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#1
11-12-2018, 03:04 PM
 kertmoore Guest Join Date: Nov 2018 Posts: 16
Article June 25, 2010: How does the earth’s rotation affect the path of a bullet?

I cannot imagine any property of physics that would account for the earth spinning at 1040 mph at the equator amounting to 1,525 ft. per second would cause a targeting adjustment of only a few inches. A formula can be created; however the contradictions that occur keeps the formula at a stage of unproven theory. Following
is my analysis regarding the Coriolis effect and gravity:

I. If you stand dead center of the North Pole, there is no centrifugal force. It takes 24 hours for one to rotate 360°.
 Weights and measures have to be determined where there is no centrifugal force; at dead center of the North or South Pole?
II. An airplane and/or helicopter that is disconnected from earth can land on an aircraft carrier that is a moving target as fast as 1040 mph.
 A Cessna 172 that is disconnected from earth at flight speed of 161 mph can adjust for earth’s rotation speed of 1040 mph at the equator?

Two tests of theory & path of totality.

Subject: Gravity makes one gallon of water appear to weigh 8.34 pounds per gallon, regardless of centrifugal force generated by multiple rotation speeds of the earth.

Centrifugal force at the perimeter of the object is determined by how fast the perimeter is traveling and the diameter of the object. For the earth, the centrifugal force is variable as to its diameter; its latitude location. In Dallas, Texas latitude 32.7767 N° the speed of rotation is 863.8 mph compared to 537.4 mph at Olympia, Washington latitude 47.0379° N. These different earth rotation speeds of 863.8 mph and 537.4 mph generate a different centrifugal force on an object. In our case, the object is one gallon of water that appears to weigh 8.34 pounds per gallon both in Dallas, Texas and Olympia, Washington.

In fact, one gallon of water appears to have the same static weight as it runs in a river from north to south across multiple earth diameters (180° of latitudes) and earth’s corresponding multiple rotation speeds. The standard weight of one US liquid gallon of water is about 8.34 pounds at 62° F. There is no adjustment for the weight of water as to its latitude location. Apparently, gravity makes this adjustment for weights and measures.

If one were to believe the conclusion that gravity makes an adjustment for weights and measures to be accurate and that the conclusion can be proven to be true then you have a fact. Otherwise, if there is some other factor(s) that causes one gallon of water at 62° F to be 8.34 lbs. but it cannot be proven, then you have a theory. Is the theory about how to calculate centrifugal force, or is the theory about whether the earth rotates 360° in a 24 hour time frame?

This is quickly becoming complicated. At what latitude is the one gallon of water at 8.34 pounds at 62° F determined? It has to be dead center of the North or South Pole where there is no centrifugal force. At dead center of the poles, there is no diameter or travel speed to consider for the math components of centrifugal force. For the math of Centrifugal Pounds Force, a gallon of water in Olympia, Washington is 12.415 lbs. as opposed to Dallas, Texas at 19.915 lbs. Otherwise, gravity has made all of the math work out to be that water weight is 8.34 lbs. per gallon regardless of its location in lakes, ponds, and rivers running from the north to the south.

There are “Earth does not move experiments”: The Michelson–Morley experiment was performed over the spring and summer of 1887 by Albert A. Michelson and Edward W. Morley …an attempt to detect the relative motion of matter through the stationary luminiferous aether ("aether wind"). The result was negative…the direction of movement through the presumed aether, and the speed at right angles, was found not to exist; this result…eventually led to special relativity, which rules out a stationary aether. The experiment has been referred to as "the moving-off point for the theoretical aspects of the Second Scientific Revolution". Michelson–Morley type experiments have been repeated many times with steadily increasing sensitivity…recent optical resonator experiments confirmed the absence of any aether wind at the 10−17 level.

Bradley's (in 1725) so-called ”aberration of star light” gave the first experimental disproof of the heliocentric hypothesis. A hypothesis is the starting point for further investigation.
http://www.geocentricity.com/ba1/no066/vdkamp.html

Science cannot be wrong, there surely must be an equation that resolves the conundrum that gravity makes one gallon of water weight be 8.34 lbs. per gallon regardless that earth’s multiple, surface speeds range from 1040 mph at the equator to zero mph at the poles.

Subject: How does a Cessna 172 at flight speed of 161 mph adjust for earth’s rotation speed of 1040 mph at the equator?

In Dallas, Texas latitude 32.7767° N the speed of earth’s rotation is 863.8 mph compared to 537.4 mph at Olympia, Washington latitude 47.0379° N. These different earth rotation speeds are determined from the fact that earth rotates 360° in a 24 hour time frame corresponding to the diameter of the earth at these respective latitudes. In fact, there are 180° of latitudes resulting in earth’s multiple surface speeds ranging from 1040 mph at the equator to zero at the poles.

These multiple speeds are mathematically described as the Coriolis effect (Rotating Earth) for an object that becomes disconnected from earth. Described in “Long Range Shooting Handbook” by Ryan M Cleckner, military snipers have to adjust their sight targeting based on their latitude location as well as the angle of the sniper’s shot relative to his latitude location In an email exchange I had with Mr. Cleckner regarding one parachuting from an airplane, he states “Although I’ve jumped out of a plane many times, I was never in charge of determining the jump point for the drop zone - I can’t speak to how those calculations are made.”

It gets complicated. Degrees of latitude are parallel so the distance between each degree remains almost constant. Each degree of latitude is approximately 69 miles (111 kilometers) apart. Coriolis effect takes into account 180° of latitudes then each half mile, each quarter mile and so on with 360° of angles off each latitude. Does a sniper’s rifle that requires precision targeting have precision GPS? A .30-06 hunting rifle fired due north at the equator has to adjust for earth spin at 1,525 ft. per second. Usually these type rifles are used for taking 350 to 500 yard shots.

The Coriolis effect requires that airline pilots must adjust flight path to reach their target destination. The instrumentation used by airlines is Honeywell: Air Data Inertial Reference System (ADIRS) which includes an Alignment Calibration Module (ACM) for the Coriolis effect (e.g. multiple rotation speeds).

The Coriolis effect is further described and illustrated as an animation on Exploring Earth stating that pilots take the Coriolis effect into account so they do not miss their targets. Exploring Earth states that the analysis is a collaboration with NASA/Goddard Space Flight Center and others (identified at the bottom of the animation web site). View the animation and pay attention, it is misleading. For the animation of south to north - Tierra del Fuego, Argentina directly toward Rio de Janeiro, Brazil – a flight taking an estimated time of five hours, the surface speed of the target destination is 948.5 mph. The target destination would move 4,743 miles; the Earth's diameter is 7,918 miles.
https://www.classzone.com/books/eart...1904page01.cfm

How did WWII vintage aircraft land on an Aircraft Carrier; how is it done with modern day aircraft? ADIRS supposedly provides calibration for point to point navigation. What happens when a Navy pilot (the aircraft is disconnected from earth) arrives at his Aircraft Carrier destination that is on the equator and the target landing “spot” is moving at 1,525 ft. per second?

