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  #1141 (permalink)  
Old 16-March-2005, 10:37 PM
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Quote:
Originally Posted by TravisM
Billions of dollars. 10,000, 100,000 and 1,000,000 atoms of carbon 12, for even more clarity.

Also, predict the precession of Mecury. Show your math.

[edit to add atomic #]
Jerry? Lunatik?

<sound of crickets chirping>

I really think we should avoid responding to their posts until they address this. The fact that they cannot use their theory to predict anything demonstrates that their theory is worthless.

if the theory is predictive, then they could devise an experiment to test it.
  #1142 (permalink)  
Old 17-March-2005, 12:34 AM
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An aside: I believe any theory that describes nature more acurately than GR will most likely be "proportionally" more complex. 8)
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  #1143 (permalink)  
Old 17-March-2005, 02:17 AM
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Quote:
Originally Posted by pghnative
Quote:
Originally Posted by TravisM
Billions of dollars. 10,000, 100,000 and 1,000,000 atoms of carbon 12, for even more clarity.

Also, predict the precession of Mecury. Show your math.

[edit to add atomic #]
Jerry? Lunatik?

&lt;sound of crickets chirping>

I really think we should avoid responding to their posts until they address this. The fact that they cannot use their theory to predict anything demonstrates that their theory is worthless.

if the theory is predictive, then they could devise an experiment to test it.
Not to mention-how in the heck did Huygens get to Titan at the predicted time if it's velocity at entry into Titan's atmosphere was twice it's predicted velocity, as Jerry has claimed?

Or how about the Enceladus flyby? Predicted closest approach 313 miles, closest approach 311 miles. Seems like the effect Jerry predicted at Titan should have messed up Cassini's flyby of Enceladus as well. Of course, it was two miles off after a 7 year journey....

edited to add the Enceladus flyby.
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  #1144 (permalink)  
Old 17-March-2005, 02:34 AM
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Or how, using one of several available orbit calculators on the web, if the desired gravitational boost from any of the fly-bys was off by even 0.1% then the ensuing trajectories would not have lined the probe up for subsequent ones, thereby not allowing the craft to even come close to the saturn system to begin with...

[edited factor of %]
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  #1145 (permalink)  
Old 17-March-2005, 07:30 AM
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PLANETARY SPIN RATIOS PER AXIOMATIC EQUATIONS AS A FUNCTION OF OF VARIABLE E &amp; G:

Taking planetary black-body heat in Kelvin and planet's orbit Energy, per Axiomatic, we can arrive at a spin ratio, SR, which will yield each planet's spin as measured in earth days. The resulting equation is:

(PK/PE) :/: (13.36E-16) x (PK/231.7K) :/: (planetary orbit/365) :/: (planet spin) x (AU)^1/2 = SR

where:
SR = Spin Ratio for planets
PK = Planet Kelvin, black-body heat
PE = Planet orbit solar Energy, per Axiomatic
VK = Venus Kelvin, base planet spin (245 days retro spin, per 244 days orbit), here as base spin "zero" (where Kelvin and Energy nearly equal)
VE = Venus Energy, per Axiomatic
planetary orbit = in Earth days
planet spin = in Earth days (or fraction of Earth day)
AU = G'/G, ratio of planet's G' and Earth's G, linear deltaG = ~7.2E-11 per AU
(AU)^1/2 = same as square root of (G'/G)
Spin Ratio = SR, a Kelvin and Energy dependent number, which when Earth's SR (2.32) is divided by planet's SR yields daily planet spin.
x = times
:/: = divided by
VK/VE = (2.317K/17.33E+16 J) = 13.36E-16 , which is a ratio, used in SR equation above, as a base for Venus (near zero) parity.

The above equation for Earth yields:

(254.3K/9E+16J) :/: (13.36E-16) x (254.3K/231.7K) :/: (365/365) :/: (1) x (1) = 2.32

Tables of above values, where orbit and spin are in Earth days, and SR is as calculated per equation above:

