Randy Yates wrote:> Jerry Avins <jya@ieee.org> writes: > >> gyansorova@gmail.com wrote: >>> On Apr 24, 8:33 am, Jerry Avins <j...@ieee.org> wrote: >>>> Michael K. O'Neill wrote: >>>> >>>> [very nice summary snipped] >>>> >>>> It might interesting to know that the first trial of an airborne INS >>>> took place from Bedford Air Force Base around 1953. I >> ... >> >>> surely the V2 German rockets pre-date this? >> I don't know what they had on board. Marine torpedoes were steered by >> gyroscopes, but as far as I know, neither they nor the V2s had stable >> platforms. > > What's a stable platform? A type of woman's shoe?A stable platform is an inertial reference surface, one of Draper's inventions and the bit of art that made inertial navigation at least conceivable. Imagine a plane plate with two gyros fastened to it. Their axes are parallel to the plate and each is perpendicular to the another. Mount the plate in frictionless gimbals and if it is forced horizontal and released it will always be horizontal. Right? Wrong! Aside from the difficulty of getting frictionless gimbals, the earth is turning. Friction can be overcome by nulling the motions caused by the gyro torques with counter torques and measuring (and integrating) the forces that need to be applied. It's hard to stop the earth. Draper's stable platform works on a rotating earth and continues to work as latitude and longitude are changed. Avins's stable platform is simpler than Draper's, but unlike Draper's, it can't be built. For the sale of a simple explanation, let's suppose that it can. The first idea is simply a platform with a pendulous weight, like a gyrocompass. The problem with that is that starting or stopping suddenly will cause the pendulum to oscillate. My fictitious pendulum doesn't oscillate; not because it isn't free to, but because there's no way to excite it. The idea is simple: put the bob at the center of the earth. When the platform moves suddenly, the pendulum bob gets left behind, as always. In this special case, it doesn't try to catch up because it's still where gravity wants it to be. My scheme is made impractical by high temperature and the solidity of rock, but let's see how it behaves. Suppose the bob were moved to one side by an external force. What would it do? Well, it's a pendulum and the interesting thing is that an observer on the surface would calculate the period as if gravity at the bob were that at the surface. The period of a pendulum is 2pi*sqrt(L/g), and that works out to 88 minutes (the same, not coincidentally, as the orbital period of a satellite grazing an airless earth). So all Draper has to do is adjust the servo that nullifies friction on his gyro-stabilized platform to have a free resonance of 88 minutes. A pair of poles smack dab on the jw axis. Easy, no? Wait! It only works on the ground. The period lengthens as altitude increases (high satellites have longer periods), so the period needs to be made inverse to sqrt(radial-distance-from-center). Altitude is measures by doubly integrating vertical acceleration -- after all, we do know which way up is -- and then it works like a charm. It's really easy when you know how! :-) And it's all built around a stable platform. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Rate Gyros and Accelerometers
Started by ●April 22, 2007
Reply by ●April 24, 20072007-04-24
Reply by ●April 25, 20072007-04-25
"Jerry Avins" <jya@ieee.org> wrote in message news:A_qdnb-T-Kb6MrPbnZ2dnUVZ_qarnZ2d@rcn.net...> Randy Yates wrote: > > Jerry Avins <jya@ieee.org> writes: > > > >> gyansorova@gmail.com wrote: > >>> On Apr 24, 8:33 am, Jerry Avins <j...@ieee.org> wrote: > >>>> Michael K. O'Neill wrote: > >>>> > >>>> [very nice summary snipped] > >>>> > >>>> It might interesting to know that the first trial of an airborne INS > >>>> took place from Bedford Air Force Base around 1953. I > >> ... > >> > >>> surely the V2 German rockets pre-date this? > >> I don't know what they had on board. Marine torpedoes were steered by > >> gyroscopes, but as far as I know, neither they nor the V2s had stable > >> platforms. > > > > What's a stable platform? A type of woman's shoe? > > A stable platform is an inertial reference surface, one of Draper's > inventions and the bit of art that made inertial navigation at least > conceivable. > > Imagine a plane plate with two