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The opening section of this article has been copied from the article that is refererenced I have corresponded with Anders Persson via email, and I have his permission to use text of him that is publicly accessable for Wikipedia. -- Cleonis | Talk 22:27, 11 January 2006 (UTC)
The standard approach as I learned, is not to use centripetal force, but centripetal acceleration. The weight depends on the difference g-a. Harald88 22:54, 11 January 2006 (UTC)
[copied from personal page:]
[Cleaonis wrote:] The poleward force, that is introduced in the Eötvös effect article, is obvious if the thinking is anchored on inertial space. Some people have argued to me that the resultant poleward force cannot actually exist. For, they said, if it would exist then all oceans would flow to the poles. It's tricky.
That is what is stated in the article: the atmospheric layer of the earth is equally thick everywhere. The situation is comparable to the layer of mercury on the dish of a mercury mirror as is used in astronomy. (Currently the biggest mercury mirror has a diameter of 6 meters.) The shape of the dish is parabolic. The closer you get to the rim, the stronger the force on the mercury to flow to the center. When the dish is rotating at the designed angular velocity, the mercury redistributes itself into an even layer.--Cleonis | Talk 18:59, 15 January 2006 (UTC)
There is an equilibrium at play. This equilibrium is not an equilibrium of forces, for in motion along a circular trajectory there is an unbalanced force at work: the centripetal force.
The rotation is around the Earth's axis, of course, so strictly speaking the blue arrow should have been drawn perpendicular to the Earth's axis, pointing towards the Earth's axis. But in general, most of the resulting motion is parallel to the surface. When a zeppelin is moving in east-to-west direction then there is a surplus of centripetal force, and this surplus of centripetal force will pull the zeppelin towards the nearest pole.
If I would have drawn the blue arrow in the correct direction, I would have had to decompose it into a component perpendicular to the local surface, and a component parallel to the local surface.
Generally I try to make pictures as uncluttered as possible. Generally, I try to use as few arrows as possible. Here, the blue arrow represents that a centripetal force is present and working. -- Cleonis | Talk 23:54, 15 January 2006 (UTC)
Historically, Eötvös noticed the effect in relation to ships with a small velocity relative to the Earth. And only a supersonic aircraft can go fast enough to be at rest with respect to the inertial frame. The situations we are interested in are the situations with just a small velocity relative to the Earth -- Cleonis | Talk 00:44, 16 January 2006 (UTC)
There is the large Zenith telescope, currently the largest murcury mirror telescope, with a diameter of 6 meters.
Assuming the rotationn rate of the mirror is one revolution every 10 seconds, then the incline of the parabolic dish at the rim is about 7 degrees away from the horizontal. The rim is then a couple of decimeters higher than the center.
Do you agree with the following statements:
Due to the incline of the surface, there is at all points a centripetal force.
If the dish would slow down, that centripetal force would cause the mercury to flow to the center.
The centripetal force that is exerted on the mercury is unbalanced. The mercury exerts a reaction force on the dish, but that does not affect the motion of the mercury. --
Cleonis |
Talk 00:09, 16 January 2006 (UTC)
An even simpler case than the mercury mirror is two balls connected by a spring. The common center of mass of the two balls is in free-motion, let's say the circling balls are onboard a space-station. The motion of the two balls can be accounted for with a single force: a centripetal force that is exerted by the spring on the balls. -- Cleonis | Talk 13:45, 16 January 2006 (UTC)
I copy and paste from above:
I'm pretty sure by now that we actually don't disagree. I think any apparent dissent is babylonian confusion.
In the case of motion over the Equator, parallel to the equator, you were correct in pointing out that when a zeppelin is cruising at 490 m/s with respect to inertial space, instead of 465 m/s, that it is the centripetal acceleration that has changed.
The most refined gravito-meters are a tube that is pumped vacuum, and then a test mass is dropped, and the acceleration of the test mass is measured.