AU; PLANET; P-KELVIN; P-ENERGY; ORBIT; SPIN; (AU)^1.2; SPIN RATIO

0.39 ; Mercury; 442K ; 60.55E16 J; 88 days; 58.8 ; 0.624 ; 0.0456
0.72 ; Venus ; 231.7K ; 17.33E16 J; 245 ; -244 ; 0.850 ; 0.005 (this is is way off - retro) ?
1.0 ; EArth ; 254,3K ; 9E16 J ; 365 days; -1- ; 1 ; 2.32 (base SR)
1.5 ; Mars ; 210.1K ; 3.86E16 J; 687 ; 1.03 ; 1.225 ; 2.246
5.2 ; Jupiter ; 110K ; 0.335E16 J; 4329 ; 0.415 ; 2.28 ; 5.41
9.5 ; Saturn ; 81.1 K; 0.1004E16J; 10753 ; 0.455 ; 3.08 ; 4.97
19.2; Uranus ; 58.1K ; 0.024E16J; 30660 ; 0.718 ; 4.38 ; 3.21
30 ; Neptune ; 46.6K ; 0.01E16J ; 60225 ; 0.673 ; 5.48 ; 3.69
39.5 ; Pluto ; ~37.5K ; 0.006E16J ; 90520 ; 6.3 ; 6.28 ; 0.304

Divide Earth's SR = 2.32 (which is our1 day spin) by any of the above planetary SR's, and you will find a close match to actual spin in Earth days. Here's the lineup:

Merc: 2.32/0.0456 = ~51 (vs. actual 58.8 )
Ven: 2.32/ 0.005 = ~464 (way off by ~x2! vs. actual -245)?
Ear: 2.32/2.32 = 1 day (our home planet's spin)
Mar: 2.32/ 2.246 = ~1.033 (vs. 1.03 actual)
Jup: 2.32/ 5.41 = ~0.428 (vs. 0.415 actual)
Sat: 2.32/ 4.97 = ~0.467 (vs. 0.455 actual)
Uran: 2.32/3.21 = ~0.723 (vs. 0.718 actual)
Nept: 2.32/ 3.69 = ~0.629 (vs. 0.673 actual)
Plut: 2.32/ 0.304 = ~7.63 (vs. 6.3 actual) ?

You can see from the above that Venus is way out of line (it has retro spin) and Pluto are off by more than 20%, Mercury is off by 12%, but the other planets are a fairly close fit. Consering I am working with rough estimates, taken from Nasa Planet tables, and used mean distances for the planets, the spin ratios as a function of planetary heat and orbit Energy come in fairly close to actual spin.

Coincidence? Numerology? Handwaving? Woowoo? Against the Mainstream? Some will think it odd, others will dislike intensely, or not believe it, and call it gibberish. Fine error? Gross error? But these are the numbers as they work out. They clearly show that there seems to be a relationship between planetary interior heat and the Energy levels in which their orbits reside, factoring in a variable G, as per the equation above. I think it interesting enough to think about some more, maybe get better K numbers, especially Mercury and Pluto, and work it out with better orbital numbers. New physics? Maybe... I am sure "careful eyes will be narrowly watching"...

Does this prove the Axiomatic Equation? Not to my mind. But it offers one more reason to go out there and look for a variable Newton's G 'proportional'. What say you now? Hogwash? I'll be curious to see your reactions. I suggest we start a different thread, since this one's run pretty deep already. This above is not a formal paper, which I will write at another time with full explanations.


Sorry guys, you won't find any of this in any textbook, it's a BABB first!

[Edited to fix Venus orbit &amp; spin days.]
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  #1146 (permalink)  
Old 17-March-2005, 09:07 AM
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Quote:
Originally Posted by Jerry
...but no one understands why, so they all put on a smiley face and say everything is nominal.
You're kidding, aren't you? This is rocket science after all, if NASA (or any other other country/space agency who has the capabilities to put something in Earth orbit) will do something like this when they have a problem (and you're saying that they do have a lot of troubles) then they'll be unable to lift a rocket in space, not talking about more complex missions! [/quote]
  #1147 (permalink)  
Old 17-March-2005, 09:26 AM
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Quote:
Originally Posted by Lunatik
You can see from the above that Venus is way out of line (it has retro spin) and Pluto are off by more than 20%, Mercury is off by 12%, but the other planets are a fairly close fit. Consering I am working with rough estimates, taken from Nasa Planet tables, and used mean distances for the planets, the spin ratios as a function of planetary heat and orbit Energy come in fairly close to actual spin.
It's the spin ratio as a function of planetary heat, orbit energy, and spin itself, that come close to actual spin in your calculations. You use spin to calculate spin. What happens to your equations if you take the result out of the function?

