This page discusses the April 12, 2011 Fun Physics Facts for the Family Demonstration...
The following video may be downloaded and played. It is a "tongue-in-cheek" trailer for the Fun Physics Facts April 2011 Demonstration: (Gyrocompass-SpyTrailer-1080p.mov). [Instead of uploading this "tongue-in-cheek" video to YouTube, I will show it during the introduction to the demonstration as an advertisement for The Most Marvelous Mechanical Machine Demonstration.]
This is my favorite mechanical device of all time, partly because
- It is exact and points to the true North Pole from anywhere in the world. A magnetic compass seldom points to the true North Pole and it does not work over large regions of the Earth's surface nor inside ferromagnetic submarines and ships.
- Instead of trying to minimize friction as is typically required for most mechanical devices, this one actually requires friction to be added back to its bearings for proper operation. In other words, friction does not cause a degradation in its operation, unlike most mechanical devices.
- In addition, it is quite intriguing that a pendulum exists for which the pivot may be accelerated without perturbing the bob.
- A Schuler-tuned horizontal platform bounds the error for an inertial guidance system, unlike a 3D platform whose error increases without bound and thus requires frequent recalibration to known locations on the Earth's surface. The Schuler-tuned platform does not require such recalibration -- a remarkable feature!
Notice how at this close proximity these two magnetic compasses are not pointing in the same direction. This is because the magnetism of one compass is affecting the needle of the other, and vice versa.
In the Jerry Bruckheimer film, Pirates of the Caribbean: The Curse of the Black Pearl (2003), directed by Gore Verbinski and based upon the Disney theme park ride, I quote Commodore James Norrington ridiculing Captain Jack Sparrow's compass at around 19 minutes from the beginning of the film, "A compass that doesn't point North -- humph!" The following video has a couple excerpts regarding Captain Jack Sparrow's compass from the movie: (SchulerPendulum-Pirates_1_Compass-1080p.mov). [This video contains about 15 seconds of excerpts from the movie and thus I will not upload it to YouTube for copyright reasons, rather I will show these excerpts during the lecture in an educational classroom setting in order to discuss why a magnetic compass does not point true North and the other limitations of magnetic compasses.]
We begin with a discussion of an "unperturbable pendulum"...
Perhaps this is not even possible!?And why would we even care?By watching a gyroscope, we notice that almost any movement produces small torques on the axle of the gyroscope that lead to slow precessional motions, called polhode motions, of its axis (see the photograph of a gyroscope resting on a rotating platform in the photograph below.
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Thus the gyroscope's axis does not continue to point in a stable direction in the inertial frame as it moves. In fact, no matter how the gyroscope is designed and manufactured, it is impossible to eliminate these polhode motions completely. A case in point is the Gravity Probe B satellite containing four of the most perfect gyroscopes ever manufactured (see the "Gravity Probe B" topic in the PDF below for a description of this experiment). These gyroscopes are nearly perfect quartz spheres, coated with a nearly uniform superconducting niobium film, suspended by an electric field, spinning in a nearly perfect vacuum, and measured by SQUID magnetometers. But even these nearly perfect gyroscopes have experienced polhode motions that have caused much more analysis and numerical modeling than was initially expected and consequently these polhode motions have delayed the publication of the experimental Lense-Thirring results testing Einstein's General Relativity theory for over three years.Here is a description of the Gravity Probe B experiment:
So, apparently it is nearly impossible to make a perfect gyroscope, at least perfect enough to provide a stable platform against which to measure any movements relative to the inertial frame for long periods of time (e.g., the Gravity Probe B's gyroscopes were used to measure General Relativity's geodetic and Lense-Thirring effects against the inertial frame for a period of 10 months).
Most long term navigational requirements, however, occur on or near the surface of the Earth, and thus typically only require 2-dimensional positional information, not 3-dimensional data. And it turns out that the inertial navigational errors in 3 dimensions stemming from system errors are unbounded and must be adjusted by external positional aids over long times, while in 2 dimensions these errors can be bounded and thus accounted for and controlled providing an unaided inertial navigational system. This bounding of the 2-dimensional errors relies primarily on the concept of a Schuler Pendulum, discovered by Maximilian Schuler in 1923.
In two dimensions, accelerometers are mounted on a platform. If this platform is horizontal, then the accelerometers measure the accelerations in the directions of travel directly, and their outputs can be integrated directly to calculate the vehicle's velocity and position. If the platform is not horizontal, however, then angle-dependent scale factor errors enter into the accelerometer calculations and drift errors enter into the orientation calculations which causes systematic errors to creep into the navigational computations. In fact, it can be shown that in 3 dimensions, the drift errors are unbounded and will cause the navigational system to seriously miscalculate its location after a relatively short time. In 2 dimensions where the 2-dimensional plane remains locally horizontal so that the accelerometers mounted on this plane measure horizontal accelerations that can be integrated without the need for scale factors, the drift errors then remain bounded and thus the navigational system will be accurate for a much longer time.
Hence we wish to maintain a stable horizontal platform on which to mount our gyroscopes and accelerometers in order to minimize the navigational errors. But this platform is aboard a ship, submarine, airplane, or land rover, and consequently is buffeted by movements in all directions, that is, the platform experiences random accelerations along any direction as the vehicle travels. If we wish to have the platform remain horizontal, then it must be stable in the face of these random accelerations. Is this possible? How is this possible?A Little History First: Maximilian Schuler, Hermann Anschultz-Kaempfe, Albert Einstein, Elmer Sperry
Let's begin with just a brief historical vignette, shall we? Hermann Anschutz-Kaempfe (1872-1931) founded a German company in 1905 (Anschutz & Co. GmbH, Kiel, now the Raytheon Marine GmbH, Kiel) to manufacture navigational devices based upon the gyroscope. His cousin, Maximilian Schuler (1882-1972), joined the company in 1906. One of the chief problems that plagued the company was how to establish a stable platform on a rolling, pitching, and turning vessel. Maximilian Schuler discovered the solution to this problem which he published in 1923.
