Vertical projectile motion from a perfect sphere rotating at constant angular velocity ignoring friction

Purpose to compare and contrast geostationary flat earther claim that object on a round spinning earth when throwing vertically and not horizontally should not land in the same place it started relative to the thrower because the earth is moving.  If the differential equations provide a certain solution than the object on a round spinning earth would land in the same place it started relative to the throwing observer due to the initial horizontal velocity of the observer caused by the rotation of the earth being identical to the initial horizontal velocity of the ball caused by the rotation of the earth from a different reference frame in which the earth is rotating and the thrower is initially moving horizontally but not initially vertically in Cartesian coordinates at the initial starting time and the thrown object is moving horizontally at the same velocity as the observer and moving vertically at it's maximum positive vertical velocity which is it's initial vertical velocity identical to a positive value of the initial speed the thrower throws the object from the throwers frame of reference.  The two reference frames share the same origin, but one reference frame is rotating relative to the other at a constant rotational velocity and air friction is ignored.  If the result shows that the object would land where the throwing observer initially threw it from in the observers frame of reference this would fail to refute the claim of a rotating spherical earth without proving a flat geostationary earth, however if the result is different by a large enough margin this would refute a spinning spherical earth without proving a flat geostationary earth.  Provided of course I set up the differential equations correctly.  The refutation or lack thereof is presuming the standard equation for calculating the force vector from gravity using the universal constant for gravitation is correct if the earth is a rotating sphere which does not necessarily logically follow but is none the less used because it is the current standard physics model which most rotating globe shaped earthers claim.

R=Radius=Radius of sphere earth

Mobject = Mass of object
MSphere = Mass of sphere earth

Xinitial = 0 for both thrower and thrown object

Yinitial = R for both thrower and object

THETAinitial=0 radians

Vxinitial = Vx0
Vx0 = Radius*dtheta/dtime

dV/dtime= -1*G*Msphere*Mobject/(Mobject*[|X|^2+|Y|^2])

dVx/dtime = -|dV/dtime|*X / (|X|^2+|Y|^2]) ^ 0.5

dVy/dtime = -|dV/dtime|* Y  / (|X|^2+|Y|^2]) ^ 0.5

dVx/dtime =-1*G*Msphere*Mobject*X/(Mobject*[|X|^2+|Y|^2]^1.5)

dVy/dtime =-1*G*Msphere*Mobject*Y/(Mobject*[|X|^2+|Y|^2]^1.5)

Vfinalobject and TIMEfinal occurs when |Xfinalobject|^2+|Yfinalobject|^2 = |R|^2 for a X value other than zero


Xfinalthrower = R*sin(Thetafinal)=R*[(timefinal-timeinitial)*(dtheta/dtime)]
And

Yfinalthrower = R-R*cos[(timefinal-timeinitial)*(dtheta/dtime)]

If
Xfinalobject=Xfinalthrower and Yfinalobject=Yfinalthrower

Or in other words the thrower and the thrown object both exist at
(R, Thetafinal) = (R, (dtheta/dtime)*(Timefinal-Timeinitial))

Then

An observer throwing an object vertically standing at Xinitial, Yinitial at TIMEinitial would observe the object landing where he stands at Xfinal, Yfinal, TIMEfinal which would be equivalent to R, THETAfinal

Which would mean from the throwers point of view throwing an object vertically (and not horizontally) while standing on a rotating sphere would result in the object landing in the same place it is thrown from failing to refute a rotating spherical earth.

Else if


XfinalObject does not equal Xfinalthrower or if Yfinal object does not equal Yfinalthrower

Then

if the difference is large enough this would not fail to refute a rotating spherical earth



Now for a simplification by rounding which might not hold correctly you can use

dVx/dtime =-1*g*|X|/([|X|^2+|Y|^2]^0.5)

dVy/dtime =-1*g*|Y|/([|X|^2+|Y|^2]^0.5)




If you wish to round further yet the following can be used however it will result in a different position between the thrower's final position and the landed thrown object 's final position none the less if the difference is small enough a strong case can be made for failure to refute a rounded globe earth.

