INERTIA

Both friction and RE attractive gravity are INVERSELY PROPORTIONAL to r (distance).

It must have some link to provide the torque to continue the coordinated rotation of the Earth with its wrapper of air.

Moreover, the damping effect of the frictional layer should have PUT A STOP TO THE EARTH'S ROTATION A LONG TIME AGO.

"The law of conservation of angular momentum applies to rigid bodies. Not to liquids and not to gases. The reason for this necessity is that imparting a torque to a molecule in a rigid body affects the whole body, which is not the case with the other two states.

Consider the World, without an atmosphere, spinning in a vacuum. If we then wrap a non-moving atmosphere around it, that atmosphere will serve to damp the spin of the World."

The supposed frictional force, inversely proportional to altitude, would have dampened the very rotation of the Earth, from the very start.

Both friction and RE attractive gravity are INVERSELY PROPORTIONAL to r (distance).

That is, the higher the altitude, the lower the magnitude of the force.

Friction can only be invoked for a few hundred meters. After the boundary layer, NO FRICTIONAL FORCES ARE INVOLVED.

None that can explain how all of the layers of the atmosphere rotate along at the very same speed as the first layer.

Remember, the frictional force is inversely proportional to the altitude: after the few hundreds of meters, it cannot be invoked anymore.

This is a fact of science.

What is needed is a NEW FORCE: A LATERAL GRAVITATIONAL FORCE WHICH IS DIRECTLY PROPORTIONAL TO THE ALTITUDE, in order to explain the rotation of the atmosphere. This is called the restoring forces paradox.

Thermal effects cannot be brought into the discussion.

<< Irrelevant!  >>

Quote from: sceptimatic on January 31, 2018, 02:37:51 AM

For those who cannot see the video, Ill go through what Brian Cox says as he gets into the Typhoon and explaining what's happening from his comments, then I'll explain why the globe is killed off by his words.
Of course, the globe can always be kept alive by MAGIC. Anything can be kept alive by MAGIC.
That is down to the people who prefer to deal with MAGIC.

I'm dealing with logic and common sense in as simple as way as is necessary for people to see how we are duped.

Ok, at 755 Brian says: Turning towards the SETTING sun the Typhoon accelerates to CATCH up with the Earth's SPIN.
Brian says: Beneath us, a 6 thousand billion billion tonne rock is spinning at 650 mph. Match that speed and something interesting happens to the suns motion across the sky.

The pilot says: We have reached 650 mph, so we are travelling at PRECISELY at the speed of the Earth's rotation. So we STOP the sun as we can see it's about two thirds down.

Brian Cox says: So it should just STAY there now because we are going EXACTLY the same speed as the Earth.

Brian Cox says: But travel faster than the planets SURFACE and the normal passage of the day, is REVERSED.
As the jet accelerates it begins to OVERTAKE the spin of the Earth. Causing the setting sun to rise again.

The pilot says: It's starting to GROW A LITTLE. (THE SUN).

Brian Cox replies: It is, I can see it; we are beating the Earth.


So for all you people that didn't see nor hear Brian Cox - go through what he said and think about what I said and forget about those people who try to use the passenger in the train scenario to dupe you.
If you take the time to understand what I'm saying, you'll see that Brian Cox and his script crew have killed off the rotating globe.

https://www.bbc.co.uk/iplayer/episode/b07kxdr9/forces-of-nature-with-brian-cox-2-somewhere-in-spacetime#

Exactly correct.

Good to see you back posting scepti...

The vertical mixing, thermal effects, and the gravitational effects are totally defied on a cosmic scale by the barometer pressure paradox.

"It has been known now for two and a half centuries, that there are more or less daily variations in the height of the barometer, culminating in two maxima and two minima during the course of 24 hours. The same observation has been made and puzzled over at every station at which pressure records were kept and studied, but without success in finding for it the complete physical explanation."


First, the correct station pressure data as it is measured all around the world.

First reference.

NATIONAL WEATHER SERVICE DATA:


The most basic change in pressure is the twice daily rise and fall in due to the heating from the sun. Each day, around 4 a.m./p.m. the pressure is at its lowest and near its peak around 10 a.m./p.m. The magnitude of the daily cycle is greatest near the equator decreasing toward the poles.

http://oceanservice.noaa.gov/education/yos/resource/JetStream/atmos/pressure.htm

Each day, around 4 a.m./p.m. the pressure is at its lowest and near its peak around 10 a.m./p.m.