But let's back up even a little more, for there is yet further connections leading all the way back to Foucault. You see, Foucault's 1851 gyroscope was employed in an 1885 patent by the Dutchman Marinus Gerardus van den Bos in a navigational device called the gyrocompass (the amazingly ingenious gyrocompass is the subject of another Example). It never really worked properly for van den Bos, however. In 1903 the German engineer Hermann Anschutz-Kaempfe constructed a working gyrocompass for which he obtained both a German and an American patent in 1908. The American Elmer Ambrose Sperry filed an American patent for his improvements on the gyrocompass, specifically the suspension and oscillation damping mechanisms, in 1911 which was later issued, with amendments, in 1918. When Sperry attempted to sell his gyrocompass to the German Navy in 1914, Hermann Anschutz-Kaempfe sued for patent infringement. Elmer Sperry's claim was that the Anchutz-Kaempfe patent was not a significant improvement on van den Bos's 1885 patent and thus was invalid. Albert Einstein (yup, the one and only) was called to testify in the court case, initially siding with Sperry. But, as was to become a recurring theme with Einstein, he reversed his opinion and sided with Anschutz-Kaempfe saying that the Sperry patent infringed upon the Anschutz-Kaempfe patent's method of damping. Anschutz-Kaempfe won the lawsuit in 1915. The Franklin Institute in Philadelphia, which houses a very large Foucault's pendulum by the way, has a very intriguing "Case File #2524" reporting details of the Sperry/Anscutz-Kaempfe conflict. This arose because Elmer A. Sperry was nominated for the prestigious John Scott Legacy Medal. Upon the nomination of Sperry for this award, Hermann Anschutz-Kaempfe sent a letter to Hugo Bilgram, the Chairman of the Franklin Institute's Committee on Sciences and the Arts which determined the winners of the Scott Medal, objecting to and critical of Sperry's nomination. The Case File includes a number of letters between Anschutz-Kaempfe, Hugo Bilgram of the Franklin Institute, and Sperry. These letters show that Sperry, by his own admission, would pinpoint the weaknesses of other inventions and then attempt to make improvements. This is precisely what he did for his suspension and oscillation damping improvements for the gyrocompass, originally patented in the U.S. by Anschutz-Kaempfe in 1906. But Anschutz-Kaempfe argued that Sperry's gyrocompass was not superior to his own and thus the Sperry patent did not constitute valid improvements.On to the Problem:
In any case, and without attempting to navigate the muddied waters (pun intended) between Sperry and Anschutz-Kaempfe, we return to our original story concerning Max Schuler who in fact did make a significant improvement to the gyrocompass for Anschutz & Co. GmbH, Kiel. As mentioned in an earlier paragraph, the development of an accurate navigational tool necessitated the establishment of a stable platform. Difficulties arise on ships that are turning, pitching, and rolling in the turbulent seas. Each acceleration of the ship would tend to destabilize any stable platform. So, what is the engineer to do? Max Schuler solved this problem in 1923 by the expedient use of a pendulum, not just any pendulum, however, but a very specific compound pendulum.Say we start with the pendulum shown in the above Figure. If the pendulum in Part A is 1 meter in length, then it has a period of 2 seconds. What happens if we were to quickly shove the support horizontally to the right with an acceleration of a drawn as the blue arrow in Part B? If the impulse from a is large enough to produce its displacement in a time short compared to the pendulum's 2 second period, then initially the bob would attempt to stay in its place, producing a pendulum bob offset from its equilibrium position. But then gravity would take over and the bob would swing to its new equilibrium position and oscillate around it. How long would it take to swing to the new equilibrium position vertically below the new position of its support? Yup, that's right, it would take a half-second to swing into its vertical orientation, based upon its two second period and assuming a linear harmonic oscillator model for the pendulum. On the other hand, Part C illustrates the situation where the acceleration a is much smaller (green arrow), so small in fact that the displacement due to the impulse occurs in a time interval long compared with the period. The bob then does not lag begin its pivot, but it just keeps up and moves over simultaneously with the pivot.Thus we see that if the pivot's displacement is fast compared to the pendulum's period, the bob does not keep up and is offset from its vertical equilibrium position (the cable does not remain locally vertical, in other words). The bob then swings back to equilibrium in a time determined by the pendulum's period. If the pivot's displacement is slow compared to the period, the bob's movement does keep up with the pivot and the bob remains in its vertical equilibrium position during the displacement. The bob does not have to swing back to equilibrium.
Slow perturbation relative to restoring force -- the pendulum bob keeps up with the acceleration of the pivot.
Fast perturbation relative to the pendulum's restoring force -- the pendulum bob does not keep up and lags behind.