And there is still the basic question of TravisM and many others, a few posts above this one...
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  #1148 (permalink)  
Old 17-March-2005, 01:36 PM
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Quote:
Originally Posted by Lunatik
Quote:
Originally Posted by papageno
As usual, you run away and stick your head in sand, instead of addressing criticism to your ideas.
You have not addressed the points I made in this post.
Do you still think that you are doing good science?
I am presenting an idea. Testing that idea is good science. Your "stick you head in sand" is an editorial comment; I do not respond to those, your opinion only, be advised. Come back to me with your calculations, and we'll talk.
You do not respond to my physical arguments.
From your posts, it seems that you ignore them, because you do not actually understand and you would not know how to address them.
Why do you expect me to do calculations?
I gave references to tests of Newtonian gravity and General Relativity.
These tests put constraints on non-Newtonian gravity that are much tighter than your variable G "speculation" requires.
In response you claimed without backup that those papers did not actually test variable G theories. This clearly shows that you do not understand what you are talking about.

And you still have not addressed the points I made here.
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  #1149 (permalink)  
Old 17-March-2005, 01:36 PM
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Quote:
Originally Posted by TravisM
Billions of dollars. 10,000, 100,000 and 1,000,000 atoms of carbon 12, for even more clarity.

Also, predict the precession of Mecury. Show your math.

[edit to add atomic #]
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  #1150 (permalink)  
Old 17-March-2005, 02:53 PM
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Quote:
Originally Posted by Lunatik
VK/VE = (2.317K/17.33E+16 J) = 13.36E-16 , which is a ratio, used in SR equation above, as a base for Venus (near zero) parity.

The above equation for Earth yields:

(254.3K/9E+16J) :/: (13.36E-16) x (254.3K/231.7K) :/: (365/365) :/: (1) x (1) = 2.32

Tables of above values, where orbit and spin are in Earth days, and SR is as calculated per equation above:
Lunatik, don't you find it odd that the planet (Venus) that you used to create a base line spin ratio is the only planet that is not even close to the actual spin in your formula?

If the formula that you used to create the baseline figures don't match the the object that you used to create the base line figures in the first place, all other numbers in your table is circumspect, regardless of how close the other numbers match up.
  #1151 (permalink)  
Old 17-March-2005, 03:09 PM
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Quote:
Originally Posted by Metricyard
Quote:
Originally Posted by Lunatik
VK/VE = (2.317K/17.33E+16 J) = 13.36E-16 , which is a ratio, used in SR equation above, as a base for Venus (near zero) parity.

The above equation for Earth yields:

(254.3K/9E+16J) :/: (13.36E-16) x (254.3K/231.7K) :/: (365/365) :/: (1) x (1) = 2.32

Tables of above values, where orbit and spin are in Earth days, and SR is as calculated per equation above:
Lunatik, don't you find it odd that the planet (Venus) that you used to create a base line spin ratio is the only planet that is not even close to the actual spin in your formula?

If the formula that you used to create the baseline figures don't match the the object that you used to create the base line figures in the first place, all other numbers in your table is circumspect, regardless of how close the other numbers match up.
And isn't Lunatik's formule a bit top heavy? I mean, you end up with (for Earth), 2.32, which has no special meaning, I believe (please correct me someone if I'm wrong). You can remove from your equation pK (which is didvided by itself), and all constants. You end up with
(1/PE) :/: planetary orbit :/: planet spin * (AU^1/2),
if I simplify it correctly and keep using your notation. Is there any reason why you make it more complicated than this? It doesn't change the correctness or incorrectness at all, but it's a lot easier to read and calculate, and it gets a lot more visible that you use planet spin to calculate your planet spin...
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  #1152 (permalink)  
Old 17-March-2005, 05:52 PM
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Quote:
Originally Posted by Metricyard
Lunatik, don't you find it odd that the planet (Venus) that you used to create a base line spin ratio is the only planet that is not even close to the actual spin in your formula?
Yes, I do. Venus represents as close an approximation to "zero spin" where its spin is in sync with its orbit, except slightly retro, 245/244. If Venus had a real spin, I'd expect it to fall into line. I used it for a "base spin" where hypothetically the Kelvin temp is on par with its orbit Energy. I will address this issue when I write a formal paper on this, so don't know more for now. It does not change the overall relationships, as shown with the results.

Quote:
Originally Posted by Fram
And isn't Lunatik's formule a bit top heavy? I mean, you end up with (for Earth), 2.32, which has no special meaning, I believe (please correct me someone if I'm wrong). You can remove from your equation pK (which is didvided by itself), and all constants. You end up with
(1/PE) :/: planetary orbit :/: planet spin * (AU^1/2),
if I simplify it correctly and keep using your notation.
I had tried this simplification, but the numbers don't work. Do you get the same results in yours? Let me know if you do, because I think simplifying it would be grand.

