tag:blogger.com,1999:blog-8800134045308911566.comments2023-04-09T17:08:07.218-07:00Greg's Space CalculationsGreg Hullenderhttp://www.blogger.com/profile/16720604327299886491noreply@blogger.comBlogger27125tag:blogger.com,1999:blog-8800134045308911566.post-38930712957971453072022-12-06T22:19:03.693-08:002022-12-06T22:19:03.693-08:00여수출장안마
여수출장안마
통영출장안마
괴산출장안마
통영출장안마
전북출장안마
전북출장안마<a href="https://www.ssculzang.com/yeosuculzang" rel="nofollow">여수출장안마</a><br /><a href="https://www.gogocallgirl.net/yeosucallgirl" rel="nofollow">여수출장안마</a><br /><a href="https://www.cpanma.com/55" rel="nofollow">통영출장안마</a><br /><a href="https://www.wpwz77.com/55" rel="nofollow">괴산출장안마</a><br /><a href="https://www.cpcz88.com/55" rel="nofollow">통영출장안마</a><br /><a href="https://www.ssculzang.com/jeonbukculzang" rel="nofollow">전북출장안마</a><br /><a href="https://www.gogocallgirl.net/jeonbukcallgirl" rel="nofollow">전북출장안마</a><br /><br /><br /><a href="https://www.ssculzang.com/">심심출장마사지</a> <a href="https://www.cpanma.com/">총판출장마사지</a>https://www.blogger.com/profile/00966232747634195271noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-83808896315141117932021-12-01T05:42:53.012-08:002021-12-01T05:42:53.012-08:00Any chance you can provide a downloadable version ...Any chance you can provide a downloadable version of this website? I've been coming back for a few years to run calculations from time to time and find this blog to be the most user friendly. I miss the little rocket animation though, Can't find that one any more and would hate to lose this one to the inevitable encroachment of temporal change.Uncle A Was Righthttps://www.blogger.com/profile/16606449259792223101noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-2266209382097202242021-11-05T07:00:40.737-07:002021-11-05T07:00:40.737-07:00For anyone here from reading Project Hail Mary, th...For anyone here from reading Project Hail Mary, the ship has a constant acceleration of 1.5G over 12 light years, flipping from boosting to decelerating in the middle. It will reach 0.9953c, take 3.90 years in ship time and 13.22 years on earth. Could someone calculate the beatles' travel time?gockhttps://www.blogger.com/profile/04652916516266303078noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-48460809606297794382021-02-24T20:46:51.694-08:002021-02-24T20:46:51.694-08:00If a ship travels 25 light years with a constant a...If a ship travels 25 light years with a constant acceleration at 1 g, does the ship's engine consume 26.9 years (Earth time) or 6.4 years (ship time) of energy?Anonymoushttps://www.blogger.com/profile/09115375451846493022noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-6626572962556611622018-02-07T10:43:12.948-08:002018-02-07T10:43:12.948-08:00Glad you liked it! I keep meaning to do a few more...Glad you liked it! I keep meaning to do a few more for things like planets, orbits, eclipses, etc. but I've never got around to it.Greg Hullenderhttps://www.blogger.com/profile/16720604327299886491noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-37525762585379880042018-02-07T08:36:08.743-08:002018-02-07T08:36:08.743-08:00What a fantastic resource! Thank you so much for c...What a fantastic resource! Thank you so much for creating and posting this calculator.Chris Merrillhttps://www.blogger.com/profile/09536931912462625066noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-55197333119471605012017-08-24T09:19:43.604-07:002017-08-24T09:19:43.604-07:00I should pay more attention to this page. If you&#...I should pay more attention to this page. If you're still interested, it's just 167 days--not 3.7 years. But that assumes 1g acceleration to the half-way point and then 1g deceleration from there.Greg Hullenderhttps://www.blogger.com/profile/16720604327299886491noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-54048554718740367532017-06-15T07:57:54.717-07:002017-06-15T07:57:54.717-07:00I want a calculator that will tell me what relativ...I want a calculator that will tell me what relativistic acceleration and velocity a rocket will get if it ejects 1 mg. per second at .38 C and weighs 15 metric tons. INVENTORhttps://www.blogger.com/profile/08390550080354960538noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-35625664655922396412017-06-05T20:28:30.216-07:002017-06-05T20:28:30.216-07:00Worrying about gravity (General Relativity) is, I ...Worrying about gravity (General Relativity) is, I think, messing you up, since the entire Twin Paradox can be explained using only Special Relativity.