Neil deGrasse Tyson has even commented that the spinning earth affects football games; that the Coriolis effect does have to be considered regarding the flight of the football on a field goal attempt. http://www.nfl.com/news/story/0ap300...lped-cincy-win

Regarding these supposedly known facts, then for the objects that are disconnected from earth, the following must be true:

In Dallas Texas, latitude 32.78° N, rotation speed is 863.8 mph:
1. Hover in a helicopter and you will arrive at your (same latitude as Dallas, Texas) Artesia, NM destination in 31 minutes or your San Diego California destination in one hour and 22 minutes.
2. A parachute drop of 5,280 ft. (1 mile) descending at an average of 20 mph requires elapsed time of three minutes. The target football stadium moves 43.2 miles.
3. A field goal kicked from east to west that has a one second flight time has an increased distance of 422 yards.

In a Cessna 172 at 161 mph for a 300 mile flight for the same longitude of 102.07° W:
1. Earth’s rotation speed at Stratford, Texas is 835.4 mph
 Flying South to North, from Midland Texas to Stratford, TX; the target destination has moved 1,557 Miles.
2. Earth’s rotation speed at Midland, Texas is 884.2 mph
 Flying North to South, from Stratford, TX to Midland, TX; the target destination has moved 1,647 Miles.

In reality we know that light aircraft, such as a Cessna 172 that has air speed of 161 mph does not have to adjust its flight path to be on target to its destination, a helicopter cannot hover and reach its distant destination and a parachutist or airplane pilot does not have to hit a moving target. So, do these examples represent the utilization of Einstein's Theory of Relativity to maintain a fixed relationship to the rotation of earth?

Science cannot be wrong, there surely must be an equation that reconciles between certain disconnected objects that are - or are not - subject to the Coriolis effect. If you decide to search for the formula or create the formula, remember precision is required:
the earth’s multiple rotation speeds range from 1040 mph at the equator to zero mph at the poles and there are innumerable angles off each of the innumerable latitudes.

For this exercise, do not include the curvature of earth. That can be worked on at a later time.

Nikola Tesla explains the preceding conflicts as follows: “Today's scientists have substituted mathematics for experiments, and they wander off through equation after equation, and eventually build a structure which has no relation to reality.”

Sources:
Honeywell: Air Data Inertial Reference System (ADIRS)
https://aerospace.honeywell.com/en/p...ference-system
https://aerospace.honeywell.com/en/s...alignment-tool

Determine point to point time and distance
https://sunrise-sunset.org/
http://tjpeiffer.com/crowflies.html

Subject: What is to be gained by those who promote a spherical earth and cover up flat earth?

Perhaps at its initial conception of earth as a sphere there was no conscious decision to gain anything. Now that we have the ability to attempt space travel, governments and academia have so much of their power structure based on solar centric science, no one wants to risk what happens when the truth becomes known.

The August 2017 solar eclipse path of totality was 70 miles wide. A shadow (the path of totality) cannot be smaller than the object (the moon) that causes the shadow. Therefore, the moon is less than 70 miles wide.

And, science acknowledges the moon and sun is the appearance of being the same size. High altitude balloon (130,000 ft.) videos indicate close proximity of the sun to the earth, likely less than 3,000 miles. The sunspot that shows on the clouds is relatively small; the appearance is that the sunspot is the same size as the sun

It is very likely the sun and moon are at the same distance from earth.
#2
11-12-2018, 03:44 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 80,133
The equations say that a bullet will be deflected only a very small amount. The experiments also show that a bullet is, in fact, only deflected a small amount. If the experiment and equations both agree, but it is your opinions which differ with both, then it is your opinions which must be revised, not the equations or experiments.
#3
11-12-2018, 05:05 PM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 18,157
Quote:
 Originally Posted by kertmoore The August 2017 solar eclipse path of totality was 70 miles wide. A shadow (the path of totality) cannot be smaller than the object (the moon) that causes the shadow. Therefore, the moon is less than 70 miles wide.
So, you know have a staggering lack of knowledge about basic Physics and you want to criticize Scientists?

Wow, just wow.

Read up about eclipses. Esp. penumbras and umbras. Get a big light. Put something smaller in front of it. Put a piece of paper after it. Move the paper farther and farther away. What happens?

This is grade school stuff. Literally.

If you don't know grade school Science, the best course of action is to trust people who know a billion times more than you.
#4
11-12-2018, 08:11 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,990
Quote:
 Originally Posted by kertmoore A shadow ... cannot be smaller than the object ... that causes the shadow.
I'm trying to figure out why you'd assert something that can be disproved by anyone in a few seconds.

I haven't had any success thus far. Can you help me?
#5
11-12-2018, 08:32 PM
 snfaulkner Guest Join Date: May 2015 Location: 123 Fake Street Posts: 6,943
Quote:
Too long and full of bullshit so I only skimmed it.

Planes might be disconnected from the earth, but they aren't disconnected from the atmosphere. And the atmosphere moves at more or less the same speed as the earth.

And if the sun and moon are at the same distance, then they should collide during an eclipse. Since that has never happened, then they must not be at the same distance.

And, I'm already exhausted by this tediousness. Read a damn text book instead on loony CT websites.
#6
11-12-2018, 11:01 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,990
Quote:
 Originally Posted by kertmoore The August 2017 solar eclipse path of totality was 70 miles wide. A shadow (the path of totality) cannot be smaller than the object (the moon) that causes the shadow. Therefore, the moon is less than 70 miles wide.
It may possibly be worth noting that the width of the umbra (the area of totality) varies a lot. It can reach 1000 km at high latitudes, and of course can be zero (in the case of an annular eclipse). (Could this be an indication that the moon regularly shrinks and expands?)

Also notable is the extreme accuracy of eclipse predictions when these are based on the standard model of the solar system. Many hundreds of thousands of people in the US saw the August 2017 eclipse, and in every case the timing they observed matched the prediction, to a fraction of a second.

I'd love to see the formulas that can produce comparable results from a flat earth model.
#7
11-12-2018, 11:05 PM
 cochrane Guest Join Date: Jun 2006 Location: The Nekkid Pueblo Posts: 20,946
Link to the column so it can be quickly found :

https://www.straightdope.com/columns...h-of-a-bullet/
#8
11-13-2018, 09:52 AM
 Jasmine Member Join Date: Jul 1999 Location: Chicagoland Posts: 1,574
The key factor here is that it is not just the earth spinning, it is also the earth and everything on it that is spinning as well. So, even when the bullet is fired, it is still subject to the same effects as everything else that is spinning at over 1,000 MPH along with the earth.

