flat earth and physics
« on: June 24, 2019, 08:36:14 PM »
You might have gone through but i would suggest reading again Reply #35 and 41 in the following link for the two possibilities - if interested

https://www.theflatearthsociety.org/forum/index.php?topic=76422.msg2075952#msg2075952


8
Flat Earth General / Re: Flat earth and physics
« on: June 24, 2019, 06:48:58 PM »
And here is a possible glitch in the solar system - need more voters if true

https://www.theflatearthsociety.org/forum/index.php?topic=76422.0

also, read Reply #36 to the aforesaid question

Both axial rotation and orbital revolving of earth independent of each other - Right

In quantum mechanics, the analogue of Newton's law (F=ma) is Schrödinger's equation - Wikipedia (atomic theory)

Mass of all electrons are the same, Mass of all protons are the same and Mass of all neutrons are also the same

i dont have in-depth knowledge of the big bang theory but how did all above happen when there were more chances of different masses of electrons, protons, and neutrons in randomness during early natural selection. 

similarly, heavy nuclei which require fusion (high temperature and isolation) were formed after the high temperature - please ignore it if im wrong

9
Flat Earth General / Re: Flat earth and physics
« on: June 23, 2019, 07:34:07 PM »
Sorry, i posted in hurry so my apologies for the mistakes that I did in my previous post especially the diameter which should be the radius of gravitating mass. Here is a clear version

Gravitational acceleration of any spherical mass = GM/R^2, where R= Radius of the sphere, G= Gravitational constant, M = Mass of a sphere

This means

The "g" of Earth = GM/R^2 where R= Radius of the earth, G= Gravitational constant, M= Mass of an earth

The "g" of an apple = GM/R^2 where R= Radius of the apple, G= Gravitational constant, M =Mass of an apple

The "g" of a sphere of 1 kg = GM/R^2 where R= Radius of the said sphere, G= Gravitational constant, M =Mass of the said sphere = 1kg

Gravitating mass can be any mass. Falling mass can also be any mass

Earth - Apple case
when the earth is gravitating mass then the apple is falling mass and vice versa

As you can see in the figure, the "g" of the earth on its surface or near to its surface = GM/R^2  BUT


The "g" of the earth above its surface at any height h, g = GM/(R+h)^2

This means we can calculate "g" of any falling mass on earth at any height h as it doesn't depend upon the mass of falling object, therefore, we don't even need to plug the mass falling in the equation as all masses fall at the same rate on earth- we all know - True

Quote
But please only use g= GM/d^2 for cases when a small object is near a massive object (such as a large moon, planet or sun).
It is rather meaningless otherwise.
Watch this at 6:51 https://www.youtube.com/watch?time_continue=411&v=hV2MwaMApZw

CRUX
Both masses are falling and gravitating at the same time as per the Universal Law of Gravitation while Gravitating mass is stationary in the split analysis of F=GMm/d^2; F=Mg1 and F=mg - WHY BECAUSE ALL OBJECTS DON'T FALL AT THE SAME RATE AS EXPLAINED IN GALILEO SECTION
hope this may help

shell thereom


Inside a Shell
https://en.wikipedia.org/wiki/Talk:Shell_theorem#Inside_a_Shell

Application of Shell Theorem
https://en.wikipedia.org/wiki/Talk:Shell_theorem#Application_of_Shell_Theorem

here is in detail




10
Flat Earth General / Re: Flat earth and physics
« on: June 22, 2019, 06:37:00 PM »
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No they are not in any "higher type of motion when fall towards each other as explained in the link".
It's just the you can no longer use the simple case of a constant acceleration, g = G Me/Re2.

Now, because both masses move and the distance changes significantly you must use the full Fg = G m1 m2/d2 to solve the problem.

It is a common problem, treated in planetary science and covered in many places, including Gravitational two-body problem.
These treatments often present a simplification called the Reduced mass.

But all these assume a fairly good understanding of mathematics often including vector algebra and calculus.

And none raises question about Newton's Law of Universal Gravitation.

OK here is another try,

We know that F = GMm/ d ^2, where g= GM/d^2. This means gravitational acceleration of gravitating mass (g= GM/d^2) doesn’t even depend upon falling mass - Right

Thus if aforementioned is true then why would gravitational force between two point masses be directly proportional to the product of their masses when the gravitational acceleration of either mass can be produced in the absence of falling mass.