Of course this rounding results in the typical equations for a thrown projectile from a flat surface onto the same flat surface without the direction or magnitude of forces from gravity changing possibly because physicists assume a flat earth model when simplifying projectile problems although they actually believe in a round earth and this is not necessarily contradictory because the difference over small distances is presumed negligible.  Some of these rounded points might not actually exist on the Sphere but could be very close to points on the Sphere.

only do rounding for small enough values of (tfinal-tinitial)

dVx/dtime = 0

dVy/dtime = - 1 * g

Xfinalthrower= Radius*sin[(dtheta/dtime)*(Timefinal-Timeinitial)]

Yfinalthrower = Radius*cos[(dtheta/dtime)*(Timefinal-Timeinitial)]

Vy0 = Vyinitial

Vx0 = Radius*dtheta/dtime

Rounded Timefinal-timeinitial=2*|Vy0/g|

Rounded Xfinalobject =  Rounded (Timefinal-timeinitial) * Vx0
Rounded Xfinalobject = 2*|Vy0/g|* Vx0

Rounded Xfinalobject = 2*|Vy0/g|* Radius*dtheta/dtime

rounded Xfinalthrower= Radius*sin[(dtheta/dtime)*(2*|Vy0/g|)]

Xfinalthrower / Xfinalobject =
= (sin[(dtheta/dtime)*(2*|Vy0/g|)])/[(dtheta/dtime)*(2*|Vy0/g|)]

Limit as theta approaches 0+ of (sin[theta]) / theta is 1-

Limit as Vy0 approaches 0+ of Rounded Xfinalthrower/RoundedXfinalobject is 1-

For small values of Vy0
Rounded Xfinalthrower is approximately equal to rounded Xfinalobject

Conclusion
When throwing objects vertically for small enough vertical heights with a small enough duration of time between throwing and landing a rotating earth should have approximately the same results as a stationary earth and a flat earth approximately the same results as a globe shaped earth.  There is a failure to refute the round earth claim with rounded simplified equations, failure to refute a round rotating earth does not refute a flat geostationary earth.  Perhaps if the differential equations were used without rounding the object might not be calculated to land in the same place but typically physicists round so much that a flat earth and round earth produce the same results with projectile motion for objects thrown over short distances for a short period of time.


Copyright Carl Janssen 2020 March 26

Arbitrary dates and stories imagined by evolutionists and creationists

People who believe in creation origins or evolutionary origins both make up different stories, neither produced practical knowledge for the current time period.  Even if you could produce a new species by breeding and such evolution was proved to happen in the present that would not prove that the previous species came about by evolution instead of being created in such a way that they could evolve into new species.  There are multiple potential pasts that could lead to the same present.  Any timeline imagined through extrapolation could be imagined to start at a earlier or later point and lead up to the same present set of conditions from that starting point.  If you must extrapolate back past thousands of years than you must extrapolate back past billions or trillions or any number of years no matter how long to mathematically "approach the infinite" and not towards a big bang a finite number of years ago.  To say that the laws of physics were different before the big bang billions of years without observing different laws of physics is no less absurd than to say the laws of physics were different thousands of years ago as a result of creation without observing different laws of physucs.  Life could very well always existed eternally and never came from non life.  There very well could be an eternal universe without a big bang.