Here is a primitive example: A quarterback is flushed out of the pocket and is sprinting horizontally towards the sidelines when he releases a long pass. The football, even though released, is still subject to that "sideways" momentum that was produced from the quarterback's sprint towards the sidelines.
#9
11-13-2018, 10:12 AM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by kertmoore I The August 2017 solar eclipse path of totality was 70 miles wide. A shadow (the path of totality) cannot be smaller than the object (the moon) that causes the shadow. Therefore, the moon is less than 70 miles wide.
2200 years ago, Aristarchus estimated the size of the Moon with considerably more accuracy than you have done (https://en.wikipedia.org/wiki/On_the...hus)#Half_Moon). Please recheck your work.
#10
11-13-2018, 11:40 AM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by Andy L 2200 years ago, Aristarchus estimated the size of the Moon with considerably more accuracy than you have done (https://en.wikipedia.org/wiki/On_the...hus)#Half_Moon). Please recheck your work.
In addition, for the solar eclipse in February 1999 (https://en.wikipedia.org/wiki/Solar_...ruary_16,_1999) the area of totality was zero miles wide. What conclusion about the size of the moon can be drawn from that fact?
#11
11-13-2018, 12:11 PM
 Czarcasm Charter Member Charter Member Join Date: Apr 1999 Location: Portland, OR Posts: 58,488
Quote:
 Originally Posted by Andy L In addition, for the solar eclipse in February 1999 (https://en.wikipedia.org/wiki/Solar_...ruary_16,_1999) the area of totality was zero miles wide. What conclusion about the size of the moon can be drawn from that fact?
The moon doesn't exist, which proves that the moon landings were faked!
#12
11-13-2018, 12:15 PM
 Channing Idaho Banks Guest Join Date: Mar 2015 Location: beautiful Idaho Posts: 2,585
Quote:
 Originally Posted by Jasmine The key factor here is that it is not just the earth spinning, it is also the earth and everything on it that is spinning as well. So, even when the bullet is fired, it is still subject to the same effects as everything else that is spinning at over 1,000 MPH along with the earth. Here is a primitive example: A quarterback is flushed out of the pocket and is sprinting horizontally towards the sidelines when he releases a long pass. The football, even though released, is still subject to that "sideways" momentum that was produced from the quarterback's sprint towards the sidelines.
Once the football is free it is no longer connected to the q b. He has to throw harder to overcome his own momentum.
#13
11-13-2018, 01:04 PM
 kertmoore Guest Join Date: Nov 2018 Posts: 16
Five theories

Supposedly there is navigation equipment that commercial and military aircraft use for calibrating adjustments to account for the spinning earth; the Coriolis effect. This type equipment is not on a Cessna 172 that flies at 161 mph which has no capability to adjust for earth spinning at 1040 mph at the equator.

I cannot imagine any property of physics that would account for the earth spinning at 1040 mph at the equator amounting to 1,525 ft. per second would cause a targeting adjustment of only a few inches. A formula can be created; however the contradictions that occur keeps the formula at a stage of unproven theory. There is very limited information about Inertial Navigation Systems (INS), because it is a cover up; it is fake. Supposedly, Inertial Navigation Systems deal with the inertia of the spin of the earth speed at the takeoff location point so that the required flight adjustments occur for destination targeting.

Air Data Inertial Reference System (ADIRS) with digital gyros provides automatic re-alignment
https://aerospace.honeywell.com/en/p...ference-system

Dual Portable Alignment Tool applications include alignment of flight critical surfaces, boresight alignment for weapons systems and calibration of inertial navigation systems.

Our solution makes field level calibration possible by filtering out ambient motion caused by wind, personnel, or other environmental factors. The inertial reference technology eliminates the need for line of sight while maintaining +/- 0.5 mRad accuracy for pitch, roll and yaw.
https://aerospace.honeywell.com/en/s...alignment-tool

Supposedly, Coriolis effect must be accounted for in long range missile and artillery systems. The authors (E. Linacre and B. Geerts) http://www-das.uwyo.edu/~geerts/cwx/...artillery.html indicate that targeting errors were prior to rifling of the cannon bore. It appears to me that Gustave Coriolis' observations became the basis of a formula and formulas causing long range targeting adjustments of only a few inches regardless that the earth supposedly spins ranging from zero to 1,525 ft. per second at the equator.

I think that Einstein recognized the problems with Coriolis effect and he then developed the theory of relativity and theory of special relativity. Yet, the Coriolis effect continues to be used by science to quantify the effect of a spinning earth on objects that become disconnected from earth. Science also uses the Eötvös effect which is "the change in perceived gravitational force caused by the change in centrifugal acceleration resulting from eastbound or westbound velocity."

So, we have three theories: Coriolis effect, Eötvös effect, Gravity and then the "everything moving with the planet" theory of relativity and theory of special relativity; all of these theories are used to explain the forces associated with the spinning earth. And then somehow these theories defy the properties of physics related to the centrifugal force for spin speeds ranging from zero at the North and South Poles to 1040 mph at the equator.

Nikola Tesla explains the contradictions as follows:

“Today's scientists have substituted mathematics for experiments, and they wander off through equation after equation, and eventually build a structure which has no relation to reality.”

I believe the properties of physics and centrifugal force are correct. I do agree with you that Coriolis effect and Eötvös effect is wrong. In addition, I believe the theory of relativity and theory of special relativity is wrong and that the theory of gravity is still just a theory.

At the present, I have not found any literature as to how any adjustments are made for the curvature of the earth for long range artillery targeting.

All aircraft utilize gyroscopes to maintain level flight; the gyroscope maintains level orientation to indicate turns, ascent and descent. Especially for instrument flying, the gyroscope is essential when the pilot cannot see the horizon. As to level flight, the gyroscope spins up (calibrates) at ground level before takeoff and there is no feature to adjust for the curvature of the earth for any distance of flight. When an aircraft is enclosed in clouds and the pilot loses sight of the horizon, the pilot is maintaining level flight via the gyroscope which was calibrated at the location of takeoff. Obviously, the distance can be thousands of miles for military aircraft that can be refueled in the air. There is no mechanism to maintain inflight calibration of the gyroscope.
#14
11-13-2018, 01:17 PM
 Czarcasm Charter Member Charter Member Join Date: Apr 1999 Location: Portland, OR Posts: 58,488
#15
11-13-2018, 01:25 PM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 18,157
Quote:
 Originally Posted by kertmoore At the present, I have not found any literature as to how any adjustments are made for the curvature of the earth for long range artillery targeting.
Translation: You don't know how to use Google.

This is the very first hit for "naval guns curvature of the earth": Link.

Again, all you are achieving here is displaying a deep lack of basic knowledge.

Once a plane leaves the ground the motion of the air takes over. If there's a 200mph wind from the North, the pilot corrects for that. Why the air is moving doesn't matter much. Just how fast and what direction. How can you not understand this at all. Not at all!

What are you going to do next, debunk that Foucault pendulums don't exist?

Think about this. The ancient Greeks knew the world was round. How does a conspiracy effort perpetuate itself for over 2500 years? There are many thousands of amateur astronomers out there (all of which know much, much more than you apparently do) and they are all magically in on it?

How many excuses are you allowed to perpetuate such a ridiculous idea before you have to face up that you really don't understand basic facts?

You don't know how light works, you don't understand the idea of what an atmosphere is, etc.

Also, what column is that about???

Last edited by ftg; 11-13-2018 at 01:26 PM.
#16
11-13-2018, 01:49 PM
 cochrane Guest Join Date: Jun 2006 Location: The Nekkid Pueblo Posts: 20,946
#17
11-14-2018, 07:23 PM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by Chronos The equations say that a bullet will be deflected only a very small amount. The experiments also show that a bullet is, in fact, only deflected a small amount. If the experiment and equations both agree, but it is your opinions which differ with both, then it is your opinions which must be revised, not the equations or experiments.
I had a similar reaction to the fellow who thought that the speed of gravity must be faster than light, even though the predictions of general relativity (in which the speed of gravity is equal to the speed of light) match observations of both the solar system and of orbiting neutron stars, just because when he thought about gravity effects being at light speed, the results he imagined didn't match observations.
#18
11-16-2018, 01:35 PM
 kertmoore Guest Join Date: Nov 2018 Posts: 16
The theory that "the Earth and all its atmosphere and weather are all moving with the planet" is sourced from: Special relativity explains motion only if you’re traveling in a straight line at a constant speed and Einstein’s general theory of relativity explains the general case of any sort of motion.

