| Topic: |
Science > Physics |
| User: |
"W. Watson" |
| Date: |
08 Sep 2006 08:41:16 PM |
| Object: |
The Shadow of a Pole and Azimuth |
Suppose I take perfectly round pole about 2' long and 1/2" wide out onto a
very smooth surface, and let it cast a shadow from the Sun onto the surface.
Now I take out a perfectly straight metal ruler and lay it along the shadow
of the pole, and draw a line along one edge. Suppose I do this when the Sun
crosses the local meridian and about 2 hours later. Will the angular
distance agree with the computed azmithual angle as computed accurately from
an ephemeris?
I suspect not because the dark (umbra) shadow is elongated and wider near
the base of the pole, and narrows to the top of the shadow; thereby
producing an incorrect azimuthal line. Even with one rod, I suspect the
drawn line will be a few degrees off the true azimuth line. It may be
possible that the line from each time is off by a dissimilar amount, so the
difference is also incorrect. Perhaps my observation about the shadow is wrong.
I've done this a few times, and get about a two degree difference when using
the computed azimuth at, say, 2 pm or so. That is, one might expect the line
at the meridian to be accurate on N-S, but when working back from the 2 pm
measurement, the angle to the N-S is off by 2 degrees.
Any comments?
Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
--
"I have made a ceaseless effort not to ridicule, not to bewail,
not to scorn human actions, but to understand them."
-—Baruch Spinoza
Web Page: <home.earthlink.net/~mtnviews>
.
|
|
| User: "Sorcerer" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
08 Sep 2006 09:57:07 PM |
|
|
"W. Watson" <wolf_tracks@invalid.com> wrote in message
news:0ZoMg.16827$Qf.3077@newsread2.news.pas.earthlink.net...
| Suppose I take perfectly round pole about 2' long and 1/2" wide out onto a
| very smooth surface, and let it cast a shadow from the Sun onto the
surface.
| Now I take out a perfectly straight metal ruler and lay it along the
shadow
| of the pole, and draw a line along one edge. Suppose I do this when the
Sun
| crosses the local meridian and about 2 hours later. Will the angular
| distance agree with the computed azmithual angle as computed accurately
from
| an ephemeris?
|
| I suspect not because the dark (umbra) shadow is elongated and wider near
| the base of the pole, and narrows to the top of the shadow; thereby
| producing an incorrect azimuthal line. Even with one rod, I suspect the
| drawn line will be a few degrees off the true azimuth line. It may be
| possible that the line from each time is off by a dissimilar amount, so
the
| difference is also incorrect. Perhaps my observation about the shadow is
wrong.
|
| I've done this a few times, and get about a two degree difference when
using
| the computed azimuth at, say, 2 pm or so. That is, one might expect the
line
| at the meridian to be accurate on N-S, but when working back from the 2 pm
| measurement, the angle to the N-S is off by 2 degrees.
|
| Any comments?
What you've created is a sundial, and what you've discovered is
a combination of the sidereal day, you own latitude, the tilt of
the Earth and the eccentricity of the Earth's orbit.
Let's take those one at a time, but imagine they are extreme.
1) The sidereal day.
Imagine the Earth rotated on its own axis once a year, always
keeping the same face to the sun, just as the moon keeps the same
face toward the Earth. Your shadow would always be due south
or due north, no matter what time your clock said.
In reality a sidereal day can be measured by your method, but
with two poles instead of one of because stars don't make
much of a shadow. Pick a star you'll recognise again, then align
the two poles so that you can sight along them and form a
straight line. Note down the exact time when star and poles
are aligned. Now repeat the following night, again noting the time.
Do NOT move the poles. You will find there is roughly a 4
minute discrepancy, because 24 hours * 60 minutes = 1440,
and 1440/4 = 360 degrees and the Earth has a 365 day year.
In other words it rotates on its own axis 366 times a year,
but appears to lose a day by going around the sun.
2) Tilt and latitude.
If you were at the north pole on 21 June, the sun circles
the horizon, right? The further north you are, the longer
summer days are and the shorter winter days are. Take
an Alaskan cruise and see for yourself, but this is the
wrong time of year, we are fast approach equinox.
3) Eccentricity of the Earth's orbit.
Again let's exaggerate and imagine that we graze the
surface of the sun at perihelion and go out as far as
(say) Mars at aphelion, still taking a year for the orbit.