<br /><br />Earth's gravity has nothing to do with it -- the difference in age of the two twins would be the same if the starting point was a zero-g space station. (There might be a very tiny factor caused by Earth's gravity, but it's negligible for the purpose of this discussion, I believe.)<br /><br />So what is the difference between the two twins?<br /><br />Twin #1, who stays on the Zero-G space station, or on the Earth, never changes inertial reference frame, because he never experiences any acceleration. He stays in a single frame.<br /><br />Twin #2 (the one on the rocket) can measure his acceleration using an accelerometer, or even just by feeling it. Especially when he changes direction. And if he had a telescope pointed at his brother's house, where there was an accelerometer setup with a giant display, he would see the whole time that his brother is not undergoing acceleration.<br /><br />Therefore, this situation is not symmetrical.<br /><br />If you use instant acceleration, it is symmetrical up until the turn-around point, at which point there is a kind of "jump".<br /><br />I found this discussion helpful, both the accepted answer and additional answer my Muphrid: https://physics.stackexchange.com/questions/66926/does-it-matter-in-which-direction-i-travel-in-relativity-theoryMike Lucashttps://www.blogger.com/profile/14909340252314100715noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-65756284143028220392017-06-05T13:39:09.893-07:002017-06-05T13:39:09.893-07:00Hello Mike. Thank you very much for trying to help...Hello Mike. Thank you very much for trying to help me understand. Since my previous posts, I did more research. The instant direction change was not a good scenario. Better stick to standard 1g acceleration example. In the past when I did read basic relativity book, Google did not exist. I made in my mind fixed idea that gravity and acceleration due to principle of equivalence has exactly same influence on speed coz the "gravity feel" and it made perfect sense in my brain. Im not psychist, just dumb curious person. Since original post I did realize current mainstream science consider Earth and Rocket not equivalent reference frame, by principles I do not understand yet and maybe I will never do. But the main problem that broke my theory was finding that there is no experiment on the time dillatation using centrifuge. This experiment would support my idea, but i did not find any. Do you know of such experiment? I will tell you why this idea still cannot leave my mind easily. PPL usually imagine that scenario in real universe. Lot of stars and galaxies. From point of time dilatation they should not need to exists. So if I imagine the Universe containing JUST an Earth and a Rocket and nothing else, then how the reference frame knows Rocket did change direction from receding to approaching and not Earth? Gravity feel always 1g. There are no other objects to tell so. Maybe principle of equivalence do not work 100% due to... presence of gravitons? I don't know but I did submit that I will not find satisfying answer in current life period, maybe never. Still thank you for your effort! :)Castanedahttps://www.blogger.com/profile/15275690021003586479noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-8573104217649580452017-06-05T09:16:06.839-07:002017-06-05T09:16:06.839-07:00Hi Siloton, you're correct that it's not s...Hi Siloton, you're correct that it's not specifically due to acceleration that there's a time difference in the Twin Paradox. Rather, it's because the twin that leaves earth and comes back, even if he accelerates instantly, cannot remain in the SAME reference frame on his whole trip because he has to change direction to come back. Changing direction, even if done instantaneously, is a change to a DIFFERENT inertial reference frame.Mike Lucashttps://www.blogger.com/profile/14909340252314100715noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-90130915665159951472017-04-17T15:32:41.529-07:002017-04-17T15:32:41.529-07:00Sorry to bother you. I would like to pay only for ...Sorry to bother you. I would like to pay only for a chance to discuss a problem with you. If there is a discrepancy we should find it quickly. But if you're sure your calculations are right even if principles cannot be formulated... well its up to you Sir, thank you for your time anyway...Castanedahttps://www.blogger.com/profile/15275690021003586479noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-50407112112110103102017-04-17T15:23:16.496-07:002017-04-17T15:23:16.496-07:00I didn't answer your previous question because...I didn't answer your previous question because you're getting out of my depth. I was hoping someone who knows more would answer.Loren Pechtelhttps://www.blogger.com/profile/08348494458707790769noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-78852417533059585192017-04-17T15:21:28.026-07:002017-04-17T15:21:28.026-07:00Sir, I will pay you for an explanation. Is it acce...Sir, I will pay you for an explanation. Is it acceptable? Thank youCastanedahttps://www.blogger.com/profile/15275690021003586479noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-78439259169845270162017-04-16T05:46:24.390-07:002017-04-16T05:46:24.390-07:00And maybe I can make myself even more clear, you w...And maybe I can make myself even more clear, you wrote: "Since you can't change velocity without an accelerated reference frame". Well, in a case of 1g on an Ship same as on a Earth, how the "accelerated reference frame" knows i am the Ship and not the Earth? Those observers cannot tell a difference.Castanedahttps://www.blogger.com/profile/15275690021003586479noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-36287534394386890922017-04-16T02:06:12.593-07:002017-04-16T02:06:12.593-07:00Oh you have patience with me man, If you explain t...Oh you have patience with me man, If you explain to me, I will send you a donation I think :). OK exactly as you write, those clocks "are same" until one of those subjects go through acceleration / deceleration so they can be compared. You say you cannot take velicity without acceleration, yes, but if we make an mental experiment, considering one can reach speed of light without acceleration and back also, there should be zero time shift I suppose? So if yes then acceleration is the only cause or if no, can you please direct me to some equations on Wiki or somewhere that sums how much time you would skip during the same travel "without acceleration"? Thank you ones more.Castanedahttps://www.blogger.com/profile/15275690021003586479noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-2563718901113925852017-04-15T20:22:05.323-07:002017-04-15T20:22:05.323-07:00Both the ship and Earth see each other's clock...Both the ship and Earth see each other's clocks as going slow. However, that only applies so long as nobody accelerates. Once you introduce an accelerated reference frame that fixes the paradox, the observer that went to the stars and came back will always be younger than his twin that stayed home. Since you can't change velocity without an accelerated reference frame there are no paradoxes.Loren Pechtelhttps://www.blogger.com/profile/08348494458707790769noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-90406572023474837372017-04-15T10:08:07.903-07:002017-04-15T10:08:07.903-07:00Ahh thank you Sir. I appreciate your effort! I tho...Ahh thank you Sir. I appreciate your effort! I thought speed itself has no effect on time speed. In example of ship and Earth consider this, how the "Time" knows if the Ship recedes from Earth, or Earth recedes away from Ship and so on what "side" be "slower"? Now, if I take acceleration as an only cause for time dilatation now it makes perfect sense for me coz now time knows exactly how fast on each side he exactly runs :) Of course I would love to inspect another explanations. Thank you ones more...Castanedahttps://www.blogger.com/profile/15275690021003586479noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-5386589771624717472017-04-15T09:50:04.424-07:002017-04-15T09:50:04.424-07:00@siloton You are mixing up the fact that gravity s...@siloton You are mixing up the fact that gravity slows time with the fact that speed slows time. Furthermore, what counts is the gravitational field, not what you actually feel. Being in orbit doesn't change the slowing due to gravity (but adds slowing due to speed.)