Now what would happen if we make the pendulum's period infinitely long? For instance, take the compound pendulum shown in Part D of the following Figure. This pendulum has an infinite period since the pivot resides at precisely the center of mass of the pendulum. Notice that if we quickly shove, or accelerate, the pivot of this pendulum as shown in Part E, the pendulum's bobs both move over simultaneously with the movement of the pivot. Thus the orientation of the pendulum remains vertical. Part F shows what happens when the pivot is slowly shoved to the right: both bobs once again move over simultaneously with the pivot. Even when the bobs are not vertically aligned, as shown in Part G, the bobs move in conjunction with the pivot and without changing their angular orientation.Now you may be thinking that this is exactly what is needed to maintain a stable platform, but upon careful analysis we see that in fact we do not want a platform that have an infinite period of oscillation since then a horizontal displacement caused by a horizontal acceleration will mean that the platform is no longer locally horizontal once it is displaced from its initial location (remember the Earth's surface is curved). The infinite period pendulum was horizontal in its initial location, but once it has moved over it is no longer horizontal in the new location. Why is this a problem? Well, we have alluded to the fact that not having a locally horizontal platform means that the errors stemming from the need to integrate the accelerometer's output leads to an unbounded error in the position coordinate. On the other hand, when the platform maintains a locally horizontal orientation, then the positional error arising from integration of the accelerometer's output is bounded. We will examine just how this happens in greater detail in a moment. But for now, let's summarize: a 2 seconds oscillation period is too short, it does not keep the platform horizontal, while an infinite period is too long but for a different reason, the platform after the displacement is no longer locally horizontal. But perhaps this implies that for some value between 2 seconds and infinity there might be a period T_oscillation that is capable of keeping the platform locally horizontal? Let's see if this is indeed the case and if we can find this period.
A pendulum with an infinite period does not need to "swing back" into position, but then its platform does not remain horizontal either.
Here is a video (SchulerPendulum-AltT_AgtT-1080p.mov) explaining and demonstrating the short period and infinite period pendulum behaviors under the action of perturbing accelerations.Here is a link to the video on YouTube: http://youtu.be/7qUPY6GZWL8(YouTube makes available various resolutions of this video.)
Description of the video's experiment: In this video a pendulum having less than a second period is perturbed by accelerations at it pivot. When the accelerations are small, the pendulum bob keeps up with the acceleration. When the perturbation is large, the bob cannot keep up and thus lags behind. So a pendulum with a 1 second period would not keep a platform horizontal in the face of perturbing accelerations. At the other extreme, we consider a pendulum having an infinite period. Now any perturbation leaves the pendulum in its present orientation. This is good and bad, it is good because the pendulum does not have to "catch up" to the perturbation of the pivot, but it is bad because once the platform has moved sideways under the influence of the perturbing acceleration, it is no longer horizontal because of the curvature of the Earth's surface. So, an infinite period pendulum does not maintain the horizontal platform. So, is there a pendulum of some special period that can keep a horizontal platform horizontal in the face of random perturbing accelerations of its pivot?Puzzler 1: Consider the following experiment: Accelerate a Horizontal Platform, measure the acceleration, and use this information to drive the platform back to horizontal. Write the necessary equations for the motion, and compute the resulting period.
Do you recognize the period you computed in the first Puzzler query? Can you give a physical argument for why your calculation must have found this period?
Hint: Consider how long a simple pendulum would have to be in order to oscillate with this period.
(The mathematics involved in this calculation requires some Calculus, but when the details are all worked out the results are identical to the mathematics that describe the swing of a pendulum, the up and down motion of a weight suspended from a spring, or the oscillation of the quartz crystal in a Timex watch -- the Harmonic Oscillator.)If you wish, you may skip the following mathematics, and go right to the result below, that is: The Solution that keeps a platform stable is a pendulum whose period is 84.4 minutes!We begin with an informal argument as to what to expect. Say we have a platform that is initially horizontal, and we accelerate it in a northerly direction. Immediately, upon acceleration, the platform will no longer be horizontal (see the platform at its initial location labelled 1 being accelerated to location 2 in the following Figure). But let's say that we measure the acceleration of the platform by an accelerometer mounted on the platform itself. We then wish to use this measurement to somehow activate a mechanism to drive the platform back to the new horizontal plane at its new farther north location 2. Remarkably, this can be accomplished thereby keeping the platform horizontal in the face of a northward acceleration. Let's see how this might be done.Since an angle equals the intercepted arc length divided by the radius, for short times \Delta t, the arc length S is estimated by the straight line length along the z axis in the North direction. Thenand after taking two time derivatives, signified by the double dot notations, we findIf the platform measures an acceleration of a at point 1 and it continues this magnitude of acceleration in the horizontal plane at position 2, then the acceleration that the platform would measure at 2 would be given by a_z=a\cos\phi. In addition to this acceleration, gravity would also produce a component g_z=g\sin\phi along the same direction as the acceleration a_z. Hence the total acceleration, \alpha, measured by an accelerometer along the z axis of the platform would be the sum of these two components [it is the sum and not the difference because the accelerometer measures a force in the backwards (south) direction due to the acceleration in the north direction while the component of the gravity force due to the misalignment angle \phi is also in the backwards (south) direction, hence the sum],
and soFor small \phi, then \cos\phi \approx 1 and \sin\phi \approx \phi, thusWe now recognize that the homogeneous version of the above second-order differential equation is the differential equation for the undriven harmonic oscillator. Its solution is the oscillationwhose graph is the familiar back and forth wave plotted belowThus the differential equation, given above, for the platform drive mechanism is just a harmonic oscillator with a nonzero driving force. From the theory of differential equations, its solution is thus a combination of the general solution of the homogeneous differential equation, plus a specific solution of the nonhomogeneous differential equation. In fact, we can solve this nonhomogeneous equation by inspection from the solution to the homogeneous equation. We find that the drive mechanism will have a period ofThis we recognize as the same period as a simple pendulum whose length is roughly the radius of the Earth, R_{\oplus}, since R=R_{\oplus}+h where h is the height of the platform above the ground and therefore h << R_{\oplus} and R\approx R_{\oplus}. This makes a certain amount of sense since a pendulum the length of the radius of the Earth would have its bob sitting at the center of the Earth (of course, this is physically impossible), which is an equilibrium position. Any motion of the pivot of the pendulum would leave the bob still at the center of the Earth which means that the pendulum's cable points vertically downwards, towards the center of the Earth, at all times even under random accelerations of the pivot. Let's compute this period using the mean radius of the Earth for R:Whoa! Now this is a familiar number, let's see, it is the time it would take a spacecraft to orbit the Earth at the Earth's surface if it didn't run into anything and there was no air resistance! In other words, it is roughly the time it takes for a low-orbit satellite to orbit the Earth. [Low-orbit satellites are about 200 miles high instead of right at the Earth's surface which makes their orbital periods a little longer.]