As worked out by Kepler, we travel much faster
going around the sun when close to it than we do
when we are furthest away.
http://home.cvc.org/science/kepler.htm
The earth doesn't speed up its rate of rotation, though, and
so when close to the sun the sun will be overhead for longer
than it will when further away, and this will not agree with
your wrist watch. In other words we'd be keeping the same
face toward the sun when closest to it.
You can see the same effect from lunations:
http://antwrp.gsfc.nasa.gov/apod/ap051113.html
The guys who built this were building a calendar/clock, doing
what you are doing.
http://witcombe.sbc.edu/earthmysteries/EMStonehenge.jpg
I can see from the shadows that the photograph was taken
in mid-afternoon, and on 21 July the sun rises over the heel
stone (bottom right), which of course is why they put it there
3,500 years ago. They needed big stones so that nobody
would move them and they needed them tall for better
accuracy. Remember those guys knew how to smelt copper
and tin to make bronze for axes and clothing clasps, they were
pretty smart. Your average Joe Sixpack today wouldn't have
a clue.
Androcles
.
|
|
|
| User: "W. Watson" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
08 Sep 2006 11:53:10 PM |
|
|
Sorcerer wrote:
"W. Watson" <wolf_tracks@invalid.com> wrote in message
news:0ZoMg.16827$Qf.3077@newsread2.news.pas.earthlink.net...
| Suppose I take perfectly round pole about 2' long and 1/2" wide out onto a
| very smooth surface, and let it cast a shadow from the Sun onto the
surface.
| Now I take out a perfectly straight metal ruler and lay it along the
shadow
| of the pole, and draw a line along one edge. Suppose I do this when the
Sun
| crosses the local meridian and about 2 hours later. Will the angular
| distance agree with the computed azmithual angle as computed accurately
from
| an ephemeris?
|
| I suspect not because the dark (umbra) shadow is elongated and wider near
| the base of the pole, and narrows to the top of the shadow; thereby
| producing an incorrect azimuthal line. Even with one rod, I suspect the
| drawn line will be a few degrees off the true azimuth line. It may be
| possible that the line from each time is off by a dissimilar amount, so
the
| difference is also incorrect. Perhaps my observation about the shadow is
wrong.
|
| I've done this a few times, and get about a two degree difference when
using
| the computed azimuth at, say, 2 pm or so. That is, one might expect the
line
| at the meridian to be accurate on N-S, but when working back from the 2 pm
| measurement, the angle to the N-S is off by 2 degrees.
|
| Any comments?
What you've created is a sundial, and what you've discovered is
a combination of the sidereal day, you own latitude, the tilt of
the Earth and the eccentricity of the Earth's orbit.
Let's take those one at a time, but imagine they are extreme.
1) The sidereal day.
Imagine the Earth rotated on its own axis once a year, always
keeping the same face to the sun, just as the moon keeps the same
face toward the Earth. Your shadow would always be due south
or due north, no matter what time your clock said.
In reality a sidereal day can be measured by your method, but
with two poles instead of one of because stars don't make
much of a shadow. Pick a star you'll recognise again, then align
the two poles so that you can sight along them and form a
straight line. Note down the exact time when star and poles
are aligned. Now repeat the following night, again noting the time.
Do NOT move the poles. You will find there is roughly a 4
minute discrepancy, because 24 hours * 60 minutes = 1440,
and 1440/4 = 360 degrees and the Earth has a 365 day year.
In other words it rotates on its own axis 366 times a year,
but appears to lose a day by going around the sun.
2) Tilt and latitude.
If you were at the north pole on 21 June, the sun circles
the horizon, right? The further north you are, the longer
summer days are and the shorter winter days are. Take
an Alaskan cruise and see for yourself, but this is the
wrong time of year, we are fast approach equinox.
3) Eccentricity of the Earth's orbit.
Again let's exaggerate and imagine that we graze the
surface of the sun at perihelion and go out as far as
(say) Mars at aphelion, still taking a year for the orbit.
As worked out by Kepler, we travel much faster
going around the sun when close to it than we do
when we are furthest away.
http://home.cvc.org/science/kepler.htm
The earth doesn't speed up its rate of rotation, though, and
so when close to the sun the sun will be overhead for longer
than it will when further away, and this will not agree with
your wrist watch. In other words we'd be keeping the same
face toward the sun when closest to it.
