# NavList:

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

**Re: Free Gyroscope**

**From:**Peter Hakel

**Date:**2009 Nov 15, 19:33 -0800

Note that cotangent = cosine / sine, so the sin Lat term cancels out in the drift equation, which then becomes:

drift rate = sin Lat - cos Lat * cos Az * tan Tilt

With Lat = 0 this simplifies to,

drift rate = - cos Az * tan Tilt

In your case this rate is indeed initially zero, but that's because you have chosen Tilt = 0, not due to Lat = 0. According to the tilt equation, however,

the tilt will increase with the rate sin Az. Since you have not oriented your gyroscope N/S, its axis will then rise straight up (at constant azimuth), then will deviate towards North (with initial Az=45 the drift rate is negative), thus following your "gyro star," just as Frank has explained. After the axis passes the meridian, the tilt rate becomes negative (sin Az < 0) so the axis begins to dip back towards the horizon.

Peter Hakel

drift rate = sin Lat - cos Lat * cos Az * tan Tilt

With Lat = 0 this simplifies to,

drift rate = - cos Az * tan Tilt

In your case this rate is indeed initially zero, but that's because you have chosen Tilt = 0, not due to Lat = 0. According to the tilt equation, however,

the tilt will increase with the rate sin Az. Since you have not oriented your gyroscope N/S, its axis will then rise straight up (at constant azimuth), then will deviate towards North (with initial Az=45 the drift rate is negative), thus following your "gyro star," just as Frank has explained. After the axis passes the meridian, the tilt rate becomes negative (sin Az < 0) so the axis begins to dip back towards the horizon.

Peter Hakel

**From:**Subhendu Hati <subhenduhati---.com>

**To:**navlist@fer3.com

**Sent:**Fri, November 13, 2009 11:03:13 PM

**Subject:**[NavList 10679] Re: Free Gyroscope

Dear Sir,

Thanks for the reply! It was just as I thought if I have to stick to the theory of "following a gyro star".

But the point which is bugging me is this:

The book from where I have studied behavior of free gyroscope say that

the rate of drift of the spin axix of a free gyroscope is = sin lat ( 1 - cot lat.Cos.Az tan Tilt) per hour

and the rate of tilt is = Cos Lat. Sin Az per hour.

Assuming that the above two are correct and that movement about the verticle axix is drift (movement in Azimuth) and movement about the horizontal axis is tilt (rising & dipping),

there should not be any movement about the verticle axix at the equator as sin lat (or sin zero) will make the entire equation for drift as zero.

But to follow the "Gyro star", as mentioned by you "... that the axis would climb first vertically from the horizon

**then arc over towards the meridian**..." it will have to move about the verticle axix which means it would drift.How do I explain this?

Thanks & Best Regards

On 11/14/09,

**frankreed{at}historicalatlas.com**<frankreed{at}historicalatlas.com> wrote:

Kumar, you wrote:

"How will a free gyroscope move when initially set horizontal at the equator with a azimuth of say 045 degs (Not East-West)"

You can think about this in two ways. The easiest if you're just starting out is to think in terms of an inertial frame of reference (fixed relative to the distant stars). Then you can us the idea that you mentioned of a "gyro star". The gyroscope, free of all other forces, will continue to point towards whatever "gyro star" it is initially aimed at while the Earth turns underneath it. So in your case, the gyroscope will track a point in the sky 45 degrees away from the celestial pole. The real star Deneb (alpha Cygni) has a declination of just about 45 degrees, so if you set a (completely) free gyroscope pointing towards Deneb it would just about match your conditions and the motion of Deneb in the sky would match the motion of the axis of your gyroscope. If you're on the equator, that means that the axis would climb first vertically from the horizon and then arc over towards the meridian. The axis would cross the meridian travelling horizontally at the an altitude of 45 degrees six hours later. It would then descend towards the western horizon which it would reach six hours later, cross the lower branch of the meridian six hours after that and finally return to its original direction pointing NE and horizontal six hours after that (24 hours after you started).

When you've had more experience with these things, and especially when you need to deal with other forces acting on the gyroscope, then it becomes more useful to work in a frame of reference which is rotating with the Earth. In this frame of reference, there is a Coriolis acceleration which causes precession in any gyroscope about an axis parallel with the Earth's axis. It's two different ways of looking at the same thing.

-FER

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Capt. Subhendu Hati

Vice Principal

R.L. Institute of Nautical Sciences

Madurai

Cell: +91 98657 06616

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