Click for full-size image
A quick interlude...
You've heard about "Riding the Gravy Train" [A slang phrase used in reference to having an easy life, typically the life of leasure for someone who has inherited great wealth and thus does not have to work for a living. For example, in the 1995 remake of the original 1954 film "Sabrina" (starring Audrey Hepburn, Humphrey Bogart, and William Holden), the heartless and ruthless businessman Linus Larrabee (acted by Harrison Ford in the 1995 film) is not "riding the gravy train", while his baby brother playboy, the handsome and debonair David Larrabee (played by the newcomer Greg Kinnear in 1995) could easily be said to be "riding the gravy train". (Now there is another, recent, urban slang use of this phrase to describe someone having a long bout of diarrhea sitting on the commode -- but this is not the meaning that I am referring to here.)], but have you heard about "Riding the Gravity Train"? While most of us do not ride on the gravy train, many of us have actually ridden on a gravity train before, you probably have too. [Technically, all modern roller coasters are gravity trains, but this is taking us too far afield for our purposes this evening.]Extraordinary Fact and Fanciful Application: Eighty-four minutes is also the time that it takes a ball to fall through a hypothetical evacuated pipe extending from the surface through the center of the Earth to a diametrically opposite point and then return to its starting position! A science fiction application of this would be a magnetic levitating vehicle that can travel to any location on the Earth's surface in just 42 minutes without the need of any energy source for propulsion!?! In fact, Robert Hooke in a letter to Isaac Newton proposed this type of motion for a body moving inside the Earth, and Lewis Carroll's Professor Mein Herr discusses just such a train on frictionless tracks running through a tunnel between cities in his Sylvie and Bruno Concluded book (MacMillan and Co., London, Vol. 2, Ch. 7, pp. 106-108, 1893).To quote Mein Herr from pages 107-108 of Lewis Carroll's book: "Easily", said Mein Herr. "Each railway is in a long tunnel, perfectly straight: so of course the middle of it is nearer the centre of the globe than the two ends: so every train runs half-way down-hill, and that gives it force enough to run the other half up-hill."
If you dropped a ball down an evacuated tube drilled through the center of the Earth, it would take 84 minutes to return (from the North Pole down through the center up to the South Pole, and then returning).
(As long as a straight evacuated tube could be built through the Earth connecting the two locations, say New York City and Paris, and the vehicle could be made frictionless, say through magnetic levitation, then gravity itself accelerates the vehicle during the first half of the trip from New York City and then slows it during the second half of the trip so that it comes to rest at Paris, 42 minutes travel time and no fuel required for propulsion beyond that needed to maintain the vacuum in the tube and levitate the vehicle above the tube's surface so that there is no friction between the vehicle and the tube. Pretty neat trick, huh? If only we could build such a tube through the Earth's mantle, but, of course, this is not possible, at least using today's technology.)
It can be shown that this 84 minutes round trip is also valid for any straight line evacuated tube and a frictionless vehicle, including a tunnel drilled from NYC to Paris.
The following PDF file derives the time periods for gravity trains thereby proving that the gravity train from NYC to Paris takes just 42 minutes:
Now notice that our above mathematical argument for the time period of the oscillation did not depend upon the initial acceleration being in the northward direction in any inherent way, in fact, the acceleration could have been in any direction and the argument would have been the same. Thus we see that a servomechanism that drives the platform around an axis perpendicular to the measured horizontal acceleration of the platform, \alpha, by an angle given by -\phi determined from the above Equation will keep the platform locally horizontal. This is accomplished by recognizing that the torque required to slew the angle of the platform is proportional to the angular velocity which is likewise proportional to the velocity since \dot{\phi}=\dot{z}/R. The velocity \dot{z} is determined by integrating the acceleration, \ddot{z}, measured by the accelerometer. This means that the accelerometers [Historically, these sensors have been called accelerometers, but, in fact, they do not measure the acceleration, rather they measure the force and should be called force sensors instead of accelerometers. We can see that they measure forces and not accelerations by considering a vertically oriented accelerometer. If it is dropped the accelerometer would measure zero even though it is obviously falling at the acceleration of gravity. So one must take care when interpreting the meaning of the measurements that an accelerometer is determining; one must be certain to adjust for gravity.] mounted on the platform will continue to measure horizontal accelerations and the errors that build from integrating the accelerometer outputs for long time intervals will remain bounded [a fact that we will derive later] --- in other words, the navigational system will be much more accurate over longer time intervals if the platform containing the accelerometers is locally horizontal.While it is not physically possible to make a simple pendulum at the surface of the Earth whose length is the radius of the Earth, can we build a compound pendulum whose period is equal to this period, T_{oscillation}? The short answer is "Yes and No". The long answer is that we can build a motor-driven servo mechanism to accomplish this Schuler Pendulum stabilization for a horizontal platform -- and this is what is done in practice.But instead of getting into this now, let's consider an analogy to the above calculations, one that is more intuitive and perhaps more readily understood.
