A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Bruce J. Pennino
Date: 2013 Feb 7, 07:00 -0500
----- Original Message -----From: Brad MorrisSent: Thursday, February 07, 2013 1:42 AMSubject: [NavList 22272] Re: Re: Mars - Mercury Question
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: £10,000 for a method that could determine longitude within 60 nautical miles (111 km); £15,000 for a method that could determine longitude within 40 nautical miles (74 km); £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!
BradOn Feb 3, 2013 3:20 PM, "Greg Rudzinski" <gregrudzinski---com> wrote:
Near earth object 2012DA14 link:
[NavList] Re: Mars - Mercury Question
From: Greg Rudzinski
Date: 3 Feb 2013 12:05
Nasa near earth object link:
[NavList] Re: Mars - Mercury Question
From: Frank Reed
Date: 3 Feb 2013 11:30
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. :)
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