Are these trick questions?

How many times will the Sun cross the southern meridian at Oxford in
2007?

How many times will Sirius cross the southern meridian at Oxford in
2007?

How many times will the Moon cross the southern meridian at Oxford in
2007?

Themos Tsikas wrote:
How many times will the Sun cross the southern meridian at Oxford in
2007?

How many times will Sirius cross the southern meridian at Oxford in
2007?

How many times will the Moon cross the southern meridian at Oxford in
2007?


Given that the answer to each question is different, it is most likely
not a trick question; you may learn something from the answers.

Best,
Stephen

Remove footfrommouth to reply

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On Mar 14, 6:17?pm, Stephen Tonkin
wrote:
Themos Tsikas wrote:
How many times will the Sun cross the southern meridian at Oxford in
2007?

How many times will Sirius cross the southern meridian at Oxford in
2007?

How many times will the Moon cross the southern meridian at Oxford in
2007?

Given that the answer to each question is different, it is most likely
not a trick question; you may learn something from the answers.

Whilst the moon loses something just under an hour each day compared
to the sun as a tidal day is about 55 minutes longer than the solar
day, the number of stellar transits will be the same will they not?

When does 2007 start and when does it end? Or is that still a matter
of choice?

wrote:
On Mar 14, 6:17?pm, Stephen Tonkin
wrote:
Themos Tsikas wrote:
How many times will the Sun cross the southern meridian at Oxford in
2007?

How many times will Sirius cross the southern meridian at Oxford in
2007?

How many times will the Moon cross the southern meridian at Oxford in
2007?

Given that the answer to each question is different, it is most likely
not a trick question; you may learn something from the answers.

Whilst the moon loses something just under an hour each day compared
to the sun as a tidal day is about 55 minutes longer than the solar
day, the number of stellar transits will be the same will they not?

Stellar transits will be the same was lunar, or solar transits? Well,
anyway, the three answers are different. Without trying to do all the
homework, remember that Earth is orbiting the sun, but not orbiting
distant stars.

When does 2007 start and when does it end? Or is that still a matter
of choice?

The place (Oxford) and year (2007) have been given. The beginning and
end of that year in that place are well defined.


Tim

This is not homework, although I think it could be a good homework
question. We'll take "2007" to mean what the BBC says it is.

The answer to the first one I am confident of: 365

The second one: Viewed from above (north) of the ecliptic, the Earth
is going anti-clockwise. So the Sun appears to move to the east. That
makes the answer 366.

The third one: The moon rises later every day, it moves to the east.
So it takes longer to cross the meridian. Say it takes 28 solar
crossings for the moon's 27 crossings. Then it will cross 27*365/28
times, that's around 352. To get the right answer, we need a precise
figure for the lunar period and the precise location of the Moon when
the New Year comes in.

On Mar 15, 12:34 pm, "Themos Tsikas"
wrote:
This is not homework, although I think it could be a good homework
question. We'll take "2007" to mean what the BBC says it is.


The answer to the first one I am confident of: 365


Try 2008 which is a leap year with 366 days.


The second one: Viewed from above (north) of the ecliptic, the Earth
is going anti-clockwise. So the Sun appears to move to the east. That
makes the answer 366.


Here is where John Flamsteed jumped the tracks by using the return of
Sirius to a meridian to justify the motions of the Earth -

"... our clocks kept so good a correspondence with the Heavens that I
doubt it not but they would prove the revolutions of the Earth to be
isochronical..." Flamsteed

That is a grave error and completely at variance with the principles
which create the equable 24 hour day which in turns creates the 1461
day calendrical cycle.

The correlation between clocks and axial rotation at precisely 4
minutes for 1 degree of rotation never requires a reference to any
external motion,not the local Sun,not the distant stars,nothing.That
they introduced both the Sun and celestial sphere geometry (Fixed
stars) in the late 17th century as a gauge for the Earth's motions
remains of the greatest astronomical catastrophe imaginable,that
being no overstatement.

The clock/axial rotation correlation falls out of the original
Copernican resolution for retrogrades by invoking an orbital motion
for the Earth leaving axial rotation to explain the daily cycle.The
timekeepers then overlayed the Equation of Time system on terrestrial
geography and treated axial rotation as a constant even if it is not
hence the correlation always remains 15 degrees of rotation per hour
making exactly 24 hours for each cycle.

The 'trick ' question may be explaining these things using a calendar
system for the additional day would complicate matters for you whereas
I can keep things seperate and assign heliocentric principles to the
pre-Copernican observations.





The third one: The moon rises later every day, it moves to the east.
So it takes longer to cross the meridian. Say it takes 28 solar
crossings for the moon's 27 crossings. Then it will cross 27*365/28
times, that's around 352. To get the right answer, we need a precise
figure for the lunar period and the precise location of the Moon when
the New Year comes in.

On Mar 15, 5:34 am, "Themos Tsikas"
wrote:
This is not homework, although I think it could be a good homework
question. We'll take "2007" to mean what the BBC says it is.

The answer to the first one I am confident of: 365

The second one: Viewed from above (north) of the ecliptic, the Earth
is going anti-clockwise. So the Sun appears to move to the east. That
makes the answer 366.