You can see the same effect from lunations:
http://antwrp.gsfc.nasa.gov/apod/ap051113.html
The guys who built this were building a calendar/clock, doing
what you are doing.
http://witcombe.sbc.edu/earthmysteries/EMStonehenge.jpg
I can see from the shadows that the photograph was taken
in mid-afternoon, and on 21 July the sun rises over the heel
stone (bottom right), which of course is why they put it there
3,500 years ago. They needed big stones so that nobody
would move them and they needed them tall for better
accuracy. Remember those guys knew how to smelt copper
and tin to make bronze for axes and clothing clasps, they were
pretty smart. Your average Joe Sixpack today wouldn't have
a clue.
Androcles
Thanks for the information packed response. Yep, smart is right. There were
some clever people way back in our history.
As I mentioned in my response above to another poster, my eventual interest
in doing this is to establish true N-S quickly and easily. I think this is
working pretty well. I plan to set up markers (direction arrows) at places
like our science museum entrance and fields where we sometimes have sky
shows. It just helps our astro participants line up their scopes reasonably
without having to use some other method. Of course, the bonus to this is
that it's done during the day, which is a big help if you're driving stakes
in the ground or errecting some marker.
Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
--
"I have made a ceaseless effort not to ridicule, not to bewail,
not to scorn human actions, but to understand them."
-—Baruch Spinoza
Web Page: <home.earthlink.net/~mtnviews>
.
|
|
|
| User: "Sorcerer" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
09 Sep 2006 08:05:53 AM |
|
|
"W. Watson" <wolf_tracks@invalid.com> wrote in message
news:WMrMg.8989$bM.334@newsread4.news.pas.earthlink.net...
| Sorcerer wrote:
|
| > "W. Watson" <wolf_tracks@invalid.com> wrote in message
| > news:0ZoMg.16827$Qf.3077@newsread2.news.pas.earthlink.net...
| > | Suppose I take perfectly round pole about 2' long and 1/2" wide out
onto a
| > | very smooth surface, and let it cast a shadow from the Sun onto the
| > surface.
| > | Now I take out a perfectly straight metal ruler and lay it along the
| > shadow
| > | of the pole, and draw a line along one edge. Suppose I do this when
the
| > Sun
| > | crosses the local meridian and about 2 hours later. Will the angular
| > | distance agree with the computed azmithual angle as computed
accurately
| > from
| > | an ephemeris?
| > |
| > | I suspect not because the dark (umbra) shadow is elongated and wider
near
| > | the base of the pole, and narrows to the top of the shadow; thereby
| > | producing an incorrect azimuthal line. Even with one rod, I suspect
the
| > | drawn line will be a few degrees off the true azimuth line. It may be
| > | possible that the line from each time is off by a dissimilar amount,
so
| > the
| > | difference is also incorrect. Perhaps my observation about the shadow
is
| > wrong.
| > |
| > | I've done this a few times, and get about a two degree difference when
| > using
| > | the computed azimuth at, say, 2 pm or so. That is, one might expect
the
| > line
| > | at the meridian to be accurate on N-S, but when working back from the
2 pm
| > | measurement, the angle to the N-S is off by 2 degrees.
| > |
| > | Any comments?
| >
| >
| > What you've created is a sundial, and what you've discovered is
| > a combination of the sidereal day, you own latitude, the tilt of
| > the Earth and the eccentricity of the Earth's orbit.
| >
| > Let's take those one at a time, but imagine they are extreme.
| > 1) The sidereal day.
| > Imagine the Earth rotated on its own axis once a year, always
| > keeping the same face to the sun, just as the moon keeps the same
| > face toward the Earth. Your shadow would always be due south
| > or due north, no matter what time your clock said.
| > In reality a sidereal day can be measured by your method, but
| > with two poles instead of one of because stars don't make
| > much of a shadow. Pick a star you'll recognise again, then align
| > the two poles so that you can sight along them and form a
| > straight line. Note down the exact time when star and poles
| > are aligned. Now repeat the following night, again noting the time.
| > Do NOT move the poles. You will find there is roughly a 4
| > minute discrepancy, because 24 hours * 60 minutes = 1440,
| > and 1440/4 = 360 degrees and the Earth has a 365 day year.
| > In other words it rotates on its own axis 366 times a year,
| > but appears to lose a day by going around the sun.
