PDA

View Full Version : Re: Tidal gererators


Oz
Aug4-04, 01:25 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>[Posted to sci.physics.research AND uk.business.agriculture]\n\n[Could any technical posters note the level of explanation likely to be\nrequired for a fairly non-technical crosspost.]\n\n\nGerald L R Stubbs &lt;stubbs@zetnet.co.uk&gt; writes\n&gt;Is that not exactly what I was postulating? That whatever the state of\n&gt;the tides, worldwide, it is the centre of gravity of the earth that\n&gt;exerts an\n&gt;influence on other heavenly bodies, or to put it another way, it is\n&gt;the total\n&gt;mass of the earth that exerts influence, which is not related to the shape\n&gt;of the earth from day to day.\n\nOK, stubbsy, I have a go.\n\nI will take the sun and the earth, dispensing with the moon which we\nremove for the sake of simplicity.\n\nThe earth sun system is balanced. That is the CENTRE of the earth feels\nno acceleration just like an orbiting spaceman. This is because gravity\nis precisely strong enough to exactly balance the centrifugal force.\n\nNote that centrifugal force is a fictitious force that we have to bring\nin if we are to force something into a circular orbit. Its only seen by\nbodies forced into a circular orbit, in this case by gravity. Since\ngravity acts on every particle of our bodies, no net force is seen,\nwhich is why astronauts in orbit feel neither gravitational nor\ncentrifugal forces despite travelling in a circular path.\n\nNow, crudely, the bit of the earth on the surface facing the sun is\ncloser than the centre. Its thus imbalanced. It feels a stronger\nattraction than the centre and so tries to fall in towards the sun.\nEqually the far side of the earth is further away, and feels a lesser\npull from the sun and so tries to fly away from the sun.\n\nThere is a similar effect for the centrifugal force, acting in the SAME\ndirection. That is the tighter orbit inside is going too slowly and\nwants to drop to a lower orbit and vice-versa for the far side.\n\nThis is all symmetrical, so you get symmetrical tides.\nThe height of the tide turns out to depend on the cube of the sun-earth\ndistance.\n\nNow lets think what happens if the tides (rotating once in 24 hours) get\nslowed by continents. Well, its just like a pair of brake shoes, slowing\nthe earth\'s rotation. The sunward tide, which is water trying to orbit\nslower than the earth rotates, is \'behind itself\'. Instead of being\nexactly facing the sun its a bit round the side. The force is thus ever\nso slightly unbalanced and doesn\'t quite point directly along the sun-\nearth axis. There is thus a small resultant force pushing the earth\nalong the orbital direction, changing its orbital speed and thus\naltering its orbital distance.\n\nI\'ve never actually done this calculation, so I don\'t know if rotation\none way will push the orbit away, and the other direction pulling it in.\nI\'m not about to work it out, one hires astrophysicists for this. My\nguess is that the direction matters.\n\nNote that the earth pulls tides from the sun, so there is an effect on\nthe sun, from the earth and the tides on the earth.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n&gt;&gt;Use oz@farmeroz.port995.com&lt;&lt;\nozacoohdb@despammed.com still functions.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>[Posted to sci.physics.research AND uk.business.agriculture]

[Could any technical posters note the level of explanation likely to be
required for a fairly non-technical crosspost.]


Gerald L R Stubbs <stubbs@zetnet.co.uk> writes
>Is that not exactly what I was postulating? That whatever the state of
>the tides, worldwide, it is the centre of gravity of the earth that
>exerts an
>influence on other heavenly bodies, or to put it another way, it is
>the total
>mass of the earth that exerts influence, which is not related to the shape
>of the earth from day to day.

OK, stubbsy, I have a go.

I will take the sun and the earth, dispensing with the moon which we
remove for the sake of simplicity.

The earth sun system is balanced. That is the CENTRE of the earth feels
no acceleration just like an orbiting spaceman. This is because gravity
is precisely strong enough to exactly balance the centrifugal force.

