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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?

The Earth has a radius of approximately 3965 miles. Using the Pythagorean theorem, that calculates to an average curvature of 7.98 inches per mile or approximately 8 inches per mile (squared) – source Internet

As shown in the diagram

OB=OD=R=Radius of earth

ABDC is a continuous structure – we haven’t reached at that level so just ignore it

The world's longest and highest continuous bridges over land or water might help in finding AC and BD

Let AB and CD are the two high-rise vertical structures.
Height of AB = CD and length AC and BD must conspicuous (say at least one limes) for curvature
Two stable helicopters or hot air balloons at height A and C can also be used instead of high-rise vertical structures.

An object x falls and strikes at B if drops from height A, similarly
An object x falls and strikes at D if drops from height C

B and D can be found via a continuous laser beam from A to B and C to D respectively

Measure AB, BD, DC, and AC with modern equipment of engineering survey

If BD = AC then the earth is flat – case closed

If BD < AC then we can measure not only the correct curvature of earth but also its radius via geometry as shown in the figure

We need extra reinforcement (steel) in the lower portion of very high-rise building if the earth is round.

Someone removed his post in reference to laser experiment. Anyway
Here is the laser experiment but another way (not close to the surface of the water)

if a laser beam hit point Q at a distance greater than one mile then the earth is flat
if a laser beam hit above point Q at a distance greater than one mile then the earth is round
if a laser beam hit below point Q at a distance greater than one mile then the earth is ....

According to Einstein, gravity is not a force but earth is accelerated frame, therefore, is it possible for the earth to accelerate in two opposite direction simultaneously, for example, falling of antipodal apples from the same elevation at the same time.

Technology, Science & Alt Science / Am I correct?
« on: April 04, 2019, 02:23:57 PM »
For any object in space– omitting gravity, IMPOV

An object of any mass (continuous) irrespective of size can be pushed or pulled without any resistance - Right?

It is the gravitating mass, due to which a falling mass shows resistance because of its weight against another force. For Example an object on earth

If this is true then shouldn’t the definition or the concept of inertia, which means resistance, needs revising.

No idea if someone agrees with the following but

An object is said to be in a state of a well-balanced condition if its centroid or center of mass (c.o.m) locates itself at its spontaneous strategic position if left on its own accord. An undisturbed mass at rest is always in well-balanced condition.

An aforementioned object is said to be in a state of an unbalanced condition if its c.o.m off its strategic position due to any means. The course of the shifting of the centroid, which is under duress due to an unbalanced condition, not only moves the object forward but also guides the direction of its motion. A disturbed mass is always at unbalanced condition - Motion. Thus

An object is said to be at rest if its c.o.m remain at its original position.

An object is said to be in state of motion if its c.o.m off its original position.

An object may spin about its undisturbed c.o.m in the direction of applied force when only a part of the mass is disturbed.

A mass may spin and move if its c.o.m and a part of the mass are disturbed.

An object may also be found at fully or partially disturbed and undisturbed conditions

Let A and B are two spherical objects of masses M and m at rest such that A > B and therefore M > m. Ca and Cb are the centre of masses of A and B respectively.

Both A and B maintain their state of rest if Ca and Cb maintain their original position – Undisturbed state

Both A and B lose their state of rest if Ca and Cb loss their original position respectively - Disturbed state

Now, assume A is disturbed and moves with velocity V while B is undisturbed. Following is one the possible conditions when A collides with B.

A pushes B in front of it in its original direction of momentum until both gains V1.

Here B never offers any resistance to A rather it takes the momentum from A – total momentum of the system still remains the same.  Let a and b are the jitters/impulses (shock waves) produced within A and B respectively due to their collision. Cb shifts away from its original position when “b” passes through it, which makes B unstable. Similarly, Ca also shifts towards its original position when “a” passes through it, which makes A less unstable than before.

Although both A and B osculate (juxtaposed) each other but exert no further force on each other after when both masses attains V1. Both Ca and Cb off center, therefore, A is under reduced momentum of while B gained a momentum.

A or B spins if the line action of a and b are truncated – not passes through the centroid.

Push or pull is considered a force. Since A pushes B due to its momentum, therefore, momentum is a Force F.

So force which is push or pull depends upon on both moving mass and its final velocity, not acceleration.

For any object in space – considering gravity

As said, it is the gravitating mass, due to which a falling mass shows resistance because of its weight against another force, therefore the heavier the mass the greater will be its gravity and resistance or the greater the mass the greater the force will be required to displace its c.o.m, therefore, a tiny apple can’t change the strategic position of c.o.m of earth.