Yet science states for objects that become disconnected from earth, flight path adjustments are required to account for Coriolis effect and Eotvos effect to enable the object to hit the destination, hit the target for aircraft and bullets, e.g. the earth spins at 1,525 ft. per second at the equator as opposed to zero mph at the North and South Poles.
.
Nikola Tesla stated: "Einstein's relativity work is a magnificent mathematical garb which fascinates, dazzles and makes people blind to the underlying errors."

#19
11-16-2018, 01:37 PM
 Czarcasm Charter Member Charter Member Join Date: Apr 1999 Location: Portland, OR Posts: 58,488
Quote:
 Originally Posted by kertmoore The theory that "the Earth and all its atmosphere and weather are all moving with the planet" is sourced from: Special relativity explains motion only if you’re traveling in a straight line at a constant speed and Einstein’s general theory of relativity explains the general case of any sort of motion. Yet science states for objects that become disconnected from earth, flight path adjustments are required to account for Coriolis effect and Eotvos effect to enable the object to hit the destination, hit the target for aircraft and bullets, e.g. the earth spins at 1,525 ft. per second at the equator as opposed to zero mph at the North and South Poles. . Nikola Tesla stated: "Einstein's relativity work is a magnificent mathematical garb which fascinates, dazzles and makes people blind to the underlying errors." Multiple earth speeds: https://www.youtube.com/watch?v=2-vI...ature=youtu.be
#20
11-16-2018, 05:08 PM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 18,157
Quote:
 Originally Posted by kertmoore The theory that "the Earth and all its atmosphere and weather are all moving with the planet" is sourced from: Special relativity explains motion only if you’re traveling in a straight line at a constant speed and Einstein’s general theory of relativity explains the general case of any sort of motion.
The basics of atmospheric motion doesn't have a single bleeping thing to do with relativity. Nothing. Nada.

Furthermore, you clearly do not understand SR or GR at all. How could you? You don't even understand Newton's Laws!

This displays an astonishing lack of knowledge of the simplest concepts of Science.

Again. If your ignorance is this astonishing, why are you questioning people who actually know stuff?
#21
11-16-2018, 08:33 PM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by ftg The basics of atmospheric motion doesn't have a single bleeping thing to do with relativity. Nothing. Nada.
Good point. All the calculations involved with Coriolis motion could have been done by Newton centuries before Einstein was born.
#22
11-16-2018, 10:48 PM
 John W. Kennedy Charter Member Join Date: Apr 1999 Location: Chatham, NJ, USA Posts: 5,048
Quote:
 Originally Posted by kertmoore Nikola Tesla stated: "Einstein's relativity work is a magnificent mathematical garb which fascinates, dazzles and makes people blind to the underlying errors."
Tesla wasn’t worthy to apply talc to Einstein’s slide rule.
__________________
John W. Kennedy
"The blind rulers of Logres
Nourished the land on a fallacy of rational virtue."
-- Charles Williams. Taliessin through Logres: Prelude
#23
11-20-2018, 09:36 AM
 kertmoore Guest Join Date: Nov 2018 Posts: 16
Centrifugal Force on the moon is same as Centrifugal Force at the North and South Poles of Earth

Gravity on the Moon is about 17% what it is on the Earth. A person whose weight is 200 pounds on Earth would weigh 34 pounds on the Moon.

The moon does not spin. In its orbit around earth it rotates one time each 27 days. Therefore there is no discernible centrifugal force. The 200 lbs. person will weigh 34 lbs. on every point of the moon's surface. There are two places on earth where there is no centrifugal force. It takes 24 hours for one to rotate 360° at dead center of the North and South Pole.

Centrifugal force at the perimeter of the object is determined by how fast the perimeter is traveling and the diameter of the object. For the earth, the centrifugal force is variable as to its diameter; its latitude location.

At the equator which travels at 1040 mph, there is centrifugal force amounting to 69.1% of the weight of the object. This same person who weighs 200 lbs. at the North and South Pole will weigh 62 lbs. on the equator.

Regarding the spin of earth, the table below identifies the effect of centrifugal force:

Measured weight of a 200 pounds person (object) at each latitude.

Miles Rotational Pounds Measured
89.75° 17 4.5 0.3% 199.4
75.00° 1,025 269 17.9% 164.2
60.00° 1,982 520 34.6% 130.9
45.00° 2,805 736 48.9% 102.2
30.00° 3,433 901 59.9% 80.2
15.00° 3,830 1005 66.8% 66.4
Equator 3,963 1040 69.1% 61.8

I. Pounds force is sourced from:
https://www.omnicalculator.com/physi...trifugal-force
Note: Tangential velocity is earth's rotation speed. With entries for Mass (weight), Radius and Tangential velocity, the calculator will solve Force.

II. Rotation speed each latitude:
https://www.vcalc.com/wiki/MichaelBa...ed+at+Latitude
Note: Rotation speed can be located by determining as the crow flies distance http://tjpeiffer.com/crowflies.html between two cities on the same latitude , determining sunrise https://sunrise-sunset.org/ time stamp at each of the two locations then calculating speed of the sun based on the time interval between the sunrise (or sunset) timestamps. For San Diego, California using the manual method, the sun speed calculates to be 864 mph. The calculator speed is 875 mph. The nominal variance between the two methods validates the accuracy of the time stamp as compared to the calculator.

III. Earth Radius by Latitude Calculator
The calculator states: Earth radius at sea level is 6378.137 km (3963.191 mi) at the equator. It is 6356.752 km (3949.903 mi) at the poles and 6371.001 km (3958.756 mi) on average. Regardless of what latitude you enter, the radius will be an answer of between 3,950 miles to 3,963 miles. The calculator represents a ball with a flat top and flat bottom, I kid you not! The variance in radius between the poles to the equator is 13 miles.

In the rotation calculator it does calculate 1040mph at the equator. For the analysis, latitude 89.75° is selected which is 17 miles radius; 17 miles from dead center of the North or South Poles. The speed of rotation at 89.75 latitude is 4.5 mph. Interpolation of the radius appears to be reasonable for estimating Centrifugal Force at each latitude.

Why would the radius calculator be misleading? Perhaps the model had to be forced to account for the Coriolis effect https://www.omnicalculator.com/physics/coriolis-effect and/or the Eötvös effect. These two science theories claim that flight path of aircraft and targeting for bullets and long range artillery has to be adjusted for the so called spin of the earth.

If you stand dead center of the North Pole, there is no centrifugal force which is the same as for the moon. For any location on earth other than dead center of the North and South Poles, to obtain accurate weights and measures, one must consider centrifugal force.