<br /><br />Both factors are minute at the scale that we experience here on Earth. The only time the average person will encounter the effects is with GPS (and even then you don't know it unless you dig into the equations that run your system.)<br /><br />To get a noticeable slowing from speed requires velocities that are an appreciable fraction of lightspeed. On Earth that means the realm of the particle accelerator.<br /><br />To get a noticeable effect from gravity requires getting very close to a neutron star or black hole--exceedingly dangerous places to be.Loren Pechtelhttps://www.blogger.com/profile/08348494458707790769noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-40037298801031885572017-04-15T09:14:33.811-07:002017-04-15T09:14:33.811-07:00Hello, I don't understand one point, when ther...Hello, I don't understand one point, when there is acceleration = 1g on the top of the form, it means that ship and crew suffer the same "gravity feel" as those people, that did stay on Earth. In this case, ship time and earth time should be the same even after return, no matter final speed the ship reaches. In your calculator they differs. How is it possible? Thank you very muchCastanedahttps://www.blogger.com/profile/15275690021003586479noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-38453641278587021122017-03-25T10:10:17.423-07:002017-03-25T10:10:17.423-07:00I've thought about it, but that turns out to b...I've thought about it, but that turns out to be really hard to do in a systematic way. The trouble is that some entries force other ones, and even figuring out which ones can't be changed is a challenge.Greg Hullenderhttps://www.blogger.com/profile/16720604327299886491noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-4667055786750123642017-03-07T14:20:30.331-08:002017-03-07T14:20:30.331-08:00Or if you want to go in another direction:
On the...Or if you want to go in another direction:<br /><br />On the evening of 2040/09/09 at 00:00:00 UTC (2040/09/08 at 8 PM local time), a spaceship takes off from Cape Canaveral accelerating at 1 g. <br /><br />Encounter Time Distance Vis. EDT ETA UTC<br /><br />Moon 0.1036 d 392.8 Mm Visible 09/09-02:29<br />Mercury 2.216 d 179.7 Gm Visible 09/11-05:11 <br />Venus 2.429 d 216.0 Gm Visible 09/11-10:18<br />Mars 3.130 d 358.6 Gm Visible 09/12-03:07<br /><br />Jupiter 5.119 d 959.3 Gm Visible 09/14-02:51<br />Saturn 6.571 d 1.581 Tm Visible 09/15-13:42<br />Uranus 8.858 d 2.872 Tm R 3:57 AM 09/17-20:36<br />Neptune 10.891 d 4.341 Tm R 9:56 PM 09/19-21:23<br /><br />Sirius 2.888 a 8.60 ly R 3:30 AM 2043/07/30<br />Procyon 3.143 a 11.46 ly R 3:35 AM 2043/10/31<br />Capella 4.366 a 42.9 ly R 11:06 PM 2045/01/20<br />Aldebaran 4.764 a 65.3 ly R 12:07 PM 2045/06/15<br /><br />Achernar 5.490 a 139 ly R 1:27 AM 2046/03/07<br />Canopus 6.264 a 310 ly R 5:29 AM 2046/12/14<br />Pleiades 6.611 a 444 ly R 10:56 PM 2047/04/20<br />Betelgeuse 6.969 a 643 ly R 1:47 AM 2047/08/29<br /><br />Rigel 7.251 a 860 ly R 1:40 AM 2047/12/10<br />M42 (Orion) 7.683 a 1.344 kly R 1:55 AM 2048/05/16<br />C106 (47 Tucanae) 10.12 a 16.7 kly Not visible 2050/10/24<br />C80 (Omega Centauri) 10.07 a 15.8 kly S 7:16 PM 2050/10/05<br /><br />LMC 12.30 a 158 kly Not visible 2052/12/28<br />SMC 12.52 a 199 kly Not visible 2053/03/19<br />M83 (Southern Pinwheel) 16.73 a 15.21 Mly S 8:38 PM 2057/06/01<br />C77 (Centaurus A) 16.53 a 12.4 Mly S 7:38 PM 2057/03/21<br /><br />Fornax (NGC 1399) 17.42 a 62 Mly R 1:15 AM 2058/02/08<br />Centaurus (NGC 4696) 19.07 a 170 Mly S 7:09 PM 2059/10/03<br />Shapely (NGC 5124) 20.37 a 650 Mly S 8:24 PM 2061/01/20<br />Horologium (IC 1933) 20.73 a 950 Mly R 2:31 AM 2061/06/03<br /><br />3C 273 21.648 a 2.0 Gly Visible 2062/05/03<br />3C 48 22.102 a 3.9 Gly R 8:26 PM 2062/10/16<br />3C 47 22.196 4.3 Gly R 8:58 PM 2062/11/20<br />3C 147 22.362 5.1 Gly R 11:17 PM 2063/01/19<br /><br />3C 9 23.014 10.0 Gly Visible 2063/09/14<br /><br />Edge of Universe 23.326 a 13.8 Gly Visible 2064/01/06<br /><br />Anonymoushttps://www.blogger.com/profile/16233289383754544838noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-19486706293367735732017-03-02T13:15:45.116-08:002017-03-02T13:15:45.116-08:00This is great. Here's an example of what can ...This is great. Here's an example of what can be done with it.