Although, the presence of G means the presence of two masses but the presence of G in the equation of g= GM/d^2 still means the presence of only gravitating mass M as rest of masses would be canceled out (F is directly proportional the product of masses) after plugging the value of G. So same as above or we can just ignore G as it is a constant. Hence

Can a lone gravitating mass generate a gravitational acceleration when it is the gravitational force (requires two point masses) due to which two masses come closer to each other with acceleration called gravitational acceleration?

As we know g= GM/d^2 is the gravitational acceleration of any gravitating mass, therefore “g” is indirectly proportional to the square of height d [or (d+h) as don't be confused with diameter d of M]. This means “g” decreases when d or d+h increases and vice versa. Remember the said mass who’s g= GM/d^2 is at rest when we play with height d or d+h in order to decrease or increase the gravitational acceleration. This is the power of gravitating mass due to which all objects fall at the same rate on it when it is at stationary i.e. the center of the said mass is at one fixed point. But this is not the case as the on-center distance between two point masses decreases from both sides instead of only from one side as explained earlier therefore aforementioned two point masses of the subject go into higher types of motion forthwith as soon as they feel the gravitational force.

OR simply the "g" of earth = GM/d^2 where d is the diameter of the earth of mass M. The "g" of earth above the earth at height h will be g = GM/(d+h)^2 . This means "g" of earth decreases when h increases as d is constant


Anyway, again why not gravitational force is directly proportional to the addition of point masses instead of their multiplication if not then any genuine reasons that could satisfy.

Again, F = GMm/ d ^2, where g= GM/d^2 (adddendum d = radius of gravitating mass; pls see next post for clarity)

Therefore F = mg = W OR simply weight of an object of mass m on gravitating mass M is W = mg

Calculating the weight of the building is the basic requirement in order to design its many different parts such as foundation beams girded columns slabs etc. As F = mg is wrong theoretically, therefore, W = mg is also wrong. Although it might help indirectly in designing engineering structures but it is the greater factors of safety due to which we are on the safe side.

The basic unit of mass is gram or kg which is itself not only derived from W = mg (wrong in theory) but also their incongruent square of the on-certre distances elicits two different gravitational accelerations as explained in the following link.

Is the definition of UNIT [gram, kilogram] of MASS absolute or iffy?
https://en.wikipedia.org/wiki/Talk:Gram#Is_definition_of_UNIT_[gram,_kilogram]_of_MASS_absolute_or_iffy?

hope all above may help

11
Flat Earth General / Re: Flat earth and physics
« on: June 21, 2019, 07:46:31 PM »
acceleration happens when one mass is stationary and the other is falling i.e on-center distance d decreases from one side but since both masses are falling towards each other therefore the on-center distance decreases from both sides at every moment. this means if one mass moves the other also moves at the same time equally so which on-center distance d would you chose in order to plug in F = Gm^2/d^2 , g=Gm/d^2 , v= gt etc . for calculating the acceleration or falling time of either mass (or if the masses are different say m1= 1kg m2=1.2kg)

both masses are in higher type of motion when fall towards each other as explained in the link, not just in acceleration g=Gm/d^2 so the striking time will be much less as compared to calculated by the universal law of gravitation. this automatically negate G - Right


12
Flat Earth General / Re: Flat earth and physics
« on: June 21, 2019, 06:12:22 PM »
Quote
Galileo is done. Now

2- Newton law of gravitation.
Please read carefully - Does "g =GM/d^2" best in situ in F= GMm/d^2? - in the following link
https://en.wikipedia.org/wiki/Talk:Gravity_of_Earth/Archive_1#Does_%22g_=GM/d^2%22_best_in_situ_in_F=_GMm/d^2?

anyone either RE and FE, please explain which part of the explanation in the aforementioned link doesn't disprove the newton's universal law of gravitation if disagree -or otherwise, this is my last post 

try calculating the striking time of two identical spherical masses of 1 kg each separated by an on-center distance of 1 meter in space using universal law of gravitation - thsi may help

13
Flat Earth General / Re: Flat earth and physics
« on: June 19, 2019, 02:56:18 PM »
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If there is something in there that you think disproves Newton's Law of Universal Gravitation () please explain it to us!

The following might help what i expound in the link

try calculating the striking time of two identical spherical masses of 1 kg each separated by an on-center distance of 1 meter in space using universal law of gravitation

14
Flat Earth General / Re: Flat earth and physics
« on: June 17, 2019, 08:00:51 PM »
Quote
Please read carefully - Does "g =GM/d^2" best in situ in F= GMm/d^2? - in the following link
https://en.wikipedia.org/wiki/Talk:Gravity_of_Earth/Archive_1#Does_%22g_=GM/d^2%22_best_in_situ_in_F=_GMm/d^2?
no idea how did you go through the aforementioned link but it's quite clear even for high school students that it is wrong. Anyway, again its freedom of acceptance. there is no compulsion

3- time dilation and relativity

plaese read carefully
Is the Einstein’s light clock shown in article standard?
https://en.wikipedia.org/wiki/Talk:Time_dilation/Archive_2018#Is_the_Einstein%E2%80%99s_light_clock_shown_in_article_standard?