Second reference.

GRAPHS SHOWING THE DAILY SEMIDIURNAL BAROMETRIC PRESSURE CHANGES AT 10:00 AM/10:00 PM (MAXIMUMS) AND 4:00 PM/4:00 AM (MINIMUMS):

http://www.geografia.fflch.usp.br/graduacao/apoio/Apoio/Apoio_Elisa/flg0355/textos/Ahrens_cap9.pdf (PG. 211)


Third reference.

A remarkable characteristic of the semi-diurnal barometric variation is the regularity of the occurrence of the maxima and minima and their uniformity in time of day in all latitudes. While the amplitude of these waves may vary greatly with latitude, with elevation, and with location, whether over the sea or over the land, the local times of maxima and minima are very constant.

http://www.archive.org/stream/bulletinobserv06terruoft/bulletinobserv06terruoft_djvu.txt
(Bulletin of Applied Physical Science)


A remarkable characteristic of the semi-diurnal barometric variation is the regularity of the occurrence of the maxima and minima and their uniformity in time of day in all latitudes.

ALL LATITUDES, no exception recorded.

EVER.


Fourth reference.

It has been known now for two and a half centuries, that there are more or less daily variations in the height of the barometer, culminating in two maxima and two minima during the course of 24 hours. The same observation has been made and puzzled over at every station at which pressure records were kept and studied, but without success in finding for it the complete physical explanation. In speaking of the diurnal and semidiurnal variations of the barometer, Lord Rayleigh says: ‘The relative magnitude of the latter [semidiurnal variations], as observed at most parts of the earth’s surface, is still a mystery, all the attempted explanations being illusory.



Fifth reference.

The atmospheric pressure is greatest at about 10:00 a.m. and 10:00 pm. and least at about 4:00 a.m. and 4:00 p.m. The variations are primarily the result of the combined effects of the sun's gravitational attraction and solar heating, with solar heating being the major component.

http://ufdc.ufl.edu/UF00001262/00001


THIS REFERENCE EVEN HAS A GRAPH ATTACHED WHICH DOES SHOW THE 10:00 AM AND 10:00 PM MAXIMUMS (PAGE 569).


The best reference from Soil Engineering.

The atmospheric pressure is greatest at about 10:00 a.m. and 10:00 pm. and least at about 4:00 a.m. and 4:00 p.m.


Sixth reference.

The barometric pressure curve shows a portion of the normal twice-daily oscillation that occurs due to solar and lunar gravitational forces (atmospheric tides), with high pressures at approximately 10:00 AM and PM, and low pressures at 4:00 AM and PM.

http://info.ngwa.org/gwol/pdf/930158405.PDF


Seventh reference.


http://www-das.uwyo.edu/~geerts/cwx/notes/chap01/diurnal.html

Surface pressure measurements in Taiwan (at 25 deg. N) are least around 4am and (especially) 4 pm Local Standard Time, and most around (especially) 10am, and 10pm LST; the amplitude of the semidiurnal cycle is about 1.4 hPa.


Eighth reference.


http://books.google.ro/books?id=vNkZAQAAIAAJ&pg=RA1-PA217&lpg=RA1-PA217&dq=barometer+pressure+semidiurnal+change+10+am+4+pm&source=bl&ots=zgQHfJMC_w&sig=NMbmgLuqwPVwEfGVp3WuSu8Mdgg&hl=ro&sa=X&ei=-As4UqWRL4qp4ATI2ICIBA&ved=0CEAQ6AEwAQ#v=onepage&q=barometer%20pressure%20semidiurnal%20change%2010%20am%204%20pm&f=false

THIS IS REAL SCIENCE: DAILY SEMIDIURNAL CHANGES IN THE BAROMETER PRESSURE READING.

Maximums at 10:00 am and 10:00 pm, and minimums at 4:00 am and 4:00 pm.



Ninth reference.

Humboldt carried a barometer with him on his famous South American journeys of 1799-1804. In his book Cosmos he remarked that the two daily maxima at about 10 a.m. and 10 p.m. were so regular that his barometer could serve somewhat as a clock.

http://www-eaps.mit.edu/faculty/lindzen/29_Atmos_Tides.pdf



U.S. Weather Bureau, “Ten-Year Normals of Pressure Tendencies and Hourly Station Pressures for the United States,”
Technical Paper No. 1, Washington, D.C. 1943.

