New Moon "conspiracy." Lunar gravity is not 1/6th that of the Earth, but 64% and NASA is c

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>... Debate the physics, not the opinions....Posted by derekmcd</DIV></p><p>Good luck.&nbsp; Casting pearls before swine.&nbsp; Pearls before swine.<br /></p> <div class="Discussion_UserSignature"> </div>

 

[Smersh]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>33800 nautical miles = 38896 mihttp://history.nasa.gov/SP-4029/Apollo_08a_Summary.htmAt 055:38:40 the crew were notified that they had become the first humans to travel to a place where the pull of Earth&rsquo;s gravity was less than that of another body. The spacecraft was 176,250 n mi from Earth, 33,800 n mi from the Moon, and their velocity had slowed to 3,261 ft/sec. Gradually, as it moved farther into the Moon&rsquo;s gravitational field, the spacecraft picked up speed.http://history.nasa.gov/ap08fj/09day3_green.htmPublic Affairs Officer - "This is Apollo Control, Houston at 55 hours, 38 minutes into the flight and we have been asked for a reaction here in the Control Center during that television passage. I think the remark from Lovell that got the most reaction was in his description of the blue and brown Earth and not being sure of whether he would land on it. This triggered a tremendous spike of laughter, the likes of which I can't recall, which immediately settled down to business. And in general, the room the - there was just zero talking going on in the room at the time, except for what we all heard from Mike Collins. In an exchange which the crew - And as we have been talking, the Apollo 8 has passed the - into the Moon's Sphere of Influence; and quite literally this is a historic landmark in space flight because, for the first time, a crew is literally out of this world. They are under the influence of another celestial body, the Moon, from which the Earth - 33,820 straight line nautical miles. We indicated earlier our space digital chart, at some point - not yet completely clear - will switch over and start giving us Moon-related values. That switch just took place and we immediately have configured. Velocity is now 3,989 feet per second in relation to the Moon and the last value, in relation to the Earth, was 3,261 feet per second in relation to the Earth. We'll see this number go down off the Moon related figure over the coming period. ...Apollo 8 ... now presently 33,681 [nautical] miles [62,377 km] from the Moon and moving in a Moon related velocity [of] 3,989 feet per second [1,216 m/s]. At 55 hours, 42 minutes into the mission, this is Apollo Control, Houston."This is, however, from a NASA website which will of course be corrupted information.&nbsp; The only better source would be to get a hold of the magnetic data telemetry tapes with the information on it.&nbsp; And those too, would be, of course, manipulated.&nbsp; It's called "moving the goal posts".&nbsp; Loony lunar conspiracy 'theorists' will move the goal posts so far, the only evidence possible to prove it to them would be to fly them up there so they can look at the artifacts left behind with their own eyes.&nbsp; Somehow, I think if this were to happen, they would find a way to still deny it.&nbsp; If they refuse to understand the math, then there is no hope of enlightening them. <br /> Posted by derekmcd</DIV></p><p>Thanks and good find Derek but as you say, it's NASA data so therefore will undoubtedly be corrupt. Also, it's data from Apollo 8, which was the mission that was quoted by Zorgon when he originally raised the neutral point issue. Mind you, I'm not sure we could get the data from anywhere other than a NASA source, other than the original tapes, as you said. Maybe Shuttle_guy can get us the originals ... LOL ! </p><p>I could still post it anyway though I guess. It does directly address the example he quoted originally, so maybe I will post it. </p> <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1><p><font color="#800000"><em>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea."<br />Churchill: "Nancy, if you were my wife, I'd drink it."</em></font></p><p><font color="#0000ff"><strong>Website / forums </strong></font></p> </div>

 

[Smersh]