Time dilation Triangle
https://en.wikipedia.org/wiki/Talk:Time_dilation/Archive_2017#Time_dilation_Triangle
The base of the time dilation triangle is taken in dilated time which is wrong while analyzing it by Pythagoras theorem. The lower mirror moves with v and covers a distance d in time t, not in dilated time t' - right

Relativity postulates/ Frame of references/ Escape velocity
https://en.wikipedia.org/wiki/Talk:Time_dilation/Archive_2011#Relativity_postulates/_Frame_of_references/_Escape_velocity

Second diagram from top
https://en.wikipedia.org/wiki/Talk:Relativity_of_simultaneity#Second_diagram_from_top

Relativity of Simultaneity (Experiment)
https://en.wikipedia.org/wiki/Talk:Relativity_of_simultaneity/Archive_2#Relativity_of_Simultaneity_(Experiment)_2

15
Flat Earth General / Re: Flat earth and physics
« on: June 17, 2019, 09:05:15 AM »
Quote
The mass of the earth vs a 5kg ball and a 100kg will show you there is no significant differnece between the reactions.

The significant difference depends upon the chosen falling masses. its all about the understanding of the concept of notion. Galileo might be right if i add " unless it is the natural tendency of all smaller objects to drive toward the massive object at the same rate in the region" - it will be easy to understand this if you read the newtons law of gravitation in the aforementioned link first.

16
Flat Earth General / Re: Flat earth and physics
« on: June 17, 2019, 05:55:43 AM »
Quote
I don't know about "goat tree" unit, but I could understand the "goat per tree".

i said s = vt or v=d/t is ok which means all other statements where per is involved presumably. I mentioned it in my original statement, not in this forum. Your idea about the units will also be cleared but later when its turn comes. Thank you and nice a day as well

17
Flat Earth General / Re: Flat earth and physics
« on: June 17, 2019, 05:42:32 AM »
Quote
But you are being ridiculous! Galileo had no accurate timepieces anyway!
- Here, I'm not making fun of someone. it seems you didn't read my post properly because I told I would provide you with proofs one by one starting from Galileo. Second, you wouldn't talk about the timepieces if you had reached the crux of the theory. Anyway, since we believe in personal freedom, therefore, its your personal choice whether you agree or not. Although it's very simple to understand what i said in reference to the Galileo's statement  but still i can help if you have any question left

Galileo is done. Now

2- Newton law of gravitation.

Please read carefully - Does "g =GM/d^2" best in situ in F= GMm/d^2? - in the following link
https://en.wikipedia.org/wiki/Talk:Gravity_of_Earth/Archive_1#Does_%22g_=GM/d^2%22_best_in_situ_in_F=_GMm/d^2?

Addendum:
Further reading if interested

"No physical connection in Newton's idea" in the link
https://en.wikipedia.org/wiki/Talk:Cavendish_experiment#No_physical_connection_in_Newton's_idea

Radial Falling and Density of Lead in the following link
https://en.wikipedia.org/wiki/Talk:Cavendish_experiment#Radial_Falling_and_Density_of_Lead

Does the gravitational constant “G” which ensconced in both classical and modern physics subliminal? in the following link
https://en.wikipedia.org/wiki/Talk:Gravitational_constant/Archive_1#Does_the_gravitational_constant_%E2%80%9CG%E2%80%9D_which_ensconced_in_both_classical_and_modern_physics_subliminal?

18
Flat Earth General / Re: Flat earth and physics
« on: June 16, 2019, 05:44:15 PM »
Quote
No! That "both Einsteinian and Newtonian mathematical models of the gravity" are accurate is based on very strong evidence.
Newton's Laws of Motion and Universal Gravitation are quite accurate within regions where velocities and gravitational fields are "not too hig
h".

ok ill provide you with proof one by one. Let's start from the Galileo in the following link

*Is Galileo's statement correct, theoretically?*
https://en.wikipedia.org/wiki/Talk:Equations_for_a_falling_body#Is_Galileo.27s_statement_correct.2C_theoretically.3F

repeat the Galileo experiment with two different masses say 5 kg and 100kg but don't drop them simultaneously.
drop 5kg from height h and note the striking timing of earth and 5 kg, then
drop 10kg from the same height h and note the striking timing of earth and 100kg
you will find two different striking times - gravity (g) of 100 kg mass is greater than the gravity (g) of 5kg mass - agree.
so the earth will accelerate faster towards 100 kg mass as compared to 5kg mass

Quote
In physics it is incorrect to add or subtract quantities with different units but is is quite legitimate to multiply or divide them when the units are correct for the whole equation.