Semidiurnal variations: maximums at 10:00 am/10:00 pm and minimums at 4:00 pm/4:00 am



Surface pressure exhibits a remarkably stable semidiurnal oscillation with maxima at 10 a.m. and 10 p.m. and minima at 4 a.m. and 4 p.m. local time. This semidiurnal oscillation in surface pressure is a universal phenomenon observed worldwide and can be identified even in disturbed weather conditions.

http://amselvam.webs.com/SEN1/bio2met.htm



NATIONAL WEATHER SERVICE DATA:


The most basic change in pressure is the twice daily rise and fall in due to the heating from the sun. Each day, around 4 a.m./p.m. the pressure is at its lowest and near its peak around 10 a.m./p.m.


A remarkable characteristic of the semi-diurnal barometric variation is the regularity of the occurrence of the maxima and minima and their uniformity in time of day in all latitudes. (Bulletin of Applied Physical Science)


ALL LATITUDES, no exception recorded.

Surface pressure exhibits a remarkably stable semidiurnal oscillation with maxima at 10 a.m. and 10 p.m. and minima at 4 a.m. and 4 p.m. local time. This semidiurnal oscillation in surface pressure is a universal phenomenon observed worldwide and can be identified even in disturbed weather conditions.


BAROMETER PRESSURE PARADOX

One maximum is at 10 a.m., the other at 10 p.m.; the two minima are at 4 a.m. and 4 p.m.

The heating effect of the sun can explain neither the time when the maxima appear nor the time of the minima of these semidiurnal variations.

If the pressure becomes lower without the air becoming lighter through a lateral expansion due to heat, this must mean that the same mass of air gravitates with changing force at different hours.


Lord Rayleigh: ‘The relative magnitude of the latter [semidiurnal variations], as observed at most parts of the earth’s surface, is still a mystery, all the attempted explanations being illusory.’



Currently, the barometer pressure paradox CANNOT BE EXPLAINED AT ALL.

Richard Lindzen tried, some 40 years ago, to include the effects of ozone and water absorption in the atmospheric tide equations; notwithstanding that in his original paper he did express some doubts, the scientific community happily concluded that the barometer pressure paradox has been solved.


Not by a long shot.

Here is S.J. Woolnough's paper detailing the gross error/omission made by Lindzen.

http://cree.rdg.ac.uk/~dynamic/index_files/papers/Woolnough_et_al_2004.pdf

While the surface pressure signal of the simulated atmospheric tides in the model agree well with both theory and observations in their magnitude and phase, sensitivity experiments suggest that the role of the stratospheric ozone in forcing the semidiurnal tide is much reduced compared to theoretical predictions. Furthermore, the influence of the cloud radiative effects seems small. It is suggested that the radiative heating profile in the troposphere, associated primarily with the water vapor distribution, is more important than previously thought for driving the semidiurnal tide.

The vertical mixing, thermal effects, and the gravitational effects are totally defied on a cosmic scale by the barometer pressure paradox.

Here is S.J. Woolnough's paper detailing the gross error/omission made by Lindzen.
http://cree.rdg.ac.uk/~dynamic/index_files/papers/Woolnough_et_al_2004.pdf

Figure 7a shows the time and zonal mean updraft
mass flux profile from the convection scheme for the
equatorial grid points.

While the surface pressure signal of the simulated atmospheric tides in the model agree well with both theory and observations in their magnitude and phase, sensitivity experiments suggest that the role of the stratospheric ozone in forcing the semidiurnal tide is much reduced compared to theoretical predictions. Furthermore, the influence of the cloud radiative effects seems small. It is suggested that the radiative heating profile in the troposphere, associated primarily with the water vapor distribution, is more important than previously thought for driving the semidiurnal tide.

Friction only applies for the first several hundreds of meters.

The Hadley cells argument cannot be used by the RE.
. . . . . .
Put another way: if the Sun's heating (insolation) at the equator provides
the thermal energy that maintains the Hadley cell N-S rotation, what
force/energy keeps the equatorial winds rotating at the same speed as the
rotating (sic) Earth?

. . . . . . . . .
The field of gravity is such that its strength at a point, s1, within the atmosphere is inversely proportional to (R + h)^2. Such rapid decrease in field strength with altitude helps to ensure that our atmosphere is not compacted into a thin layer at sea level. In contrast, the strength of the supposed new field would be directly proportional to (R + h) and thus increase with altitude.