At the equator the gravitational acceleration g is 9.780 m/sec^2
The centripetal acceleration that corresponds to a velocity of 465 m/s with respect to inertial space and a radius of 6370 km is 0.034 m/sec^2
So the true gravitation at the equator is 9,780 + 0.034 = 9.814 m/sec^2
And what a gravito-meter measures is 9.780 m/sec^2
Naturally it is the measured acceleration that is the most useful for any science and engineering.
When a zeppelin moves at 490 m/s with respect to inertial space then the corresponding centripetal acceleration is 0.038 m/sec^2.
Hence a gravitometer onboard a zeppelin cruising at 490 m/s will measure a gravitational acceleration of 9,814 - 0.038 = 9,776 m/sec^2
The mathematical derivation section has been brought into accordance with that. I will try to amend the 'explanation of the cosine' section. -- Cleonis | Talk 23:58, 16 January 2006 (UTC)
I copy and paste from above:
Here are some other examples that I have used, and stil use, to visualize the physics taking place.
A velodrome is a racetrack specifically for bicycle races. A velodrome with a track length of 250 metres usually has straights of about 60 metres long, and U-turns with an inner radius of about 20 metres, and about 65 metres in length. The banking of the turns is 42 degrees.
What would you see if someone would try to race with a little one-seater hovercraft on a velodrome track? Then you can pretty much calculate the lap times in advance, for there is only one velocity that will allow the hovercraft to negotiate the 42 degrees banked turns. Too fast and he swings wide, too slow and he'll scrape the inside
And what about the Arecibo dish? Suppose that parabolic dish is very sturdy, and some people would organize hovercraft races that go round and round on the dish. The Arecibo dish has a diameter of 300 meters. Suppose a hovercraft is going round on that dish. Assuming that the rim is 30 meter higher than the center then it takes a hovercraft about 40 seconds to complete a full lap.
What happens when a racer opens up the throttle? Well, then he will swing wide, but his angular velocity won't change. On a parabolic dish, when you increase your velocity, you automatically climb to a higher trajectory, and a higher trajectory is a longer trajectory. All trajectories on a parabolic dish take the same amount of time to complete a full rotation. That is a property of harmonic oscillation For a given harmonic oscillator, the period of one oscillation is independent of the amplitude. -- Cleonis | Talk 00:04, 17 January 2006 (UTC)
Let there be a mercury mirror dish, with a layer of mercury. A small gravito-meter is placed on the rotating mercury. (The gravito-meter is a vacuum tube, in which a test mass is dropped.) The small gravito-meter is floating on the mercury.
As the the gravito-meter is co-rotating with the dish and mercury, it will measure an acceleration perpendicular to the local mercury surface.
Let the dish slow down. Then the gravito-meter will slow down together with the mercury. The mercury will sag to the middle of the dish. In latitudinal direction, the gravito-meter will measure a deceleration. (If the test mass does not fall exactly parallel to the tube, then a sideways acceleration/deceleration is inferred.) In longitudinal direction the measured acceleration will remain perpendicular to the top surface of the mercury. -- Cleonis | Talk 09:08, 17 January 2006 (UTC)
This article currently reads like an essay, not an encyclopedia article. I propose that the text be considerably shortened, removing all the numbers relating to velocities and masses of example vessels. In addition, the mathematical deduction is unnecessary. In my opinion, it is sufficient to link to the articles on Coriolis effect and centrifugal force respectively. -- PeR 15:22, 9 November 2006 (UTC)
Clarified wording regarding the formula for acceleration along the equator derived from the more general formula - GH August 10, 2020 Ghnatiuk ( talk) 07:09, 10 August 2020 (UTC) — Preceding unsigned comment added by Ghnatiuk ( talk • contribs) 06:51, 10 August 2020 (UTC)
The statement that the second term of the correction formula for the Eötvös effect corresponds to the centrifugal term of the equation of motion for a rotating coordinate system is incorrect.