So momentum is defined as mass x velocity So in the SI System has units of kg.m.t-1.
And force is mass x acceleration [/b] So in the SI System has units of kg.m.t-2.

Where did you get these strange ideas from?

multiply 3 goats with 5 trees
3 x5 = 15 goat.tree so can pls tell me what does meant by 15 goat.tree what kind of unit is goat.tree

Quote
In physics it is incorrect to add or subtract quantities with different units but is is quite legitimate to multiply or divide them when the units are correct for the whole equation.

any reference pls

19
Flat Earth General / Re: Flat earth and physics
« on: June 15, 2019, 06:48:53 PM »
Quote
They are all wrong anyway except s=vt. i mean even for RE

Why is it?

because both Einsteinian and Newtonian mathematical models of the gravity are wrong - already proved

Guess law is derived from Newton law of gravity

Einstein's time dilation, relativity, etc are also wrong - already proved

rest are derived from Einsteinian and Newtonian mathematical models

similarly, the multiplication of goats and trees has no useful meaning in both physics and math because they can't be multiplied and the same is applied to momentum which = mass x velocity and all similar equation in physics

20
Flat Earth General / Re: Flat World From Space
« on: June 14, 2019, 07:23:26 PM »
how deep is it

21
Flat Earth General / Re: Flat earth and physics
« on: June 14, 2019, 07:00:11 PM »
They are all wrong anyway except s=vt. i mean even for RE

22
Eratosthenes measured the angle of the sun above Alexandria by measuring the shadow cast by a stick. He got 7.2° .

Assume the earth is bright while the sun doesn't have its own sunlight. As sun is 109 times bigger than the earth, therefore, keeping the distance constant (800 km between the stick and the well or b/t two cities) on the surface of sun, the measuring angle of the earth above Alexandria on the sun will be the shadow cast by a stick .i.e.   7.2° / 109 - RIGHT??

My estimated value of the circumference of the sun when extending the Eratosthenes experiment to the sun is way different than standard. Am I wrong? 

23
Quote
This is how:

Top edge of the Sun still sends enough (even too much) light to your eyes (and everywhere else).
Bottom edge of the Sun still sends enough (even too much) light to your eyes (and everywhere else).
Left edge of the Sun still sends enough (even too much) light to your eyes (and everywhere else).
Right edge of the Sun still sends enough (even too much) light to your eyes (and everywhere else).
Every other point of the Sun still sends enough (even too much) light to your eyes (and everywhere else).

sunlight takes an average of 8 min and 20 sec to reach the earth. Most of the sunlight is in the spacelike region (unable for us to see) and their direction of traveling keep them away further in the spacelike region of lightcone - true. 

24
Quote
Error that small wouldn't make Earth any less spherical. :)

how about similar cones formed by the spherical circles above, below, left and right etc of the cone as shown in the diagram

would the interference of light rays from these cones nullify the formation of shadows on earth

25
Quote
if a tree falls in the forest, does it make a sound -
i know about this philosophical thought experiment - changing of sound waves into electrical signals

I do hope you know how do our eyes allow us to see objects if not Google it for detail explanation - it's all about light. 

My question is how do our eyes see the whole image of the sun when more than 90% of sunlight rays do not even fall on the earth and ultimately into our eyes. No reflection is involved either

the spacelike region of the lightcone might help in understanding the question

26


Quote
The nearest I can guess as to your meaning is that every point on the sun's surface radiates light in every direction.

Yes, sun's surface radiates light in every direction. As said, our eyes can only things when light reflected off from it to our eyes. since earth receives a very small portion of light from sun (maybe 5% of the diameter of the earth - just guessing) and the rest of the light that radiates doesn't even fall on the earth and goes in different direction, therefore, how do our eyes see the aforesaid light rays which don't reach our eyes?

27
i have not yet gone through your mathematics but here is another question

here is how do we see an object

if the light from the sun strikes an object, the light reflected off that object moves through space and reach our eyes therefore

how come we see the full sun when earth receives a very small portion of light from the sun due to its smaller size as compared to the sun as said in the question.

28
Is it authentic

Earth receives a very small portion of light from the sun due to its smaller size as compared to the sun.