<< irrelevant >>
Conclusion
The World either rotates or it doesn't.

If the World rotates, then its atmosphere must rotate, because we do not experience lethal windspeeds as a function of latitude.

In this case, a restoring force is necessary to explain periods of local atmospheric calm. This field would have an effect on all material objects and would seriously restrict our daily motion in all but an eastwardly direction.

• The atmosphere was presumably derived from the already rotating earth, so it never had to be brought "up to speed".
• There is nothing outside the atmosphere to stop it moving with the earth, so even the smallest drag would bring it up to speed eventually.
• There is quite adequate mixing of the lower atmosphere, in your "friction zone", with the upper atmosphere, transferring any horizon velocity with it.

If the World does not rotate, then its atmosphere cannot rotate, and successive periods of local calm are caused in this case simply by decreasing kinetic energy (and linear momentum) of the air molecules as the magnitudes of their velocities are reduced by collisions. This requires the absence of any rotational field and also the absence of even a non-rotating vector field (which would make itself apparent via atmospheric damping).
Unlike the field of gravity, there exists no evidence to support the idea of a restoring vector field.

Since there is no restoring field, the World and its associated atmosphere cannot be rotating about an axis."

The exact formula for the lateral deflection of a vertically fired projectile:



g = 32ft/s²

T_E = period of rotation = 86,400 s

λ = latitude


Bedford latitude = 52.13 degrees

δ = 5.2 ft (far larger than the recorded 8 inches)

This is the best case scenario for the RE, taking into account the Coriolis force (which at the time of the publishing of Earth is not a Globe was not yet fully investigated and accounted for).

If the speed is taken into account:

http://www.damtp.cam.ac.uk/user/reh10/lectures/ia-dyn-handout14.pdf


One of the easiest experiments which can be done to find out that the Earth is stationary.

Quote from: cikljamas on February 03, 2018, 02:35:37 AM

Please, just try running certain distance in the moving (5km/h) train in a counter direction and then do the same (try running across the same length within the stationary train) and compare results. And then you can even try running the same distance in the moving (5 km/h) train in the same distance of train's motion and compare all (three) measured times. Any rational person don't even need to carry out such an experiment because (solely on the basis of our thought experiment) it is already more than obvious that all three measured times would be quite different.

You started out so well, then crashed and burned.
It is already more than obvious that all three times will be the same, that the motion of the train wont magically make it harder for you to walk to the front or back.

Quote from: cikljamas on February 03, 2018, 02:35:37 AM

Quote from: Alpha2Omega on February 02, 2018, 06:02:42 PM

Quote from: Alpha2Omega on February 02, 2018, 01:17:02 PM

Have you attempted the simple dropped-ball-in-a-moving-train experiment yet. What [you keep insisting] says no, you haven't. Please do so before continuing with this "thought experiment". The simple physical experiment suggested would clearly show you that your premise is wrong, and, thus, any thought experiment that assumes it is correct is meaningless.

Please, just try dropping something (a ball, a book, a beanbag, anything like that) on a somewhat rapidly moving, but not accelerating, train. If you are right, it will land on the floor several meters behind the point on the floor directly below where it was dropped.

Why this doesn't happen has already been explained many times. You obviously don't believe it won't happen, but that should be easy enough to check for yourself. Please do.

Please, just try running certain distance in the moving (5km/h) train in a counter direction and then do the same (try running across the same length within the stationary train) and compare results. And then you can even try running the same distance in the moving (5 km/h) train in the same distance of train's motion and compare all (three) measured times. Any rational person don't even need to carry out such an experiment because (solely on the basis of our thought experiment) it is already more than obvious that all three measured times would be quite different.

Unfortunately, there's no passenger train service within a couple hundred miles of where I live. Even more important, your proposed experiment has too many variables that are hard to control in practice. For instance, you're suggesting timing three separate relatively short runs, but have no way to ensure that your pace was the same all three times. Accurate and repeatable timing could also be an issue.

“So far as hypotheses are concerned, let no one expect anything certain from astronomy, which cannot furnish it, lest he accept as the truth ideas conceived for another purpose, and depart from this study a greater fool than when he entered it.” - *NICOLAS COPERNICUS*

Modern astronomers have lengthened the sun's distance by nearly a hundred millions of miles, which has necessarily increased the earth's supposed orbit more than 300 000 000 of MILES!!! But this extreme alteration is neither acknowledged nor permitted to detract from the great name of Kepler, lest it might also reflect upon the "science" of astronomy ; FOR IN THIS EXACT "SCIENCE" THE ALTERATION OF MILLIONS OF MILES IS "A MERE DETAIL!"