Baseball Analogy:
Well, if you didn't like the above mathematical treatment for the pendulum approach, there is another way of understanding the horizontal platform. If you play baseball, you have probably batted the ball from the bat's "sweet spot", and you have probably also hit the ball from a spot that is not the "sweet spot". When the bat connects with the ball at its "sweet spot", it feels like a solid hit and you have the sense of a well hit ball. But when the bat connects with the ball somewhere but the "sweet spot", it sends tingles in your hands and shivers up your arms. If you are very far from the "sweet spot", it can actually cause pain. You sense that the ball is not well hit. Physically, what is the difference? What is happening?The bat's "sweet spot" is the point along the bat where an applied impulse causes the bat the rotate around the hand grip. In other words, when the ball strikes the "sweet spot", the grip of the bat remains instantaneously stationary. There is no tendency to jerk the grip out of the hands. You feel like you have a solid hit on the ball. When the ball strikes the bat at some other point, the grip is not the point around which the bat rotates and thus the grip is jerked by the ball's impulse. This quick jerk causes tingles, even pain, in the hands and sends what feels like vibrations up the arms. It does not feel like a solid hit.Physically, whenever an impulse is applied to a rigid body, the impulse causes two things to happen. First of all, the impulse is a force that accelerates the body in the direction of the impulse. This means that the center of mass of the body follows Newton's Second Law, F=ma. But if the impulse is "off-center", that is, if the impulse vector does not lie on a line passing through the center of mass of the body, then the impulse also produces a torque that causes the body to rotate, following Newton's Second Law for rotation, Gamma=I\alpha (torque, Gamma, equals the moment of inertia, I, times the angular acceleration, \alpha). Since the torque from the impulse causes rotation, the body has to be rotating about some axis. This axis of rotation is called the Instantaneous Center, it is the position that is instantaneously stationary when the impulse is applied. In other words, it is the position that the impulse causes the rigid body to rotate around. For the bat, if the Instantaneous Center is located at the hand grip, the batter feels a solid hit. If the Instantaneous Center is not located at the hand grip, the grip jerks when the ball is struck causing a tingling sensation in the batter's hands.
The impulse from the baseball on the bat occurs at point P known as the Center of Percussion. This impulse causes the Center of Mass C to accelerate in the direction of the impulse. At the same time the impulse causes the bat to rotate around a point O known as the Instantaneous Center. If the Instantaneous Center is located at the grip, then the batter's hands do not feel any jerking of the bat (no tingling sensation, in other words).
The following video (SchulerPendulum-Instantaneous_Center_NoGershwin-1080p.mov) demonstrates the Instantaneous Center for a baseball bat.
Here is a link to the video on YouTube: http://youtu.be/0FcjqHpnpFQ(YouTube encodes various alternative resolutions, so download the highest resolution you can.)Description of the video's experiment: In this video a baseball bat is struck at three different locations along its length by a rubber mallet. When the bat is struck near to the hand grip, the grip quickly moves to the side out of the Sandy's fingers. A slow motion repeat of the video clearly shows the grip moving horizontally after the mallet strike. But when the bat is struck at its "sweet spot", the grip does not move horizontally, rather the bat rotates around the point at the grip. This is the instantaneous center, the point about which the bat rotates when struck. Again, slow motion clearly illustrates this rotation without horizontal motion. A third mallet strike at an intermediate point produces both motion and rotation.
Click for full-size image
Note 1: Okay, there are actually two different things going on with the baseball bat. One is the Center of Percussion/Instantaneous Center thing discussed above. The other is the fact that the bat has natural vibration modes and if the bat is struck at a node of a particular vibration mode, then this vibration is not stimulated by the ball strike and the bat does not vibrate in this mode. So, there are really two (or more) so-called "sweet spots" for a bat. One is the Center of Percussion, others are the nodes for the various vibration modes. Both of these mechanisms, the Center of Percussion/Instantaneous Center and the vibration modes will cause tingling in the hands if the ball strikes the bat far from the "sweet spots".Note 2: Striking the ball at the bat's "sweet spot" constitutes a good hit. Why? Well, if the ball strikes at a spot not the "sweet spot", then the impulse from the ball will tend to rotate the bat out of the batter's hands. The batter must compensate for this by applying extra hand force on the grip in order not to loose control of the bat. The energy required to do this is ultimately robbed from the energy delivered to the ball by the bat, thus the ball ends up with less energy and hence travels a shorter distance. In other words, when the ball is struck at the bat's "sweet spot", more energy goes into the motion of the ball and less energy goes into holding onto the bat and the vibrations of the bat. In other words, you really can "feel" a home run hit, as any batter will tell you![Note that these "sweet spots" for a bat do not constitute the precise spots for maximum performance (fastest ball off the bat, or maximum energy transferred to ball, or the point of maximum coefficient of restitution, depending upon how you wish to define maximum performance). The reason is because of the complex interplay between the various sweet spots. Additionally, metal bats also have transverse vibration modes which cause them to act like a trampoline, producing yet a faster rebound of the ball from the bat. All of these mechanisms interact to yield a maximum performance point that differs slightly from any of the bat's Center of Percussion or vibration node "sweet spots". The Center of Percussion/Instantaneous Center position is typically quite near to the maximum performance point for the bat and thus it is usually considered to be the bat's "sweet spot".]
Note 3: If you would like to see another problem related to the Center of Percussion / Instantaneous Center, see the Billiard Bumper Bounce Puzzler on the Physics Puzzler page: Puzzlers: 5 (Physics)With this introduction to the Center of Percussion/Instantaneous Center, let's apply this concept to the horizontal platform. If the platform is to remain horizontal under the impulses caused by the vehicle's accelerations, then the Instantaneous Center from those impulses should be located at the Earth's center. If the center of the Earth is the Instantaneous Center, then the platform rotates about the center of the Earth. This means that the acceleration of the platform leave the point at the center of the Earth stationary, which, if you think about it, means that the platform remains horizontal during the acceleration!