Copyright Carl Janssen 2020 March 25

Early Masonic Manuscripts Freemasons Freemasonry free mason masons

http://web.archive.org/web/20200318043530/https://en.m.wikipedia.org/wiki/Masonic_manuscripts

http://web.archive.org/web/20200318060424/https://en.m.wikipedia.org/wiki/Ahiman_Rezon


http://web.archive.org/web/20200318055229/http://www.freemasons-freemasonry.com/regius.html

http://web.archive.org/web/20200318055406/https://en.m.wikisource.org/wiki/Matthew_Cooke_Manuscript

http://web.archive.org/web/20200318061311/https://www.sacred-texts.com/mas/gar/index.htm

http://web.archive.org/web/20200318062102/http://www.themasonictrowel.com/Articles/Manuscripts/manuscripts/edinburgh_register_house_ms_1696.htm

http://web.archive.org/web/20200318061918/https://archive.org/stream/EdinburghRegisterHouseMS1696/Edinburgh%20Register%20House%20MS%201696_djvu.txt

http://web.archive.org/web/20200318061905/http://theoldcharges.com/chapter-21.html

http://web.archive.org/web/20200321053541/http://en.m.wikipedia.org/wiki/History_of_Freemasonry#Anderson's_Constitutions

http://web.archive.org/web/20200318043336/http://freemasoninformation.com/masonic-education/books/andersons-constitutions-of-1723/amp/

Construction workers build houses not landlords

Government does not build the roads construction workers do
Landlords do not build the houses construction workers do
Landlords do not help the construction workers financially they charge them rent to live in houses built by construction workers
 bankers loan the landlords to pay the construction workers
Bankers don't loan the landlords with money they earned they create money by ledger entry

Copyright Carl Janssen 2020 March 20

1965 scientist claims the moon is plasma - UNCUT | RetroFocus ABC News (Australia)

Published on May 8, 2019

In 1965 this scientist tried to turn conventional science on its head. Among R Foster’s theories was that the moon was in fact made of plasma, not rock, and that landing on it would not be possible. Mr Foster claimed that once his ‘profound and decisive’ investigations were proven, a complete re-investigation of the laws of nature would be necessary. This is the uncut version of the interview posted on 28 March 2019. The ABC has been unable to confirm Mr Foster’s identity beyond the entry in the production notebook from 1965: “People – Int Tasmanian Professor (FOSTER)”. We have also been unable to find any documentation of his work. For more from ABC News, click here: http://www.abc.net.au/news/ Subscribe to us on YouTube: http://ab.co/1svxLVE You can also like us on Facebook: http://facebook.com/abcnews.au Or follow us on Instagram: http://instagram.com/abcnews_au Or even on Twitter: http://twitter.com/abcnews

1965 scientist claims the moon is plasma - UNCUT | RetroFocus

ABC News (Australia)

https://m.youtube.com/watch?v=XhIwZuPGfss

http://web.archive.org/web/20200119084751/https://m.youtube.com/watch?v=XhIwZuPGfss

http://web.archive.org/web/20200301132518/https://m.youtube.com/watch?v=XhIwZuPGfss

The historical Jesus Habermas

http://web.archive.org/web/20200227200706/https://www.garyhabermas.com/books/EvidenceBook/GaryHabermas_Evidence-for-the-historical-Jesus-Release_1point1.pdf

Visual border height on perfect spherical planets

Abstract

Derived formulas, drew diagrams and ran calculations for spherical earth claim

Two individuals on planet earth at the same elevation standing 6 miles apart with a body of water between them should have a visual barrier of 6 feet of water between them

A 6 foot tall barrier of water is the same as average height of European Caucasian  humans rounded to the nearest foot

If people approximately 6 feet tall standing at the same elevation (relative to sea level) can see each other's feet with binoculars while standing 6 miles apart then this would refute the claimed radius of the earth if it is a sphere.  It would not refute that the earth is a sphere because it could just be a much bigger sphere instead of flat for example which might indicate hidden land.