Coriolis effect and airplanes

Do Coriolis effect and airplanes have something in common? Of course they have! Let's say that an airplane (m = 50,000 kg) takes off from London (α = 51.50° N) and travels to North America (to the west) with the velocity v = 500 km/h. With our Coriolis effect calculator we can compute that this airplane is subjected to the Coriolis force F ≈ 800 N which means that it deflects to the north with the acceleration a = F / m = 0.016 m/s². It is almost 0.2% of the Earth's gravity! You can imagine that pilots need to plan a proper flight in advance and fly not to the west but to the southwest to reduce the effect of the Coriolis force.

Dominik Czernia, PhD student
https://www.omnicalculator.com/physi...-and-airplanes

Interval latitudes are innumerable and angles off of the innumerable latitudes are innumerable. Somehow with the theory of gravity the Coriolis effect theory converts equator spin speed of 1040 mph-1,525 ft. per second to cause earth spin to become nominal adjustments for targeting destination points.

1. How does a Cessna 172 at flight speed of 161 mph adjust for earth’s rotation speed of 1040 mph at the equator? Pilots do not adjust for the Coriolis effect. There is no commercial aircraft capable of overcoming speeds of 1040 mph when the aircraft becomes disconnected from earth.
2. How does a hunter-rifleman-sniper adjust for adjust for earth’s rotation speed of 1,525 ft. per second at the equator? GPS systems would be required for any handgun, shotgun or rifle.

Coriolis effect and Eötvös effect are direct contradictions to the theory of relativity and theory of special relativity.

Combinations of theories have become assumed to be fact.

Nikola Tesla explains the contradictions as follows:

“Today's scientists have substituted mathematics for experiments, and they wander off through equation after equation, and eventually build a structure which has no relation to reality.”
#24
11-20-2018, 09:56 AM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
1) The moon does rotate - once every 27 days.

2) Your calculation of the magnitude of the centrifugal force is off by a couple of orders of magnitude. Please show your math.

3) When a person on a plane traveling at 150 miles an hour throws a paper airplane across the aisle, how does the paper airplane magically adjust for the flight speed of the plane?
#25
11-20-2018, 11:49 AM
 Exapno Mapcase Charter Member Join Date: Mar 2002 Location: NY but not NYC Posts: 30,634
Quote:
 Originally Posted by kertmoore At the equator which travels at 1040 mph, there is centrifugal force amounting to 69.1% of the weight of the object. This same person who weighs 200 lbs. at the North and South Pole will weigh 62 lbs. on the equator.
#26
11-20-2018, 01:31 PM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 18,157
kertmoore

You have completely and utterly failed to respond to people pointing out totally incorrect statements in your posts.

I suggest the following:

Step 1: Acknowledge your many, many errors.
Step 2: Acknowledge you don't even have a grade school knowledge of Physics.
Step 3: Acknowledge that people who know more that you ... know more than you.
Step 4: Promise to learn from people who aren't as ignorant as you.
#27
11-21-2018, 12:41 PM
 kertmoore Guest Join Date: Nov 2018 Posts: 16
Quote:
 Originally Posted by Andy L 1) The moon does rotate - once every 27 days. 2) Your calculation of the magnitude of the centrifugal force is off by a couple of orders of magnitude. Please show your math. 3) When a person on a plane traveling at 150 miles an hour throws a paper airplane across the aisle, how does the paper airplane magically adjust for the flight speed of the plane?
1. On line two of my post, I did state the the moon does rotate one time each 27 days.
2. The centrifugal force math is at: https://www.omnicalculator.com/physi...trifugal-force
The data required to complete the centrifugal force calculator is cited in the post.
3. I do not know enough about inertia to respond. Commercial aircraft have INS (inertial navigation systems). Light aircraft do not have INS. INS as to the spin of the earth is a fake, a cover up. Commercial pilots may or may not think they adjust for the spin of the earth. I am a light aircraft pilot. Never did adjust for spin of the earth, never was trained to adjust for spin of the earth. Never operated INS.
#28
11-21-2018, 12:45 PM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by kertmoore 1. On line two of my post, I did state the the moon does rotate one time each 27 days. 2. The centrifugal force math is at: https://www.omnicalculator.com/physi...trifugal-force The data required to complete the centrifugal force calculator is cited in the post. 3. I do not know enough about inertia to respond. Commercial aircraft have INS (inertial navigation systems). Light aircraft do not have INS. INS as to the spin of the earth is a fake, a cover up. Commercial pilots may or may not think they adjust for the spin of the earth. I am a light aircraft pilot. Never did adjust for spin of the earth, never was trained to adjust for spin of the earth. Never operated INS.
You wrote "The moon does not spin."

Using the website you suggest produces an answer of 0.63 pounds, not 63 pounds

My question was about a paper airplane; paper airplanes do not have INS
#29
11-21-2018, 01:00 PM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 18,157
Quote:
 Originally Posted by kertmoore 3. I do not know enough about inertia to respond.
Understatement of the year.

People have found serious flaws in all of your posts. Do you acknowledge your obvious, repeated, mistakes?
#30
11-21-2018, 01:17 PM
 Czarcasm Charter Member Charter Member Join Date: Apr 1999 Location: Portland, OR Posts: 58,488
Quote:
 Originally Posted by kertmoore 1. On line two of my post, I did state the the moon does rotate one time each 27 days.
The moon does not spin-it rotates.
Are you sticking with this?
#31
11-21-2018, 02:15 PM
 kertmoore Guest Join Date: Nov 2018 Posts: 16
Quote:
I know that there is no adjustment required for weight at the pole as opposed to the weight of an object at the equator. However, science says that the earth spins 1040 mph at the equator and zero at the poles. Because the earth does spin according to science, then science has to deal with the effect of centrifugal force created by the spin of the earth at all of the innumerable interval latitudes.

The calculator states: Earth radius at sea level is 3949.903 miles at the poles. The calculator says radius at the poles is 13 miles less than at the equator. The calculator represents a ball with a flat top and flat bottom, I kid you not! The variance in radius between the poles to the equator is 13 miles.

For the equator, entry of 1040 mph and radius of 3,963 miles into the centrifugal force calculator
https://www.omnicalculator.com/physi...trifugal-force is the result of 69.1% force amounting to 138.2 lbs. force for a 200 lb. person (object). At the equator, the 200 lb. person (object) has a measured weight of 61.8 lbs.

The best solution is to interpolate to obtain a realistic radius for other latitudes:
Rotation speed https://www.vcalc.com/wiki/MichaelBa...ed+at+Latitude
each latitude is easily validated by checking sunrise timestamps of two locations on the same latitude:

For the equator, radius is 3,963 miles and speed is 1040 mph which becomes a factor of 3.811 for each 1 mph. For the latitude 89.75°, the speed calculator establishes earth spin at 4.5 mph which is multiplied times the factor of 3.811 establishes the radius to be 17 miles.

For the latitude 89.75, entry of 4.5 mph and radius of 17 miles into the centrifugal force calculator is the result of .3% force amounting to .6 lbs. force for a 200 lb. person (object). At the 89.75 latitude, the 200 lb. person (object) has a measured weight of 199.4 lbs.