~snip~

The Sun is 93 million miles away.

Too many variables? What a pathetic excuse...

You see Alpha, this simple experiment (modified in a proper manner) is all we really need to prove you and your HC believers are utterly wrong. But you know that, there is no doubt in my mind that you are perfectly aware of it. However, we can talk like this 1000 years, and you will still tell us your silly HC fairytales, that is why we really need to perform our simple experiment, don't you think so?

Property of walking is constant contact with the surface (base of the train).
Property of running is NOT constant contact with the surface (base of the train).

So, if we transformed our walker into a runner, and carried out the same kind of an experiment, should we expect a different result? Would it take less amount of time for our walker to take the whole distance of 100 m long train if he ran 10 km/h inside the moving train - in counter direction of train's motion (assuming that he had already canceled out initial inertia before he crossed the starting line in front part of the train)?

Yes. Why? Because significant part (maybe 50 %) of his journey he will spend in the air due to the property of running. Spending 50 % of it's journey in the air (and with already canceled out initial inertia) he will allow the rigid base of the train to slip/slide below his feet to a certain extent.

Quote from: cikljamas on February 04, 2018, 06:07:04 AM

Too many variables? What a pathetic excuse...

You see Alpha, this simple experiment (modified in a proper manner) is all we really need to prove you and your HC believers are utterly wrong. But you know that, there is no doubt in my mind that you are perfectly aware of it. However, we can talk like this 1000 years, and you will still tell us your silly HC fairytales, that is why we really need to perform our simple experiment, don't you think so?

Yesterday I performed the even simpler, far more practical, and more conclusive experiment I recommended.

Driving at a steady 62 mi/hr (100 km/hr) on a straight stretch of level road with the windows closed, I held a "squeeze ball" (a soft ball with vinyl cover) at the ceiling and let it drop into the passenger seat. It fell straight down to the cushion 38" (97 cm) below the drop point (see specification drawing; the trajectory was pretty much exactly along the line indicating front-seat headroom in the drawing). The diameter of the ball is about 6 cm, so the distance of the fall was about 90 cm.



Using the formula for distance traveled, s, under constant acceleration, a

s = ¹/₂ a t²

and solving for t (time), we get

t² = 2 s / a
t = sqrt (2 s / a)

Plugging in the distance of fall for s, and the standard value for acceleration of gravity,

t = sqrt (2 * 0.9 m / (9.8 m/s²))
 = sqrt (0.18 s²)
t = 0.43 seconds

Distance traveled by the car is, of course distance = velocity times time

d = vt

v = 100 km/h
 = 100 km/h * 1000 m / km / (3600 s/h)
v = 28 m/s

So, in the 0.43 seconds it took the dropped ball to fall to the seat, the car traveled

d = 28 m/s * 0.43 s
d = 12 meters.

The specifications for the car show the overall length as 189.6", or 4.816 meters, meaning the car travels 2 ¹/₂ times times its own length while the ball is in the air.

If "detaching" the ball and allowing it to free fall inside the cabin meant all its forward momentum immediately vanished, the car would be "scooting past it" until it hit something. It didn't; it dropped straight down from the FOR of the car, which means it maintained exactly the same forward speed as the car.

Your hypothesis:

Quote

Property of walking is constant contact with the surface (base of the train).
Property of running is NOT constant contact with the surface (base of the train).

So, if we transformed our walker into a runner, and carried out the same kind of an experiment, should we expect a different result? Would it take less amount of time for our walker to take the whole distance of 100 m long train if he ran 10 km/h inside the moving train - in counter direction of train's motion (assuming that he had already canceled out initial inertia before he crossed the starting line in front part of the train)?

Yes. Why? Because significant part (maybe 50 %) of his journey he will spend in the air due to the property of running. Spending 50 % of it's journey in the air (and with already canceled out initial inertia) he will allow the rigid base of the train to slip/slide below his feet to a certain extent.

says that being "in the air" and not in contact with the vehicle (steadily moving train in your case, steadily moving car in mine) will "allow the rigid base of the [vehicle] to [slide] below ... to a certain extent". To what extent? In the case of the dropped ball in the air for about 0.4 seconds, how far would you expect the rigid body of the car to "slide" past the ball? I observed zero. Therefore, your hypothesis is not supported unless the "extent" is so close to zero that it couldn't be detected.