gyros fastened to it. Their axes are > parallel to the plate and each is perpendicular to the another. Mount > the plate in frictionless gimbals and if it is forced horizontal and > released it will always be horizontal. Right? Wrong! > > Aside from the difficulty of getting frictionless gimbals, the earth is > turning. Friction can be overcome by nulling the motions caused by the > gyro torques with counter torques and measuring (and integrating) the > forces that need to be applied. It's hard to stop the earth. Draper's > stable platform works on a rotating earth and continues to work as > latitude and longitude are changed. > > Avins's stable platform is simpler than Draper's, but unlike Draper's, > it can't be built. For the sale of a simple explanation, let's suppose > that it can. The first idea is simply a platform with a pendulous > weight, like a gyrocompass. The problem with that is that starting or > stopping suddenly will cause the pendulum to oscillate. My fictitious > pendulum doesn't oscillate; not because it isn't free to, but because > there's no way to excite it. > > The idea is simple: put the bob at the center of the earth. When the > platform moves suddenly, the pendulum bob gets left behind, as always. > In this special case, it doesn't try to catch up because it's still > where gravity wants it to be. My scheme is made impractical by high > temperature and the solidity of rock, but let's see how it behaves. > Suppose the bob were moved to one side by an external force. What would > it do? Well, it's a pendulum and the interesting thing is that an > observer on the surface would calculate the period as if gravity at the > bob were that at the surface. The period of a pendulum is 2pi*sqrt(L/g), > and that works out to 88 minutes (the same, not coincidentally, as the > orbital period of a satellite grazing an airless earth). > > So all Draper has to do is adjust the servo that nullifies friction on > his gyro-stabilized platform to have a free resonance of 88 minutes. A > pair of poles smack dab on the jw axis. Easy, no? > > Wait! It only works on the ground. The period lengthens as altitude > increases (high satellites have longer periods), so the period needs to > be made inverse to sqrt(radial-distance-from-center). Altitude is > measures by doubly integrating vertical acceleration -- after all, we do > know which way up is -- and then it works like a charm. It's really easy > when you know how! :-) And it's all built around a stable platform. > > Jerry > -- > Engineering is the art of making what you want from things you can get. > ����Gravity at the center of the earth is zero. So even if you could build the "Avins stable platform", I don't think it would work the way you think. I don't see how the period of your pendulum, for example, could be 88 minutes, since the force of gravity experienced by a bob at the center of the earth is zero. In candor, however, I have no idea of what its period might be, and you might be correct when you say that "the interesting thing is that an observer on the surface would calculate the period [of the pendulum] as if gravity at the bob were that at the surface." Mike
Reply by ●April 25, 20072007-04-25
Jerry Avins wrote: (snip)> The idea is simple: put the bob at the center of the earth. When the > platform moves suddenly, the pendulum bob gets left behind, as always. > In this special case, it doesn't try to catch up because it's still > where gravity wants it to be. My scheme is made impractical by high > temperature and the solidity of rock, but let's see how it behaves. > Suppose the bob were moved to one side by an external force. What would > it do? Well, it's a pendulum and the interesting thing is that an > observer on the surface would calculate the period as if gravity at the > bob were that at the surface. The period of a pendulum is 2pi*sqrt(L/g), > and that works out to 88 minutes (the same, not coincidentally, as the > orbital period of a satellite grazing an airless earth).This reminds me of a physics quiz many years ago, calculating the period for a satellite orbiting inside a tunnel in a planet. I won't give the answer, in case anyone wants to work it out, consider a planet of mass m, uniform density, radius R, and tunnel (orbit) radius r. -- glen
Reply by ●April 25, 20072007-04-25
Michael K. O'Neill wrote:
...