<p>Ok I have posted it <strong>here.</strong></p><p><img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" /> </p> <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1><p><font color="#800000"><em>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea."<br />Churchill: "Nancy, if you were my wife, I'd drink it."</em></font></p><p><font color="#0000ff"><strong>Website / forums </strong></font></p> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Ok I have posted it here. <br /> Posted by Smersh</DIV></p><p>I see they have posted a few.</p><p>pleladian is using a formula to find the ratio between 2 masses that are in a static relation to each other when the neutral point is known.&nbsp; In other words, those 2 object are NOT moving.&nbsp; There are no other forces to account for.&nbsp; The number he input to figure the ratio of the 2 static masses is wrong.&nbsp; The number he used is derived from the formula to find lagrange points which can be seen here to be far, <em><strong>far</strong></em> more complex.</p><p>http://www.ottisoft.com/samplact/Lagrange%20point%20L1.htm</p><p>and a deeper derivation.</p><p><font size="2"><strong>Here.</strong></font>&nbsp;</p><p>In an earlier post, I used a formula to figure the nuetral point of 2 static objects.&nbsp; Using the known masses of the earth and moon, i came up with something like 24,000 miles (this is the number they should be inputing into their formula).&nbsp; The above formulae will give a more accurate description of distance of spacecraft that are actually moving under the influence of 2 bodies that are in orbit about their barycenter and involve centripital forces.&nbsp; These Lagrange point formulae are in line with what NASA and the astronauts have quoted.</p><p>They are obviously having issues distinguishing the difference of the two formulae and their purpose.&nbsp;</p><p>And as Mee_n_Mac pointed out, the quote of 43,xxx was taken from an article in Time magazine... not exactly a scientific source.&nbsp; </p><p>Math doesn't make errors, only the people doing the math and quoting figures.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Ok I have posted it here. <br />Posted by Smersh</DIV></p><p>One of the paragraphs "over there" is this:</p><p><font color="#000000"><font size="2">At 055:38:40 the crew were notified that they had become the first humans to travel to a place where the pull of Earth&rsquo;s gravity was less than that of another body. The spacecraft was 176,250 n mi from Earth, 33,800 n mi from the Moon, and their velocity had slowed to 3,261 ft/sec. Gradually, as it moved farther into the Moon&rsquo;s gravitational field, the spacecraft picked up speed.</font></font></p><p><font color="#000000">Now, this can be interpreted to mean that they are talking about a point where the <strong>magnitude</strong> of the Earth's gravitatinal attraction equals the <strong>magnitude</strong> of the Moon's gravitational attraction.&nbsp; That need not be a point at&nbsp;which the two attractions cancel out and result in zero vectorially.&nbsp; There are a lot of places where the magnitudes are equal, and&nbsp;they don't even all lie in the plane of the Moon's orbit.&nbsp;&nbsp;In particular a solution of this problem may have nothing to do with libration points.&nbsp; My point here is make sure you have the problem adequately defined before applying difficult mathematics -- you might be deriving the right solution to the wrong problem.<br /></font>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p>You make a good point.&nbsp; The solution may even be simpler in that the craft may be slowing due to the influence of earth's gravity, but increase in velocity in relation to the moon.&nbsp; This would happen if the craft is ahead of the moon and the direction of the moon's orbit is causing the closing speed to increase.</p><p>But given the statement quoted from NASA, the distance is almost precisely what the L1 point is.&nbsp; Now, with that said, the craft does not have to be precisely between the moon and earth.&nbsp; The L1 point is the point directly between the 2 masses, but the hill sphere of the moon would form and arc between the 2 masses with the L1 point being the center of the arc.&nbsp; The hill sphere wouldn't exactly be a sphere, but rather an ellipsoid similar to an egg.&nbsp; The L1 point would be on the fat part of the egg.&nbsp; The moon would be the yolk.&nbsp; The arc of the hill sphere on the near side of the moon wouldn't be semi-circular, rather the distance from the surface of the moon to the edge of the hill sphere as you move perpendicular would increase slightly as that point is further from the earth.</p><p>I'm not sure if that makes any sense, but it does in my head.&nbsp;</p><p>The quotes from NASA and the astronauts are quite in line with with what the edge of the hill sphere is... which is very close to the distance of the L1 point.&nbsp; By using the L1 point as an example, I'm trying to illustrate why the formula they are using is completely wrong in determining the moon is 64% of earth's gravity. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>You make a good point.&nbsp; The solution may even be simpler in that the craft may be slowing due to the influence of earth's gravity, but increase in velocity in relation to the moon.&nbsp; This would happen if the craft is ahead of the moon and the direction of the moon's orbit is causing the closing speed to increase.But given the statement quoted from NASA, the distance is almost precisely what the L1 point is.&nbsp; Now, with that said, the craft does not have to be precisely between the moon and earth.&nbsp; The L1 point is the point directly between the 2 masses, but the hill sphere of the moon would form and arc between the 2 masses with the L1 point being the center of the arc.&nbsp; The hill sphere wouldn't exactly be a sphere, but rather an ellipsoid similar to an egg.&nbsp; The L1 point would be on the fat part of the egg.&nbsp; The moon would be the yolk.&nbsp; The arc of the hill sphere on the near side of the moon wouldn't be semi-circular, rather the distance from the surface of the moon to the edge of the hill sphere as you move perpendicular would increase slightly as that point is further from the earth.I'm not sure if that makes any sense, but it does in my head.&nbsp;The quotes from NASA and the astronauts are quite in line with with what the edge of the hill sphere is... which is very close to the distance of the L1 point.&nbsp; By using the L1 point as an example, I'm trying to illustrate why the formula they are using is completely wrong in determining the moon is 64% of earth's gravity. <br />Posted by derekmcd</DIV></p><p>I haven't done the calculation, and given how silly their notion is I am not going to, but it the moon's gravity were 64% of the Earth's, then I am quite sure that the change in the barycenter for the Earth-Moon system would place the barycenter sufficiently far from the center of the Earth, that the motion of the Earth in orbiting that barycenter would be more than a wee bit noticeable.