The c.o.m of falling mass is below it original position while the c.o.m of flying object is above its original position

The collision effect of the aforementioned A and B depends upon the size, shape, density, and velocity, etc therefore both a and b may or may not reach the center of mass of A and B respectively.  The two outer particles of A and B at tacnode, which collide with other, start exerting a force on the neighboring particles, the said neighboring particles pass a or b to the next connected particles closely and so on and this is how shock wave passes through A or B or M and m respectively.

Both a and b depend upon the mass below, above, right, left, back and in front and how their elasticity or interlocking system, etc is.

It is said that all objects fall at the same rate if this is true then why the damaging/penetrating effect of the same mass is different if fall (at the same rate) from different heights on the ground.

This means force is directly proportional to the mass of the falling object and its final velocity (not acceleration)

Therefore Force F = MV but not F = ma or mg where g=GM/d^2 or 9.8 m/s/s.

Similarly, addition or multiplication of two or more things of different types (e.g. goats and trees) has no useful meaning in mathematics unless totaling them, therefore, I don’t understand why mass and velocity (or mg) are in the multiplication form with each other in the formula of momentum. Why not momentum = M+V if the product of MV is allowed. And the same is applied to all similar mathematical equations.

Is 3 (goats) x 5 (trees) = 15 goat-tree possible or meaningful?

Anyway, shouldn’t force be measured relative to the displacement of c.o.m of moving mass as explained above and or indirectly via relative to standard penetration on the ground or any standard surface?

Addendum #1: Neither an apple nor earth shows resistance when they feel the same amount of force pulling each other together. Similarly, if an object of any mass (continuous) irrespective of size can be pushed or pulled without any resistance then why the greater the mass the greater force required to move it? Shouldn’t the acceleration of an apple and earth be the same when they feel the same amount of force pulling each other together? 

Addendum #2: It is said all the laws of physics remain the same in every inertial frame if moving with constant speed. This is true only if the inertial frame is moving in earth’s (or any other celestial’s) atmosphere due to its smooth ride on equipotential lines (same elevation).

No celestial gravity’s atmosphere means no smooth ride on the equipotential lines of gravity and hence all the objects in a spaceship if not attached to each other are considered individual object including spaceship.

Therefore a person (if not fasten) in a spaceship and a spaceship are two different objects in space therefore initially when a spaceship accelerates from rest and then gains constant speed, all objects (if not attached to the spaceship) within the spaceship are still at rest. This means the back of the cockpit moves towards a person who is still at rest while the front of spaceship moves away from the said person.  Finally, a rear cockpit reaches a person and a person is pushed by the rear cockpit in forwarding direction of the spaceship when it catches/hit a person.

So the above statement might not hold true for space due to the lack of celestial gravity.

Technology, Science & Alt Science / Question about the calculus
« on: September 24, 2018, 09:50:08 AM »
Can someone clarify the following

The vertical change between two points is called the RISE while the horizontal change is called the RUN. Hence slope is the ratio of the vertical and horizontal changes between two points on a surface or a line. So no slope when there is no rise and run at any given point.

We all know how to draw a graph of y = f(x)=x2 curve. Take two points P and Q close to each other on y = f(x)=x2 curve.

The coordinates of P on the curve of y = f(x)=x2 are (x, y) or (x, x2).

Since it is said x changes from x to (x +dx) where dx is the symbol we use for a small change, or small increment in x while y changes from f(x) to (x+dx)2. The corresponding small change in y is written as dy.

Thus the x and y coordinates of Q are (x + dx, y + dy).

Now dy /dx = [(x + dx)2 – x2] / [x+dx-x]
dy /dx = [x2 + dx2+2xdx − x2] / dx
dy /dx = [2x + dx] / 1
dy /dx = 2x + dx
dy /dx = 2x
When dx is shrunk to zero then it is disappeared on RHS but it's not on LHS – no idea why?

Similarly, when dx heads towards zero then the coordinates of Q are also becoming equal to the coordinates of P but they are not.

Moreover, we introduce delta x or dx right at the start in order to derive the equation of slope dy/dx at any point on the curve but we ignore delta x or dx just before the derivation of dy/dx – Does it make sense?

Anyhow, this means, RISE = dy =2x while RUN = dx = 1 but is it possible for x to be present in such small change in y or dy and 1 in small change in x or dx?

Further, integration is the reverse process of differentiation. Although delta x or dx is ignored during the process of derivation of dy/dx but can it be ignored in the process of integration which makes a lot of difference in summation?