The spin ratio of Earth SR = 2.32 is a number relating Kelvin planet heat to its orbit Energy, which means there is more black body heat K than E by this ratio, as I understand it. I don't have more information than that for now.

I did check to see if perhaps these numbers are merely arbitrary, meaning that internal numbers of equation are just expressing themselves; but if that was so, all the results would be consistent, which they are not, as Pluto and Venus show. The other planets have a closer fit to their actual spin as expressed in Earth days. I keep looking for how this could occur naturally (numerologically) but have not found it. If you see it, let me know! Thanks.

About other issues, Halley's, papageno's physics, Mercury's precession, I have no answers at this time. Perhaps in the future... Rome was not built in a day. 8)
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  #1153 (permalink)  
Old 17-March-2005, 06:26 PM
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Quote:
Originally Posted by Fram
Quote:
Originally Posted by Lunatik
You can see from the above that Venus is way out of line (it has retro spin) and Pluto are off by more than 20%, Mercury is off by 12%, but the other planets are a fairly close fit. Consering I am working with rough estimates, taken from Nasa Planet tables, and used mean distances for the planets, the spin ratios as a function of planetary heat and orbit Energy come in fairly close to actual spin.
It's the spin ratio as a function of planetary heat, orbit energy, and spin itself, that come close to actual spin in your calculations. You use spin to calculate spin. What happens to your equations if you take the result out of the function? ...
I've thought about this also. My simple answer is that the above equation is a ratio of spin. If this is what it represent, a Spin Ratio, how can it be a ratio if spin is not part of it?
Quote:
Originally Posted by Lunatik
(PK/PE) :/: (13.36E-16) x (PK/231.7K) :/: (planetary orbit/365) :/: (planet spin) x (AU)^1/2 = SR
Note that I standardized those numbers that are constants, but planet spin is what this ratio is made of, so it needs to be included or no ratio results, if I understand it right.
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Old 17-March-2005, 06:56 PM
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Originally Posted by Lunatik
About other issues, Halley's, papageno's physics, Mercury's precession, I have no answers at this time.
Or in other words: "My ideas, as presented, do not work".

And don't forget to add your idea's incorrect prediction on inertial mass to the list.
  #1155 (permalink)  
Old 17-March-2005, 07:42 PM
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Quote:
Originally Posted by Lunatik
Quote:
Originally Posted by Fram
Quote:
Originally Posted by Lunatik
You can see from the above that Venus is way out of line (it has retro spin) and Pluto are off by more than 20%, Mercury is off by 12%, but the other planets are a fairly close fit. Consering I am working with rough estimates, taken from Nasa Planet tables, and used mean distances for the planets, the spin ratios as a function of planetary heat and orbit Energy come in fairly close to actual spin.
It's the spin ratio as a function of planetary heat, orbit energy, and spin itself, that come close to actual spin in your calculations. You use spin to calculate spin. What happens to your equations if you take the result out of the function? ...
I've thought about this also. My simple answer is that the above equation is a ratio of spin. If this is what it represent, a Spin Ratio, how can it be a ratio if spin is not part of it?
Quote:
Originally Posted by Lunatik
(PK/PE) :/: (13.36E-16) x (PK/231.7K) :/: (planetary orbit/365) :/: (planet spin) x (AU)^1/2 = SR
Note that I standardized those numbers that are constants, but planet spin is what this ratio is made of, so it needs to be included or no ratio results, if I understand it right.
You first use it to calculate Spin Ratio, but then you use that to calculate the Spin, as you know well enough, as that is the crux of your numerological trick. A bit of circular reasoning...
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Old 17-March-2005, 07:49 PM
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Quote:
Originally Posted by Lunatik
Quote:
Originally Posted by Fram
And isn't Lunatik's formule a bit top heavy? I mean, you end up with (for Earth), 2.32, which has no special meaning, I believe (please correct me someone if I'm wrong). You can remove from your equation pK (which is didvided by itself), and all constants. You end up with
(1/PE) :/: planetary orbit :/: planet spin * (AU^1/2),
if I simplify it correctly and keep using your notation.
I had tried this simplification, but the numbers don't work. Do you get the same results in yours? Let me know if you do, because I think simplifying it would be grand.
What do you mean, the numbers don't work? You divide by 365. That does nothing, that doesn't change the result one bit, it just gives you a different number in the end, but the different planet numbers relate to each other just the same as they did before. The same with all the other constants. They are useless there. If you don't get the same results, try again. It does cast a doubt on your first results, if you cannot calculate them good enough. You have brought in a lot of things that are not relevant and only muddle the equation.
I have not calculated with your equation and again with my equation, as that is not necessary. The simplification is correct or is not correct. But don't expect to get 2.32 again, as that will not happen, and isn't necessary.
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Old 17-March-2005, 08:33 PM
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Jerry wrote:

Quote:
No, well yes and no, I don't see how you could differentiate between acceleration caused by wind, and acceleration caused by gravity - the problem is very complex, because there are so many planes of motion to factor in - there is a rotational acceleration as well.
Gravity is the restoring force, wind forces would be the impulse.