The centrifugal term is proportional to the square of the angular velocity of the rotating coordinate system with respect to the inertial coordinate system. On the other hand, the second term in the Eötvös correction formula is proportional to the squared velocity of the object with respect to the Earth.
Example: an object located at the equator that is stationary with respect to the Earth is circumnavigating the Earth's axis with a tangential velocity of about 465 meters per second. In the case of the equation of motion with respect to a coordinate system that is co-moving with the Earth, the centrifugal term is proportional to the square of 465.
In the examples given in the article, with an airship with cruising velocity of 25 m/s, it is clear that the value of 25 m/s does not correspond to the centrifugal term. -- Cleonis | Talk 15:51, 9 November 2006 (UTC)
The linking of the Eötvös effect with the Coriolis force is fundamentally wrong. The Eötvös effect is only an effect of change of the local centrifugal acceleration (inertia = opposition to change of direction), because the local rotation speed of a ship (sum of Earth's rotation speed and speed of a ship vary eastwards or westwards) vary. Gyrogravitation ( talk) 08:50, 18 May 2010 (UTC)
The new diagram on the right contains a duality : "Coriolis effect" is written twice, whereas the vector sum should be "resultant vector" or so. Gyrogravitation ( talk) 08:41, 18 May 2010 (UTC)
For meteorology, the factor that matters is the component that acts parallel to the local surface. Hence the Coriolis effect as taken into account in meteorology is described with a sine law. At the equator, the component parallel to the local surface is zero, hence at the equator meteorology does not take any Coriolis effect into account. (Advanced atmospheric models do need to take the Eötvös effect into account of course.)
To my knowledge, there is no standard name for the vector sum of the Coriolis effect and the Eötvös effect, which is why that arrow remains unnamed in the diagram. It would be convenient if there would be such a standard, but unfortunately there isn't. -- Cleonis | Talk 16:09, 9 November 2006 (UTC)
I copy and paste from above:
What is stated in the opening paragraph is that Eötvös noticed peculiarities in the gravimetric readings. Usually, a gravimeter that is taken along on journeys operates with a spring with a weight underneath. Stronger Earth gravity causes longer elongation of the spring. The spring provides the required amount of normal force. Depending on the amount of velocity with respect to the Earth, a different amount of normal force is required, and that is what is registered by the gravimeters. It would be very cumbersome to think of the Eötvös effect in terms of (apparently) altered gravity.
I deviated from Anders Persson's article to make the wikipedia article compliant with modern standards of physics education. Invoking centrifugal force as explanation is recommended against. -- Cleonis | Talk 14:08, 12 November 2006 (UTC)
I am no physicist but I find most articles concerning newtonian physics especially with regards to circular motion and gravity oddly contorted , It seems "invoking" a centrifugal force is now a cardinal sin which requires ungodly (in the newtonian sense of the word ) rhetorical contraptions to somehow retrieve a centripetal force from an homo-directional gravitational force as if modern physics had made everyone even less comfortable with the concept of a fictitious force . I'd contend that dear Issac considered both the centrifugal and gravitational forces as fictitious and would be rather flabbergasted that in light of Einstein's contributions such semantical confusion should still prosper . Such strenuous efforts seem also to have for consequence poorly worded assertions such as "the weight of a given object is reduced by x % on the equator " relative to what absolute weight exactly? (I'd haveto guess that would be relative to "polar" weight with zero "fictitious" centrifugal force ) but given all the additional parameters at play here such as the varying r value (oblate spheroid ) and the irregular mass distribution which might be to a point be reciprocally cancelling I find those physics rather confusing and not quite up to "modern" standards . 88.190.24.84 ( talk) 13:33, 7 July 2019 (UTC)
According to the text in the section, by increasing the eastward speed the downward force should increase. However the graph shows the opposite of this, with greater eastward speed the downward force decreases. Something doesn't add up. — Preceding unsigned comment added by 165.225.200.212 ( talk) 11:32, 11 August 2020 (UTC)
![]() | This article is rated B-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||
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The opening section of this article has been copied from the article that is refererenced I have corresponded with Anders Persson via email, and I have his permission to use text of him that is publicly accessable for Wikipedia. -- Cleonis | Talk 22:27, 11 January 2006 (UTC)
The standard approach as I learned, is not to use centripetal force, but centripetal acceleration. The weight depends on the difference g-a. Harald88 22:54, 11 January 2006 (UTC)
[copied from personal page:]
[Cleaonis wrote:] The poleward force, that is introduced in the Eötvös effect article, is obvious if the thinking is anchored on inertial space. Some people have argued to me that the resultant poleward force cannot actually exist. For, they said, if it would exist then all oceans would flow to the poles. It's tricky.