The simple pendulum corresponding to this compound pendulum whose Instantaneous Center is at the Center of the Earth would thus have a length equal to the radius of the Earth, just as we calculate previously.In summary, applying an impulse to the horizontal platform causes the platform to not only accelerate but also to rotate. If the Instantaneous Center (axis) for the impulse-produced rotation is located at the center of the Earth, then the rotation is such that the platform remains horizontal as it accelerates since it is rotating about a point at the center of the Earth.We thus see that it is possible to design a platform that remains horizontal under the influence of random accelerations! The mechanism for accomplishing this is known as a Schuler Pendulum, or sometimes it is called Schuler-tuning, since Maximilian Schuler discovered this concept in 1923.
Having a platform that remains horizontal in the face of random impulses (accelerations) has two important consequences. The first is the subject of Puzzler 2, the second consequence is the gyrocompass that we discuss below.
- The drift errors in a inertial guidance system on a Schuler-tuned platform are bounded. They do not increase over time but remain finite. If the guidance system were not Schuler-tuned, then the errors are unbounded and increase continuously towards infinity. (See the solution PDF file for Puzzler 2 for why this is the case.)
- A gyroscope whose axle is restricted to a horizontal plane acts like a perfect compass, always pointing to true North.
Puzzler 2: Why does a stable horizontal platform bound the drift errors of a navigational system while a platform allowed to move in 3 dimensions leads to an unbounded drift error?
Puzzler 2 is a much harder question to answer, but it is also a much more important result. It means that an inertial guidance system can be manufactured so that its navigational error is finite and bounded for all time. This means that the navigational system can be used for long periods of time without recalibration to a known location, for instance, out on the open sea where there are no known locations (there are no street signposts in the middle of the ocean, in other words). On the other hand, a 3 dimensional inertial system has an unbounded error and thus must be either a much more accurate and stable mechanism or it must be recalibrated to a known position more often.
Hint: Propagate the errors for both the locally horizontal stable platform and for the platform allowed to move in 3 dimensions. [This is a nontrivial problem requiring not only familiarity with error propagation but also second-order differential equations.]
The second part of the following solution PDF includes this error propagation calculation:This PDF file gives alternative physical explanations for the reason a pendulum having the proper period will provide a stable, locally horizontal, platform. It also derives the error analysis that shows that a locally horizontal platform leads to bounded drift errors while a platform in 3 dimensions allows for the drift errors to be unbounded.With this introduction, let's move on to the Most Amazing Mechanical Device (in my humble opinion)...
Afer all of this discussion of horizontal platforms and Schuler Pendula, let's review why a gyroscope is not a perfect compass...Naively, we might think that just a simple gyroscope would constitute a compass...meaning that if we align the axis of the spinning rotor with the polar axis of the Earth, then as the gimbal housing is moved, rotated, and accelerated by the vehicle, the gyroscope's axis will stay aligned with the Earth's polar axis. While this is very clever, unfortunately it doesn't work for precisely the reason we have already studied: precession! What causes the precession, you ask? Torques, of course, cause precessions. But what if we eliminate the torques entirely? Won't the gyroscope's axis then be stable and point in one fixed direction? The problem is that we can't eliminate the torques completely. As we have mentioned earlier, even the most perfect rotors ever manufactured, the four quartz spheres spinning and suspended electrostatically in a vacuum and employed for the gyroscopes in the Gravity Probe B satellite experiment, did not entirely eliminate the torques. And any non-zero residual torques, no matter how small, lead to precessional motions of the rotor's axis. (See the following PDF file for further details of the Gravity Probe B experiment: )
Click for full-size image |
So, if a simple gyroscope won't work, then perhaps nothing will work? This is where Schuler's pendulum enters the picture. Say that the axle of a spinning gyroscope is connected to a Schuler pendulum so that the gyroscope's axle always resides in the horizontal plane since the pendulum itself always points vertically no matter how its pivot is perturbed. In other words, a Schuler pendulum is used to keep the spin axis of a gyroscope always in a horizontal plane even as the gimbal mounts are accelerated and rotated by the motion of the vehicle. A schematic of this is shown below.
Now that we know that the Schuler pendulum mechanism will keep the gyroscope's axle in a horizontal plane, let's see how this happens and what its consequences might be. First consider a gyroscope with attached Schuler pendulum located at the equator with its axle pointing in an East/West direction (see the top drawing of the gyroscope in the figure below). As the Earth turns about its polar axis, the Schuler pendulum will try to keep the gyroscope's spin axis in a horizontal plane, perpendicular to the radius line from the Earth's center. Since the radial line is rotating the horizontal plane must also "rotate" with the Earth as it spins. This rotation of the horizontal means that the Schuler pendulum must apply a torque to the gyroscope's spin axis in order to "rotate" it around with the daily rotation of the Earth. This torque thus produces its attendant precession, as shown in the left drawing of the gyroscope below.
Another way of visualizing this is to consider the 3-dimensional drawing below. The two magenta disks denote the horizontal planes to which the gyroscope's spin axis is restricted by the Schuler pendulum. As you can see, the horizontal plane is twisted around as the Earth spins, and this twist produces a torque on the spin axis of the gyroscope, causing the spin axis to precess around the horizontal plane.
On the other hand, what would happen if the gyroscope's axle were initially pointing parallel to the Earth's polar axis? Now, as you can see from the diagram below, the Schuler pendulum does not produce any net torque on the gyroscope's axle. With no net torque, there is no precession.
From the following figure, we see that when the gyroscope's spin axis is pointing North, parallel to the Earth's polar axis, then there is no net torque on the gyroscope's axle and thus no precession.
So, what does this mean? It means that if the gyroscope is pointing North, then there is no tendency to rotate its spin axis, but if the gyroscope is not pointing North, then the gyroscope's axle will precess around until it is pointing North! In other words, a Schuler pendulum stabilized gyroscope constitutes a true North pointing compass.
