Conundrum

5.9k · March 20, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

1.3k · March 20, 2004

Okay, I got 3.7283 repeating. What'd you get?

24.7k · March 20, 2004

Crap. I've been doing it wrong. That's the very thing I'm thinking of when I fall to sleep. It's thinking about women that keeps me awake!

3.8k · March 20, 2004

Edit:

Edited March 20, 2004 by sept1c_tank

March 20, 2004

N.C.T.W.W.A.S.B.E. - Always sorts it out for me, well close enough for jazz and geocaching anyhow.

March 20, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

Uh, I would say the answer is definitely B.

Or maybe the answer is True.

Oh forget it, I'll just take whatever is in the box!!! No no, wait, door number 2!

2.2k · March 20, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

Uh, I would say the answer is definitely B.

Or maybe the answer is True.

Oh forget it, I'll just take whatever is in the box!!! No no, wait, door number 2!

can't i just by a vowel?

March 20, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

Uh, I would say the answer is definitely B.

Or maybe the answer is True.

Oh forget it, I'll just take whatever is in the box!!! No no, wait, door number 2!

can't i just by a vowel?

At this time of night? No, just take one...

March 20, 2004

NO THE ANSWER IS ALWAYS

NONE OF THE ABOVE

or is it

ALL OF THE ABOVE

now I'm going to be awake trying to figure out this answer

I wish these perplexing questions wouldn't come on line so late

:blink:

2.2k · March 20, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

Where's Urkel when you need him.

March 20, 2004

The answer is YES .

Well Maybe... err... it's no... OKAY it's NO!

OH heck I forgot the constant for RHO..

Oh never mind... I thought you said "INTERSTITIAL" when you said "INTERSPATIAL"

OOPS... you did say "Interstitial".... so OK...

I HATE quantum mechanics (until next year).

3.1415926583587979727...

I hafta get back to ya...........................................

(edit: fixed INTERSTITIAL .. INTERSPATIAL .. thingy)

darn it's Saturday and you're making me think?!?!?

Edited March 20, 2004 by davwil

March 20, 2004

Just tell him that the answer is "42" and don't lose any more sleep. :blink:

March 20, 2004

Wait a minute.. that's the answer I've been looking for!

March 20, 2004

Well, what you have is incomplete informtion.

In order to get your answer down to the millimeter level you need to apply the inverse number of the Darwinian endpoint where B is divided by S (B/S). This number can be squared again into a rectangle of 8 x 4 where the expansion of gas (per Boyle's Law) allows you the configuration to permeate the entire answer in a few short breaths.

March 20, 2004

They don't teach you words like that at the school I went to.

Edited March 20, 2004 by Team Lyons

March 20, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

The problem here is, you have forgotten the unladen traveling velocity of he African swallow vs. the european swallow.

2k · March 20, 2004

Okay, I got 3.7283 repeating. What'd you get?

You forgot to carry a 1. It's 3.8283 repeating ...

March 20, 2004

I can’t sleep until I figure this out.

As long as you're awake, go out and find a night cache. :blink:

March 20, 2004

Well, first you're going to have to deviate your variances. Then you'll need to variate your deviances. Next, determine whethera free and democratic institution such as our could eventually consider evolutionary humanism as a prerequisite to the insipidness that prevails today. After you've done this, run it through cmconverter, GPX Spinner, Watcher, then cmconverter again. I checked my answer with my broken magic 8-ball and got the same results.

The answer is no. With a handheld HP1295 the best you're going to get (with any accuracy) is 1/4 centimeter.

1.5k · March 20, 2004

or hide one.

38.2k · March 20, 2004

It is my feeling that the critical isomers in the potassium ions will cause the unsaturated hydroxy steroid keytone, formed by the oxidation of the malto-dextrose and neurochemical proteins to yield progesterone upon dehydrogenation, thus negating the effect of the pre-positive charge. This is because the critical isomers will cause an un-solicited lactate steroid reaction resulting in the lubricious elements forming an oxidized dextrose compound, which will bond the ions. Similarly, the lactate steroid reaction will only progenerate if the compounding of the lubricious elements convert molecular zygotes into a bi-particle isomeric isomer before precipitating the suspended silicate and undergoing a re-generation of the asynchronous potassium ions.