Intro

My calculatiions could be wrong

If my calculations are correct the ideal distance to test the alleged shape and size of the earth is two people standing 6 miles apart from each other at the same elevation 

A way to try to stay the same elevation might be to look at each other across a body of water with their feet standing near where the water meets the land at the coast of the same body of water

The reason for that distance is that at a path of 6 miles between them there should be a visual barrier of water between them that is 6 feet tall

6 feet tall is the average height of Caucasian humans of European ancestry rounded to the nearest foot

If the earth is a sphere of the radius claimed they should be able to see more of their bodies as they get closer to each other than 6 miles and less of their bodies (at a certain point none of their bodies) as they stand farther away than 6 miles

I checked for the alleged radius of the earth at the equator at the highest mountain on the equator where the radius is supposed to be the largest and the alleged radius between the north and south poles where the earth's radius is supposed to be the smallest and in all cases two individuals standing 6 miles apart as you would walk the curve would have a visual barrier approxiamately 6 feet tall between them (between 5.989 feet and 6.016 feet) this is approximately the standing height of a person rounded to the nearest foot

The following is an explanation for how to calculate the height of a visual barrier of water between two people standing on a spherical planet followed by diagrams that go with the explanation for how to get the formula then followed by calculations using the radius of the earth according to sources listed later below,  screenshots of calculations and diagrams were created at websites listed in links below in the last section, along with various websites by other people related to the same topic for additional reading, the information in some of these websites do not give the same results as one another and it is important to understand the difference between a visual barrier due to curvature and a visual drop due to curvature I am hoping by viewing these other websites along with my own explanation of how to calculate a visual barrier this will become more clear to the reader.

Method of Formulation

If two points A and B are selected on a outer surface of a sphere with less than one fourth of the circumference of a circle with an equal radius to the sphere apart from each other as traveled by the shortest arc path along the surface of the sphere the following can be calculated using a diagram of a two dimensional circle instead of a more complicated diagram of a three dimensional sphere because only two orthogonal dimensions are needed to represent the shortest path of travel in Cartesian coordinates.

Let us call the direction of the shortest straight line between the two points (A and B) horizontal and the other direction perpendicular to it vertical.

If one were to try to imagine eyes being at the one point and looking at the other point (at a distance of zero units from the surface) the eyes could not see the other point because there would be an opaque barrier between the points and the vertical height and horizontal width of the barrier could be calculated.

Let R be the radius of the sphere and P be the shortest arc distance between the two points (A and B) that one could travel walking on the surface of the sphere to get from one point to the other.  Let H be the shortest straight line distance between the two points (A and B) which would be the horizontal width of the visual barrier at it's thickest in the horizontal direction or the horizontal width of the visual barriers base.  Let V be the height of the opaque barrier at it's highest in the vertical direction which would be the distance measured between a point called W halfway between the two points (A and B) to a point called T which shall be located at the top of the sphere halfway between the two points horizontally but not vertically.  Let C be a point at the center of the sphere.

Let the angles measured be in radians.  Let us call the angle ACB the name Theta.  The angle ACW  has an equal value to the angle  WCB both being one half of Theta.  Two symmetric triangles with right angles are formed one contains points A, C and W and the other containing points W, C and B.  Using those two triangles the following can be calculated.

P = length of curved Path between points A and B

V = height of Visual barrier = length of line segment WT

Length of WT + Length of CW = V + Length of CW = R

Length of CW = R * cos(Theta/2)

V = R - Length of CW

V = R-R*cos(Theta/2) = R*[1-cos(Theta/2)] = R*[1-cos(P/2R)]

H = width of base of visual barrier

H = the straight Horizontal distance between points A and B

H = length of line segment AB  = length of AW + length of WB

length of AW = length of WB = R * sin(theta/2)

H = 2R*sin(Theta/2) = 2R*sin(P/2R)

R = the Radius of the planet

Theta is an angle in radians

Theta = P/R

Theta/2 = P/2R

Theta/2 = inverse sine of H/2R

The straight line length from A to T is equal to the straight line length from T to B is equal to the square root of V squared plus (H/2) squared.