That is what is stated in the article: the atmospheric layer of the earth is equally thick everywhere. The situation is comparable to the layer of mercury on the dish of a mercury mirror as is used in astronomy. (Currently the biggest mercury mirror has a diameter of 6 meters.) The shape of the dish is parabolic. The closer you get to the rim, the stronger the force on the mercury to flow to the center. When the dish is rotating at the designed angular velocity, the mercury redistributes itself into an even layer.--Cleonis | Talk 18:59, 15 January 2006 (UTC)
There is an equilibrium at play. This equilibrium is not an equilibrium of forces, for in motion along a circular trajectory there is an unbalanced force at work: the centripetal force.
The rotation is around the Earth's axis, of course, so strictly speaking the blue arrow should have been drawn perpendicular to the Earth's axis, pointing towards the Earth's axis. But in general, most of the resulting motion is parallel to the surface. When a zeppelin is moving in east-to-west direction then there is a surplus of centripetal force, and this surplus of centripetal force will pull the zeppelin towards the nearest pole.
If I would have drawn the blue arrow in the correct direction, I would have had to decompose it into a component perpendicular to the local surface, and a component parallel to the local surface.
Generally I try to make pictures as uncluttered as possible. Generally, I try to use as few arrows as possible. Here, the blue arrow represents that a centripetal force is present and working. -- Cleonis | Talk 23:54, 15 January 2006 (UTC)
Historically, Eötvös noticed the effect in relation to ships with a small velocity relative to the Earth. And only a supersonic aircraft can go fast enough to be at rest with respect to the inertial frame. The situations we are interested in are the situations with just a small velocity relative to the Earth -- Cleonis | Talk 00:44, 16 January 2006 (UTC)
There is the large Zenith telescope, currently the largest murcury mirror telescope, with a diameter of 6 meters.
Assuming the rotationn rate of the mirror is one revolution every 10 seconds, then the incline of the parabolic dish at the rim is about 7 degrees away from the horizontal. The rim is then a couple of decimeters higher than the center.
Do you agree with the following statements:
Due to the incline of the surface, there is at all points a centripetal force.
If the dish would slow down, that centripetal force would cause the mercury to flow to the center.
The centripetal force that is exerted on the mercury is unbalanced. The mercury exerts a reaction force on the dish, but that does not affect the motion of the mercury. --
Cleonis |
Talk 00:09, 16 January 2006 (UTC)
An even simpler case than the mercury mirror is two balls connected by a spring. The common center of mass of the two balls is in free-motion, let's say the circling balls are onboard a space-station. The motion of the two balls can be accounted for with a single force: a centripetal force that is exerted by the spring on the balls. -- Cleonis | Talk 13:45, 16 January 2006 (UTC)
I copy and paste from above:
I'm pretty sure by now that we actually don't disagree. I think any apparent dissent is babylonian confusion.
In the case of motion over the Equator, parallel to the equator, you were correct in pointing out that when a zeppelin is cruising at 490 m/s with respect to inertial space, instead of 465 m/s, that it is the centripetal acceleration that has changed.
The most refined gravito-meters are a tube that is pumped vacuum, and then a test mass is dropped, and the acceleration of the test mass is measured.