Don't believe me? Conduct your own experiment. In the meantime, please save everyone's time (especially yours) and cease with the wordy speculation based on your flawed premise.

Quote from: EvolvedMantisShrimp on February 04, 2018, 07:33:11 AM

The Sun is 93 million miles away.

I am the president of USA.

But the fool on the hill
Sees the sun going down
And the eyes in his head
See the world spinning round

The exact formula for the lateral deflection of a vertically fired projectile:

Within HC theory (rotating earth), when flying or rolling (ThrustSSC) 1000 km/h (which is roughly the alleged speed of the earth at 52 degrees N) WESTBOUND, that is to say : in counter direction of earth's rotation, we counteract (ENTIRELY - 100 % - cancel out) initial inertia (impetus), so that - if we carried out the same kind of an experiment (shooting the ball upwards) from the cannon which is attached to the moving frame of 1000 km/h fast object - we should expect the ball to come down much closer to the muzzle of the gun than in the case when the ball was discharged from a non-moving object (local frame of reference).
Why?

Within HC theory a non-moving object (local FOR) is in fact moving object (inertial FOR).

Well, Jack, do i really have to explain that to you?

hence the ball that would be discharged from our APPARENTLY stationary cannon would have very significant impetus.

How HC believers are going to explain that?

All that they can call upon is "air drag"

Sandokhan provided for us very compelling explanations on which basis we can discard even that

The motion of the train wont make it harder or easier for you to WALK to the front or back of the train

when our walker becomes our runner then we are going to see the differences in the final results of his running (not walking) race comparing his results after running in counter direction of train's motion vs running in the same direction of train's motion.

Detecting the Aether Wind: the Michelson-Morley Experiment

I hope this is going to be our decisive thought experiment:

So, with such gradual acceleration and with the speed which is 4 times slower than the speed of our runner (in counter direction) we have provided for our experiment two very important conditions :

The final result of our experiment will be the faster arrival

Care to carry out such an experiment in reality and see for yourself if (even by conducting such a simple experiment) we could very easily determine (only if we wanted to) whether the earth is in motion or not!!!

Too many variables? What a pathetic excuse...

You see Alpha, this simple experiment (modified in a proper manner) is all we really need to prove you and your HC believers are utterly wrong.

we counteract (ENTIRELY - 100 % - cancel out) initial inertia (impetus),

impetus
noun
the force or energy with which a body moves.
     "hit the booster coil before the flywheel loses all its impetus"
synonyms: momentum, propulsion, impulsion, impelling force, motive force, driving force, drive, thrust, continuing motion.

The exact formula for the lateral deflection of a vertically fired projectile:



g = 32ft/s²

T_E = period of rotation = 86,400 s

λ = latitude


Bedford latitude = 52.13 degrees

δ = 5.2 ft (far larger than the recorded 8 inches)

This is the best case scenario for the RE, taking into account the Coriolis force (which at the time of the publishing of Earth is not a Globe was not yet fully investigated and accounted for).

If the speed is taken into account:

http://www.damtp.cam.ac.uk/user/reh10/lectures/ia-dyn-handout14.pdf


One of the easiest experiments which can be done to find out that the Earth is stationary.

That is why the vertically fired projectile (cannon ball, bullet, tennis ball) is the most direct experiment to prove that the Earth is stationary: we have a very precise formula which shows exactly what the lateral deflection should be in the case of a rotating Earth.

Quote from: cikljamas on February 04, 2018, 06:07:04 AM

Within HC theory (rotating earth), when flying or rolling (ThrustSSC) 1000 km/h (which is roughly the alleged speed of the earth at 52 degrees N) WESTBOUND, that is to say : in counter direction of earth's rotation, we counteract (ENTIRELY - 100 % - cancel out) initial inertia (impetus), so that - if we carried out the same kind of an experiment (shooting the ball upwards) from the cannon which is attached to the moving frame of 1000 km/h fast object - we should expect the ball to come down much closer to the muzzle of the gun than in the case when the ball was discharged from a non-moving object (local frame of reference).
Why?