Similarly, we are integrating all the dy(s) not the y-coordinates when we integrate both sides of the equation of dy = 2xdx but it turns into the area under the graph – no idea how? -

As y-coordinate of the chosen point Q is (x+dx)2 therefore

Aren’t both y=x2 and y =(x+dx)2 two different types curves no matter how small delta x is which is a portion of the distance of known value of x. Since we don’t know the limit of the distance of the x, therefore, can we fit delta x within the distance of x?

Both y=x2 and y =(x+dx)2 works fine for all numerical values in which we set the limit of x but since we don’t know the maximum limit of x, therefore, isn’t it wrong to say y=f(x)=x2=(x+dx)2? Aren’t we actually started graphing another equation of y=(x+dx)2 (unbeknownst) instead of y=x2 after point P by choosing another point Q at a distance of delta x from P on the curve of y=f(x)=x2 in which the maximum limit of x is unknown?

We can see the difference if we start graphing both y=(x+dx)2 and y=x2 right from the origin at the same time.

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

The following diagram (not to the scale) depicts the path traced by earth in its orbit (either circular or elliptical) around the sun.

Any point on the outer circle represents solar midnight while on inner circle solar noon. The length of an outer circle is greater than the length of the inner circle (or Arc I > Arc II). And hence

Anti noon (midnight points) are more than solar noon (middays points) when the earth revolves around the sun in its orbit – Any reason

Spherical Earthers:

As the earth orbits the sun, the moon circles the earth and hence both the earth and moon orbit around the sun. Moon just travels around the earth with only one side that always faces the earth. Moon’s rotation period is the same as its revolution period - synchronous rotation

The earth not only rotates about its axis but also moves around the sun in its orbit. Earth rotates once in about 24 (rotates extra approx one degree) hours with respect to the sun in order to solar noon reoccur –  (no need to go into detail of the solar day)

Since the moon doesn't rotate (spin) like the earth does on its axes, therefore, shouldn’t moon change side for sunlit for us due to the effect of the synchronous rotation when both the earth and the moon are imagined together at any two opposite positions in their orbits around the sun?

Let in the following image (not to the scale)

Relative to observer A
The direction of the axis of the north pole of sun or moon is facing upward
The direction of the axis of the south pole of sun or moon is downward

Relative to observer B
The axis of the north pole of sun or moon is downward
The axis of the south pole of sun or moon is upward

Spherical Earthers: Would observer B see the letters "sun or moon" in the picture upside down?

Both Flat and Spherical Earthers: Is there any authentic picture or video of the moon or sun available which is taken from the south pole of the earth in order to see if the spots on the moon or sun look upside down / inverted.

Flat Earth General / Light clock and Time Dilation
« on: April 05, 2018, 11:21:37 AM »
Is time dilation real?

Would time dilate at the same rate in the following light clocks if moving at the same speed in the same direction relative to the stationary observer?

Path of a light pulse is ABA or BAB in the light clock which consists of two mirrors
Path of a light pulse is ABCA or ACBA in the light clock which consists of three mirrors

The tendency of an object to resist changes in its state of motion varies with mass. Force is needed to overcome the inertia of an object. An object of a mass, therefore, depends upon its inertia. More mass means more inertia and more gravity and vice versa. A more massive object offers more resistance and therefore requires more force to change its state of still or constant motion. Gravitational force exists between any pair of masses. For Example; An Earth and an Apple

How does an Apple overpower the inertia of an Earth if both gravity and inertia are real?

How much is a force other than gravity needed to overcome the inertia of an earth in the absence of all other forces?

Flat Earth General / Are all masses like black holes?
« on: February 11, 2018, 07:54:07 PM »
Does gravity become infinity where the center of gravity of a mass lies?

Gravity pulls towards the center of mass. Simple Example;

Acceleration due to the gravity of earth on its surface; g = GM/R^2
Acceleration due to gravity of earth at its center; g = GM/R^2 = INFINITY, where R=0

How are two objects attracted to their center of mass with F=GMm/d^2 which is less than infinity?

Both earth and an apple (smaller objects) accelerate toward each other due to the force of gravitation but an apple appears a lot to the earth due to its greater acceleration as compared to the earth toward an apple, which is so minuscule to be distinguished.

Since the difference in masses is mammoth therefore it seems that not only the earth is stationary as compared to an apple but also the reduction in on-center distance "d" occurs due to falling of an apple ONLY (in its acceleration mode), but verily, both masses are changing their positions and hence on-center distance decreases due to the falling of both masses in their higher derivatives of motion (complex motion) before they strike each other. 

