# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Coriolis and gyros (second attempt)(typos corrected)**

**From:**Gary LaPook

**Date:**2009 Aug 25, 10:51 +0200

It's still just a "fictitious" force used to explain to an observer on
earth what he thinks he is seeing. Think about this one. Say you are in
a space ship in the coasting phase of an interstellar mission moving at
a constant velocity, not accelerating or rotating, not subject to any
real forces--an inertial reference frame. You have a gyroscope on board
and you point one end of its axis of rotation at Sirius. Due to the
gyroscopic property of "rigidity in space" it will stay pointed at
Sirius. Now do the same experiment on our rotating earth. Point the
axis of a gyroscope at Sirius and it will continue to point at Sirius
due to the same "rigidity in space" while the earth turns under it. No
need for a force to keep it pointed at Sirius in the space ship and no
need for a force to keep it pointing at Sirius on earth either.

gl

P H wrote:

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gl

P H wrote:

<!-- DIV {margin:0px;} --> 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

From:Gary LaPook <glapook@pacbell.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@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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