On Fri, 19 Jul 1996, Roderick Wall wrote: > > HI all > Thanks to all who replied to my question regarding the equation > of time correction. Sorry for the late reply but my E-mail has not been > working after upgrading my system to a Pentium 100. > > My question was: > > > > > My question is with regard to the tilt of the earth axis effect. > > > > The SUNDIAL AUSTRALIA ISBN 0 646 22200 7 book indicates that (for the > > tilt of the earth axis effect) there is no difference between local > noon > > and solar noon at dates corresponding to the Equinoxes and Solstices > (4 > > times per year), but produces a difference of up to +/- 9.86 minutes > > between each Equinox and Solstice (4 times per year). > > > > Can someone please explain and describe the above effect. > > > > Thanks in advance: Roderick Wall > > > >-- End of excerpt from Roderick Wall > > All the replies that I have received has helped me a little, but I am > still having problems understanding the earth axis effect. > > In the British Sundial Society BULLETIN No. 93.3 October 1993 page > figure 1 it shows A diagram of the Celestial Equator and its > relationship to the Ecliptic. > > On page 9 it says: The first problem arises from the fact that time is > measured with respect to the celestial equator. and the motion of the > sun along the ecliptic, when projected in the direction parallel to the > celestial equator, changes throughout the year. > > I can see that because the sun is also moving away from the Celestial > Equator ( as well as along it) that for one unit of movement (say 15 > deg) then the movement along the celestial equator will be less (less > than 15 deg). > > What I don't understand is that: At the summer solstice (and winter > solstice) the motion of the sun is parallel to the Celestial Equator and > I would think that one unit of motion of the sun would equal the same > amount on the Celestial Equator. But Figure 2 (at O) shows that the rate > of change is still high (The point half way between the vernal equinox > and the autumnal equinox). Also if the distance of movement along the > Celestial Equator is less than one unit then at some other time the > movement must be more than one unit to catch up with time. > > What am I missing in trying to understand this? > > Thanks in advance again. and sorry for the late reply. > > Roderick Wall >
The difficulty is that you are trying to visualise the problem in terms of plane geometry but is it necessary to think of the motions and positions in terms of spherical geometry. The mean Sun gets ahead of the true Sun until halfway from the Equinox to the Solstice. Then the true Sun gradually catches up by the Solstice. Viewed another way, the speed of both along their respective arcs is 1/365.25 of a circle every day, so both have to go the same distance along their respective arcs between the time of the Equinox and the time of the Solstice, when both have RA = 6 hrs. Thus, there must be a maximum difference about halfway between these points. If you have a globe you don't mind drawing on, try marking out the angles with bits of string. I would agree that it is hard to visualise these relationships.