Some believe that when neutral heloiogenic compounds are progenerated, it causes the pseudo-hydroid enzymes to bond incongruously with the potash silicate to form convergent hydrogen silicates, but what they don’t realize is that if these silicates bond, the effect will not occur when there is more that one synthesized molecular reaction. Furthermore, the neutral heliogenic compounds will not exhibit any random variable areocentric reaction without first undergoing a polymer synthesis. This is largely because the amnio derivatives of the linear di methyl alkaloids contain recessive compounds. The effect here is the distillates becoming a stabilized, heliocentric compound which manifests it itself in a fibrous, columnoid form. But lets not forget that the hydroxy keytone doesn’t form any bi-partical isogenic isomers when the complex compound elements convert tetrogenic zygotes into congruously bonded hydrostatic isomers linked with the bonding of the crystalline linear isomers. Now in the case where there is a thermostatic charge, the columnoid elements have a predilection towards oxidation throughout the poly-nucletide syntheses. In no case does this mean that acceleration of these particles will incur a crypto-genic peptide bond unless there is a prior concentration of the sub-atomic particles and a quarkite convergence with the distichous compounds. If so, it holds that of the napthol strands are stabilized by synthesized purine elements in a pyramadine base, which is effectively, the phagocytic destruction of the pathogens. This begs for a our reclassifying the bonding of the bi-particle oxides and mono-nuclear solinoids thatl manifest themselves in an anamapohorous fusible substance, the exact form of which has yet to be determined. So sleep well.

1.6k · March 20, 2004

The answer to EVERYTHING is definately 42...

But then you have to wonder, what is the question?

March 20, 2004

Where's my seizure medication when I need it the most? :blink:

March 20, 2004

Okay, I've read all of the posts. Then I printed them and ran them through a Troybilt Chipper/Shredder that I have installed a 454 V8 with 4-barrel carb in to get a workable average. Using that average I have come to the conclusion that all of the caches in the world are off by 272.35 meters at 47.89 degrees.

I am going to go out and start moving the caches to the right coordinates. :blink:

March 20, 2004

Okay, I've read all of the posts. Then I printed them and ran them through a Troybilt Chipper/Shredder that I have installed a 454 V8 with 4-barrel carb in to get a workable average. Using that average I have come to the conclusion that all of the caches in the world are off by 272.35 meters at 47.89 degrees.
I am going to go out and start moving the caches to the right coordinates.

They are THAT far out? WOW, thanks for all your work, will you please start by moving mine? Thanks so much.....

March 20, 2004

... can I narrow the total corrected search field to millimeters?

So basically, If we apply Einstein’s quantum mechanical theory to satellites with respect to earth, can we get better result?

Quick answer: no.

All of the appreciable error is introduced by ionic distortion, inaccurately knowing satellite position (speed will no help as much as position), reflected waves, etc. Some relativistic calculations are considered in your little GPS, but even the added precisions there is far greater than would be added by Einstein’s 10th.

Einstein's quantum theory works great for determining minority carrier drift in semi-conductors, but when considering thousands of miles of distance, the small deviations are negligible.

My interpretation is it comes down to this: There are a lot of unknowns that we would have to know in order to get sub millimeter accuracy. Einstein’s 10th is not going to help you.

see RELATIVITY

But I only have a TI-85. :blink:

Okay did i sound geeky enough, or do I need to add some more words so most people can't read it?

Edited March 20, 2004 by geckoee

1.1k · March 20, 2004

I have forwarded your question to my friend Will Hunting. He is pretty good at this stuff.

5.9k · March 20, 2004

It occurred to me, that if one could additionally determine the satellite pitch, yaw, and roll, relative to the center of gravity after establishing a true datum line, one could, in theory, make manual drift corrections. Further, by measuring density altitude and applying the corrections for total atmospheric particulate in percent of tonnage for the given atmosphere between the sending satellite transmitter and the receiving unit, one could apply sub millimeter rectification to the WAAS corrected signal, thus realizing atomic level field placement for any given location on or below the surface of the earth.

Thoughts?

2.5k · March 20, 2004

Reminds me of when this lawn supervisor was out on a sprinkler maintenance job and he started working on a Findlay sprinkler head with a Langstrom 7" gangly wrench. Just then, this little apprentice leaned over and said, "You can't work on a Findlay sprinkler head with a Langstrom 7" wrench." Well this infuriated the supervisor, so he went and got Volume 14 of the Kinsley manual, and he reads to him and says, "The Langstrom 7" wrench can be used with the Findlay sprocket." Just then, the little apprentice leaned over and said, "It says sprocket not socket!"

2.2k · March 20, 2004

It occurred to me, that if one could additionally determine the satellite pitch, yaw, and roll, relative to the center of gravity after establishing a true datum line, one could, in theory, make manual drift corrections. Further, by measuring density altitude and applying the corrections for total atmospheric particulate in percent of tonnage for the given atmosphere between the sending satellite transmitter and the receiving unit, one could apply sub millimeter rectification to the WAAS corrected signal, thus realizing atomic level field placement for any given location on or below the surface of the earth.

Thoughts?

thoughts? we don't need no stinkin' thoughts. we're geocachers. :blink:

March 20, 2004

Everyone except pater is missing the most important part here:

can I narrow the total corrected search field to millimeters?