Sandy demonstrates the precession of the wheel's axle when the turntable is rotated. (The following video shows this experiment.)
In the following video (Gyrocompass-Demonstration_Sandy-1080p.mov), Sandy demonstrates how the gyrocompass's axle will precess whenever a torque is applied to the axle by rotation about a vertical axis. But when the gyrocompass's axle is already vertical and thus aligned with the rotation axis, then there is no torque on the axle and hence no precession of the gyrocompass's axle.
Here is a link to the video on YouTube: http://youtu.be/nErU-4cBO-8
(YouTube offers several alternative resolutions of this video.)
Description of the video's experiment: In this video Sandy supports the weight of the bicycle wheel rotor. She does not keep it from moving, in other words, she allows the wheel to precess if it wishes to do so. Sandy first stands on the lazy susan turntable with the bicycle wheel not rotating. As I spin the lazy susan turntable, the non-rotating wheel has no tendency to precess, so its axle remains horizontal as I rotate the turntable. Next the wheel is spun up to a high angular velocity and Sandy holds its axle in a horizontal direction. Now when I rotate the turntable, the wheel attempts to precess about a perpendicular axis. Sandy allows this precession to occur (she only supports the weight of the wheel), as we clearly can see. When I reverse the direction of the turntable, what happens? Watch the video to find out. And lastly Sandy holds the spinning wheel with its axle in a vertical direction. Now the wheel's axis and the turntable's axis are parallel, thus there is no net torque applied to the wheel's axle by the rotation of the turntable and hence the wheel does not precess in this situation.
While the mathematics is somewhat more complicated, the mechanism is the same even when the Schuler pendulum stabilized gyroscope is located at a latitude besides the equator. In this case the gyroscope does not reach an equilibrium position but rather it continues to precess after reaching its North pointing position. The precession rate exactly matches the Earth's spin rate (23h54min4.09s) and thus the gyroscope continues to point to the true North direction.
And here I demonstrate this North-pointing behavior by placing the gyroscope on a tilted spinning platter with its spin axis initially misaligned with the platter's spin axis. After a few revolutions, however, the gyroscope's spin axis is parallel to the platter's spin axis, and it remains so as the platter continues to rotate. This demonstrates precisely the behavior of a Schuler-tuned gyrocompass located on the Earth and pointing in the direction of the Earth's polar axis.
Problem: What happens if the rotation direction of the platter is reversed?
Here is a video (Gyrocompass-Marvelous_Machine-1080p.mov) providing the answer.
Here is a link to the video on YouTube: http://youtu.be/2SENzmIrQNM
(YouTube provides various alternative resolutions for the video.)
Description of the Video's Experiment: Notice that at the start of the video the gyroscope's rotor is not spinning, then as the platter is rotated the rotor's axis rotates around with the platter. It does not point in a fixed direction, rather it is fixed relative to the platter and thus rotates with the platter. When the gyroscope's rotor is spinning, however, the rotor axis appears to point in a fixed direction as the platter is rotated. But a careful observation shows that the rotor's axis is in fact precessing around some fixed axis. The question is, what fixed axis? Friction dampens the amplitude the precession so that it disappears and when the precession's amplitude vanishes the rotor's axis is pointing in a direction parallel to the platter's rotation axis. If there were no friction, then the rotor would continue to precess around the platter's rotation axis.
What happens when the platter rotation direction is reversed? As the video shows, the gyroscope's rotor axis reverses direction (the gyroscope turns over) and friction settles it once again parallel to the platter rotation axis but now pointing in the opposite direction.
You have to imagine in your mind that the lazy susan platter in the above video is the Earth spinning around its polar axis. You have to imagine that the tilt of the platter resting on the book and the additional tilt induced by the Altoids can represents the tilt of the gyrocompass's horizontal platform caused by its latitude position on the Earth's surface. You then recognize that the gyroscope's axis precesses until it points in the same direction as the spin axis of the platter -- this is analogous to the gyrocompass pointing true North on the Earth's surface (in other words, in the direction of the Earth's spin axis). This is the mechanism for how a gyrocompass operates. There is a difference, however, and that is the fact that the gyroscope's axle in the gyrocompass is restricted to a horizontal plane by Schuler tuning. This restriction does not allow the gyrocompass's axis to align with the Earth's spin axis (excepting at the Equator), rather it produces a precessional motion whose period is precisely the Earth's spin rate (23h 56m 4.09s) thus keeping the gyrocompass's axis pointing in the true North direction!!!!
Summary Concepts:
- Maximilian Schuler invented a very clever device to stabilize a gyrocompass. Without this stabilization, the gyrocompass's axis continously precesses due to small torques (there is no such thing as a perfect gyroscope - see the file), and so you never would know the precise true North direction. But with Schuler tuning, the gyrocompass's axis settles into a precessional motion with a angular speed exactly equal to the spin rate of the Earth and consequently its axis always points in the true North direction!
- Besides allowing for the gyrocompass function to work, Schuler tuning also bounds the error for an inertial guidance system. Without Schuler tuning, the error would be unbounded and thus would keep getting larger over time. (See the for details of why this is the case.)
- Instead of the need to minimize the friction, as is typical for most mechanical devices, the gyrocompass actually requires friction to be added back to the gimbal mount.
The following link goes to a page containing all FPF Demonstration video YouTube links: FPF Videos
And now for this month's Puzzler solution...
Coconuts
And now for the First Coconut Problem from the Ultimate Puzzler page. (By the way, a $20 reward is offered for the first verified solution to the Third Coconut Problem. Here is a link to the Ultimate Puzzler: Ultimate Puzzler .)