| >
| > 2) Tilt and latitude.
| > If you were at the north pole on 21 June, the sun circles
| > the horizon, right? The further north you are, the longer
| > summer days are and the shorter winter days are. Take
| > an Alaskan cruise and see for yourself, but this is the
| > wrong time of year, we are fast approach equinox.
| >
| > 3) Eccentricity of the Earth's orbit.
| > Again let's exaggerate and imagine that we graze the
| > surface of the sun at perihelion and go out as far as
| > (say) Mars at aphelion, still taking a year for the orbit.
| > As worked out by Kepler, we travel much faster
| > going around the sun when close to it than we do
| > when we are furthest away.
| > http://home.cvc.org/science/kepler.htm
| > The earth doesn't speed up its rate of rotation, though, and
| > so when close to the sun the sun will be overhead for longer
| > than it will when further away, and this will not agree with
| > your wrist watch. In other words we'd be keeping the same
| > face toward the sun when closest to it.
| > You can see the same effect from lunations:
| > http://antwrp.gsfc.nasa.gov/apod/ap051113.html
| >
| >
| > The guys who built this were building a calendar/clock, doing
| > what you are doing.
| > http://witcombe.sbc.edu/earthmysteries/EMStonehenge.jpg
| > I can see from the shadows that the photograph was taken
| > in mid-afternoon, and on 21 July the sun rises over the heel
| > stone (bottom right), which of course is why they put it there
| > 3,500 years ago. They needed big stones so that nobody
| > would move them and they needed them tall for better
| > accuracy. Remember those guys knew how to smelt copper
| > and tin to make bronze for axes and clothing clasps, they were
| > pretty smart. Your average Joe Sixpack today wouldn't have
| > a clue.
| >
| > Androcles
| >
| >
|
| Thanks for the information packed response. Yep, smart is right. There
were
| some clever people way back in our history.
|
| As I mentioned in my response above to another poster, my eventual
interest
| in doing this is to establish true N-S quickly and easily. I think this is
| working pretty well. I plan to set up markers (direction arrows) at places
| like our science museum entrance and fields where we sometimes have sky
| shows. It just helps our astro participants line up their scopes
reasonably
| without having to use some other method. Of course, the bonus to this is
| that it's done during the day, which is a big help if you're driving
stakes
| in the ground or errecting some marker.
If that were ALL you need, Polaris is the answer.
http://www.onr.navy.mil/Focus/spacesciences/observingsky/constellations3.htm
http://antwrp.gsfc.nasa.gov/apod/ap050218.html
http://antwrp.gsfc.nasa.gov/apod/ap050714.html
It sounds to me as if you wish to be involved in education,
teaching youngsters some basic principles.
This is just a suggestion, but if you can sell the idea, have
access to some land on a semi-permanent basis and you
could raise some funds from parents (avoid government grants)
you could enlist the aid of youngsters to create your own
stone or wooden circle along the lines of Stonehenge
before it became a ruin.
http://www.this-is-amesbury.co.uk/woodhenge.html
That way you'd spark interest not only in astronomy and
celestial mechanics but in history, mathematics and
construction techniques as well. Get them involved in a
project they can be proud of, and always remember:
when one teaches, two learn.
Androcles
.
|
|
|
|
|
|
| User: "" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
08 Sep 2006 09:02:19 PM |
|
|
W=2E Watson wrote:
Suppose I take perfectly round pole about 2' long and 1/2" wide out onto a
very smooth surface, and let it cast a shadow from the Sun onto the surfa=
ce.
Now I take out a perfectly straight metal ruler and lay it along the shad=
ow
of the pole, and draw a line along one edge. Suppose I do this when the S=
un
crosses the local meridian and about 2 hours later. Will the angular
distance agree with the computed azmithual angle as computed accurately f=
rom
an ephemeris?
I suspect not because the dark (umbra) shadow is elongated and wider near
the base of the pole, and narrows to the top of the shadow; thereby
producing an incorrect azimuthal line. Even with one rod, I suspect the
drawn line will be a few degrees off the true azimuth line. It may be
possible that the line from each time is off by a dissimilar amount, so t=
he
difference is also incorrect. Perhaps my observation about the shadow is =
wrong.
I've done this a few times, and get about a two degree difference when us=
ing
the computed azimuth at, say, 2 pm or so. That is, one might expect the l=
ine
at the meridian to be accurate on N-S, but when working back from the 2 pm
measurement, the angle to the N-S is off by 2 degrees.