Quote:
At least to a couple of significant figures this may be true. Still, you must have a complete profile of all accelerations, and you must know the kinetic energy of Huygens at all times -
Rubbish! you do not need to know the KE of an adult or a child on a swing or a Marine on a parachute to be able to determine the natural period. It is purely dependant on the effective lenght of suspension and local gravity. I thought I had made that clear.
According to the ESA, the wind velocity was always changing. If the rate of change was constant, you are correct, but if the wind has a periodic function, you will not be able to separate the wind force from the natural period. Since the probe was rotating and is not completely symmetric aerodynamically, this has the potential of imposing a harmonic on the wind deflections. This is not a simple problem.
Quote:
Quote:
remember, before Huygens started experiencing any atmospheric drag, the accelerometers would not detect acceleration during free-fall.
The accelerometers measured acceleration in xyz real time. How do you think they could determine that Huygens rocked at all during its descent? I also chose my words carefully and said "Now, if I could find the data for the effective lenght of the parachute tether I could give you a reasonable figure for the expected frequency of oscilation of the Huygens probe imediately prior to touch down. " (My bold added)
Yes! If you can pull this information together, and assume the parachute is in equalibrium, this should work, or at least provide a reasonably good number. Good luck finding the data.

Quote:
Originally Posted by frog
Quote:
There are at least three descent profiles that can be constructed: Doppler based, acceleration based, and temperature/pressure based; and variations using ground radar and sonar. All of these profiles must be in reasonable agreement with each other, and trace a reasonable path all the way from release from Cassini to the landing.

I think they are running into the same family of inconsistencies trying to map a complete descent profile for Cassini that they have run into trying to map the entry, descent and landing of Spirit, Opportunity and every other Mars probe. What happened in the upper atmosphere of Titan is going to be under scrutiny for a very long time - this is now a siege thread.
You are trying to deflect from the fact that Huygens landed safely where you predicted disaster. You were just plain wrng on the first page of this thread and rather than admit your eror and seek truth you keep up this pretence.
I predicted a rapid descent, and all of the evidence released to date is consistent with that prediction, except for the partial reconstruction of the descent profile.

Quote:
Others will have their own views but personally I don't like to see nonsense propagated as fact when there are possibly students reading this board who may get erronious ideas from you. There is no seige or agenda on my part, only a desire to have a clearer understanding of the universe and her mysteries. So far I have learnt none from you, however, the thread is worthwhile because the systematic refution of all your claims by some very able members of this board who have given unstintingly of their time, have provided me with a far greater insight into the cosmos than I had previously. I heartily thank them for that =D> =D> =D>
All my claims? The ESA has only released preliminary descent profiles without enough information to fully characterize the descent, and the images released are completely inconsistent with that model.

Elias insists the images are not duplicates, yet anyone willing to take the time to look at them can clearly tell that they are.

No one has even tried to offer another explanation for the Mooning Titanian. If this is not the heat shield, what is it? Why is there no explanation for duplicate images and the curious images of a very round object? I don't know how anyone could look at these images, (on up through page 23), and say the camera taking these pictures was on a rotating platform: They don't rotate. What's going on?

If I am wrong, the Pressure-verses-time data during the descent should be consistent with known gas laws, increasing at a predictable rate through the entire descent. If I am right, the gas pressure will show precious little variance after the first twenty to thirty minutes of descent. Simple test. Where is this data?
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Old 17-March-2005, 10:37 PM
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Quote:
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I predicted a rapid descent, and all of the evidence released to date is consistent with that prediction, except for the partial reconstruction of the descent profile.
LOL! None of the data released to date agree with your prediction! We may not have the complete descent profile, but we do know that:
Code:
1) The parachute sequence occurred within seconds of predicted time.
2) We know how long it took Huygens to reach the surface.
3) We know how long Huygens transmitted data after it landed.
4) We have the preliminary radio telescope data which showed they got the signal from Huygens at the predicted time and could even detect the stabilization parachute deployment. They also detected when Huygens landed.
There is more, but this is enough to show that you predictions of a rapid descent are false. In fact, the last 40+ pages on this topic have shown your predictions do not hold water.

