Editing Schuler tuning (section)

==Principle==
As first explained by German engineer [[Maximilian Schuler]] in a 1923 paper,<ref>{{cite journal
 |last        = Schuler
 |first       = M.
 |title       = Die Störung von Pendel und Kreiselapparaten durch die Beschleunigung des Fahrzeuges
 |journal     = Physikalische Zeitschrift
 |volume      = 24
 |issue       = 16
 |year        = 1923
 |url         = https://apps.dtic.mil/dtic/tr/fulltext/u2/b806120.pdf
 |accessdate  = 2008-12-02
 |url-status  = live
 |archiveurl  = https://web.archive.org/web/20170901013138/http://www.dtic.mil/cgi-bin/GetTRDoc?Location=U2&doc=GetTRDoc.pdf&AD=ADB806120
 |archivedate = 2017-09-01
}}</ref> a pendulum that has a period that equals the [[orbit]]al period of a hypothetical [[satellite]] orbiting at the surface of [[Earth]] (about 84.4 minutes) will tend to remain pointing at the center of Earth when its support is suddenly displaced.  Such a pendulum (sometimes called a ''Schuler pendulum'') would have a length equal to the radius of Earth.  Consider a simple gravity [[pendulum]], whose length to its center of gravity equals the [[radius]] of Earth, suspended in a uniform gravitational field of the same strength as that experienced at Earth's surface.  If suspended from the surface of Earth, the center of gravity of the pendulum bob would be at the center of Earth.<ref name="Cannon">[http://accessscience.com/content/Schuler-pendulum/606900 Schuler Pendulum] by Robert H. Cannon, Accessscience.com</ref>  If it is hanging motionless and its support is moved sideways, the bob tends to remain motionless, so the pendulum always points at the center of Earth.  If such a pendulum were attached to the inertial platform of an inertial navigation system, the platform would remain level, facing "north", "east" and "down", as it was moved about on the surface of the Earth.

The '''Schuler period''' can be derived from the classic formula for the period of a [[pendulum]]:
:<math>T = 2\pi \sqrt\frac{L}{g} \approx 2\pi \sqrt\frac{6371000}{9.81} \approx 5063 \ \text{seconds} \approx 84.4 \ \text{minutes}</math>
where '''''L''''' is the [[Earth radius#mean radius|mean radius of Earth]] in meters and '''''g''''' is the local [[Gravity of Earth|acceleration of gravity]] in [[metres per second per second]].