> Gravity at the center of the earth is zero. So even if you could build the
> "Avins stable platform", I don't think it would work the way you think. I
> don't see how the period of your pendulum, for example, could be 88 minutes,
> since the force of gravity experienced by a bob at the center of the earth
> is zero.
>
> In candor, however, I have no idea of what its period might be, and you
> might be correct when you say that "the interesting thing is that an
> observer on the surface would calculate the period [of the pendulum] as if
> gravity at the bob were that at the surface."
I thought I made "as if" clear. I will put the argument another way to
put your mind at ease about the period. Imagine an unbalanced sphere
mounted in frictionless gimbals that when disturbed oscillates with a
period of 2pi*sqrt(r/g), where r is the earth's radius and g is the
acceleration of gravity at the surface. (The period, if you work it out,
is around 5280 seconds. That the number is familiar _is_ a coincidence.)
A horizontal displacement will start it swinging like any other
pendulum, but the change of latitude or longitude will redefine "down"
in such a way that it always "likes" where it is, so it won't swing
back. Now, servo the gimbals to behave frictionlessly and replace the
pendulous sphere with a pair of gyros that can provide the signals
needed to close that servo loop and you have Draper's stable platform.
Gravity at the earth's center isn't an issue.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●April 25, 20072007-04-25
glen herrmannsfeldt wrote:> Jerry Avins wrote: > > (snip) > >> The idea is simple: put the bob at the center of the earth. When the >> platform moves suddenly, the pendulum bob gets left behind, as always. >> In this special case, it doesn't try to catch up because it's still >> where gravity wants it to be. My scheme is made impractical by high >> temperature and the solidity of rock, but let's see how it behaves. >> Suppose the bob were moved to one side by an external force. What >> would it do? Well, it's a pendulum and the interesting thing is that >> an observer on the surface would calculate the period as if gravity at >> the bob were that at the surface. The period of a pendulum is >> 2pi*sqrt(L/g), and that works out to 88 minutes (the same, not >> coincidentally, as the orbital period of a satellite grazing an >> airless earth). > > This reminds me of a physics quiz many years ago, calculating the > period for a satellite orbiting inside a tunnel in a planet. > > I won't give the answer, in case anyone wants to work it out, > consider a planet of mass m, uniform density, radius R, and > tunnel (orbit) radius r.The period depends on the local gravity. That's what causes a departure from the free-space straight line. The force due to gravity is inverse with the square of the radius outside a spherical body, and decreases linearly from the surface to the center IFF the body is of uniform density. Assuming uniform density for the earth (it has an iron core) will lead to error. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●April 26, 20072007-04-26
Jerry Avins wrote: (snip)> The period depends on the local gravity. That's what causes a departure > from the free-space straight line. The force due to gravity is inverse > with the square of the radius outside a spherical body, and decreases > linearly from the surface to the center IFF the body is of uniform > density. Assuming uniform density for the earth (it has an iron core) > will lead to error.And it is difficult to tunnel through molten rock, too. Maybe moons would be a better choice. So yes, the gravitational force is proportional to r (tunnel radius). How does the period depend on r? (On the quiz I got the right equation and the wrong conclusion.) -- glen
Reply by ●April 26, 20072007-04-26
"Michael K. O'Neill" <MikeAThon2000@nospam.hotmail.com> writes:> "Jerry Avins" <jya@ieee.org> wrote in message > news:A_qdnb-T-Kb6MrPbnZ2dnUVZ_qarnZ2d@rcn.net... >>....... > > Gravity at the center of the earth is zero. So even if you could build the > "Avins stable platform", I don't think it would work the way you think. I > don't see how the period of your pendulum, for example, could be 88 minutes, > since the force of gravity experienced by a bob at the center of the earth > is zero.And this tells us what? The total force on the bop at equilibrium position of _any_ pendulum will be zero; that's just the definition of equilibrium. The interesting thing is: what will be the force when the bob is pulled out of equilibrium by a distance x. Then, iff the force is proportional to this distance, we have a harmonic oszillation. -- hw