Quote:
All my claims?
Please indicate one specific claim that has been borne out by the data.

Quote:
The ESA has only released preliminary descent profiles without enough information to fully characterize the descent, and the images released are completely inconsistent with that model.
Who have you contacted at the ESA to get this data? The preliminary descent profile is enough to show your prediction is incorrect.

Quote:
No one has even tried to offer another explanation for the Mooning Titanian. If this is not the heat shield, what is it?
I don't see anything resembling a heat shield in these images and I've yet to see anyone else make that peculiar claim.

Quote:
If I am wrong, the Pressure-verses-time data during the descent should be consistent with known gas laws, increasing at a predictable rate through the entire descent. If I am right, the gas pressure will show precious little variance after the first twenty to thirty minutes of descent. Simple test.
"After the first twenty to thirty minutes of descent". I thought you said it was rapid. Didn't you previously claim that poor Huygens wasn't able to eject the heat shield until right near the surface? Didn't you claim the heat shield hit the surface, bounced and then tagged the probe? If the heat shield didn't deploy until right near the surface then there'll be no "Pressure-verses-time data" to look at. The instruments require the heat shield to be gone before they can work.

Quote:
Where is this data?
It's being analyzed by people who know what they're talking about. Who have you contacted at the ESA to get this data?
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  #1159 (permalink)  
Old 17-March-2005, 11:06 PM
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Quote:
Originally Posted by TravisM
Billions of dollars. 10,000, 100,000 and 1,000,000 atoms of carbon 12, for even more clarity.

Also, predict the precession of Mecury. Show your math.

[edit to add atomic #]
[quote="Jerry"]Morris Anderson has posted his "the universe in wave equations" book on the web.

This includes wave equations that predict the precession, page 77 of Time, Matter, and Gravity.

There are a 101 mathematical models out there that look at GR from wave mechanics prospectives. Morris varies the speed of light, as I do, rather than time.
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  #1160 (permalink)  
Old 17-March-2005, 11:11 PM
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Quote:
Originally Posted by Jerry
Quote:
Originally Posted by frog
You are trying to deflect from the fact that Huygens landed safely where you predicted disaster. You were just plain wrng on the first page of this thread and rather than admit your eror and seek truth you keep up this pretence.
I predicted a rapid descent, ...
Um....no Jerry, you did in fact predict disaster. From the original post (remember that???), with emphasis added.

Quote:
Originally Posted by Jerry
Normally, I post a topic like this on the “Against the Mainstream” thread. This is because virtually everything I do is highly speculative. This time, I believe there is a serious threat to the Huygen mission, &lt;snip>.

I have concluded the same failure mode that lead to the demise of Beagle, and indirectly two other Mars probes is inherent in the orbital mechanics of the Huygens probe scheduled to separate from Cassini December 26, 2004. ...

This past Summer, the Beagle expedition ended as another of a long line of Mars failures &lt;snip> What is most curious about this finding is that the successful landings of both Spirit and Opportunity still experienced flight anomalies that are consistent with the Beagle scenario: .&lt;snip>

I believe these anomalies occurred because of an intrinsic pressure differential in the Martian atmosphere, not an anomalous event. I will further demonstrate the root cause of the failure of Polar Orbiter, the Global Surveyor and at least four other Martian missions is the same, and that Huygens will almost certainly fail unless new physical considerations are included in the descent profile.

&lt;snip> If these predictions are correct, the actual density of Titan is ~ 4.42g/cc, more than twice the current theoretical value (1.88g/cc)
edited to add: The phrases "serious threat" and "will almost certainly fail" are pretty close to "disaster" in the minds of most reasonable people. Certainly they are closer to meaning "disaster" than "I predicted a rapid descent". Note that the adjective "rapid" is meaningless. Any velocity could be "rapid".
  #1161 (permalink)  
Old 17-March-2005, 11:14 PM
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Oh, and by the way, both Jerry and Lunatik continue to ignore this simple request:

Quote:
Originally Posted by TravisM
Quote:
Originally Posted by TravisM
Billions of dollars. 10,000, 100,000 and 1,000,000 atoms of carbon 12, for even more clarity.
&lt;snip>
If you can formulate a theory, you can formulate a test of the theory. The fact that they refuse to formulate a test indicates that their theory does not hold water.
  #1162 (permalink)  
Old 17-March-2005, 11:27 PM
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Jerry wrote:

Quote:
...

