> Jerry Avins <jya@ieee.org> writes:
>
>> glen herrmannsfeldt wrote:
>>> Jerry Avins wrote:
>> ...
>> I shouldn't have had to think about your problem, but I did. In the
>> 50s, there was talk of long-distance transportation tunnels that would
>> be bored on a straight path. A tunnel from New York to San Francisco,
>> for example, would be far below the surface at its midpoint. Despite
>> the tunnel's being straight, it would seem to be slanted downward at
>> each end. If evacuated to eliminate air resistance, the gravity-driven
>> end-to-end time would be 88 minutes regardless of the separation of
>> the destinations. "I shoulda known!"
>
> I was not aware of this wonderful property of rotational symmetric
> harmonic oszillators up to this moment. It's so easy to see:
>
> a) For any harmonic oszilator period is independent of amplitude.
>
> b) Given an n-dimensional harmonic oszillator, then you can define a
> function of a single variable, by giving (n-1) variables a fixed
> value. Now the functions defined this way for various choices of the
> fixed coordinates obviously differ only by a constant. Geometrically:
> if you have a rotational parabola, then intersections with a plane
> parallel to the axis of rotation will show _congruent_ parabolas.
> That's it.
But the straight tunnel is neither the fastest nor the slowest
gravity-powered path between two cities. Steeper initial and final
grades can yield shorter times. The curve degenerates to a cycloidal
path in a uniform gravitational field.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by Jerry Avins●April 27, 20072007-04-27
Heinrich Wolf wrote:
> Jerry Avins <jya@ieee.org> writes:
>
>> Heinrich Wolf wrote:
>>> Jerry Avins <jya@ieee.org> writes:
>>> ...
>>>> ...
>>> 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?
>
> See the carrets above. Please correct me, if this doesn't answer
> your question. (No native english speaker.)
Careless reading. Sorry!
>>> 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.
>
> Parody--- I understood it this way. Schuler's work seems to be
> related to the gyro-compass; have no good refernce yet, but if I find
> more I will ping you in this NG.
Thank you.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by Heinrich Wolf●April 27, 20072007-04-27
Jerry Avins <jya@ieee.org> writes:
> glen herrmannsfeldt wrote:
>> Jerry Avins wrote:
> ...
> I shouldn't have had to think about your problem, but I did. In the
> 50s, there was talk of long-distance transportation tunnels that would
> be bored on a straight path. A tunnel from New York to San Francisco,
> for example, would be far below the surface at its midpoint. Despite
> the tunnel's being straight, it would seem to be slanted downward at
> each end. If evacuated to eliminate air resistance, the gravity-driven
> end-to-end time would be 88 minutes regardless of the separation of
> the destinations. "I shoulda known!"
I was not aware of this wonderful property of rotational symmetric
harmonic oszillators up to this moment. It's so easy to see:
a) For any harmonic oszilator period is independent of amplitude.
b) Given an n-dimensional harmonic oszillator, then you can define a
function of a single variable, by giving (n-1) variables a fixed
value. Now the functions defined this way for various choices of the
fixed coordinates obviously differ only by a constant. Geometrically:
if you have a rotational parabola, then intersections with a plane
parallel to the axis of rotation will show _congruent_ parabolas.
That's it.
--
hw
Reply by Heinrich Wolf●April 27, 20072007-04-27
Jerry Avins <jya@ieee.org> writes:
> Heinrich Wolf wrote:
>> Jerry Avins <jya@ieee.org> writes:
>>...
>>>...
>
>> 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?
See the carrets above. Please correct me, if this doesn't answer
your question. (No native english speaker.)
>
>> 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.
Parody--- I understood it this way. Schuler's work seems to be
related to the gyro-compass; have no good refernce yet, but if I find
more I will ping you in this NG.
--
hw
Reply by Jerry Avins●April 27, 20072007-04-27
glen herrmannsfeldt wrote:
> Jerry Avins wrote:
> (snip of orbits inside a tunnel)
>
>> 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.
>
> Yes. I ended up with an equation with an R**3 in it,
> that is, the planet radius, the right equation, and concluded
> that it was just like Kepler, not noticing R vs. r.
> (Well, it was late at night. The fun of take home closed
> book quizzes.)
I shouldn't have had to think about your problem, but I did. In the 50s,
there was talk of long-distance transportation tunnels that would be
bored on a straight path. A tunnel from New York to San Francisco, for
example, would be far below the surface at its midpoint. Despite the
tunnel's being straight, it would seem to be slanted downward at each
end. If evacuated to eliminate air resistance, the gravity-driven
end-to-end time would be 88 minutes regardless of the separation of the
destinations. "I shoulda known!"
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by glen herrmannsfeldt●April 26, 20072007-04-26
Jerry Avins wrote:
(snip of orbits inside a tunnel)
> 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.
Yes. I ended up with an equation with an R**3 in it,
that is, the planet radius, the right equation, and concluded
that it was just like Kepler, not noticing R vs. r.
(Well, it was late at night. The fun of take home closed
book quizzes.)
-- glen
Reply by Jerry Avins●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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by Jerry Avins●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 Heinrich Wolf●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 Heinrich Wolf●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