The sun is 109 times the diameter of the earth. So imagine a spherical circle (convex) of size earth on the surface of the sun. The intensity of light decrease as it increases its distance from the sun, therefore, the intensity diagram of light can be a shape of frustum cone which starts from a spherical type convex base on the surface of the sun. So does this mean sun rays located at the center of the cone would be perpendicular to the earth while resting at an angle?

there would be no shadow of the stick if light rays reach earth in form of wave

Is aforementioned true?

29
Here is the diagram (not to the scale) of two major pieces of plus (+) type container with lil bit more detail than before


Preparation of container

Make pieces/parts of the glass container of desired dimensions in desired units  (say length= 1meter, height=1', width = 1’) in a factory. The middle part is different than the other as shown in the figure. 

Each unit is marked with lines at regular interval for measurement in desired units as shown for measuring, leveling and plumbing reasons.

The height of 7.8” is also marked all the way on 2 sides of the container for curvature reasons as shown

Preparation of ground

Level the ground where the container has to be placed – it shouldn’t settle after placing the full container.   

The radius of the earth is constant in the area where the ground is leveled

Construction

Place the middle piece of the container first on the aforementioned-leveled ground– check its level and then plumb. This means both leveling and plumbing of the container are done at the same time if they are Ok.

Since the height of the middle portion of the container on a perfectly leveled ground is very small therefore it would be hard to notice discrepancies in plumbing of leveled container on either side if there is curvature of earth, therefore add next pieces of the container on all four sides  - Make sure each piece, which is added, perfectly leveled and plumbed. Keep adding units in each known direction and check its leveling and plumbing till a point would come where we wouldn’t be able to plumb the leveled piece of the container if the there is a curvature of the earth. In simple words, if we plumb the container then its level is disturbed and vice versa. This means the leveling and adjustment of the plumbing of the container can’t be done simultaneously.

For Flat Earth
Both leveling and plumbing of the container go together during assembling of the container over the entire length of one / two miles on a perfectly leveled ground

For Round Earth
Both leveling and plumbing of the container DON’T go together during assembling of the container over the entire length of one or two miles on a perfectly leveled ground. This means we need to change the design of the pieces in order to make them parallel to the curvature of the earth if there is an issue with simultaneous leveling and plumbing at any point during assembling of the container. This requires more materials – Another check

Additional check
Volume of water that will be added to the container = 7.98” x 12" x 63360" x 4 = 193669926.9 cubic inches =112077.504 cft after assembling ( 1 mile=63360 inches) =

The earth is round if the container requires water more than the pre-calculated volume of 112077.504 cft up to the marking level of 7.98” (all the way) or otherwise flat.

Since parts of the container can be transported and assembled easily and chosen ground upon agreement can be leveled as well, therefore, rechecking of the curvature can be conducted anywhere on the dry as well as frozen land.

Apropos funding –GoFundCurvature, if there is no theoretical issue with the sanity check. 

I can donate first if either flat/round earther or anyone else shows interest in initiating.

Note: design of the container can be modified for efficacy and accuracy reasons
Quote
All of these methods end up needed optical alignment and eliminating all chance of refraction without it being in a vacuum as was done in the gamma ray "accidental" case I mentioned.

no refraction involved
Quote
At least in the European and Arabian countrues, including Persia (Iran), relativity few have doubted that the is a Globe for over 2000 years.
there is no solid proof that the earth is round or flat

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You ideas might be quite correct "in principle" but almost impossible to implement in practice because the earth is so large.
In a one mile long trough, assuming no breeze, the water at the ends is only 2 inches below a horizontal tangential to the centre.

7.98 inches is the minimum height requirement of water level for this experiment. Height of the water level can be increased if we increased the height of the container. The greater the thickness of water level and height of the container the greater will be the accuracy of the experiment. Similarly, the container is detachable and has 4 sided (covered) so breeze and its transportation to the other different part of the world for testing shouldn't be a problem. Also, we can build the container like a plus (+) sign or similar type in four different directions - 2 miles in each direction. I can design it. we don't even need water, it's just for the third check as the simultaneous check of the plumbing and leveling @ any point during assembling of the container on the leveled ground is enough as explained earlier

No problem in simultaneous checking of plumbing and leveling @ any point on the aforesaid container means the earth is flat as shown in Fig A otherwise the shape of the container will not be rectangular as shown in Fig B if there is curvature.

Moreover, the materials used in the container as shown in Fig B > materials used in the container as shown in Fig A

Note: The radius of the earth is constant for leveled ground



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