At the equator the gravitational acceleration g is 9.780 m/sec^2
The centripetal acceleration that corresponds to a velocity of 465 m/s with respect to inertial space and a radius of 6370 km is 0.034 m/sec^2
So the true gravitation at the equator is 9,780 + 0.034 = 9.814 m/sec^2
And what a gravito-meter measures is 9.780 m/sec^2
Naturally it is the measured acceleration that is the most useful for any science and engineering.
When a zeppelin moves at 490 m/s with respect to inertial space then the corresponding centripetal acceleration is 0.038 m/sec^2.
Hence a gravitometer onboard a zeppelin cruising at 490 m/s will measure a gravitational acceleration of 9,814 - 0.038 = 9,776 m/sec^2
The mathematical derivation section has been brought into accordance with that. I will try to amend the 'explanation of the cosine' section. -- Cleonis | Talk 23:58, 16 January 2006 (UTC)
I copy and paste from above:
Here are some other examples that I have used, and stil use, to visualize the physics taking place.
A velodrome is a racetrack specifically for bicycle races. A velodrome with a track length of 250 metres usually has straights of about 60 metres long, and U-turns with an inner radius of about 20 metres, and about 65 metres in length. The banking of the turns is 42 degrees.
What would you see if someone would try to race with a little one-seater hovercraft on a velodrome track? Then you can pretty much calculate the lap times in advance, for there is only one velocity that will allow the hovercraft to negotiate the 42 degrees banked turns. Too fast and he swings wide, too slow and he'll scrape the inside
And what about the Arecibo dish? Suppose that parabolic dish is very sturdy, and some people would organize hovercraft races that go round and round on the dish. The Arecibo dish has a diameter of 300 meters. Suppose a hovercraft is going round on that dish. Assuming that the rim is 30 meter higher than the center then it takes a hovercraft about 40 seconds to complete a full lap.
What happens when a racer opens up the throttle? Well, then he will swing wide, but his angular velocity won't change. On a parabolic dish, when you increase your velocity, you automatically climb to a higher trajectory, and a higher trajectory is a longer trajectory. All trajectories on a parabolic dish take the same amount of time to complete a full rotation. That is a property of harmonic oscillation For a given harmonic oscillator, the period of one oscillation is independent of the amplitude. -- Cleonis | Talk 00:04, 17 January 2006 (UTC)
Let there be a mercury mirror dish, with a layer of mercury. A small gravito-meter is placed on the rotating mercury. (The gravito-meter is a vacuum tube, in which a test mass is dropped.) The small gravito-meter is floating on the mercury.
As the the gravito-meter is co-rotating with the dish and mercury, it will measure an acceleration perpendicular to the local mercury surface.
Let the dish slow down. Then the gravito-meter will slow down together with the mercury. The mercury will sag to the middle of the dish. In latitudinal direction, the gravito-meter will measure a deceleration. (If the test mass does not fall exactly parallel to the tube, then a sideways acceleration/deceleration is inferred.) In longitudinal direction the measured acceleration will remain perpendicular to the top surface of the mercury. -- Cleonis | Talk 09:08, 17 January 2006 (UTC)
This article currently reads like an essay, not an encyclopedia article. I propose that the text be considerably shortened, removing all the numbers relating to velocities and masses of example vessels. In addition, the mathematical deduction is unnecessary. In my opinion, it is sufficient to link to the articles on Coriolis effect and centrifugal force respectively. -- PeR 15:22, 9 November 2006 (UTC)
Clarified wording regarding the formula for acceleration along the equator derived from the more general formula - GH August 10, 2020 Ghnatiuk ( talk) 07:09, 10 August 2020 (UTC) — Preceding unsigned comment added by Ghnatiuk ( talk • contribs) 06:51, 10 August 2020 (UTC)
The statement that the second term of the correction formula for the Eötvös effect corresponds to the centrifugal term of the equation of motion for a rotating coordinate system is incorrect.
