A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Brad Morris
Date: 2013 Feb 7, 09:15 -0500
In your first answer, you wrote that the apparent angular rate of Mercury, as viewed from Earth, is approximately 1/4 degree per day. I took this as an average rate, not a constant rate.
Around maximum elongation, the apparent angular rate of Mercury must be low, as Mercury is either coming directly towards or going directly away from us.
If the angular rate can be low and average 1/4 degree per day, it follows that there are times when the apparent angular rate is higher. When is it higher? When Mercury's motion is across our field of view. When is it highest? When the motion is perpendicular to our field of view. That occurs (and is visible to us), during a Transit.
I then calculated this maximum apparent angular rate at just over 6 arc minutes per hour or just over 2.5 degrees per day. Is this wrong?
I then asserted that the maximum sensitivity of an observer to be 0.1 arc minutes. This is based upon the micrometer drum and sextant verniers. This is clearly the weakest portion of my technical argument. If accepted, we can then relate the apparent angular rate and measurement sensitivity to yield an absolute time. I estimated 1 minute of time to note the minimum threshold of sensitivity, using the maximum apparent angular rate of Mercury. Is the arithmetic wrong?
I then calculated the displacement on the Earth's surface for 1 minute of time. I calculated 15 nautical miles. Is this wrong?
I then compared the displacement against the Longitude Prize rules. The most accurate rule is determination to within 30 nautical miles. Since 15 is less than 30, I expected that this qualifies under the rules. Did I get this wrong?
Brad, In my very first answer on Mercury, I gave a rough estimate of the speed of Mercury motion which shows that the speed is not sufficient for the determination of longitude, even at the time of transit. Of course the expression of "determination of longitude" is somewhat ambiguous. I took the specifications of the Longitude prize as the definition of this expression. But telling whether you are in Atlantic or in Pacific can be also called a "determination of longitude". Alex. > The conversation about the orbit of Mercury around the Sun, and how at > maximum elongation the apparent angular change is at minimum, made sense. > The angular rate of change is too small then for the determination of > longitude. > > A few referred to not being able to see Mercury when at maximum angular > rate of change, because then you are blinded by the Sun. I accepted this, > at first, as correct on inspection. There are exceptions to this general > rule of course. What about a Transit of Mercury? Mercury appears as a > black dot on the face of the sun, quite visible. > > Some simple research shows this occurring 13 or 14 times per century The > Transit of Mercury takes about 5 hours. Its clear that the angular rate > of > Mercury around the sun is nearly constant, so the variation in transit > period would be WHERE on the sun it transits. Would this be enough to > determine longitude, albeit of limited utility? > > Assume the Sun's S.D. to be 16 arc minutes. So the diameter on the day of > the transit (next one in 2016) would be 32 arc minutes. Assume further > that Mercury transits on the Sun's equator. The rate of change then is 32 > arc minutes / 5 hours / 60 minutes per hour / 60 seconds per minute * 60 > arc seconds per arc minute = 0.10666 arc seconds per second. > > I find therefore, that in 60 seconds of time, the angular difference is > 6.4 > arcseconds or approximately 0.1 arc minutes. This is at the thresh hold > of > measurement for a hand held sextant. > > Given that the circumference of the earth is 24901 miles or 21638 nautical > miles and that there are 24 time zones of 60 minutes each, one minute of > time is about 15 nautical miles. > > The main longitude prizes were: Â£10,000 for a method that could determine > longitude within 60 nautical miles (111 km); Â£15,000 for a method that > could determine longitude within 40 nautical miles (74 km); Â£20,000 for a > method that could determine longitude within 30 nautical miles (56 km). > > To summarize then, the angular rate of change for a Transit of Mercury is > approximately 0.1 arc minutes per minute. This is measurable using a hand > held sextant. The distance equivalence of one minute of time is > approximately 15 nautical miles. The top prize was for a method that > could > determine longitude to 30 miles. > > I think this actually can determine longitude (okay: only 13 or 14 times > per century, in 5 hour bursts!). Did I go wrong? > > In looking up some facts for this, I found that the Transit of Venus and > the Transit of Mercury occur simultaneously only once in the next 250,000 > years! How's that for an oddball fact! > > Regards > Brad > > > > > On