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Re: Coriolis and gyros (second attempt)(typos corrected)
From: Peter Hakel
Date: 2009 Aug 24, 20:29 -0700
From: Gary LaPook <glapook---.net>
To: navlist@fer3.com
Sent: Monday, August 24, 2009 10:29:08 AM
Subject: [NavList 9614] Re: Coriolis and gyros (second attempt)(typos corrected)
There is a problem with your analysis.
Since the period of the pendulum is fixed by its length then it is
necessarily true that as the amplitude of the swing diminishes during
the day, due to air resistance, that its maximum velocity and its
average velocity also slows down. Try this yourself with any pendulum
and observe how slowly it moves near the end as it slows to a stop. The
formula for coriolis force includes a term for velocity across the
turning reference frame so that coriolis force and coriolis acceleration
is proportional to this velocity over the ground. So if your analysis is
correct then the rate of change of the azimuth of the pendulum's swing
should vary throughout the day, changing at a more rapid rate in the
morning and more slowly later in the day as the pendulum slows down. But
the rate of change of azimuth is constant, 11.32 degrees per hour which
disproves your analysis.
Now looking at the case of the gyroscope's undergoing earth rate
apparent precession. If this apparent precession is caused by coriolis
due to the speed of the rotating flywheel moving in one direction at the
point at the bottom and in the opposite direction at the top, (as you
claim) then, without doing the diagram of the precessional forces (I
leave that to you), it must also be true that gyroscopes spinning in
opposite directions would also precess in opposite directions. One
rotating clockwise would precess toward the east and one turning
counterclockwise would precess toward the west yet all gyroscopes, no
matter which way they are spinning, precess toward the east, again
disproving your analysis.
A further disproof of your analysis of earth rate precession of a
gyroscope is due to that pesky term in the coriolis formula that makes
the coriolis force proportional to velocity over the ground. If your
analysis were correct then slowly turning gyroscopes would precess more
slowly than rapidly spinning gyros yet they all show the same apparent
precession due to the earth rate of 15.04 degrees per hour times the
sine of the latitude.
You have not addressed the movement of a gyrocompass at the equator
where coriolis is zero since the sine of zero degrees of latitude is
also zero, there's that pesky coriolis formula again.
Also consider the earth rate apparent precession of a gyroscope at the
north pole which changes at a rate of 15.04 degrees per hour since the
sine of 90 degrees of latitude is 1. Yet a gyroscope located exactly at
the pole is not moving at all (ignoring the earth's movement around its
orbit) but is remaining at one fixed spot on the earth so it has no
velocity across the ground so, again by the formula, there should be no
coriolis available to cause the apparent precession.
gl
frankreed{at}HistoricalAtlas.com wrote:
> "Coriolis is not involved since the Pantheon has not moved across the surface of the earth since its foundations were laid in 1758 so its ground speed is zero."
>
> LOL. Gary the pendulum bob is itself in motion, right? Its ground speed is most certainly NOT zero. Likewise in a gyro-compass, its individuaL mass elements are moving at very high speed. If they weren't moving, it wouldn't work. The Coriolis acceleration causes the precession in both cases --in the rotating frame of reference in which the Earth is fixed.
>
> -FER
>
>
>
> >
>
>
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From: Peter Hakel
Date: 2009 Aug 24, 20:29 -0700
Gary,
First, let's agree that the pendulum has a "small" amplitude (say within 5 degrees), so that the idealized pendulum model is adequate. Then its natural oscillations have a single frequency which is independent of the amplitude. The maximum amplitude, velocity, and acceleration of the bob are all directly proportional to one another. So, when you double the amplitude scale (within those 5 degrees...), you also double the velocity scale.
Now add a small perturbation in the form of Earth's rotation underneath the pendulum. As you correctly point out, the Coriolis force/acceleration is linearly proportional to velocity. This small correction has to be added to the unperturbed velocity. The outcome of this addition is the Foucault rotation. Both the unperturbed velocity and the Coriolis correction to it will scale together in magnitude, hence halving one will halve the other. The resulting rotation angles are therefore independent of the pendulum's amplitude and its velocity. Since the oscillation time scale is unaffected by amplitude, you get the independence of the Foucault rotation rate as well. Everything checks out.