And the answer is no. I have found that most people miss the tree for the forest and thats what you are doing here.

March 20, 2004

Everyone except pater is missing the most important part here:

can I narrow the total corrected search field to millimeters?

And the answer is no. I have found that most people miss the tree for the forest and thats what you are doing here.

Even the Tadpoles have me on ignore. That's it. Forumiside is the only answer!

jk.

3.9k · March 20, 2004

It occurred to me, that if one could additionally determine the satellite pitch, yaw, and roll, relative to the center of gravity after establishing a true datum line, one could, in theory, make manual drift corrections. Further, by measuring density altitude and applying the corrections for total atmospheric particulate in percent of tonnage for the given atmosphere between the sending satellite transmitter and the receiving unit, one could apply sub millimeter rectification to the WAAS corrected signal, thus realizing atomic level field placement for any given location on or below the surface of the earth.

Thoughts?

I have spent days researching this very thing. Thanks for clearing it all up!! :blink:

Now if you could just get me the coordinates to the Keebler Cookie factory (I just have to know if they are really made by elves.) my life will be complete.

El Diablo

3.8k · March 20, 2004

Now this whole question hinges on whether you are using a Garmin or Magellan GPS to determine the number of satellites that are reporting corrected interstellar position. Any 5th grade science student would know that. :blink:

If you are using a Garmin reciever the answer is yes you can get millimeter corrected coordinates. With a Magellan receiver you are limited to 14.743 centimeters. (Assuming that you have fresh batteries of course.)

Edited March 20, 2004 by webscouter.

2.2k · March 20, 2004

The problem here is, you have forgotten the unladen traveling velocity of he African swallow vs. the european swallow.

Or the force and velocity required to cut down the mightiest tree in the forest with a herring.

March 20, 2004

I got

3.1415926535897932384626433832795028841971693993751058209749445923078164062862

089986280348253421170679821480865132823066470938446095505822317253594081284811

174502841027019385211055596446229489549303819644288109756659334461284756482337

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420199561121290219608640344181598136297747713099605187072113499999983729780499

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002501262285941302164715509792592309907965473761255176567513575178296664547791

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291042487761825829765157959847035622262934860034158722980534989650226291748788

202734209222245339856264766914905562842503912757710284027998066365825488926488

025456610172967026640765590429099456815065265305371829412703369313785178609040

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393610310291616152881384379099042317473363948045759314931405297634757481193567

091101377517210080315590248530906692037671922033229094334676851422144773793937

517034436619910403375111735471918550464490263655128162288244625759163330391072

253837421821408835086573917715096828874782656995995744906617583441375223970968

340800535598491754173818839994469748676265516582765848358845314277568790029095

170283529716344562129640435231176006651012412006597558512761785838292041974844

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033346697499235630254947802490114195212382815309114079073860251522742995818072

471625916685451333123948049470791191532673430282441860414263639548000448002670

496248201792896476697583183271314251702969234889627668440323260927524960357996

469256504936818360900323809293459588970695365349406034021665443755890045632882

250545255640564482465151875471196218443965825337543885690941130315095261793780

029741207665147939425902989695946995565761218656196733786236256125216320862869

222103274889218654364802296780705765615144632046927906821207388377814233562823

608963208068222468012248261177185896381409183903673672220888321513755600372798

394004152970028783076670944474560134556417254370906979396122571429894671543578

468788614445812314593571984922528471605049221242470141214780573455105008019086

996033027634787081081754501193071412233908663938339529425786905076431006383519

834389341596131854347546495569781038293097164651438407007073604112373599843452

251610507027056235266012764848308407611830130527932054274628654036036745328651

057065874882256981579367897669742205750596834408697350201410206723585020072452

256326513410559240190274216248439140359989535394590944070469120914093870012645

600162374288021092764579310657922955249887275846101264836999892256959688159205

60010165525637568

Sorry about all those spaces, when I uploaded the result from my HP 1295 handheld it put them all in.

I wish I could learn how to do that stuff..

1.2k · March 21, 2004

Well, what you have is incomplete informtion.
In order to get your answer down to the millimeter level you need to apply the inverse number of the Darwinian endpoint where B is divided by S (B/S). This number can be squared again into a rectangle of 8 x 4 where the expansion of gas (per Boyle's Law) allows you the configuration to permeate the entire answer in a few short breaths.

You forgot the part about reversing the polarity of the dilithium crystals accessed via the vertical, then horizontal Jefferies tubes and then rebooting the whole ship. :blink:

Edited March 21, 2004 by Sparrowhawk

March 21, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

did a n00b really ask you that? :blink:

5.3k · March 21, 2004

March 21, 2004

I can't believe I read each and every post on this topic. I guess I should be out caching in the rain instead. :blink:

March 21, 2004

Has ANYONE thought of checking the gingling pin on the overhead hectonators? Also, it may affect the poo-poo valves and the laughing pins.