</p><p>Your problem is that you are dealing with people who want to believe that there is a conspiracy to cover up the Moon's gravity.&nbsp; Logic and physics are not going to convince them of anything.&nbsp; They have already dumped Newton.&nbsp; A proper orbital mechanics calculation will fall on deaf ears.&nbsp; Remember, you are dealing with people who believe that NASA would actually want to cover up the&nbsp;"real" gravitational force of the Moon.&nbsp; I guess that by doing so they can people from discovering the snow-capped mountains on the dark side.&nbsp; You would have an easier time, a much easier time, convincing them to buy property on the dark side from Century 21&nbsp; -- say a nice ski chalet.&nbsp; Those people ought to have to pay for air -- but they think it will be free on the Moon !&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I haven't done the calculation, and given how silly their notion is I am not going to, but it the moon's gravity were 64% of the Earth's, then I am quite sure that the change in the barycenter for the Earth-Moon system would place the barycenter sufficiently far from the center of the Earth, that the motion of the Earth in orbiting that barycenter would be more than a wee bit noticeable.</p><p>Posted by DrRocket</DIV></p><p>I addressed the orbit in the 2nd post of this thread.&nbsp; I'm suprised I didn't follow through with that.&nbsp; Thanks for the reminder.</p><p>If the moon was 4 times as massive at it's current distance, it would be orbiting the earth considerably less than 29.5 days.&nbsp; However, since it is observed to be 29.5 days <em><strong>and</strong></em>&nbsp; assumed to be 4 times as massive it must be much further away.&nbsp; If it is, indeed 4 times as massive and much further away, this, of course, would make the diameter of the moon much larger given it's angular size in the sky.&nbsp; </p><p>Given the fact that the greeks had the size and distance figured out 2000 years ago and those same methods they used can be used today, I tend to lean towards basic physics versus unsupportable speculation to get my statitistics on the moon. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[Mee_n_Mac]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I'd be willing to bet that when WvB was writing that article in 1961, he used his cocktail napkin to do a back of the envelope calculation based on two static masses to come up with a relatively close figure. The figures he quotes there is not taking into consideration of both the moon's and earth's hill spheres nor the centripital forces the craft is under.&nbsp; You factor in those numbers, you find the lagrange point which is more related to the number you see NASA and the astronauts quoting.When johnlear input the 43,000 mi figure into a formula that describes two static masses, of course his figures will be wrong.&nbsp; 23,000 some miles is the correct figure for 2 static objects.&nbsp; 38,000 is the correct figure when incorporating many other variables in the n-body problem that involves velocity, orbits, centripital forces, etc, etc.{snip} <br />Posted by <strong>derekmcd</strong></DIV></p><p>I just wanted to note that Mr Lear keeps quoting WvB as coming up with the number the Times article quotes when it says the Moon's gravity = the Earth's gravity.&nbsp; Again I found the article (link in my prior post) but found no mention of WvB having contributed to it.&nbsp; If he did it was in a different sense that Mr Lear implies because when WvB did (in 1961) the same math Mr Lear does (today), WvB came up with the 24kmi we all know is correct (for that analysis, limited as it is).&nbsp; Why does Mr Lear ignore this ?&nbsp; Did WvB really contribute at all to the July '69 Times article or is Mr Lear just saying so ?</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[Mee_n_Mac]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Now, this can be interpreted to mean that they are talking about a point where the magnitude of the Earth's gravitatinal attraction equals the magnitude of the Moon's gravitational attraction.&nbsp; That need not be a point at&nbsp;which the two attractions cancel out and result in zero vectorially.&nbsp; There are a lot of places where the magnitudes are equal, and&nbsp;they don't even all lie in the plane of the Moon's orbit.&nbsp;&nbsp;In particular a solution of this problem may have nothing to do with libration points.&nbsp; My point here is make sure you have the problem adequately defined before applying difficult mathematics -- you might be deriving the right solution to the wrong problem.&nbsp; <br />Posted by <strong>DrRocket</strong></DIV></p><p>I wonder if it isn't the opposite.&nbsp; That&nbsp;is it's where the component of gravity from each source, lying along the spacecraft's path, just cancel out.&nbsp; Since they were on a free return trajectory and did minimal correction burns, I might think a simplified calculation (no burns,&nbsp;circular orbits, etc)&nbsp;for such a flightpath might rule in, or out, this possibility.&nbsp; Did someone post a link to the equations to do such ?</p><p>Or it is just as people have said, it's the point where the Moon's gravity is large enough to have to worry about and the writers didn't really understand what they were writing.</p><p>&nbsp;</p><p>EDIT : Just had a chance to read replies "over there". I'm curious why people think an accelerometer in an Apollo spacecraft would decide the point.&nbsp; It and the spacecraft are in free fall. It's going to read zero (if properly calibrated) as a result. The question is what did the accel's read on the LEMs after they landed.&nbsp; But of course this would have to come from NASA and so ..... <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" /></p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I wonder if it isn't the opposite.&nbsp; That&nbsp;is it's where the component of gravity from each source, lying along the spacecraft's path, just cancel out....Posted by mee_n_mac</DIV></p><p>So you want a point where the vector sum of the gravitational forces from Earth and the Moon is normal to a tangent line to the trajectory of the spacecraft ?&nbsp; And probably&nbsp;in some inertial reference frame.&nbsp; At first blush this does not strike me as an easy calculation.&nbsp;</p><p>You know, as I think it about this it occurs to me that the point of these proposed calculations is to figure out what "they" are talking about.&nbsp; Since it is becoming increasing clear that "they" have no idea what they are talking about,l the calculations are probably futile.&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[Mee_n_Mac]