Any comments?
Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39=B0 15' 7" N, 121=B0 2' 32" W, 2700 feet
--
"I have made a ceaseless effort not to ridicule, not to bewail,
not to scorn human actions, but to understand them."
--Baruch Spinoza
Web Page: <home.earthlink.net/~mtnviews>
The main consideration in all this is that the top of the shadow
appears narrower because the shadow is farther from the pole there and
the 0.5 degree disk of the sun is making the edge indistinct. You
would have to find a way to mark this blurry edge at a point which
corresponds to the distinct bottom edge. That would be where it
becomes as dark as at the bottom. Also, the pole would need to be
perfectly vertical and the base perfectly flat and horizontal.
.
|
|
|
| User: "W. Watson" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
08 Sep 2006 11:46:03 PM |
|
|
wrote:
W. Watson wrote:
Suppose I take perfectly round pole about 2' long and 1/2" wide out onto a
very smooth surface, and let it cast a shadow from the Sun onto the surface.
Now I take out a perfectly straight metal ruler and lay it along the shadow
of the pole, and draw a line along one edge. Suppose I do this when the Sun
crosses the local meridian and about 2 hours later. Will the angular
distance agree with the computed azmithual angle as computed accurately from
an ephemeris?
I suspect not because the dark (umbra) shadow is elongated and wider near
the base of the pole, and narrows to the top of the shadow; thereby
producing an incorrect azimuthal line. Even with one rod, I suspect the
drawn line will be a few degrees off the true azimuth line. It may be
possible that the line from each time is off by a dissimilar amount, so the
difference is also incorrect. Perhaps my observation about the shadow is wrong.
I've done this a few times, and get about a two degree difference when using
the computed azimuth at, say, 2 pm or so. That is, one might expect the line
at the meridian to be accurate on N-S, but when working back from the 2 pm
measurement, the angle to the N-S is off by 2 degrees.
Any comments?
Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
--
"I have made a ceaseless effort not to ridicule, not to bewail,
not to scorn human actions, but to understand them."
--Baruch Spinoza
Web Page: <home.earthlink.net/~mtnviews>
The main consideration in all this is that the top of the shadow
appears narrower because the shadow is farther from the pole there and
the 0.5 degree disk of the sun is making the edge indistinct. You
would have to find a way to mark this blurry edge at a point which
corresponds to the distinct bottom edge. That would be where it
becomes as dark as at the bottom. Also, the pole would need to be
perfectly vertical and the base perfectly flat and horizontal.
Yep, the vertical and base were oriented as you indicated. It seems like
using the midpoint of the top and bottom is the way to go for simple and
quick measurements. When I first started this, I had a short pole, and then
eventually noticed the distortion, which is now altogther so obvious.
Actually, my interest is finding quickly and easily where true north is
located. I want to put a marker, possibly a sight device with cross hair or
the like, on a field to indicate it. A few degrees won't matter much, but I
would like to get it as good as I can without spending hours at it. It's
pretty easy for me to pump out the azimuth of the Sun for my location.
Simply working back to 180 degrees is easily done with a piece of cardboard,
pencil, ruler and a protractor. The pole and (double) level are in my tool kit.
.
|
|
|
| User: "Richard Herring" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
11 Sep 2006 11:00:12 AM |
|
|
In message <fGrMg.9377$xQ1.3217@newsread3.news.pas.earthlink.net>, W.
Watson <wolf_tracks@invalid.com> writes
Actually, my interest is finding quickly and easily where true north is
located. I want to put a marker, possibly a sight device with cross
hair or the like, on a field to indicate it. A few degrees won't matter
much, but I would like to get it as good as I can without spending
hours at it. It's pretty easy for me to pump out the azimuth of the Sun
for my location. Simply working back to 180 degrees is easily done with
a piece of cardboard, pencil, ruler and a protractor. The pole and
(double) level are in my tool kit.
Unless I've missed a post, nobody seems to have suggested the
alternative method: just plot the locus of the shadow of the tip at
several times during the day. It should give a reasonable approximation
to an E-W line, and you don't need to know the azimuth or any other any
almanac data.