Note that centrifugal force is a fictitious force that we have to bring
in if we are to force something into a circular orbit. Its only seen by
bodies forced into a circular orbit, in this case by gravity. Since
gravity acts on every particle of our bodies, no net force is seen,
which is why astronauts in orbit feel neither gravitational nor
centrifugal forces despite travelling in a circular path.

Now, crudely, the bit of the earth on the surface facing the sun is
closer than the centre. Its thus imbalanced. It feels a stronger
attraction than the centre and so tries to fall in towards the sun.
Equally the far side of the earth is further away, and feels a lesser
pull from the sun and so tries to fly away from the sun.

There is a similar effect for the centrifugal force, acting in the SAME
direction. That is the tighter orbit inside is going too slowly and
wants to drop to a lower orbit and vice-versa for the far side.

This is all symmetrical, so you get symmetrical tides.
The height of the tide turns out to depend on the cube of the sun-earth
distance.

Now lets think what happens if the tides (rotating once in 24 hours) get
slowed by continents. Well, its just like a pair of brake shoes, slowing
the earth's rotation. The sunward tide, which is water trying to orbit
slower than the earth rotates, is 'behind itself'. Instead of being
exactly facing the sun its a bit round the side. The force is thus ever
so slightly unbalanced and doesn't quite point directly along the sun-
earth axis. There is thus a small resultant force pushing the earth
along the orbital direction, changing its orbital speed and thus
altering its orbital distance.

I've never actually done this calculation, so I don't know if rotation
one way will push the orbit away, and the other direction pulling it in.
I'm not about to work it out, one hires astrophysicists for this. My
guess is that the direction matters.

Note that the earth pulls tides from the sun, so there is an effect on
the sun, from the earth and the tides on the earth.

--
Oz
This post is worth absolutely nothing and is probably fallacious.

BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.

J. B. Wood
Aug5-04, 03:23 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article &lt;PEbMZlB1UHEBFwvA@farmeroz.port995.com&gt;, Oz\n&lt;oz@farmeroz.port995.com&gt; wrote:\n\n&gt; Note that centrifugal force is a fictitious force that we have to bring\n&gt; in if we are to force something into a circular orbit. Its only seen by\n&gt; bodies forced into a circular orbit, in this case by gravity.\n\nHello, and I\'m not sure the above is entirely correct. We should obtain a\nreaction force whenever we force a body to follow other than a\nstraight-line path. IOW, the body is undergoing a change of momentum (a\nvector quantity) when we cause it to deviate from a straight-line\ntrajectory. A more general statement of Newton\'s second law is that force\nis equal to the rate of change of momentum with respect to time. Whether\nthe terms "centripetal" and "centrifugal" for the force and its reaction,\nrespectively, can apply to non-circular, but not straight-line, motion is\na matter of definition. Sincerely,\n\nJohn Wood (Code 5550) e-mail: wood@itd.nrl.navy.mil\nNaval Research Laboratory\n4555 Overlook Avenue, SW\nWashington, DC 20375-5337\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <PEbMZlB1UHEBFwvA@farmeroz.port995.com>, Oz
<oz@farmeroz.port995.com> wrote:

> Note that centrifugal force is a fictitious force that we have to bring
> in if we are to force something into a circular orbit. Its only seen by
> bodies forced into a circular orbit, in this case by gravity.

Hello, and I'm not sure the above is entirely correct. We should obtain a
reaction force whenever we force a body to follow other than a
straight-line path. IOW, the body is undergoing a change of momentum (a
vector quantity) when we cause it to deviate from a straight-line
trajectory. A more general statement of Newton's second law is that force
is equal to the rate of change of momentum with respect to time. Whether
the terms "centripetal" and "centrifugal" for the force and its reaction,
respectively, can apply to non-circular, but not straight-line, motion is
a matter of definition. Sincerely,

John Wood (Code 5550) e-mail: wood@itd.nrl.navy.mil
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337