Reply by ●April 26, 20072007-04-26
Jerry Avins <jya@ieee.org> writes:> .... The force due to gravity > is inverse with the square of the radius outside a spherical body, and > decreases linearly from the surface to the center IFF the body is of > uniform density.Right. Thus the potential inside the earth would be a perfect harmonic oszillator, where the period is independent of the amplitude--- other than with the ordinary pendulum, where this is true only to the approximation sin(x)=x.> Assuming uniform density for the earth (it has an > iron core) will lead to error.Iron core ??? Anyone ever proved this? AFAIK the only thing that can be said is that the _average_ density is close to that of iron--- ``weighing the earth''; Cavendish IIRC. But probably they have measured some multipole moments in the recent years? But the reason why I reply is a related problem which you might be interested in: provided the earth were a homogenous sphere, what would be the pressure in the center of the earth? Many books on geology seem to get this wrong: they don't consider the reduction of gravity when moving inwards. I did the (rather simple) calculation and got an even greater result. Though baffled at the first moment, I saw why: when doing it right, the load to a surface element at the center of the earth isn't a prisme, but a cone. BTW Jerry: Ever heared about M.Schuler? This guy might have proposed something similar to Avin's stable platform already before 1940. -- hw
Reply by ●April 26, 20072007-04-26
glen herrmannsfeldt wrote:> Jerry Avins wrote: > > (snip) > >> The period depends on the local gravity. That's what causes a >> departure from the free-space straight line. The force due to gravity >> is inverse with the square of the radius outside a spherical body, and >> decreases linearly from the surface to the center IFF the body is of >> uniform density. Assuming uniform density for the earth (it has an >> iron core) will lead to error. > > And it is difficult to tunnel through molten rock, too. > Maybe moons would be a better choice. > > So yes, the gravitational force is proportional to r (tunnel radius). > How does the period depend on r? (On the quiz I got the right > equation and the wrong conclusion.)The heavy core distorts the nice theoretical linear decline. In fact, gravity will continue to increase with depth until the core is nearly reached if the core is dense enough and the rest is fluffy enough. The period is 2*pi*sqrt(r/g) When g = kr (k constant) the period is independent of r. Like turtles, it's 88 minutes all the way down. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●April 26, 20072007-04-26
Heinrich Wolf wrote:> Jerry Avins <jya@ieee.org> writes: > >> .... The force due to gravity >> is inverse with the square of the radius outside a spherical body, and >> decreases linearly from the surface to the center IFF the body is of >> uniform density. > > Right. Thus the potential inside the earth would be a perfect > harmonic oszillator, where the period is independent of the > amplitude--- other than with the ordinary pendulum, where this is true > only to the approximation sin(x)=x. > >> Assuming uniform density for the earth (it has an >> iron core) will lead to error. > > Iron core ??? Anyone ever proved this? AFAIK the only thing that can > be said is that the _average_ density is close to that of iron--- > ``weighing the earth''; Cavendish IIRC. But probably they have > measured some multipole moments in the recent years?Iron core seems to be good hypothesis on more than one count, but it is indeed not proven.> But the reason why I reply is a related problem which you might be > interested in: provided the earth were a homogenous sphere, what would > be the pressure in the center of the earth? Many books on geology > seem to get this wrong: they don't consider the reduction of gravity > when moving inwards. I did the (rather simple) calculation and got an > even greater result. Though baffled at the first moment, I saw why: > when doing it right, the load to a surface element at the center of > the earth isn't a prisme, but a cone.did your calculation account for the likelihood that the core is much denser than the mantle?> BTW Jerry: Ever heared about M.Schuler? This guy might have proposed > something similar to Avin's stable platform already before 1940.I don't know of Schuler's work. Avins's stable platform is a parody of Draper's dreamed up to explain on physical grounds what he worked out with equations. I think Draper actually wrote a paper called "The Eighty-eight Minute Pendulum". The parody didn't require deep thought. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