Each latitude radius can be interpolated by multiplying the latitude speed times the factor of 3.811 for each 1 mph. Interval latitude results may not be an exact accurate result. However the equator radius is known and the latitude radius that 89.75° latitude is 25% of 1 degree distance from dead center of the poles. The distance between 1 degree of latitude is 69 miles. At 25% of 1 degree, the distance is 17.25 miles. The factor of 3.811 for each 1 mph is a very accurate estimate the radius of 89.75° latitude.
#32
11-21-2018, 02:28 PM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by kertmoore For the equator, entry of 1040 mph and radius of 3,963 miles into the centrifugal force calculator https://www.omnicalculator.com/physi...trifugal-force is the result of 69.1% force amounting to 138.2 lbs. force for a 200 lb. person (object). At the equator, the 200 lb. person (object) has a measured weight of 61.8 lbs.
The centrifugal force calculator reports to me that the centrifugal force at the equator is 0.6331 pounds, meaning that a 200 pound person has a measured weight of 200 pounds minus 0.63 pounds or 199.37 pounds.

I note that if you accidentally put in a tangential velocity of 10,000 miles per hour instead of 1000 miles per hour, you get a centrifugal force of 63 pounds, which would mean that a person weighing 200 pounds normally would weight 137 pounds at the equator.
#33
11-21-2018, 02:35 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 80,133
Quote:
 I know that there is no adjustment required for weight at the pole as opposed to the weight of an object at the equator.
Yet one more of the many incorrect things you "know".
#34
11-21-2018, 04:50 PM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
An interesting fact by the way is that variations of the Earth's gravity from location to location were first detected in 1673, by Jean Richer https://books.google.com/books?id=Jr...ayenne&f=false
#35
11-22-2018, 11:51 AM
 Exapno Mapcase Charter Member Join Date: Mar 2002 Location: NY but not NYC Posts: 30,634
Quote:
 Originally Posted by kertmoore I know that there is no adjustment required for weight at the pole as opposed to the weight of an object at the equator. However, science says that the earth spins 1040 mph at the equator and zero at the poles. Because the earth does spin according to science, then science has to deal with the effect of centrifugal force created by the spin of the earth at all of the innumerable interval latitudes.
Once again, go to Ecuador and step on a scale. Take your own so you know it hasn't been adjusted for local conditions.

#36
11-22-2018, 01:46 PM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by Exapno Mapcase Once again, go to Ecuador and step on a scale. Take your own so you know it hasn't been adjusted for local conditions. When a one-step experiment is enough to disprove your claims, your reasoning has failed badly.
To be fair, I think that's his point. He's trying to disprove conventional science with a reductio ad absurdum argument, showing that conventional science predicts a ridiculous amount of weight reduction in Ecuador and using that to show that conventional science must be wrong.

The immediate problem with this is that he's done his calculations wrong (in two ways - apparently by using the wrong velocity for the rotation of the Earth, and by interpreting the amount of reduction of weight by centrifugal force as the amount of weight after reduction).

The underlying problem is that when he sees a disconnect between theory and observation that would have been obvious almost four centuries ago, he doesn't think "maybe I've misunderstood the theory or made an incorrect calculation" - he thinks rather "I've uncovered something that everyone has ignored!"
#37
11-22-2018, 02:18 PM
 Exapno Mapcase Charter Member Join Date: Mar 2002 Location: NY but not NYC Posts: 30,634
Is that what he's saying? My eyeballs keep spinning around when I try to follow.

Let me try a direct question then.

Does the atmosphere move with the earth at the same local speed of rotation?
#38
11-23-2018, 12:05 AM
 dropzone Member Join Date: May 2000 Location: Cloud Cuckoo Land Posts: 28,874
Oh joy, a flat earther! Santa, this is early. Are you saving an Armenian Genocide truther for Christmas?
#39
11-23-2018, 12:34 AM
 Johnny L.A. Charter Member Join Date: Jan 2000 Location: NoWA Posts: 59,987
Where does the OP get the idea that a Cessna 172 can fly 161 mph?
#40
11-23-2018, 01:16 AM
 Johnny L.A. Charter Member Join Date: Jan 2000 Location: NoWA Posts: 59,987
Quote:
 Originally Posted by Johnny L.A. Where does the OP get the idea that a Cessna 172 can fly 161 mph?
(Actually, I know where he got the number. I just want to hear him say it.)
#41
11-23-2018, 10:04 AM
 kertmoore Guest Join Date: Nov 2018 Posts: 16
Quote:
 Originally Posted by Andy L The centrifugal force calculator reports to me that the centrifugal force at the equator is 0.6331 pounds, meaning that a 200 pound person has a measured weight of 200 pounds minus 0.63 pounds or 199.37 pounds. I note that if you accidentally put in a tangential velocity of 10,000 miles per hour instead of 1000 miles per hour, you get a centrifugal force of 63 pounds, which would mean that a person weighing 200 pounds normally would weight 137 pounds at the equator.
You are correct, at the equator pounds force for a 200 lbs. person (object) is .691 pounds and the measured weight of the 200 lbs. object becomes 199.309 lbs. Calculating weight for a gallon of water which weighs 8.34 lbs. per gallon at the poles has a weight of 8.31118 lbs. at the equator as well as different weight results for each latitude. Standard worldwide weights would indicate the centrifugal force formula is faulty.

However, the only speed in the calculator that generates zero pounds force is a speed of zero. With centrifugal force, I do not know how there can be precise standard worldwide weights of a specific object. Anyone who has ridden in a car making a hard turn knows there is centrifugal force.

A formula problem https://rechneronline.de/earth-radius/ The calculator states: Earth radius at sea level is 3949.903 miles at the poles. The calculator says radius at the poles is 13.29 miles less than at the equator. The calculator represents a ball with a flat top and flat bottom. The variance in radius between the poles to the equator is 13.29 miles. This certainly cannot be true. At the formula's radius for the poles of 3949.9 miles, a spin speed of 1038.1 mph generates .691 pounds force. A reasonable conjecture is the formula is contrived to force out a desired result in another calculation; possibly seeking a desired result for the Coriolis effect as well as the Eötvös effect.

Check out Halley's Comet named after Edmund Halley, who calculated its orbit. He determined that the comets seen in 1531 and 1607 were the same object that followed a 76-year orbit.

The comet's pass in 1910 was about 13.9 million miles and in 1986, 39 million miles away from Earth. https://www.space.com/19878-halleys-comet.html

Earth's orbit speed of the sun is 67,000 mph; the sun is traveling at 45,000 mph. Reasons gravity is a theory:
1. How does earth's gravity act like a retractor beam to pull in Halley's comet that has a 55 million mile orbit?
2. How does the moon overcome Halley's Comet retractor beam power of earth's gravity?
§ Halley's Comet Mass 2.2×1014 kg vs. Moon mass 7.35 x 1022 kg, about 1.2 percent of Earth's mass.
3. How does gravity lock in the moon so that the moon does not spin? Why do we expect that the moon should spin? Why does the moon rotate only one time each 27days? Certainly, the moon is a timepiece.
4. If you say centrifugal force of the moon orbit maintains the earth to moon distance, is there any math formula to prove it? Video Link: The formula would have to take into account distance, speed and mass of the object.
5. Why does gravity of the sun affect earth but does not affect the moon?