The third one: The moon rises later every day, it moves to the east.
So it takes longer to cross the meridian. Say it takes 28 solar
crossings for the moon's 27 crossings. Then it will cross 27*365/28
times, that's around 352. To get the right answer, we need a precise
figure for the lunar period and the precise location of the Moon when
the New Year comes in.

Gentlemen,

The coords of the Moon for Jan 1, 2008 a

12h 57m 55.2s S 10d 20m 35.9s

The orbital longitude for that epoch is 277.3947 deg
indicating that the moon is on the wane.

The instant of the previous New Moon is
Dec 9.737217 (2007) followed by the next New Moon
on Jan 8.485031 (2008).

There is always one day in every month which is
deprived of a transit, moonrise and moonset. For example
the transit here in E Arkansas for Mar 2 was 11:35:26.6 pm
and the following night was 12:25:44.7 am *on Mar 4*.
There was no transit on Mar 3.

So 365 - 12 = 353 is a reasonable assumption.

I'm curious about Sirius. What's the longitude of Oxford?

Ben, 90.126 n 35.539

On Mar 15, 5:34 am, "Themos Tsikas"
wrote:
This is not homework, although I think it could be a good homework
question. We'll take "2007" to mean what the BBC says it is.

The answer to the first one I am confident of: 365

The second one: Viewed from above (north) of the ecliptic, the Earth
is going anti-clockwise. So the Sun appears to move to the east. That
makes the answer 366.

The third one: The moon rises later every day, it moves to the east.
So it takes longer to cross the meridian. Say it takes 28 solar
crossings for the moon's 27 crossings. Then it will cross 27*365/28
times, that's around 352. To get the right answer, we need a precise
figure for the lunar period and the precise location of the Moon when
the New Year comes in.

Gentlemen,

The coords of the Moon for Jan 1, 2008 a

12h 57m 55.2s S 10d 20m 35.9s

The orbital longitude for that epoch is 277.3947 deg
indicating that the moon is on the wane.

The instant of the previous New Moon is
Dec 9.737217 (2007) followed by the next New Moon
on Jan 8.485031 (2008).

There is always one day in every month which is
deprived of a transit, moonrise and moonset. For example
the transit here in E Arkansas for Mar 2 was 11:35:26.6 pm
and the following night was 12:25:44.7 am *on Mar 4*.
There was no transit on Mar 3.

So 365 - 12 = 353 is a reasonable assumption.

I'm curious about Sirius. What's the longitude of Oxford?

Ben, 90.126 n 35.539

wrote:
On Mar 14, 6:17?pm, Stephen Tonkin
wrote:
Themos Tsikas wrote:
How many times will the Sun cross the southern meridian at Oxford in
2007?
How many times will Sirius cross the southern meridian at Oxford in
2007?
How many times will the Moon cross the southern meridian at Oxford in
2007?
Given that the answer to each question is different, it is most likely
not a trick question; you may learn something from the answers.

Whilst the moon loses something just under an hour each day compared
to the sun as a tidal day is about 55 minutes longer than the solar
day, the number of stellar transits will be the same will they not?
(as a layman in this area) I expect, that because Earth orbits the Sun,
the Sun gets one more or looses (I suppose the latter is the actual case
as both rotations are counter-clockwise when looking from the North
pole) one transition per year compared to the Stars.

Claudio Grondi

When does 2007 start and when does it end? Or is that still a matter
of choice?

In uk.sci.astronomy message , Wed, 14 Mar
2007 18:17:09, Stephen Tonkin posted:
Themos Tsikas wrote:

How many times will the Sun cross the southern meridian at Oxford in
2007?

How many times will Sirius cross the southern meridian at Oxford in
2007?

How many times will the Moon cross the southern meridian at Oxford in
2007?


Given that the answer to each question is different, it is most likely
not a trick question; you may learn something from the answers.

The answer may not be _quite_ what you think, if the International
Standard is followed carefully.

Since time immemorial, the idea has been to lock the year to the
climatic seasons (granted, others did not do that). Hence the use of
the Tropical Year, dependent on the motion of the Earth with respect to
that of the Equinoxes.

Pope Gregory wanted to lock the Calendar Year to the seasons, but did
not do so with perfect accuracy, being about 1 ppm in error IIRC.

ISO 8601:2004, however, describes the Gregorian Calendar Year in a
manner which can refer only to the Earth's motion with respect to the
Fixed Stars.

Therefore, there is uncertainty (of about 40 ppm) in the official length
of 2007, and maybe a little more.

Plus 0.03 ppm for the Leap Second opportunity at the end of UTC
December. Readers significantly to the East of Greenwich have no such
opportunity.

--
(c) John Stockton, Surrey, UK. Turnpike v6.05 IE 6.
Web URL:http://www.merlyn.demon.co.uk/ - w. FAQish topics, links, acronyms
PAS EXE etc : URL:http://www.merlyn.demon.co.uk/programs/ - see 00index.htm
Dates - miscdate.htm moredate.htm js-dates.htm pas-time.htm critdate.htm etc.