Five sailors and a monkey become shipwrecked on a desert island. They collect a pile of coconuts. The ship's Captain takes half of them plus a half a coconut for his stash. The First Mate absconds with half of what is left plus half a coconut. The Engineer then hoards half of what's left plus half a coconut. Next the Gunner takes half plus half a coconut. And finally, the lowly Deck Hand takes half of what remains plus half a coconut. Left over is exactly one coconut, which they toss to the Monkey. How many coconuts did the sailors initially collect? [Note: Remember that unlike the Third Coconut Problem, the solution to this First Coconut Problem is not restricted to be an integer; it can be any real number.]If you are having trouble with the algebra, go get a box of toothpicks (or some other expendable objects that you can break in half) and do a trial-and-error solution. The trial-and-error solution will guide you in your algebraic proof.The following figure draws a bunch of toothpicks being employed to represent the coconuts. The toothpicks can be readily broken in half to represent the half of a coconut, as illustrated below. We work the problem in reverse order, starting with the 1 coconut given to the Monkey...
The numbers on the left-side of the piles denote the number of coconuts that each sailor received, the numbers in red on the right-side of the piles denote the total number of coconuts before each sailor took his or her share. Below we describe the process.
The Monkey gets the last coconut, and the Deck Hand immediately preceding the Monkey took one-half of the pile plus half a coconut. Thus the Monkey's coconut plus one-half must be equal to one-half of the pile before the Deck Hand took half. (Boy, isn't this ever confusing?!) Hence 1+1/2 must be half of the pile, and consequently the pile contained 3 coconuts before the Deck Hand took half (1+1/2) plus 1/2 a coconut for a total of 2 coconuts taken by the Deck Hand. The Gunner before the Deck Hand takes one-half of the pile plus half a coconut. This means that one-half of the pile must be 3+1/2 coconuts, and therefore the pile before the Gunner takes her half must have contained 7 coconuts. The Gunner then takes one-half of 7 or 3+1/2 plus 1/2 of a coconut, or a total of 4 coconuts leaving behind a pile of 3 coconuts for the Deck Hand to split. Similarly, 7+1/2 must be one-half of the pile split by the Engineer so the pile before the Engineer must have contained 15 coconuts. The Engineer takes one-half of 15 or 7+1/2 coconuts plus 1/2 coconut, or a total of 8 coconuts leaving behind the pile of 7 coconuts for the Gunner. Continuing in this fashion, we see that 15+1/2 is one-half of the pile for First Mate to divide in half, thus the pile for the First Mate contains 31 coconuts. The First mate takes one-half of these, or 15+1/2 coconuts, plus 1/2 coconut for a total of 16 coconuts leaving behind the pile of 15 coconuts for the Engineer. And, finally, 31+1/2 must be one-half of the pile for the Captain, thus the Captain's pile contains 63 coconuts before she takes her one-half, or 31+1/2 coconuts plus 1/2 coconut for a total of 32 coconuts leaving the pile of 31 coconuts for the First Mate. Hence we see that the original pile contained 63 coconuts!
Having worked this problem in reverse, let's demonstrate that it works in the forward direction too. The sailors initially collect a pile of 63 coconuts. The Captain takes one-half of the pile plus 1/2 coconut, or 31+1/2+1/2 = 32 coconuts, leaving a pile of 63 - 32 = 31 coconuts. The First Mate takes one-half of this pile plus 1/2 coconut, or 15+1/2+1/2 = 16 coconuts leaving a pile of 31 - 16 = 15 coconuts for the Engineer. The Engineer takes one-half of this pile plus 1/2 coconut, or 7+1/2+1/2 = 8 coconuts, leaving a pile of 15 - 8 = 7 coconuts for the Gunner. The Gunner takes one-half of this pile plus 1/2 coconut, or 3+1/2+1/2 = 4 coconuts, leaving a pile of 7 - 4 = 3 coconuts for the Deck Hand. The Deck Hand takes one-half of this pile plus 1/2 coconut, or 1+1/2+1/2 = 2 coconuts, leaving a pile of 3 - 2 = 1 coconut that is given to the Monkey. It WORKS!
To understand how to convert the above toothpick demonstration into algebra, we first define some variables:
Since there are no coconuts left, we haveX = Number of Coconuts in the initial pile
C = Number of Coconuts taken by the Captain
F = Number of Coconuts taken by the First Mate
E = Number of Coconuts taken by the Engineer
G = Number of Coconuts taken by the Gunner
D = Number of Coconuts taken by the Deck Hand
M = Number of Coconuts taken by the Monkey
The Captain takes one-half of the initial pile plus one-half of a coconut. The Captain's stash therefore contains0 = X - C - F - E - G - D - M
coconuts and the remaining pile contains X - C coconuts. The First Mate takes one-half of the remaining pile plus one-half of a coconut. The First mate thus takescoconuts, leaving behind X - C - F coconuts. The Engineer takes half of these plus half a coconut, or
Click for full-size image coconuts. The pile now has only X - C - F - E coconuts remaining. The Gunner's share is one-half of the remainder plus half a coconut, so
Click for full-size image coconuts. The remaining pile has X - C - F - E - G coconuts, which the Deck Hand takes one-half of these plus half a coconut, or
Click for full-size image coconuts. Only a single coconut is left over, so X - C - F - E - G - D = 1 = M , the last coconut being taken by the Monkey. Substituting these values for C , F , E , G , D , M into 0 = X - C - F - E - G - D - M we find
Click for full-size image which we solve for X to give
Click for full-size image
Click for full-size image Aha! We find the same answer as we found experimentally using toothpicks.The following PDF file contains the algebraic solution to this First Coconut Problem. The algebra required to solve the Third Coconut Problem is analogous to this algebra:
You should now proceed to the Second Coconut Problem on your the way to the Third Coconut Problem. Here is a link to the Ultimate Puzzler page, Ultimate Puzzler , where you will find these Coconut Problems.
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