Halley's Comet
Orbital period: 76.0 years
Distance: 0.587 AU = 55 million miles
► 1 AU is 93 Million miles, distance of sun to earth
Next perihelion (appearance): 2061
Inserted from http://solarviews.com/eng/halley.htm

Artificial gravity, or rotational gravity, is thus the appearance of a centrifugal force in a rotating frame of reference https://en.wikipedia.org/wiki/Artificial_gravity There is no measurable Coriolis force at the Earth’s equator http://www.tech-faq.com/coriolis-effect.html

Supposedly, Coriolis effect must be accounted for in long range missile and artillery systems. The authors (E. Linacre and B. Geerts) http://www-das.uwyo.edu/~geerts/cwx/...artillery.html indicate that targeting errors were prior to rifling of the cannon bore. Barrel rifling was invented in Augsburg, Germany in 1498. True rifling dates from the mid-16th century, it did not become commonplace until the nineteenth century. Correspondingly, cannon targeting was improved. It appears to me that Gustave Coriolis' 1800's observations regarding cannon targeting became the basis of a formula and formulas causing long range targeting adjustments of only a few inches regardless that the earth supposedly spins ranging from zero to 1,525 ft. per second at the equator.

I am a light aircraft pilot. I never made flight path adjustment for the spin of the earth, never was trained to adjust for the spin of the earth and never piloted an aircraft to overcome speeds of anything more than 161 mph. Realize, a pilot would have to calculate for the innumerable interval latitudes and the innumerable angles off of these innumerable latitudes. Pilots seldom fly due north/south or due east/west. The use of the theory of gravity to calculate that Coriolis effect mitigates the innumerable earth spin speeds for targeting the flight destination point and for precise sniper targeting is unrealistic.

There are “Earth does not move experiments”: The Michelson–Morley experiment was performed over the spring and summer of 1887 by Albert A. Michelson and Edward W. Morley …an attempt to detect the relative motion of matter through the stationary luminiferous aether ("aether wind"). The result was negative…the direction of movement through the presumed aether, and the speed at right angles, was found not to exist; this result…eventually led to special relativity, which rules out a stationary aether. The experiment has been referred to as "the moving-off point for the theoretical aspects of the Second Scientific Revolution". Michelson–Morley type experiments have been repeated many times with steadily increasing sensitivity…recent optical resonator experiments confirmed the absence of any aether wind at the 10−17 level.
https://en.wikipedia.org/wiki/Michel...ley_experiment
#42
11-23-2018, 10:57 AM
 ftg Guest Join Date: Feb 2001 Location: Not the PNW :-( Posts: 18,157
Quote:
 Originally Posted by kertmoore 5. Why does gravity of the sun affect earth but does not affect the moon? Halley's Comet Orbital period: 76.0 years Distance: 0.587 AU = 55 million miles
The Sun's gravity does affect the Moon. Good grief! Yet another amazing display of ignorance. Ever hear of a guy named Newton?

The "distance" you give is perihelion. As per the site you link to. Aphelion is 35AU.

If you don't know and use appropriately basic terminology, maybe you're not the genius you think you are.
#43
11-23-2018, 11:16 AM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by kertmoore You are correct, at the equator pounds force for a 200 lbs. person (object) is .691 pounds and the measured weight of the 200 lbs. object becomes 199.309 lbs.
Thank you.

Quote:
 Originally Posted by kertmoore Calculating weight for a gallon of water which weighs 8.34 lbs. per gallon at the poles has a weight of 8.31118 lbs. at the equator as well as different weight results for each latitude. Standard worldwide weights would indicate the centrifugal force formula is faulty. However, the only speed in the calculator that generates zero pounds force is a speed of zero. With centrifugal force, I do not know how there can be precise standard worldwide weights of a specific object. Anyone who has ridden in a car making a hard turn knows there is centrifugal force.
Perhaps standard weights aren't as precise as you think they are. After all, the maximum variation you've described is 0.34% - and people don't live at the poles, so the observed variation over inhabited areas will be even smaller.

Quote:
 Originally Posted by kertmoore A formula problem https://rechneronline.de/earth-radius/ The calculator states: Earth radius at sea level is 3949.903 miles at the poles. The calculator says radius at the poles is 13.29 miles less than at the equator. The calculator represents a ball with a flat top and flat bottom. The variance in radius between the poles to the equator is 13.29 miles. This certainly cannot be true.
Why can't it be true? It was measured back in the 1700s.

Quote:
 Originally Posted by kertmoore At the formula's radius for the poles of 3949.9 miles, a spin speed of 1038.1 mph generates .691 pounds force. A reasonable conjecture is the formula is contrived to force out a desired result in another calculation; possibly seeking a desired result for the Coriolis effect as well as the Eötvös effect.
What makes this reasonable?

Quote:
 Originally Posted by kertmoore Check out Halley's Comet named after Edmund Halley, who calculated its orbit. He determined that the comets seen in 1531 and 1607 were the same object that followed a 76-year orbit. The comet's pass in 1910 was about 13.9 million miles and in 1986, 39 million miles away from Earth. https://www.space.com/19878-halleys-comet.html Earth's orbit speed of the sun is 67,000 mph; the sun is traveling at 45,000 mph. Reasons gravity is a theory: 1. How does earth's gravity act like a retractor beam to pull in Halley's comet that has a 55 million mile orbit? 2. How does the moon overcome Halley's Comet retractor beam power of earth's gravity? § Halley's Comet Mass 2.2×1014 kg vs. Moon mass 7.35 x 1022 kg, about 1.2 percent of Earth's mass.
Halley's comet orbits the Sun, not the Earth.

Quote:
 Originally Posted by kertmoore 3. How does gravity lock in the moon so that the moon does not spin? Why do we expect that the moon should spin? Why does the moon rotate only one time each 27days? Certainly, the moon is a timepiece.
The moon does spin - it rotates once every 27 days.

Quote:
 Originally Posted by kertmoore 4. If you say centrifugal force of the moon orbit maintains the earth to moon distance, is there any math formula to prove it? Video Link: The formula would have to take into account distance, speed and mass of the object. https://www.youtube.com/watch?v=fnVL...ature=youtu.be
V*V/R= GMm/R - this was the first significant calculation for gravity, done in the 1600s by Newton.

Quote:
 Originally Posted by kertmoore 5. Why does gravity of the sun affect earth but does not affect the moon? Halley's Comet Orbital period: 76.0 years Distance: 0.587 AU = 55 million miles ► 1 AU is 93 Million miles, distance of sun to earth Next perihelion (appearance): 2061 Inserted from http://solarviews.com/eng/halley.htm
What does Halley's comet's orbit around the Sun have to do with the Earth?
The Sun's gravity does affect the Moon.

Quote:
 Originally Posted by kertmoore Artificial gravity, or rotational gravity, is thus the appearance of a centrifugal force in a rotating frame of reference https://en.wikipedia.org/wiki/Artificial_gravity There is no measurable Coriolis force at the Earth’s equator http://www.tech-faq.com/coriolis-effect.html
The Coriolis force is not the same thing as the centrifugal force.