<p>I wuz Plucked ... my first reponse to this vanished ... arghhh</p><p>&nbsp;</p><p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>So you want a point where the vector sum of the gravitational forces from Earth and the Moon is normal to a tangent line to the trajectory of the spacecraft ?&nbsp; And probably&nbsp;in some inertial reference frame.&nbsp; At first blush this does not strike me as an easy calculation.&nbsp;Posted by DrRocket</DIV></p><p>Yup (to all questions above)&nbsp; <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" /></p><p>&nbsp;</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'></p><p>You know, as I think it about this it occurs to me that the point of these proposed calculations is to figure out what "they" are talking about.&nbsp; Since it is becoming increasing clear that "they" have no idea what they are talking about,l the calculations are probably futile.&nbsp;&nbsp; <br />Posted by DrRocket</DIV></p><p>&nbsp;</p><p>It would be futile if the purpose is to convince "them" that "they" are all wet.&nbsp; When you have to start ignoring Newtonian physics so you can believe in your "truth" ....&nbsp; <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-surprised.gif" border="0" alt="Surprised" title="Surprised" /></p><p>&nbsp;It is an interesting lunchtime problem though; what counter argument can you put forth that's simple enough so that others, not knowing who to believe, won't get suckered into the BS being posted. As Mr Lear's argument seemingly hinges on what he <u>purports</u> WvB said in a Times article, I'd think putting that in it's proper light might be fruitful.&nbsp; Again noting that a Moon w/such a surface gravity would manifest effects we don't observe is a good line though it may not be obvious to the lay person.&nbsp; I wonder if there are some more obvious effects than an increased wobble ?&nbsp; Certainly the flight path of all the Apollo's could be used as a truth but alas all data (except that supporting Mr Lear) is "suspect". </p><p>&nbsp;</p><p>For me, I'm mostly curious as to what NASA used for a criteria to demark this point in space as being worthy of mention.&nbsp; <em>Perhaps</em> there's a non-arbitrary and obvious reason for it being where it is in the flight.&nbsp; Or perhaps, as has been suggested, it's just that the convenience of ignoring the Moon's effects became "too" inaccurate in someone's opinion.&nbsp; It's somewhat fun to do the inquiry.</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>... I wonder if there are some more obvious effects than an increased wobble ?&nbsp; ...Posted by mee_n_mac</DIV></p><p>Well, if the Moon's gravity were really 6 times what it actually is, I would think that high tide might be quite an experience.<br /></p> <div class="Discussion_UserSignature"> </div>