This can easily be observed if the difference in masses is not so huge or if we consider the following two identical spherical masses (from point to celestial), which are separated by on-center distance “d”.

First Mass = M1, Second Mass = M2, M1 = M2 = Identical, Centre-to-Centre distance b/w M1 and M2 = d, d1 = d2, d1 + d2 = d, Gravitational acceleration of M1 = g1, Gravitational acceleration of M2 = g2, g1=g2 and “c” be the mid point of “d".

Although, both M1 and M2 strike each other at “c” as per universal law of gravitation but since none of the M1 or M2 is stationary at “c” therefore neither M1 covers a distance d1 with g1=g2 nor M2 covers a distance d2 with g1=g2 on their road to “c”.  Acceleration g1=g2 is only possible if either M1 or M2 is stationary at “c”.

The earliest imaginable motions of M1 and M2 towards “c” might be due to the generation of g1 & g2 and the reduction in d (reduction in "d1" and "d2" equally on both sides of “c”) as well but after that both M1 and M2 start moving toward “c” at higher types of motion (such as gravitational jerk, jounce, crackle, pop, lock, drop etcetera or complex motion) as d1 and d2 decreases equally on both sides due to the change in positions of both M1 and M2 while on their ways to "c".

Both M1 and M2 move at much faster rate due to the formation of the complex motion instead of simply with accelerations before they hit each other at “c”. The actual striking time of M1 and M2 at "c" is much less than estimated by g1 and g2 combined.

So is F=GMm/d^2 well formulated?

Addendum: As gravity is universal therefore physical objects orbits other physical objects and hence causes the motions of planets, stars, and galaxies in the universe.

Let there is a triplet of A, B, and C on an asteroid initially. A stays on an asteroid while B and C set out for a long space journey with high speed (say 0.5c and 0.9c) at the same time in the same direction relative to A. Assume each 10 years old at the time of departure. B and C are gone for 60 years relative to A. Afterward, B and C return home at the same time and reunited with A on an asteroid.

What would be the age of A relative to B and C?
What would be the age of B relative to C and A?
What would be the age of C relative to A and B?

Since A, B and C can have only one physical appearance and one age. Thus who would be right on the physical appearance w.r.t their respective time of one another especially that of A?

Make it more simple: Let there are 96 clone brothers. Clone #96 stays on an asteroid while the rest take off at the same time with the following speeds relative to Clown#96, in the same direction for their long synchronized space journey. Assume each 10 years old at the time of departure. All 95 clones gone for 60 years relative to clone 96. Afterward, 1 to 95 return home at the same time and reunited with clone 96 on an asteroid.

Speed of clone #1 is 0.01c , #2 is 0.02c, #3 is 0.03c, #4 is 0.04 c, ..., #10 is 0.1c, ......, #20 is 0.2c, ......, #90 is 0.9c, ....,#95 is 0.95c

Each clone can have only one age and one physical appearance therefore who would be right on the physical appearance of #96 clone?

Flat Earth General / Is gravity weak force or strong force
« on: February 06, 2018, 02:04:53 PM »
Imagine two similar rings A and B. Radius of A is R1 while B is R2. Both A and B share the same center. Gravity force b/w A and B is F=GMm/ d^2, where d = 0. Thus, would you be able to separate A and B?

Is the weight of a mass of 1kg of a sphere on real earth, which is equal 9.8 N, greater than the weight of identical imaginary earth on our real earth?

Weight of a mass "m" of 1 kg of a sphere on earth F =GMm/d^2 = W = mge = 1 x 9.8 = 9.8 N.  Where ge = GM/d^2 = 9.8 m/s/s, M = Mass of earth, m = 1 kg and d = o/c distance b/t earth and spherical mass on earth

Things can be ignored at this level but not at when the mass “m” of above said sphere increases gradually till it reaches the size of earth - two identical earths - An imaginary earth on our real earth.

Weight of imaginary earth on real earth = Mge
Weight of real earth on imaginary earth = Mge

Mge = Mge or ge = ge; gravitational accelerations “g” of both identical earths are the same (9.8 m/s/s/) but acting in opposite directions and hence cancel each other. Thus both the earths are weightless, however, physists say that gravities don’t cancel each other.

Since the center of gravity of matter depends upon the shape of object therefore how do we know the unknown mass we buy in market is equivalent to the known standard mass on weigh balance when theoretically, the center to center distance b/w the known mass & earth AND the center to center distance b/w the unknown mass & earth are differ and the same analogy is applied to measurements of division of all units of mass such as gm, kg etc. 

There are infinite ways to design a light clock but here is one of the simplest light clocks shown in the link.

