+Corp Of Discovery Posted March 21, 2004 Posted 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. It's actually pretty easy to do, if you have a sonic screwdriver.....and remember to reverse the polarity of the neutron flow. Quote
+C.A.K.E. UNIT Posted March 22, 2004 Posted March 22, 2004 CAN WE JUST BEAT HTE NEW GUY WITH A WET NOODLE 'TILL HE TAKES THE QUESTION BACK Quote
+clearpath Posted March 22, 2004 Posted March 22, 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? Oh Jeez, Criminal. What I wouldn't give to spend one day inside your head ... no wait, make that 5 minutes, not sure I could survive 24 hours. Quote
+The Cheeseheads Posted March 22, 2004 Posted March 22, 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. You kiss your mom with that mouth? Quote
+Lone Duck Posted March 22, 2004 Posted March 22, 2004 It occurred to me, that if one could additionally determine the satellite pitch, yaw, and roll, Satellites don't pitch, yaw, and roll in orbit, so that changes the whole theory! Quote
Colonel Mustard Posted March 22, 2004 Posted March 22, 2004 Well DUH!!!! It's obvious it's a question from a newbie. Anyone else would have known NOT to use Einsteinian physics. It is clearly a scenario for Newtonian physics. Quote
bug and snake Posted March 22, 2004 Posted March 22, 2004 It occurred to me, that if one could additionally determine the satellite pitch, yaw, and roll, Satellites don't pitch, yaw, and roll in orbit, so that changes the whole theory! Nah!..... DISCOS was the first Drag-Free Satellite ever flown. The DISCOS module was built at Stanford University in the late 1960's with the help of the Applied Physics Laboratory at Johns Hopkins University who constructed the other two modules of the TRIAD I satellite. It was built as a test vehicle to attempt to improve the accuracy of the U. S. Navy's TRANSIT navigation-satellite system. The Stanford team was led by Prof. D. B. DeBra and the APL effort by Dr. Robert E. Fischell. From a modern point of view (particularly for LISA), DISCOS demonstrated two things: 1) No field gradient at the proof mass exceeded about 10-7 / sec2 and 2) the disturbance-force calculation techniques (and in particular the method of determining the gravity field and its gradient at the proof mass by calculating the contribution from every mass element in the satellite) are viable. This means that a LISA specific force disturbance of about 10-16 meters / sec2 / Hz1/2 from internal disturbance field gradients has been demonstrated in flight assuming a modern Drag-Free controller performance of 10-9 meters / Hz1/2 Quote
+Geo Ho Posted March 22, 2004 Posted March 22, 2004 Thanks for giving me a cranial hemorrage . . . too painful to think about . . . I can guarantee that that doesn't keep me awake at night . . . but other stuff does. Happy caching and stuff! Quote
+Lone Duck Posted March 22, 2004 Posted March 22, 2004 I got 3.14 There's no Pi in this equation. Try again. Quote
+Johnnie Stalkers Posted March 22, 2004 Posted March 22, 2004 There's no Pi in this equation. Try again. No PIE! Thats it I'm out. Quote
+Geo Ho Posted March 22, 2004 Posted March 22, 2004 There's no Pi in this equation. Try again. I was hoping for apple . . . (sigh) . . . oh well. Quote
bug and snake Posted March 22, 2004 Posted March 22, 2004 I got 3.14 There's no Pi in this equation. Try again. But there is ice cream right? Pistachio - right? Quote
dutchmaster Posted March 22, 2004 Posted March 22, 2004 PARALLEL!!! Oh man, don't you know, Pi r ROUND, cornbread's squared. Sub the speed of light for the speed of dark, which is much faster (eg before light got there it was ALREADY dark!) and the numbers won't seem so intimadating because dark speed is measured in units called Moroffs. The opposite of a moroff is a moron but now your into geocaching and a new measure of SLOW! Quote
fallout4x4 Posted March 23, 2004 Posted March 23, 2004 (edited) Even the Tadpoles have me on ignore. That's it. Forumiside is the only answer! jk. Sorry geckoee. Must have missed it. This post officially gets me out of the tadpole category. edited to add last line Edited March 23, 2004 by fallout4x4 Quote
schon Posted March 23, 2004 Posted March 23, 2004 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. Truly, you have a dizzying intellect. This thread reminds me of a dream I had, about a year ago. I'm an IT contractor for a local city. In the dream, they called me because they were having problems communicating with an alien spaceship which was orbiting Earth. Every day at 8:00, the aliens would lose the ability to communicate with the city for the rest if the day, and were blaming the city's receiving equipment - and they were getting pissed off. So the city called me, and I looked at their equipment, and couldn't find a problem - so I asked if they were getting disconnected every day at the same time - and was told yes... so I then say that the problem was that the alien spaceship was in a geosynchronous orbit, and it needed to be in a geostationary orbit, to be able to communicate with a fixed point on the planet, because the planet is blocking their signals. I then wondered how aliens can have the technology to come here from another planet, but aren't smart enough to know how to position their spaceship to communicate with a fixed location on the ground. When I tell people about this dream, I always get weird stares. Quote
+4leafclover Posted March 23, 2004 Posted March 23, 2004 two words for y'all. Occam's Razor. to paraphrase: The simplest answer is usually the correct one. Therefore: 3. Quote
+CTgeocacher (CTg) Posted March 23, 2004 Posted March 23, 2004 This is simple...just ask Randy (RJFerret). I guarantee HE knows! Randy? Where are you? Quote
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.