 

[Mee_n_Mac]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Well, if the Moon's gravity were really 6 times what it actually is, I would think that high tide might be quite an experience. <br />Posted by <strong>DrRocket</strong></DIV><br /><br />Especially around the Bay of Fundy !&nbsp; But tidal heights aren't really calculable from the Earth-Moon-Sun physics.&nbsp; They depend hugely on the geography of the area and ocean bottom.&nbsp; You can calculate the tidal heights for a Waterworld and they're pretty small compared&nbsp;to what is seen in some bays.&nbsp; So how could I "prove" to a questioning person that the observed heights are due to a 0.16G Moon and not a 0.64G Moon ?</p><p>&nbsp;</p><p>(besides tides caused by gravity ?!?&nbsp; by the Sun and Moon ?!!!?&nbsp; How "Newtonian" of you&nbsp;<img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" /> )</p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Especially around the Bay of Fundy !&nbsp; But tidal heights aren't really calculable from the Earth-Moon-Sun physics.&nbsp; They depend hugely on the geography of the area and ocean bottom.&nbsp; You can calculate the tidal heights for a Waterworld and they're pretty small compared&nbsp;to what is seen in some bays.&nbsp; So how could I "prove" to a questioning person that the observed heights are due to a 0.16G Moon and not a 0.64G Moon ?&nbsp;(besides tides caused by gravity ?!?&nbsp; by the Sun and Moon ?!!!?&nbsp; How "Newtonian" of you&nbsp; ) <br />Posted by mee_n_mac</DIV></p><p>Ok if you want to calculate something, and Iagree that tidal calculation might be hard (I have no idea how to do a real one with real geography) the think about this.&nbsp; If the Moon's gravity were really 64% of Earth's then there ought to be a detectable difference in your weight between the time when the Moon is directly overhead and when it is directly on the other side of the Earth.&nbsp; I have not done the calculation, but it shoud be relatively straightforward.</p> <div class="Discussion_UserSignature"> </div>

 

[Mee_n_Mac]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Ok if you want to calculate something, and Iagree that tidal calculation might be hard (I have no idea how to do a real one with real geography) the think about this.&nbsp; If the Moon's gravity were really 64% of Earth's then there ought to be a detectable difference in your weight between the time when the Moon is directly overhead and when it is directly on the other side of the Earth.&nbsp; I have not done the calculation, but it shoud be relatively straightforward. <br />Posted by <strong>DrRocket</strong></DIV><br /><br />I had given that some thought as well.&nbsp;&nbsp;I believed instruments of such precision existed&nbsp;though not in my basement.&nbsp; Scanning the WWW it seems to have been done, usually as a&nbsp;byproduct of calibrating out the effects of the Sun and Moon on the desired experiment. <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[Smersh]