The position of the four mirrors A, B, C, and D are shown in the light clock. The clock is not to the scale but the distance AB=BC=CD=DA and theta 1/one is equal to theta 2/ two = 45 degrees. The direction of the pulse can be from A to B to C to D to A and so on or in opposite direction. A suitable/desirable scale of the clock can also be chosen and similarly its orientation as well.

The observer Oc in the clock is at point A

A pulse of light takes 4 seconds to complete a path of ABCDA if fired at point A for all stationary observers.

Let the clock starts moving at very high speed in space relative to another stationary observer “Os” on a stationary asteroid.

Would the clocks of Oc and Os be synchronized relative to each other if Oc were moving relative to Os?

My question is short in order to avoid confusion, therefore, missing information can be assumed if necessary

Flat Earth General / Does earth become jerky in its orbit
« on: January 31, 2018, 09:54:39 PM »
Inertia is the tendency of a body to resist a change in its motion or rest. Acceleration is the rate of change of velocity while jerk is the rate of change of acceleration. 

Earth moves faster when closest and slowest when farthest in its elliptical orbit around the sun, therefore, why don’t inertial forces which produced when an earth changes its orbital speed and ultimately acceleration (causing angular jerk) affect our daily life, large water bodies, internal structure of earth, sway in super high rise structures etc?

Flat Earth General / Sidereal year/ Solar year
« on: January 30, 2018, 07:22:20 PM »
Sidereal day: time taken by the earth to rotate on its axis relative to the stars - is this for the stationary earth as people were thinking back in the past that the sun is revolving around the earth.

A sidereal day: time taken by the earth to rotate on its axis so that the distant stars appear in the same position in the sky. Since observer on earth changes its position in the orbit of earth due to the revolving of earth around the sun, therefore, would the position of stars in sky be changed due to the recording of time at two different points in the orbit of earth?

Also, the sidereal year is 20 min 24.5 s longer than the mean solar year. This means it will be one hour longer in 3 years, 24 hrs in 72 solar years and 48 hours in 144 years and so on. Thus would the number of nights in 72 sidereal years in which stars were seen be equal to number of nights in 72 solar years.

Flat Earth General / Airplane / correction for curvature of earth
« on: January 28, 2018, 08:41:30 AM »
Do airplanes make a correction for the curvature of an earth when they fly at the same altitude (elevation) for long hours/ period of time? - I know surveyors do.

Flat Earth General / Who is right Galileo or Newton?
« on: January 27, 2018, 11:42:38 PM »

Let in the video

The gravitational acceleration of earth = ge
The gravitational acceleration of bowling ball =  gbb
The gravitational acceleration of feather = gf

Now according to Newton ge > gbb> gf so the striking time of earth and bowling ball will less than the striking time of feather and earth if drop separately or at the same time.

Flat Earth General / Fighter Jet and Curvature of the earth
« on: January 26, 2018, 12:07:31 PM »
The average curvature of the earth is 7.98 inches per mile. So it would be1596 inches (133’) in 200 miles.

Let we have 400 miles perfectly or ideal leveled runway. A fighter jet starts its journey from one end of the runway and increases Its speed till it reaches to a maximum speed of 2500 km/h (or increases its speed beyond the limit which requires for its flying in the air) in the first stretch of 1 or 2 miles from the starting end. So will this jet lift off the ground automatically in air due to a curvature of the earth?

The said jet takes off in a normal way. After few minutes it turns its nose upward and shoots straight vertically up into the sky. This jet is now making 90 degrees vertical angle with the tangent of curvature of the earth just below it. After some time, the pilot decides to fly on the imaginary line in an airspace, which is parallel to the above tangent. So he abruptly turns its nose down (90 degrees) and starts flying on his desired line in the airspace. So will this jet gain elevation if fly for 100 miles on the same said line? 

Flat Earth General / Question about the astronomy
« on: January 22, 2018, 10:00:31 AM »
The First and the Last Noon of the solar year don’t coincide. This means solar noon lags by about 6 hours each year and about 24 hours after every 4 years. Days and Nights flip after two years. We update our calendar after 4 years for our own convenience by adding Feb 29. Apparently, we age 365.25 x4 +1 =1462 days in 4 solar years but the actual figure is 1461 days due to the reversing of aforementioned unnoticeable one day. The earth starts its next year (5th) in the orbit but in reality, we are 24 hrs behind the earth in its orbit around the sun.

This small change in one year reverses not only days and nights but also seasons latently with the passage of time but neither we adjust our clocks for above-mentioned unnoticeable 24 hrs nor we adjust this in our calendar.

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