If I am wrong, the Pressure-verses-time data during the descent should be consistent with known gas laws, increasing at a predictable rate through the entire descent. If I am right, the gas pressure will show precious little variance after the first twenty to thirty minutes of descent. Simple test. Where is this data?
Here Results of Doppler Wind Data. A data set that makes far more sense than you have postulated.
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  #1163 (permalink)  
Old 17-March-2005, 11:45 PM
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Quote:
Originally Posted by Fram
You first use it to calculate Spin Ratio, but then you use that to calculate the Spin, as you know well enough, as that is the crux of your numerological trick. A bit of circular reasoning...
Very good point, thanks for bringing it up. When I first wrote it, just before I posted it, I decided that the post was not clear, since most people looking at it wouldn't understand what I meant by SR. This everyone should know is how planetary black body Kelvin relates to planet's orbit Energy. The resulting SR without adjusting for deltaG (via sqrt of AU) gives readings that plot on a curve, but not direct. Once I incorporated deltaG, it all fell together. That was worked out in the first part. So SR, spin ratio, resulted, a ratio of the K/E relationship divided by spin. But this meant something to me, but I feared it wouldn't communicate to others, so I wrote the second part showing how the resulting SR relates back to each planet's spin.

Now, my point: I then decided to list the second part was for illustration purposes only, to help clarify how SR works, and not as a proof of it. I hope this makes sense, but that was how I saw it in my mind. I guess it was still not as clear as I had hoped, as you brought it up, a very important point.

Quote:
Originally Posted by Fram
I have not calculated with your equation and again with my equation, as that is not necessary. The simplification is correct or is not correct. But don't expect to get 2.32 again, as that will not happen, and isn't necessary.
I wish I could see what you're getting at, but I'm missing it. Would you have an illustration to show as a numerical example? Thanks.

Quote:
Originally Posted by Tassel
Quote:
Originally Posted by Lunatik
About other issues, Halley's, papageno's physics, Mercury's precession, I have no answers at this time.
Or in other words: "My ideas, as presented, do not work".

And don't forget to add your idea's incorrect prediction on inertial mass to the list.
[/quote]
I guess you missed the second part of that same post of which you only copied the first part, which answers yours:
Quote:
Originally Posted by Lunatik
Perhaps in the future... Rome was not built in a day. 8)
Quote:
Originally Posted by pghnative
Oh, and by the way, both Jerry and Lunatik continue to ignore this simple request:

Quote:
Originally Posted by TravisM
Quote:
Originally Posted by TravisM
Billions of dollars. 10,000, 100,000 and 1,000,000 atoms of carbon 12, for even more clarity.
&lt;snip>
If you can formulate a theory, you can formulate a test of the theory. The fact that they refuse to formulate a test indicates that their theory does not hold water.
Ditto. :roll:

Quote:
Originally Posted by papageno
And you still have not addressed the points I made here.
I guess you missed my response to that here.

Well, I do appreciate your constructive criticism, and I am looking at this with a weary eye, because if there is a flaw in it, I want to know. Perhaps when I sit down to write a paper on it, something will come up, since each part of the equation will have to be explained. But for now, your alerts to possible errors are really appreciated. You are gentlemen and scholars.

-----------------------------------------------------------------------------------------
Please note, as this discussion is separating from the Huygens Mission, I thought to also list it on its own thread here, so it doesn't turn into a priveate 'chat room'. All are welcome.
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  #1164 (permalink)  
Old 18-March-2005, 12:55 AM
Metricyard Metricyard is offline
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Quote:
Originally Posted by Lunatik
When I first wrote it, just before I posted it, I decided that the post was not clear, since most people looking at it wouldn't understand what I meant by SR. This everyone should know is how planetary black body Kelvin relates to planet's orbit Energy. The resulting SR without adjusting for deltaG (via sqrt of AU) gives readings that plot on a curve, but not direct. Once I incorporated deltaG, it all fell together. That was worked out in the first part. So SR, spin ratio, resulted, a ratio of the K/E relationship divided by spin. But this meant something to me, but I feared it wouldn't communicate to others, so I wrote the second part showing how the resulting SR relates back to each planet's spin.
What I can't figure out is why you didn't use the Earth for your Spin Ratio (SR) baseline. Seems it would have been easier then having to incoroprate Venus's data into earths data. Plus your variable G, if I recall correctly, is supposed to equal 1 on Earth. Plus you would have your K/E relations exactly match the spin ratio of 1:1.