Now imagine an individual whose eye height from the bottom of their feet to their eyes was a height of Z was standing on point A and a duplicate of that individual was standing on point B.  If they were both oriented as line segments that were purely in the vertical direction with no horizontal components in order to see each other's eyes Z would need to be greater than V.  But if the line segment of the person standing on point A was parallel to the line segment AC and the line segment of the person standing on point B was parallel to the line segment BC.

Then in order for the two people to see each other's eyes

Z = Standing eye height

Z*cos(Theta/2) > V

Z*cos(P/2R) >  R*[1-cos(P/2R)]

And

Z > V / cos(Theta/2)

Z > R*[1-cos(P/2R)] / cos(P/2R)

And the horizontal distance between their eyes length N would be

N = H+2Z*sin(Theta/2) = [2R*sin(P/2R)]+[2Z*sin(P/2R)]

Z and N are not yet included in diagrams at the present time

Diagrams for formulation made with Geogebra and IMGflip

Calculations done with geogebra and duckduckgo

Links to geogebra produced diagrams modified in IMGflip

http://web.archive.org/web/20211014154142/https://imgflip.com/i/5qespv

http://web.archive.org/web/20211014154142/https://i.imgflip.com/5qespv.jpg

http://web.archive.org/web/20211014155226/https://imgflip.com/i/5qeulb

http://web.archive.org/web/20211014155208/http://i.imgflip.com/5qeulb.jpg

http://web.archive.org/web/20211014160801/https://imgflip.com/i/5qex46

http://web.archive.org/web/20211014160802/https://i.imgflip.com/5qex46.jpg

Terms of copyright

Copyright Carl Janssen 2020, 2021 

Material Above is copyrighted some of the material below is from outside sources and not intellectual property of Carl Janssen

Permission to duplicate copyrighted portion in entirety with attribution providing the link to the original article

http://web.archive.org/web/*/http://teachingthenarrowway.blogspot.com/2020/02/visual-border-height-on-perfect.html

No permission is granted to duplicate a portion without duplicating the entire article

Sources of data for radius, and calculations for Radius and double radius

The radius of Earth at the equator is 3,963 miles (6,378 kilometers), according to NASA's Goddard Space Flight Center

http://web.archive.org/web/20211008193450/space.com/17638-how-big-is-earth.html

http://web.archive.org/web/*/https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html

Double Equator Radius

7926 miles

https://duckduckgo.com/?q=%3D3963*2&ia=calculator

the highest point above Earth’s center is the peak of Ecuador’s Mount Chimborazo, located just one degree south of the Equator where Earth’s bulge is greatest. The summit of Chimborazo is 20,564 feet above sea level

https://oceanservice.noaa.gov/facts/highestpoint.html

https://web.archive.org/web/20211013024510/https://oceanservice.noaa.gov/facts/highestpoint.html