<p>I get the feeling we're losing the plot over there now [or the conspiracy theorists are.] The Pari Spolter thread is rapidly developing into a "did we really go to the Moon?" topic. Zorgon has just posted here, with a lengthy post, pointing out that NASA lost the original tapes of the Apollo 11 moonwalks. I just found an article here at space.com that does talk about this: http://www.space.com/news/060813_apollo11_tapes.html</p><p>Zorgon also posted the video of Patrick Moore asking the Apollo 11 crew if they could see the stars or not from the Lunar surface, and the crew seeming to be uncertain.&nbsp; I think I debunked that on another board once, by quoting Phil Plait on his "no stars" explanation, but I'm not sure if that would convince the HB's at Open Minds though.</p><p>I suppose all this could still be on topic with the thread, if they believe that not going to the Moon ties in with the 64% gravity claim, alleged NASA coverup, and "Spolter Physics ..."</p><p>&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1><p><font color="#800000"><em>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea."<br />Churchill: "Nancy, if you were my wife, I'd drink it."</em></font></p><p><font color="#0000ff"><strong>Website / forums </strong></font></p> </div>

 

[Smersh]

<p>Hmmm ... I just viewed that video of Patrick Moore asking the Apollo 11 crew if they saw stars again. Armstrong said he was certain they could not see stars. Aldrin made no comment, but the one who seemed unsure was Collins, who didn't land on the Moon, but remained in orbit in the CM ... </p> <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1><p><font color="#800000"><em>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea."<br />Churchill: "Nancy, if you were my wife, I'd drink it."</em></font></p><p><font color="#0000ff"><strong>Website / forums </strong></font></p> </div>

 

[Smersh]

<p>Oh hang on, Armstrong mentions first of all the daylight side, then the Solar corona, so that must mean he's referring to while they were still in orbit. I reckon Zorgon got that wrong over at OM, and thinks they are talking about when they were on the Lunar surface, but I'm not certain. </p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1><p><font color="#800000"><em>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea."<br />Churchill: "Nancy, if you were my wife, I'd drink it."</em></font></p><p><font color="#0000ff"><strong>Website / forums </strong></font></p> </div>

 

[Mee_n_Mac]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I get the feeling we're losing the plot over there now [or the conspiracy theorists are.&nbsp; {snip} I suppose all this could still be on topic with the thread, if they believe that not going to the Moon ties in with the 64% gravity claim, alleged NASA coverup, and "Spolter Physics ..."&nbsp;&nbsp; <br />Posted by <strong>Smersh</strong></DIV><br /><br />Yes the plot is being lost and no it's really not on topic.&nbsp; Standard methodology for bogus science though; never stop long enough on one thing to ascertain it's truth or falseness, keep piling&nbsp;up the BS in hopes that the sheer weight of it overwhelms any rebuttal (or tires out the rebutter).</p><p>That stars can't be seen or photographed or videoed has been answered so many times, so convincingly, I'm frankly amazed it's been brought up.&nbsp; It's this quality called dynamic range and the fact that any device, even the human perception called sight, has a DR limit.&nbsp;If the brightest object (the lunar landscape or the Sun)&nbsp;isn't going to be overexposed then the dimmest ones (other stars)&nbsp;may not be captured. It's like trying to hear a whisper at a rock concert. In his recreation of the Earth's day sky w/o an atmosphere, I really doubt that the Sun is being properly depicted; it's not as bright as it would be in real life. The composite looks like an HDR merge (photography term). Nice visual aid but not scientifically accurate (artistic picture vs simulation).&nbsp; Add that the human visual system can adjust it's dynamic range as it scans about and you find that the eye is better than the video or still camera in DR over a large area given enough time.&nbsp; If I shine a bright light in your eyes, you won't see any stars for a few minutes afterwards no matter how dark the night or bright the stars.&nbsp; Still even humans have their limits.&nbsp; Go out on a drak night and try to see some stars while a streetlight is simultaneously in your vision.</p><p>I've still not seen any proof the WvB wrote the Times article or supplied the numbers for the supposed equigrav point contained therein. I've still not seen how "they" reconcile that with WvB's much earlier article in Collier's where he did the calculation "they" like and came up with 24k miles as the equigrav point (distance).</p><p>I did look at some gravity meter data showing the Sun and Moon's effect as they passed overhead and underfoot.&nbsp; Kind of hard to see the "wheat through the chaff" so unless I can find a nice easy predigested form of the data,&nbsp;it won't help show the Moon's gravity is not 0.64G.&nbsp; Though it might be interesting to see if "they" would agree, in principal,&nbsp;that such data would be a way to determine the truth.&nbsp; The formula "they" so like to use is perfectly suitable to make the calculations and predictions and I think a 4X difference between the 2 predictions (0.16G vs 0.64G)&nbsp;should, in theory, be obvious.</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[DrRocket]