You do seem to like over complicating things.
  #1165 (permalink)  
Old 18-March-2005, 08:24 AM
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Quote:
Originally Posted by Lunatik
Quote:
Originally Posted by Fram
I have not calculated with your equation and again with my equation, as that is not necessary. The simplification is correct or is not correct. But don't expect to get 2.32 again, as that will not happen, and isn't necessary.
I wish I could see what you're getting at, but I'm missing it. Would you have an illustration to show as a numerical example? Thanks.
Well, for one, in your formula, you at one point divide by 365. You do this for all planets, not just Earth. You can just as well leave this out of the equation, as it doesn't change the comparative values between the planets. I mean, now you divide all results by 2.32, and then you would divide all results by (2.32 / 365) or (2.32 * 365), I can't bother looking up which of the two you would end up with. Those constants add nothing when calculating the spin, they are only important for the spin ratio, but I don't know if that number as such is important. It didn't come across as important in your first post about this, but that may be my misunderstanding.
And even in the spin ratio calculation, you use PK twice, and those do erase one another (PK/PK), so at least that part can be left out of it.
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  #1166 (permalink)  
Old 18-March-2005, 10:07 AM
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Quote:
Originally Posted by Lunatik
Quote:
Originally Posted by papageno
And you still have not addressed the points I made here.
I guess you missed my response to that here.
My reply to that post, was the post where I explained what a linear function is.
You acknowledged that post right there.
Does it mean that you actually did not understand my post, hence you did not realize that it was a reply to yours?

By the way, the points you did not address are these:
Quote:
Originally Posted by papageno
But your F' is the force, not the potential.
The gravitational potential is the integral of force over a path.
Now you are mixing up energy and force: did you actually read the link you gave?

[...]

"One body equation": how do you get a gravitational interaction with one body? **
(** Maybe he means a mass in an external gravitational field (for example, an apple in the Earth's field). But you still need another mass to be the source of this "external" field.)

[...]

And which G would you use? The one at the Earth or the one at Saturn?
Wouldn't it depend on the relative distance?
Would you still get Kepler's laws?


Quote:
Originally Posted by Lunatik
Well, I do appreciate your constructive criticism, and I am looking at this with a weary eye, because if there is a flaw in it, I want to know.
You actions disprove your words: you keep ignoring my criticism.
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  #1167 (permalink)  
Old 18-March-2005, 01:13 PM
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Quote:
Originally Posted by Lunatik
Quote:
Originally Posted by pghnative
Quote:
Originally Posted by TravisM
Quote:
Originally Posted by TravisM
Billions of dollars. 10,000, 100,000 and 1,000,000 atoms of carbon 12, for even more clarity.
&lt;snip>
If you can formulate a theory, you can formulate a test of the theory. The fact that they refuse to formulate a test indicates that their theory does not hold water.
Ditto. :roll:
How hard can it be to formulate a test of your theory?
  #1168 (permalink)  
Old 18-March-2005, 01:15 PM
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Quote:
Originally Posted by pghnative
Quote:
Originally Posted by Lunatik
Quote:
Originally Posted by pghnative
Quote:
Originally Posted by TravisM
Quote:
Originally Posted by TravisM
Billions of dollars. 10,000, 100,000 and 1,000,000 atoms of carbon 12, for even more clarity.
&lt;snip>
If you can formulate a theory, you can formulate a test of the theory. The fact that they refuse to formulate a test indicates that their theory does not hold water.
Ditto. :roll:
How hard can it be to formulate a test of your theory?
Maybe he meant that his theory does not hold water.
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  #1169 (permalink)  
Old 18-March-2005, 01:24 PM
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=D> It's like this smiley, but with more waving...
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  #1170 (permalink)  
Old 18-March-2005, 02:06 PM
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Regarding the "pendulum motion" of the probe underneath the parachute:

I just have to note that it actually deviates a lot from pendulum motion because the rope that attached the probe to the parachute included a decoupling device to make the motion of the probe as less dependent as possible from the motion (rotation + pendulum motion) of the parachute. So, it is quite complex. A better approximation would be two different pendulums connected, with a certain "coupling factor" (if one could name it like this). This coupling factor is different for rotation and for the typical pendulum motion...

Decoupling devices were also present I think for the MER landers (and possibly in most landers)

You can see the effects of this decoupling in the Spirit descent reconstruction video (18.8 MB):

http://marsrovers.jpl.nasa.gov/galle..._Animation.avi
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