Quote:
 Originally Posted by kertmoore Supposedly, Coriolis effect must be accounted for in long range missile and artillery systems. The authors (E. Linacre and B. Geerts) http://www-das.uwyo.edu/~geerts/cwx/...artillery.html indicate that targeting errors were prior to rifling of the cannon bore. Barrel rifling was invented in Augsburg, Germany in 1498. True rifling dates from the mid-16th century, it did not become commonplace until the nineteenth century. Correspondingly, cannon targeting was improved. It appears to me that Gustave Coriolis' 1800's observations regarding cannon targeting became the basis of a formula and formulas causing long range targeting adjustments of only a few inches regardless that the earth supposedly spins ranging from zero to 1,525 ft. per second at the equator.
Cannons do not fire from the poles to the equator - they fire a few miles. The variation in the speed of the Earth's motion over a few miles is (unsurprisingly) much smaller than a range of 0 to 1525.

In the future, could you pick one misunderstanding at a time? We'd make better progress that way, I think.
#44
11-23-2018, 12:05 PM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by kertmoore I am a light aircraft pilot. I never made flight path adjustment for the spin of the earth, never was trained to adjust for the spin of the earth and never piloted an aircraft to overcome speeds of anything more than 161 mph. Realize, a pilot would have to calculate for the innumerable interval latitudes and the innumerable angles off of these innumerable latitudes. Pilots seldom fly due north/south or due east/west.
Your aircraft starts out with the same speed as the Earth. For simplicity's sake, let's take that as the equator. So in inertial space, your craft is moving at 1526 feet per second while sitting on the runway. Once you take off, your plane retains that initial velocity in inertial space in addition to whatever velocity the plane has through the air.

Let's suppose you're traveling at 150 miles per hour, due north. In an hour, you've traveled north 2.5 degrees, so at that point, you're at 2.5 degrees north. At that latitude the Earth's rotation is 1524.5 feet per second. Not a big difference. Would you notice 1.5 foot per second motion to the east? Of course not. You'd unconsciously adjust for it, just like while walking you compensate for the slight difference in the length of your legs (typical leg length discrepancy is on the order of a centimeter).
#45
11-23-2018, 12:11 PM
 Johnny L.A. Charter Member Join Date: Jan 2000 Location: NoWA Posts: 59,987
Quote:
 Originally Posted by kertmoore I am a light aircraft pilot... and never piloted an aircraft to overcome speeds of anything more than 161 mph.
Have you ever flown a Cessna 172?
#46
11-25-2018, 08:45 AM
 Andy L Member Join Date: Oct 2000 Posts: 5,694
Quote:
 Originally Posted by Andy L Thank you. Perhaps standard weights aren't as precise as you think they are. After all, the maximum variation you've described is 0.34% - and people don't live at the poles, so the observed variation over inhabited areas will be even smaller.
I should add that people who care about precise measurement of masses will likely use balance scales which will produce the correct value at any location (because they compare masses to standard masses).
#47
11-25-2018, 10:19 AM
 kertmoore Guest Join Date: Nov 2018 Posts: 16
Quote:
 Originally Posted by Johnny L.A. Have you ever flown a Cessna 172?
Yes, I am a licensed pilot qualified VFR (Visual Flight Rules). I am not qualified IFR (Instrument Flight Rules) for low visibility environment such as fog. Most regulated aircraft have flight instruments that are grouped according to pitot-static system (air speed), compass systems, and gyroscopic instruments (level flight, turns and orientation to the horizon). My experience never included INS (Inertial Navigation System). In my time period of flight, the 1970's, I do not remember INS, I do not think I ever heard the term INS.

IFR requires total reliance on the instruments. Training for IFR is "under the hood"; the pilot wears an extended type visor and the pilot focuses only on the instrument panel.

Pilots do not consider the spinning earth in flight planning. I discovered Coriolis effect about three years ago. Possibly I heard the term, but I did not know its meaning or application.

https://www.omnicalculator.com/physics/coriolis-effect Enter 2500 lb., 161 mph and 32 latitude. A force of 6.289 is the result. I do not know how to apply the result to a flight plan. I never experienced noticeable time difference between round trip due north/due south legs of a flight. My flight times were usually 90 minutes.

Flight planning is drawing a straight line on a flat map from departure to destination. Never was there a lead distance calculated to lead the moving destination point. At latitude 36°, about 300 miles, the force becomes 6.975. I do not know how to apply the force of 6.975 to determine a compass setting to account for the destination point moving a certain distance within 90 minutes. Supposedly, for round trip due north/due south legs of a flight, the aircraft is operating under multiple inertial and gravitational forces for the entire flight; and multiple earth spin speeds as well as multiple centrifugal forces. For a due west/due east flight on the same latitude, the same Coriolis force is calculated. Gravity is theory. Spin speed of the earth 32° latitude is 882 mph and 36° latitude is 842 mph. https://www.vcalc.com/wiki/MichaelBa...ed+at+Latitude For a 90 minute south to north flight, the destination point 36° latitude has traveled 842 miles and for Coriolis effect, it appears the theory of gravity is used to determine a compass setting. For VFR flight planning, latitude and longitude were nominal considerations. Flight planning was distance, weather, elevations and straight line point to point direction.

Supposedly, INS performs all calculations for Coriolis effect. It appears INS has other calculations for other purposes, but the idea of calculating for a spinning earth is fake. World War I vintage aircraft would not have had INS equipment.

Sources for researching INS specifications and application of the equipment is very limited. Below is INS information I found:

Honeywell: Air Data Inertial Reference System (ADIRS)
https://aerospace.honeywell.com/en/p...ference-system
https://aerospace.honeywell.com/en/s...alignment-tool

Coriolis effect illustration (use Internet Explorer)
http://www.classzone.com/books/earth...1904page01.cfm

http://www.nfl.com/news/story/0ap300...lped-cincy-win
The @Bengals stadium isn’t oriented exactly North-South. And the field goal was 42-yds. Yielding a 1/3-in deflection, not 1/2

The Coriolis effect contradicts the idea that the "earth and its atmosphere" all move together.
#48
11-25-2018, 10:36 AM
 Johnny L.A. Charter Member Join Date: Jan 2000 Location: NoWA Posts: 59,987
Quote:
 Originally Posted by kertmoore Yes, I am a licensed pilot qualified VFR (Visual Flight Rules).
The question was: Where did you get the idea that a Cessna 172 flies at 161 mph?
#49
11-26-2018, 11:16 AM
 kertmoore Guest Join Date: Nov 2018 Posts: 16
I looked up the speed. If the speed is different, it changes the formula results. Only an aircraft that has speed in excess of 1040 could adjust for conditions where the earth spins at 1040 mph at the equator. Science says pilots do have to adjust for the spinning earth. The fact is, pilots do not have to adjust flight path for a spinning earth.

I do not know how to apply the formula result to planning a flight course:
https://www.omnicalculator.com/physi...-and-airplanes

Here is the illustration:
http://www.classzone.com/books/earth...1904page01.cfm

Use Internet Explorer to see the illustration.
#50
11-26-2018, 11:26 AM
 Johnny L.A. Charter Member Join Date: Jan 2000 Location: NoWA Posts: 59,987
Quote:
 Originally Posted by kertmoore I looked up the speed.
1. Where did you look up the speed?

2. How do you know the number is correct?

3. As someone who has flown Cessna 172s, how often have you flown them at 161 mph?

(The rest of your post is irrelevant.)

.

Last edited by Johnny L.A.; 11-26-2018 at 11:27 AM.

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