--
Richard Herring
.
|
|
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
09 Sep 2006 10:18:39 AM |
|
|
On 8 Sep 2006 19:02:19 -0700, wrote:
W. Watson wrote:
Suppose I take perfectly round pole about 2' long and 1/2" wide out onto a
very smooth surface, and let it cast a shadow from the Sun onto the surface.
Now I take out a perfectly straight metal ruler and lay it along the shadow
of the pole, and draw a line along one edge. Suppose I do this when the Sun
crosses the local meridian and about 2 hours later. Will the angular
distance agree with the computed azmithual angle as computed accurately from
an ephemeris?
I suspect not because the dark (umbra) shadow is elongated and wider near
the base of the pole, and narrows to the top of the shadow; thereby
producing an incorrect azimuthal line. Even with one rod, I suspect the
drawn line will be a few degrees off the true azimuth line. It may be
possible that the line from each time is off by a dissimilar amount, so the
difference is also incorrect. Perhaps my observation about the shadow is wrong.
I've done this a few times, and get about a two degree difference when using
the computed azimuth at, say, 2 pm or so. That is, one might expect the line
at the meridian to be accurate on N-S, but when working back from the 2 pm
measurement, the angle to the N-S is off by 2 degrees.
Any comments?
Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
--
"I have made a ceaseless effort not to ridicule, not to bewail,
not to scorn human actions, but to understand them."
--Baruch Spinoza
Web Page: <home.earthlink.net/~mtnviews>
The main consideration in all this is that the top of the shadow
appears narrower because the shadow is farther from the pole there and
the 0.5 degree disk of the sun is making the edge indistinct. You
would have to find a way to mark this blurry edge at a point which
corresponds to the distinct bottom edge. That would be where it
becomes as dark as at the bottom. Also, the pole would need to be
perfectly vertical and the base perfectly flat and horizontal.
Off the top of my head, the rate is really 15 degrees per hour
multiplied by the cosine of your latitude. I havent worked it out but
where did you get the idea that 2 hours equals 2 degrees?
John Polasek
.
|
|
|
| User: "Sorcerer" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
09 Sep 2006 11:31:35 AM |
|
|
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:dmm5g2ppq3rr9spkp3bh4g7en6aich15ig@4ax.com...
| On 8 Sep 2006 19:02:19 -0700, wrote:
|
| >
| >W. Watson wrote:
| >> Suppose I take perfectly round pole about 2' long and 1/2" wide out
onto a
| >> very smooth surface, and let it cast a shadow from the Sun onto the
surface.
| >> Now I take out a perfectly straight metal ruler and lay it along the
shadow
| >> of the pole, and draw a line along one edge. Suppose I do this when the
Sun
| >> crosses the local meridian and about 2 hours later. Will the angular
| >> distance agree with the computed azmithual angle as computed accurately
from
| >> an ephemeris?
| >>
| >> I suspect not because the dark (umbra) shadow is elongated and wider
near
| >> the base of the pole, and narrows to the top of the shadow; thereby
| >> producing an incorrect azimuthal line. Even with one rod, I suspect the
| >> drawn line will be a few degrees off the true azimuth line. It may be
| >> possible that the line from each time is off by a dissimilar amount, so
the
| >> difference is also incorrect. Perhaps my observation about the shadow
is wrong.
| >>
| >> I've done this a few times, and get about a two degree difference when
using
| >> the computed azimuth at, say, 2 pm or so. That is, one might expect the
line
| >> at the meridian to be accurate on N-S, but when working back from the 2
pm
| >> measurement, the angle to the N-S is off by 2 degrees.
| >>
| >> Any comments?
| >>
| >> Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
| >> (121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
| >> Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
| >> --
| >>
| >> "I have made a ceaseless effort not to ridicule, not to bewail,
| >> not to scorn human actions, but to understand them."
| >> --Baruch Spinoza
| >>
| >> Web Page: <home.earthlink.net/~mtnviews>
| >
| >The main consideration in all this is that the top of the shadow
| >appears narrower because the shadow is farther from the pole there and
| >the 0.5 degree disk of the sun is making the edge indistinct. You
| >would have to find a way to mark this blurry edge at a point which
| >corresponds to the distinct bottom edge. That would be where it
| >becomes as dark as at the bottom. Also, the pole would need to be
| >perfectly vertical and the base perfectly flat and horizontal.
| Off the top of my head, the rate is really 15 degrees per hour
| multiplied by the cosine of your latitude. I havent worked it out but
| where did you get the idea that 2 hours equals 2 degrees?
| John Polasek
Where did you get the idea anyone said it was?