Not by the amounts you are claiming, and it has nothing to do with cancelling out inertia.
The reason is purely due to removing the Coriolis effect from the situation.
However you then have the competing effect of wind resistance and I don't think a cannonball moving at 1000 km/hr through the air (relative to the air) would still have a negligible effect. I think the wind is more likely to contribute and push it over.

How high does a bullet go?

You know I like the MythBusters, right? Well, I have been meaning to look at the shooting bullets in the air myth for quite some time. Now is that time. If you didn't catch that particular episode, the MythBusters wanted to see how dangerous it was to shoot a bullet straight up in the air.

I am not going to shoot any guns, or even drop bullets - that is for the MythBusters. What I will do instead is make a numerical calculation of the motion of a bullet shot into the air. Here is what Adam said about the bullets:

    A .30-06 cartridge will go 10,000 feet (3 000 m) high and take 58 seconds to come back down
    A 9 mm will go 4000 feet and take 37 seconds to come back down.

--------------------------------------------------------------

Taking .50 cal as a Barrett M82 rifle:

velocity = 853m/s (wiki page) acceleration = 9.8m/s2

Let's consider 100 seconds needed time for a bullet to come back on the surface of the earth :

Using our formula above :

1. If we were at the North Pole our bullet should still come back right in the gun muzzle.
2. If we were at the Equator our bullet should fall 385,80 feet (117,58 m) away from our gun.

Does this happen in reality???

1. If we were at the North Pole our bullet should come back right in the gun muzzle.
2. If we were at the Equator our bullet should fall 75,27 feet (22,5 meters) away from our gun.

Now, we have to apply the same method as we did in the case of our decisive thought experiment in which we ensured 4 times greater speed of our runner (inside the 1000 m long train) with respect to the speed of the train.

We have to avoid such enormous speeds (so that nobody can complain about supposed air drag)

if HC theory were true we should have canceled out to a certain extent initial inertia (impetus) of our gun, and the ball should fall closer to the gun in accordance to such diminished degree of (non-existent) initial inertia.

That is why the vertically fired projectile (cannon ball, bullet, tennis ball) is the most direct experiment to prove that the Earth is stationary: we have a very precise formula which shows exactly what the lateral deflection should be in the case of a rotating Earth.

Very well done.

Chatbots do not have any knowledge of Mach's principle:

Ring laser gyroscopes are one of the most direct proofs of geocentrism:

On a windless day, any vertically fired projectile experiment will yield valid results.

The final result for the cannon ball experiment was 8 inches, where at least 5.2 ft should have been recorded.

Quote from: cikljamas on January 30, 2018, 03:06:20 AM

Quote from: JackBlack on January 30, 2018, 02:53:47 AM

Really?
Coming back after so long, reviving a dead thread, just to spout the same refuted nonsense.

Quote from: cikljamas on January 30, 2018, 02:12:43 AM

It means that at the very moment of taking off, concorde passengers should be able to notice (very perceptibly) rotational motion of the earth beneath them assuming that the pilot of concorde right after taking off, turns concorde to the left or to the right (it doesn't make any difference), so that their direction of flight is now perpendicular to the direction of earth's rotation. Concorde passengers should be able to see VERY DISTINCTLY AND PERCEPTIBLY how the earth is turning below them from their left side to their right side (if concorde has turned to the right), or from their right to their left (if concorde has turned to the left). Isn't that so? If you still think that it isn't so, please explain why it isn't so!

Because they don't magically turn instantly.
They gradually turn by interacting with the atmosphere, meaning their motion will be relative to it. As the atmosphere in general moves with Earth, that means they will continue moving with Earth.

Even if your hypothetical nonsense was accurate, how do you plan on having them tell the difference between Earth rotating and them moving?

Because they don't magically turn instantly? ROFLMAO
O.K., let's say they just proceed to fly in a straight line.
What is going to happen?
As soon as they took off, their speed (relative to the surface of the earth) would be 800 km/h (400 km/h due to already (before leaving the ground) gained acceleration + 400 km/h as a consequence of already (for so much) canceled out momentum).
Every experiment can easily prove that this is absolutely correct.

Remember when Professor (cough) Brian Cox went up in a fighter jet and caught up with the rotation of the Earth, he said?
He stopped the sun from setting by keeping up with the exact rotation as we were told.

But then you get these people on here saying that the atmosphere carries on dragging planes regardless of them going with or against the rotation.

Absolutely mental.
The reality is clear to see by Brian Cox and the pilot.
They simply followed the sun as it moved away and kept up with the suns movement, making it appear that the sun had stopped dead yet obviously still moving over ground at a set speed, which was 700 mph or something.