Mount Chimbazo summit approx distance in miles from center of earth

Chimbazo Summit radius = Chimbazo Elevation + Equator Radius

3963 miles + 20564 feet / 5280 feet per mile

3966.89469697 miles

https://duckduckgo.com/?q=%3D3963%2B(20564%2F5280)&ia=calculator

Double Chimbazo Summit Radius

7933.78939394 miles

https://duckduckgo.com/?q=%3D(3963%2B(20564%2F5280))*2&ia=calculator

Earth's polar radius is 3,950 miles

http://web.archive.org/web/20211008193450/space.com/17638-how-big-is-earth.html

Double Earth's polar radius

7900 miles

https://duckduckgo.com/?q=%3D3950*2&ia=calculator

Additional Reading Recommendations

https://duckduckgo.com/?q=8+inches+per+mile+squared&ia=web

https://duckduckgo.com/?q=earth+curvature+formula+in+miles&ia=web

https://duckduckgo.com/?q=earth%27s+curvature+per+mile&ia=web

https://duckduckgo.com/?q=horizon+drop+per+mile&ia=web

https://earthcurvature.com/

http://web.archive.org/web/20211007043857/https://earthcurvature.com/

http://web.archive.org/web/20200219191233/https://flatearth.ws/8-inches-per-mile

Online Tools used in calculations, graphics and diagram creation

https://geogebra.org/graphing

https://www.math10.com/en/geometry/geogebra/fullscreen.html

https://imgflip.com/memegenerator

copied and pasted angle symbol charecter from here

http://web.archive.org/web/20210307090638/wumbo.net/symbol/angle/

If Venus is between the earth and the sun it should not be possible to view venus at certain hours during the night because the observer's view would be facing away from Venus since the observer's view would  be facing away from the sun once one is deep enough into the night.  Now if the distance between Venus and the sun is less than the distance between the earth and the sun at certain times of night it should not be possible to view venus in certain parts of the sky no matter where Venus is positioned within that distance from the sun or less.

Copyright Carl Janssen 2020 February 1


http://web.archive.org/web/20200202013609/https://www.space.com/15279-venus-weekend-skywatching-tips.html

Some astronomy guidebooks will tell you that because Venus is a so-called "inferior" planet — a world  that is closer to the sun than Earth — that you can never see it in the middle of the night. But this week, for many locations, Venus will be visible at the witching hour and for some localities well beyond the stroke of midnight!

Reverse communism negates rights to a free market for it's supporters

Reverse communism where you tax the working class to give to the elite is worse than communism where you tax the rich to give to the poor.  Big corporations and the elite and many politicians support reverse communism while pretending to support a free market when talking  to right wingers and pretending to support communism or socialism when talking to to left wingers.  Big corporations are not entitled to a free market as an excuse to censor the working class on their media platforms or do many other things libertarians say it is ok for a business to do but not a government because these big corporations elite owners oppose a free market for individuals who are in the working class.

Copyright Carl Janssen 2020 January 28

Open borders on personal property by government mandate

If your required to accept people you do not want to by law for a business that is a open borders policy for your own personal or private property until those open border policies are removed for personal and private property you can defend your borders by having closed borders for the entire state you live in.  There ought to be a right to refuse clients with whatever reason you choose but the government ruled you must accept clients with hiv aids for your private business.  Barack Hussein Obama additionally removed restrictions on immigrants with hiv aids.  There used to be something called a quarantine.  Now people are getting Ebola and the plague and measles in Europe and California.  When you are not allowed to prevent people with infectious diseases from getting on your business property and people are also crossing state lines with those same infectious diseases then diseases locally eliminated  like measles or never before present in "first world" countries will come in from the "third world."  No mandatory vaccines for school students and children will stop it unless there are mandatory vaccines for immigrants but there can not be mandatory vaccines for immigrants with open borders.  And you can never achieve a state with more (compared to not letting them in) libertarian policies by voting while accepting immigrants that vote for more government power (or anti libertarian policies) who come from locations with a lower percentage of people that support libertarian ideology.

Copyright Carl Janssen 2019 December 3

What percent of the boat is visible?

If the water surface is found from a globe earth than the lower portion of a boat should gradually disappear even when observed through a telescope or binoculars or video magnification now I have seen videos where a boat disappears until it is zoomed in on through a telescope but on a round earth with a round water surface the boat would still re appear when zoomed in on by a telescope but a gradually increasing portion of the bottom should disappear until the whole boat is gone however on a flat earth with a flat water surface the same percentage of the bottom of the boat should be present when zoomed in on.  I am impressed by the videos of a boat reappearing when zoomed in on but I want to know if the percentage of the boat visible decreases as less of the bottom can be seen indicating a globe earth with a curved water surface or if the percent remains unchanged.

Copyright Carl Janssen 2019 October 5

Convex Earth - The Documentary - The Flat Earth Scientific Proof

NJS RADIO

https://m.youtube.com/watch?v=igoizIP7f-g

http://web.archive.org/web/*/https://www.youtube.com/watch?v=igoizIP7f-g&app=desktop

http://web.archive.org/watch?v=igoizIP7f-g&app=desktop