Feb 3, 2013 3:20 PM, "Greg Rudzinski" wrote: > >> ------------------------------ >> >> Near earth object 2012DA14 link: >> >> http://neo.jpl.nasa.gov/news/news177.html >> >> >> [NavList] Re: Mars - Mercury Question >> From: Greg Rudzinski >> Date: 3 Feb 2013 12:05 >> Nasa near earth object link: >> >> http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2013%20BS15;orb=1 >> >> >> [NavList] Re: Mars - Mercury Question >> From: Frank Reed >> Date: 3 Feb 2013 11:30 >> Brad, >> It helps if you draw a picture of the Solar System and look at it "from >> above". Mercury orbits at 0.4 AUs from the Sun, Venus at 0.7, the Earth >> at >> 1.0, Mars at 1.5, Jupiter at 5.2, etc. Draw the circles for the inner >> Solar >> System, and start off Mercury and the Earth in line with the Sun. >> Obviously >> you can't see Mercury if it's in line with the Sun, but we can ignore >> that >> for the moment. Now advance both planets by ten days. Mercury moves four >> degrees per day in heliocentric longitude. The Earth moves just under >> one >> degree per day. These are the angles as measured heliocentrically --from >> the SUN's center. The lines drawn from the Sun to Mercury and the Earth >> separate by thirty degrees in ten days. So how much does Mercury move >> (relative to the Sun) as seen from the Earth? If you draw this out, you >> should find that you have a simple triangle to solve: one side has >> length >> 0.4, another 1.0 and the angle between those two sides is 30 degrees. >> There's a little plane trig problem here. You should find that Mercury >> will >> have moved 17 degrees as seen from the Earth. How much does that angle >> change in the next day? I find that it increases to 18.2 degrees. So >> even >> though Mercury is moving 4 degrees per day relative to the stars as seen >> from the Sun, from the Earth it is only advancing 1.2 degrees in a day >> and >> that rate is slowing. Just six days later, it is moving only 0.5 degrees >> per day as seen from the Earth. And six days after that, the angle >> between >> Mercury and the Sun, as seen from the Earth, isn't changing at all. >> >> Although Mercury is travelling faster around the Sun than the Earth in >> terms of linear speed (velocity is inversely proportional to the square >> root of the distance from the Sun for circular orbits around the Sun), >> its >> angular speed is reduced by simple distance: if one object is twice as >> far >> away as another travelling at the same linear speed, its angular speed >> is >> half as great, and it is reduced by the relative direction it is >> travelling. For high angular rates across the sky as seen from the >> Earth, >> we need things that are zipping by the Earth at close range and on a >> path >> that is nearly perpendicular to our line of sight. There are lots of >> these, >> but we only now have the ability to predict some close passes by very >> small >> Solar System bodies in advance. There's one coming up on the afternoon >> and >> early evening of February 15. A very small asteroid (very large meteor) >> about 150 feet across, designated with minor planet ID "2012 DA14" will >> pass the Earth at a distance of only about 20,000 miles (closer then >> most >> communication satellites). It will be visible in binoculars from >> southeast >> Asia directly under the point of closest approach moving at an angular >> rate >> of about one minute of arc per second. Part of the pass will be visible >> in >> medium-size backyard telescopes from our neck of the woods in the >> northeastern US. By then the angular rate will be reduced to about 0.1' >> per >> second, but that's still a dozen times faster than the Moon. You can get >> details for a specific location at heavens-above.com. >> >> Imagine an alternate universe where objects like "2012 DA14" are a >> hundred >> times more common and ten times bigger in diameter (making them visible >> to >> the unaided eye). Explorers could have used them to find longitude on a >> fairly regular basis by comparing observations of sightings after they >> returned home. Unfortunately, that would also imply impacts capable of >> flattening forests, villages, cities and towns to a distance of a >> hundred >> miles with severe effects over thousands of miles roughly every five >> years >> somewhere on the Earth. So this is probably a better universe. :) >> >> -FER >> >> >> ---------------------------------------------------------------- >> NavList message boards and member settings: www.fer3.com/NavList >> Members may optionally receive posts by email. >> To cancel email delivery, send a message to NoMail[at]fer3.com >> ---------------------------------------------------------------- >> >> View and reply to this message: http://fer3.com/arc/m2.aspx?i=122250 >> > > > > > View and reply to this message: http://fer3.com/arc/m2.aspx?i=122272 > > >
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