If they were on a spinning ball then they would have to literally appear to be going backwards to the observer on the ground if the observer was to watch a so called sun set whilst Brin Cox and pilot saw a stationary sun.

Quote from: cikljamas on February 05, 2018, 05:00:04 AM

How high does a bullet go?

Does the change in topic mean you've given up on your "running on a train" idea? Did you try to simply drop an object and see, simply, yet definitively, that you were wrong from the outset? I hope so.

Any comment on the experiment I ran in my car? There have been no replies about that from you. Maybe the the reply by PL soon after was related to that post, but his posts are not worth reading and I've been ignoring them since he declared I am not here to respond, or debate more than two years ago, and he seems to be true to his word about that based on quoted bits.

Quote

You know I like the MythBusters, right? Well, I have been meaning to look at the shooting bullets in the air myth for quite some time. Now is that time. If you didn't catch that particular episode, the MythBusters wanted to see how dangerous it was to shoot a bullet straight up in the air.

I am not going to shoot any guns, or even drop bullets - that is for the MythBusters. What I will do instead is make a numerical calculation of the motion of a bullet shot into the air. Here is what Adam said about the bullets:

    A .30-06 cartridge will go 10,000 feet (3 000 m) high and take 58 seconds to come back down
    A 9 mm will go 4000 feet and take 37 seconds to come back down.

--------------------------------------------------------------

Taking .50 cal as a Barrett M82 rifle:

velocity = 853m/s (wiki page) acceleration = 9.8m/s2

Let's see. Using your numbers for muzzle velocity and acceleration...

From a starting velocity v₁ and constant acceleration a, the new velocity v₂ after time t will be

v₂ = v₁ + a t

v₁ = muzzle velocity =  853 m/s
a = acceleration = -9.8 m/sec²

a is negative because it's in the opposite direction when v₁ is vertically upward.

We want to find the time when vertical velocity, v₂, is zero.

0 = 853 m/s + (-9.8 m/sec²) t

Rearranging

9.8 m/sec² t = 853 m/s

t = 853 m/s / (9.8 m/sec²)

t = 87 seconds

Theoretically, it takes that same time to fall, though, so we need to double that to 174 seconds.

Plugging that value for t into the formula presented, using the other figures provided, at the equator, we get

δ = π * g * t³ * cos λ / (3 * T_E)
 = π * 32 ft/s² * (174 s)³ * cos (0) / (3 * 86,400 s)
 = π * 32 ft/s² * 5,268,024 s³ * 1.0 / (259,200 s)
 = π * 168,576,768 ft s / (259,200 s)
 = π * 650 ft
δ = 2043 feet = 623 meters

Quote

Let's consider 100 seconds needed time for a bullet to come back on the surface of the earth :

Using our formula above :

1. If we were at the North Pole our bullet should still come back right in the gun muzzle.
2. If we were at the Equator our bullet should fall 385,80 feet (117,58 m) away from our gun.

Sanity check:
(174 s / (100 s))³ = 1.74³
 = 5.28

623 m / (118 m) = 5.28

Checks!

Quote

Does this happen in reality???

Returning to the Barrett example since it has a muzzle velocity specified, let's see!

What deflection from perfectly vertical would cause the bullet to travel 623 m horizontally in 174 seconds?

d_h = v_h t

d_h is horizontal distance, v_h is horizontal component of velocity. t is, of course, time.

v_h = v_muzzle sin(z)

z is the "zenith angle", the angular deviation from true vertical.

d_h = v_muzzle sin(z) t

We want to solve for z.

sin(z) = d_h / (v_muzzle t)
 = 623 m / (853 m/s * 174 s)
 = 623 m / (148,422 m)
sin(z) = 0.0042

z = sin^-1(0.0042)
z = 0.24° = 14 arc minutes

It doesn't take much deviation from perfectly vertical aim to swamp the effect you're looking for. Also, this completely ignores the effect of wind on a bullet in flight for almost three minutes. The horizontal adjustment of telescopic rifle sights is called "windage" for a reason; bullets are typically in flight for a few seconds at most, and, for accurate shooting, even with much less time for the wind to have an effect, wind matters!

"Does this happen in reality???"

Who knows? It probably can't be rigorously tested.

Do the drop test in a moving train. It should be definitive and you can end this discussion.