Charles Francis
Aug6-04, 03:03 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In message &lt;wood-0408040714540001@jbw-mac.itd.nrl.navy.mil&gt;, J. B. Wood\n&lt;wood@itd.nrl.navy.mil&gt; writes\n&gt;In article &lt;PEbMZlB1UHEBFwvA@farmeroz.port995.com&gt;, Oz\n&gt;&lt;oz@farmeroz.port995.com&gt; wrote:\n&gt;\n&gt;&gt; Note that centrifugal force is a fictitious force that we have to bring\n&gt;&gt; in if we are to force something into a circular orbit. Its only seen by\n&gt;&gt; bodies forced into a circular orbit, in this case by gravity.\n&gt;\n&gt;Hello, and I\'m not sure the above is entirely correct. We should obtain a\n&gt;reaction force whenever we force a body to follow other than a\n&gt;straight-line path. IOW, the body is undergoing a change of momentum (a\n&gt;vector quantity) when we cause it to deviate from a straight-line\n&gt;trajectory.\n\nThe point is that before you define a straight line you have to define a\nreference frame. Newton himself only considered a non-rotating frame.\nHuyghens introduced the idea of "fictitious" forces which apply in\nrotating frames.\n\n\n&gt;A more general statement of Newton\'s second law is that force\n&gt;is equal to the rate of change of momentum with respect to time. Whether\n&gt;the terms "centripetal" and "centrifugal" for the force and its reaction,\n&gt;respectively, can apply to non-circular, but not straight-line, motion is\n&gt;a matter of definition. Sincerely,\n\nIts a matter of the choice of reference frame.\n\n\n\nRegards\n\n--\nCharles Francis\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>In message <wood-0408040714540001@jbw-mac.itd.nrl.navy.mil>, J. B. Wood
<wood@itd.nrl.navy.mil> writes
>In article <PEbMZlB1UHEBFwvA@farmeroz.port995.com>, Oz
><oz@farmeroz.port995.com> wrote:
>
>> Note that centrifugal force is a fictitious force that we have to bring
>> in if we are to force something into a circular orbit. Its only seen by
>> bodies forced into a circular orbit, in this case by gravity.
>
>Hello, and I'm not sure the above is entirely correct. We should obtain a
>reaction force whenever we force a body to follow other than a
>straight-line path. IOW, the body is undergoing a change of momentum (a
>vector quantity) when we cause it to deviate from a straight-line
>trajectory.

The point is that before you define a straight line you have to define a
reference frame. Newton himself only considered a non-rotating frame.
Huyghens introduced the idea of "fictitious" forces which apply in
rotating frames.


>A more general statement of Newton's second law is that force
>is equal to the rate of change of momentum with respect to time. Whether
>the terms "centripetal" and "centrifugal" for the force and its reaction,
>respectively, can apply to non-circular, but not straight-line, motion is
>a matter of definition. Sincerely,

Its a matter of the choice of reference frame.



Regards

--
Charles Francis

J. B. Wood
Aug7-04, 05:04 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article &lt;QNGHYcWzphEBFwsD@clef.demon.co.uk&gt;, Charles Francis\n&lt;charles@clef.demon.co.uk&gt; wrote:\n\n&gt; The point is that before you define a straight line you have to define a\n&gt; reference frame. Newton himself only considered a non-rotating frame.\n&gt; Huyghens introduced the idea of "fictitious" forces which apply in\n&gt; rotating frames.\n&gt;\n\nHello, and I\'m sure Newton was well-aware of non-inertial frames of\nreference. The "absolute" frame of reference for Newtonian mechanics is\n"the fixed stars."\nAny decent undergrad physics text discusses this. A frame rotates\nrelative to the fixed stars and objects within that frame are changing\nmomentum relative to the stars.\n\n&gt;&gt;A more general statement of Newton\'s second law is that force\n&gt;&gt;is equal to the rate of change of momentum with respect to time. Whether\n&gt;&gt;the terms "centripetal" and "centrifugal" for the force and its reaction,\n&gt;&gt;respectively, can apply to non-circular, but not straight-line, motion is\n&gt;&gt;a matter of definition. Sincerely,\n\n&gt;Its a matter of the choice of reference frame.\n\nThat wasn\'t my point. The point is giving a name to forces that arise\nwhen an object (mass) has an other than circular or straight-line\ntrajectory relative to the fixed stars. Sincerely,\n\nJohn Wood (Code 5550) e-mail: wood@itd.nrl.navy.mil\nNaval Research Laboratory\n4555 Overlook Avenue, SW\nWashington, DC 20375-5337\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <QNGHYcWzphEBFwsD@clef.demon.co.uk>, Charles Francis
<charles@clef.demon.co.uk> wrote:

> The point is that before you define a straight line you have to define a
> reference frame. Newton himself only considered a non-rotating frame.
> Huyghens introduced the idea of "fictitious" forces which apply in
> rotating frames.
>

Hello, and I'm sure Newton was well-aware of non-inertial frames of
reference. The "absolute" frame of reference for Newtonian mechanics is
"the fixed stars."
Any decent undergrad physics text discusses this. A frame rotates
relative to the fixed stars and objects within that frame are changing
momentum relative to the stars.

>>A more general statement of Newton's second law is that force
>>is equal to the rate of change of momentum with respect to time. Whether
>>the terms "centripetal" and "centrifugal" for the force and its reaction,
>>respectively, can apply to non-circular, but not straight-line, motion is
>>a matter of definition. Sincerely,

>Its a matter of the choice of reference frame.

That wasn't my point. The point is giving a name to forces that arise
when an object (mass) has an other than circular or straight-line
trajectory relative to the fixed stars. Sincerely,

John Wood (Code 5550) e-mail: wood@itd.nrl.navy.mil
Naval Research Laboratory
4555 Overlook Avenue, SW
Washington, DC 20375-5337

Tim S
Aug12-04, 08:29 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\non 07/08/2004 11:04 pm, J. B. Wood at wood@itd.nrl.navy.mil wrote:\n\n&gt; In article &lt;QNGHYcWzphEBFwsD@clef.demon.co.uk&gt;, Charles Francis\n&gt; &lt;charles@clef.demon.co.uk&gt; wrote:\n&gt;\n&gt;&gt; The point is that before you define a straight line you have to define a\n&gt;&gt; reference frame. Newton himself only considered a non-rotating frame.\n&gt;&gt; Huyghens introduced the idea of "fictitious" forces which apply in\n&gt;&gt; rotating frames.\n&gt;&gt;\n&gt;\n&gt; Hello, and I\'m sure Newton was well-aware of non-inertial frames of\n&gt; reference. The "absolute" frame of reference for Newtonian mechanics is\n&gt; "the fixed stars."\n\nThat\'s the exact opposite of Newton\'s view. Newton\'s \'bucket\'\nthought-experiment was intended precisely to demonstrate that the absolute\nframe of reference is an inherent property of the geometry of spacetime, not\na property of its relation to other pieces of matter, such as (allegedly)\nfixed stars.\n\nTim\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>on 07/08/2004 11:04 pm, J. B. Wood at wood@itd.nrl.navy.mil wrote:

> In article <QNGHYcWzphEBFwsD@clef.demon.co.uk>, Charles Francis
> <charles@clef.demon.co.uk> wrote:
>
>> The point is that before you define a straight line you have to define a
>> reference frame. Newton himself only considered a non-rotating frame.
>> Huyghens introduced the idea of "fictitious" forces which apply in
>> rotating frames.
>>
>
> Hello, and I'm sure Newton was well-aware of non-inertial frames of
> reference. The "absolute" frame of reference for Newtonian mechanics is
> "the fixed stars."

That's the exact opposite of Newton's view. Newton's 'bucket'
thought-experiment was intended precisely to demonstrate that the absolute
frame of reference is an inherent property of the geometry of spacetime, not
a property of its relation to other pieces of matter, such as (allegedly)
fixed stars.

Tim