I'm going to let the dog out. Let me know when someone gets the right answer. So far I totally agree with 42.....

March 21, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

That's a very good question, but I wouldn't lose any sleep over it. In theory one would think that the answer would be yes, but in actual field conditions the geocacher can't actually be in the same superposition from the time that the calculations are first entered into the handheld HP 1295 computer until answer is processed, due to the geosychronous orbit and spin of four or more sattelites, the rotation of the earth on its axis, the rotation of the planet around the sun, the movement of the tectonic plates, and the respirations of the living geocacher; which subsequently compound the errors of the waas reading.

To top it all off, the coordinates on the cache page are wrong anyway, just look under the pile of sticks.

March 21, 2004

What it all comes down to is, to quote Monty Python, "It's over there, in a box!"

March 21, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

To answer this question I'd have to remove the flux capacitor from the DeLorean and it's always such a hassle to reconnect it.

March 21, 2004

can I narrow the total corrected search field to millimeters?

Yes, unfortunately the cache hider misplaced the cache by 35'.

1.3k · March 21, 2004

Today a new cacher asked me the following question:

What is the interstitial relationship between the reciprocal of the square root of sigma, where sigma is the delta of RHO and RHO sub O, and the quantum of the satellites is the satellite speed divided by the altitude multiplied by the total mass, over the corrected KCAS (Knots, Calibrated Airspeed) for the given number of satellites reporting corrected interstellar position adjusted for Einstein’s (rare) tenth theory of special relativity? By computing this data in the field with the HP 1295 handheld (supergeek) computer, can I narrow the total corrected search field to millimeters?

I can’t sleep until I figure this out.

(Paraphrashing Jerry Clower)

New geocacher, as long as I've been geocaching, that's about the most simplest question I ever been asked since I've been geocaching. I'm surprised that they'd let a man that don't know no more than you buy a GPS. And just to show you how simple the question is, my chauffeur is in the back of the room, I'll ask him to stand up and answer it....

:blink:

southdeltan

March 21, 2004

Suppose that (i) p is true or p is false and (ii) not-p is true or not-p is false.

Then p is true or not-p is true.

Now suppose that in 1900 one person says that a sea-battle will take place on 1/1/2100, and another says that a sea-battle will not take place on 1/1/2100.

Then either what the first person says is true or what the second person says is true.

But, in that case, either it is necessary in 1900 that a sea-battle takes place on 1/1/2100, or it is necessary in 1900 that one does not take place.

But the date of the predictions is irrelevant, and it is irrelevant whether any prediction is actually made at all.

So it is necessary at all times that a sea-battle takes place on 1/1/2100, or that a sea-battle does not take place on 1/1/2100.

But the argument can evidently be generalised.

So, everything that happens, happens of necessity.

First of all we need to be clear about what is meant by "necessity" here. What is at issue here is not logical necessity. It is rather inevitability. When the occurrence of a sea-battle on 1/1/2100 is said to be necessary at a certain date, what is meant is that at that date nothing can prevent a sea-battle from taking place on 1/1/2100. In particular, no one has the power to prevent it. Now Aristotle accepts that "What is, necessarily is, when it is; and what is not, necessarily is not, when it is not." So, he accepts that, if a sea-battle is actually taking place on 1/1/2100, then on 1/1/2100 it is (in this sense) taking place of necessity. Nothing can then stop it happening, because it is happening. What this argument appears to establish, however, is that, if a sea-battle takes place on 1/1/2100, not only is it necessary then that a sea-battle takes place on 1/1/2100, but it was always necessary. No one could ever have prevented it. And the same applies to everything that can happen. So, in particular, no one ever has the power to do anything other than what they actually do.

So, is there anything wrong with the argument?

Do you see the parallels here? The answer lies in the soil.

March 21, 2004

I Think that your "H.P. Supergeek micro-computer (calculator)" should be put in a Cache as a First to find prize.Then you should get some sleep get up early. Try to find what the Max. 'Cache find to hour ratio' in your area should be. Then see how close you gan get to it. Or see what the Max. CITO weight to Cache find you can get. :blink:

March 21, 2004

As long as you're awake, go out and find a night cache.

How do I run a query to find night caches?

4.1k · March 21, 2004

The answer to EVERYTHING is definately 42...

But then you have to wonder, what is the question?

The Question, my good man, is.....

'What IS the meaning of life.'

To whit....'the answer is 42'

As verified by Douglas Adams & the mice.

Conundrum

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