<p>[QUOTE....I suppose all this could still be on topic with the thread, if they believe that not going to the Moon ties in with the 64% gravity claim, alleged NASA coverup, and "Spolter Physics ..."&nbsp;&nbsp; <br />Posted by Smersh[/QUOTE]</p><p>You know, in formal logic a false premise implies any conclusion whatsoever.&nbsp; So I find it hard to imagine anything that could possible be considered "off topic" over there.</p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Standard methodology for bogus science though; never stop long enough on one thing to ascertain it's truth or falseness, keep piling&nbsp;up the BS in hopes that the sheer weight of it overwhelms any rebuttal (or tires out the rebutter).&nbsp; <br /> Posted by mee_n_mac</DIV></p><p>Indeed.&nbsp; "If you can't dazzle them with your brilliance, then baffle them with your bullsh*t."&nbsp; An argument from verbocity.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[Smersh]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Indeed.&nbsp; "If you can't dazzle them with your brilliance, then baffle them with your bullsh*t."&nbsp; An argument from verbocity.&nbsp; <br /> Posted by derekmcd</DIV></p><p>Aye that does seem to be the case in that thread.</p><p>Answering the stars allegation is not a problem, and the statement by the commentator on that video that the crew seemed uncertain, or "sad, nervous and worried" when questioned by Patrick Moore is absolute nonsense imo. </p><p>I don't really know what the answer to the missing tapes is though, ie why they went missing. Incompetence by NASA staff? [I know what it's like working for a large organisation, and failings of staff and the mis-management that can occur.]<br /> </p> <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1><p><font color="#800000"><em>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea."<br />Churchill: "Nancy, if you were my wife, I'd drink it."</em></font></p><p><font color="#0000ff"><strong>Website / forums </strong></font></p> </div>

 

[Mee_n_Mac]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Aye that does seem to be the case in that thread.Answering the stars allegation is not a problem, and the statement by the commentator on that video that the crew seemed uncertain, or "sad, nervous and worried" when questioned by Patrick Moore is absolute nonsense imo. Posted by Smersh</DIV></p><p>Not only nonsense but blatantly dishonest !&nbsp; The narrator "re-interprets" (I'm being polite)&nbsp;what was said in direct response to a direct question as ... Collins never seeing any stars at any time because he didn't bother to look out the window, ... something we're not supposed to believe because it's ridiculous (coverup thusly implied [insert nefarious .wav]).&nbsp; What's ridiculous is the narrator's over-reaching !!&nbsp; Collins didn't see any stars <strong><em>at the times previously mentioned</em></strong>. </p><p>As for "sad, nervous and worried" .... I didn't see it.&nbsp; I'd say they were tired of stupid PR interviews.</p><p>&nbsp;</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I don't really know what the answer to the missing tapes is though, ie why they went missing. <u>Incompetence</u> by NASA staff? [I know what it's like working for a large organisation, and failings of staff and the mis-management that can occur.] <br />Posted by Smersh</DIV></p><p>Probably the <u>above</u>, in all the years nobody seemed to want them so I guess they got all the "care and attention" deemed necessary.</p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[Kalstang]

<p>Welp it seems that we no longer have to debate the 64% lunar gravity or any other of John Lear's claims. He has left the OM forums. </p><p>[url=http://lucianarchy.proboards21.com/index.cgi?board=johnlear&action=display&thread=1633&page=11]Link</p>[/url] <div class="Discussion_UserSignature"> <font color="#ffff00"><p><font color="#3366ff">I have an answer for everything...you may not like the answer or it may not satisfy your curiosity..but it will still be an answer.</font> <br /><font color="#ff0000">"Imagination is more important then Knowledge" ~Albert Einstien~</font> <br /><font color="#cc99ff">Guns dont kill people. People kill people</font>.</p></font><p><font color="#ff6600">Solar System</font></p> </div>