We can also recall that the mass of the bob plays no role in this analysis. This is consistent with the notion that gravitational and inertial accelerations (such as Coriolis) are independent of the mass - as they should be, since they are in fact equivalent.
Peter Hakel
First, let's agree that the pendulum has a "small" amplitude (say within 5 degrees), so that the idealized pendulum model is adequate. Then its natural oscillations have a single frequency which is independent of the amplitude. The maximum amplitude, velocity, and acceleration of the bob are all directly proportional to one another. So, when you double the amplitude scale (within those 5 degrees...), you also double the velocity scale.
Now add a small perturbation in the form of Earth's rotation underneath the pendulum. As you correctly point out, the Coriolis force/acceleration is linearly proportional to velocity. This small correction has to be added to the unperturbed velocity. The outcome of this addition is the Foucault rotation. Both the unperturbed velocity and the Coriolis correction to it will scale together in magnitude, hence halving one will halve the other. The resulting rotation angles are therefore independent of the pendulum's amplitude and its velocity. Since the oscillation time scale is unaffected by amplitude, you get the independence of the Foucault rotation rate as well. Everything checks out.
We can also recall that the mass of the bob plays no role in this analysis. This is consistent with the notion that gravitational and inertial accelerations (such as Coriolis) are independent of the mass - as they should be, since they are in fact equivalent.
Peter Hakel
From: Gary LaPook <glapook---.net>
To: navlist@fer3.com
Sent: Monday, August 24, 2009 10:29:08 AM
Subject: [NavList 9614] Re: Coriolis and gyros (second attempt)(typos corrected)
There is a problem with your analysis.
Since the period of the pendulum is fixed by its length then it is
necessarily true that as the amplitude of the swing diminishes during
the day, due to air resistance, that its maximum velocity and its
average velocity also slows down. Try this yourself with any pendulum
and observe how slowly it moves near the end as it slows to a stop. The
formula for coriolis force includes a term for velocity across the
turning reference frame so that coriolis force and coriolis acceleration
is proportional to this velocity over the ground. So if your analysis is
correct then the rate of change of the azimuth of the pendulum's swing
should vary throughout the day, changing at a more rapid rate in the
morning and more slowly later in the day as the pendulum slows down. But
the rate of change of azimuth is constant, 11.32 degrees per hour which
disproves your analysis.
Now looking at the case of the gyroscope's undergoing earth rate
apparent precession. If this apparent precession is caused by coriolis
due to the speed of the rotating flywheel moving in one direction at the
point at the bottom and in the opposite direction at the top, (as you
claim) then, without doing the diagram of the precessional forces (I
leave that to you), it must also be true that gyroscopes spinning in
opposite directions would also precess in opposite directions. One
rotating clockwise would precess toward the east and one turning
counterclockwise would precess toward the west yet all gyroscopes, no
matter which way they are spinning, precess toward the east, again
disproving your analysis.
A further disproof of your analysis of earth rate precession of a
gyroscope is due to that pesky term in the coriolis formula that makes
the coriolis force proportional to velocity over the ground. If your
analysis were correct then slowly turning gyroscopes would precess more
slowly than rapidly spinning gyros yet they all show the same apparent
precession due to the earth rate of 15.04 degrees per hour times the
sine of the latitude.
You have not addressed the movement of a gyrocompass at the equator
where coriolis is zero since the sine of zero degrees of latitude is
also zero, there's that pesky coriolis formula again.
Also consider the earth rate apparent precession of a gyroscope at the
north pole which changes at a rate of 15.04 degrees per hour since the
sine of 90 degrees of latitude is 1. Yet a gyroscope located exactly at
the pole is not moving at all (ignoring the earth's movement around its
orbit) but is remaining at one fixed spot on the earth so it has no
velocity across the ground so, again by the formula, there should be no
coriolis available to cause the apparent precession.
gl
frankreed{at}HistoricalAtlas.com wrote:
> "Coriolis is not involved since the Pantheon has not moved across the surface of the earth since its foundations were laid in 1758 so its ground speed is zero."
>
> LOL. Gary the pendulum bob is itself in motion, right? Its ground speed is most certainly NOT zero. Likewise in a gyro-compass, its individuaL mass elements are moving at very high speed. If they weren't moving, it wouldn't work. The Coriolis acceleration causes the precession in both cases --in the rotating frame of reference in which the Earth is fixed.
>
> -FER
>
>
>
> >
>
>
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