<br /><br />On the evening of 2040/09/09 at 00:00:00 UTC (2040/09/08 at 8 PM local time), a spaceship takes off from Cape Canaveral accelerating at 1 g. <br /><br />Encounter Time Distance Vis. EDT ETA UTC<br /><br />Moon 0.1036 d 392.8 Mm Visible 09/09-02:29<br />Mercury 2.216 d 179.7 Gm Visible 09/11-05:11 <br />Venus 2.429 d 216.0 Gm Visible 09/11-10:18<br />Mars 3.130 d 358.6 Gm Visible 09/12-03:07<br /><br />Jupiter 5.119 d 959.3 Gm Visible 09/14-02:51<br />Saturn 6.571 d 1.581 Tm Visible 09/15-13:42<br />Uranus 8.858 d 2.872 Tm R 3:57 AM 09/17-20:36<br />Neptune 10.891 d 4.341 Tm R 9:56 PM 09/19-21:23<br /><br />Rigil 2.317 a 4.37 ly V 4:52 PM 2043/01/03<br />Altair 3.486 a 16.73 ly Visible 2044/03/05<br />Vega 3.859 a 25.04 ly Visible 2044/07/19<br />Arcturus 4.218 a 36.7 ly Visible 2044/11/27<br /><br />Spica 6.056 a 250 ly Visible 2046/09/29<br />Acrux 6.295 a 320 ly NV 2:39 PM 2046/12/26<br />Hadar 6.486 a 390 ly V 4:16 PM 2047/03/06<br />Antares 6.818 a 550 ly Visible 2047/07/05<br /><br />M6 (Butterfly) 7.852 a 1.6 kly Visible 2048/07/16<br />M8 (Lagoon) 8.763 a 4.1 kly Visible 2049/06/14<br />M4 (Scorpio) 9.309 a 7.2 kly Visible 2049/12/31<br />M2 (Aquarius) 10.784 a 33 kly Visible 2051/06/22<br /><br />M31 (Andromeda) 14.993 a 2.54 Mly Visible 2055/09/07<br />M81 (Bode's) 16.481 a 11.8 Mly Visible 2057/03/03<br />M101 (Pinwheel) 17.035 a 20.9 Mly Visible 2057/09/22<br />M51 (Whirlpool) 17.128 a 23 Mly Visible 2057/10/26<br /><br />Virgo (M87) 17.946 a 53.5 Mly Visible 2058/08/20<br />Coma (NGC 4889) 19.642 a 308 Mly Visible 2060/05/01<br />Hercules (IC 1182) 20.128 a 509 Mly Visible 2060/10/25<br />Corona (PGC54846) 20.747 a 964 Mly Visible 2061/06/08<br /><br />3C 273 21.648 a 2.0 Gly Visible 2062/05/03<br />3C 48 22.102 a 3.9 Gly R 8:26 PM 2062/10/16<br />3C 47 22.196 a 4.3 Gly R 8:58 PM 2062/11/20<br />3C 147 22.362 a 5.1 Gly R 11:17 PM 2063/01/19<br /><br />3C 9 23.014 a 10.0 Gly Visible 2063/09/14<br /><br />Edge of Universe 23.326 a 13.8 Gly<br /><br />However, these calculations assumed that all the places visited are on a straight line from the place of launch. This is obviously not the case. How can I take into account a spaceship turning while moving? <br />Anonymoushttps://www.blogger.com/profile/16233289383754544838noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-11583361112716065002017-02-13T18:48:40.561-08:002017-02-13T18:48:40.561-08:00Nice but it could use one improvement--remembering...Nice but it could use one improvement--remembering which values were entered. I was looking at a constant end velocity with various accelerations but when I changed the acceleration it holds the first number constant rather than the one I entered.Loren Pechtelhttps://www.blogger.com/profile/08348494458707790769noreply@blogger.comtag:blogger.com,1999:blog-8800134045308911566.post-50535513730230811962016-11-07T17:00:45.546-08:002016-11-07T17:00:45.546-08:00Hi Greg. I am wondering if maybe you could guide ...Hi Greg. I am wondering if maybe you could guide me through the use of your calculator to help me with a story. I need to get my crew to the inner layer of the Oort Cloud; 2000 AU from Earth, give or take. According to one link I found, the best I could do for "hard sci-fi" tech was the fusion rocket design (http://www.universetoday.com/15403/how-long-would-it-take-to-travel-to-the-nearest-star/#,) which, based on estimated distances, would take about 3.7 years to reach that distance, provided I had the calculations right. This isn't quite the speed of light, obviously, so I am having trouble figuring out your calculator. However, I imagine that the same principle of continuous acceleration could work. I guess I'm trying to reverse engineer the calculator; I want to know how long the ship would have to accelerate and decelerate for in order to be safe but reach the requisite speed. Also, I want to know if that 3.7 years at speeds that are pushing, but not achieving, light speeds, would have relativistic effects on the crew. Does this make sense, or are the two technologies (fusion rockets and continuous acceleration) completely incompatible? Thanks for your help!Sable Aradiahttps://www.blogger.com/profile/02772874690100912042noreply@blogger.com