The centrifugal term is proportional to the square of the angular velocity of the rotating coordinate system with respect to the inertial coordinate system. On the other hand, the second term in the Eötvös correction formula is proportional to the squared velocity of the object with respect to the Earth.
Example: an object located at the equator that is stationary with respect to the Earth is circumnavigating the Earth's axis with a tangential velocity of about 465 meters per second. In the case of the equation of motion with respect to a coordinate system that is co-moving with the Earth, the centrifugal term is proportional to the square of 465.
In the examples given in the article, with an airship with cruising velocity of 25 m/s, it is clear that the value of 25 m/s does not correspond to the centrifugal term. -- Cleonis | Talk 15:51, 9 November 2006 (UTC)
The linking of the Eötvös effect with the Coriolis force is fundamentally wrong. The Eötvös effect is only an effect of change of the local centrifugal acceleration (inertia = opposition to change of direction), because the local rotation speed of a ship (sum of Earth's rotation speed and speed of a ship vary eastwards or westwards) vary. Gyrogravitation ( talk) 08:50, 18 May 2010 (UTC)
The new diagram on the right contains a duality : "Coriolis effect" is written twice, whereas the vector sum should be "resultant vector" or so. Gyrogravitation ( talk) 08:41, 18 May 2010 (UTC)
For meteorology, the factor that matters is the component that acts parallel to the local surface. Hence the Coriolis effect as taken into account in meteorology is described with a sine law. At the equator, the component parallel to the local surface is zero, hence at the equator meteorology does not take any Coriolis effect into account. (Advanced atmospheric models do need to take the Eötvös effect into account of course.)
To my knowledge, there is no standard name for the vector sum of the Coriolis effect and the Eötvös effect, which is why that arrow remains unnamed in the diagram. It would be convenient if there would be such a standard, but unfortunately there isn't. -- Cleonis | Talk 16:09, 9 November 2006 (UTC)
I copy and paste from above:
What is stated in the opening paragraph is that Eötvös noticed peculiarities in the gravimetric readings. Usually, a gravimeter that is taken along on journeys operates with a spring with a weight underneath. Stronger Earth gravity causes longer elongation of the spring. The spring provides the required amount of normal force. Depending on the amount of velocity with respect to the Earth, a different amount of normal force is required, and that is what is registered by the gravimeters. It would be very cumbersome to think of the Eötvös effect in terms of (apparently) altered gravity.
I deviated from Anders Persson's article to make the wikipedia article compliant with modern standards of physics education. Invoking centrifugal force as explanation is recommended against. -- Cleonis | Talk 14:08, 12 November 2006 (UTC)
I am no physicist but I find most articles concerning newtonian physics especially with regards to circular motion and gravity oddly contorted , It seems "invoking" a centrifugal force is now a cardinal sin which requires ungodly (in the newtonian sense of the word ) rhetorical contraptions to somehow retrieve a centripetal force from an homo-directional gravitational force as if modern physics had made everyone even less comfortable with the concept of a fictitious force . I'd contend that dear Issac considered both the centrifugal and gravitational forces as fictitious and would be rather flabbergasted that in light of Einstein's contributions such semantical confusion should still prosper . Such strenuous efforts seem also to have for consequence poorly worded assertions such as "the weight of a given object is reduced by x % on the equator " relative to what absolute weight exactly? (I'd haveto guess that would be relative to "polar" weight with zero "fictitious" centrifugal force ) but given all the additional parameters at play here such as the varying r value (oblate spheroid ) and the irregular mass distribution which might be to a point be reciprocally cancelling I find those physics rather confusing and not quite up to "modern" standards . 88.190.24.84 ( talk) 13:33, 7 July 2019 (UTC)
According to the text in the section, by increasing the eastward speed the downward force should increase. However the graph shows the opposite of this, with greater eastward speed the downward force decreases. Something doesn't add up. — Preceding unsigned comment added by 165.225.200.212 ( talk) 11:32, 11 August 2020 (UTC)