"when working back from the 2 pm measurement, the angle to the N-S is
off by 2 degrees."
The 2:00 pm measurement is 30 +/- 2 degrees, subtract 30 and
the angle to N-S is +/- 2 degrees.
You've summarised it quite well with " I havent worked it out ".
Androcles.
.
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
09 Sep 2006 01:34:23 PM |
|
|
On Sat, 09 Sep 2006 16:31:35 GMT, "Sorcerer"
<Headmaster@hogwarts.physics_b> wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:dmm5g2ppq3rr9spkp3bh4g7en6aich15ig@4ax.com...
| On 8 Sep 2006 19:02:19 -0700, wrote:
|
| >
| >W. Watson wrote:
| >> Suppose I take perfectly round pole about 2' long and 1/2" wide out
onto a
| >> very smooth surface, and let it cast a shadow from the Sun onto the
surface.
| >> Now I take out a perfectly straight metal ruler and lay it along the
shadow
| >> of the pole, and draw a line along one edge. Suppose I do this when the
Sun
| >> crosses the local meridian and about 2 hours later. Will the angular
| >> distance agree with the computed azmithual angle as computed accurately
from
| >> an ephemeris?
| >>
| >> I suspect not because the dark (umbra) shadow is elongated and wider
near
| >> the base of the pole, and narrows to the top of the shadow; thereby
| >> producing an incorrect azimuthal line. Even with one rod, I suspect the
| >> drawn line will be a few degrees off the true azimuth line. It may be
| >> possible that the line from each time is off by a dissimilar amount, so
the
| >> difference is also incorrect. Perhaps my observation about the shadow
is wrong.
| >>
| >> I've done this a few times, and get about a two degree difference when
using
| >> the computed azimuth at, say, 2 pm or so. That is, one might expect the
line
| >> at the meridian to be accurate on N-S, but when working back from the 2
pm
| >> measurement, the angle to the N-S is off by 2 degrees.
| >>
| >> Any comments?
| >>
| >> Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
| >> (121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
| >> Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
| >> --
| >>
| >> "I have made a ceaseless effort not to ridicule, not to bewail,
| >> not to scorn human actions, but to understand them."
| >> --Baruch Spinoza
| >>
| >> Web Page: <home.earthlink.net/~mtnviews>
| >
| >The main consideration in all this is that the top of the shadow
| >appears narrower because the shadow is farther from the pole there and
| >the 0.5 degree disk of the sun is making the edge indistinct. You
| >would have to find a way to mark this blurry edge at a point which
| >corresponds to the distinct bottom edge. That would be where it
| >becomes as dark as at the bottom. Also, the pole would need to be
| >perfectly vertical and the base perfectly flat and horizontal.
| Off the top of my head, the rate is really 15 degrees per hour
| multiplied by the cosine of your latitude. I havent worked it out but
| where did you get the idea that 2 hours equals 2 degrees?
| John Polasek
Where did you get the idea anyone said it was?
"when working back from the 2 pm measurement, the angle to the N-S is
off by 2 degrees."
The 2:00 pm measurement is 30 +/- 2 degrees, subtract 30 and
the angle to N-S is +/- 2 degrees.
You've summarised it quite well with " I havent worked it out ".
Androcles.
OK I didn't read it in detail. I saw
"crosses the local meridian and about 2 hours later"
and I saw "the angle to the N-S is off by 2 degrees" but did not
notice he said it was an error of 2 degrees. (So that's the downside
of speed-reading).
So, actually his angle for two hours should be 30 degs x cos 39 = 23.3
deg. but he is not stating the obvious: the ephemeris value.
Is it too much to ask what that angle was and what his reading was?
John Polasek
.
|
|
|
| User: "Sorcerer" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
09 Sep 2006 02:59:40 PM |
|
|
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:pl16g2d2glpn2ipcf36itvu1a8i97o35dl@4ax.com...
| On Sat, 09 Sep 2006 16:31:35 GMT, "Sorcerer"
| <Headmaster@hogwarts.physics_b> wrote:
|
| >
| >"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
| >news:dmm5g2ppq3rr9spkp3bh4g7en6aich15ig@4ax.com...
| >| On 8 Sep 2006 19:02:19 -0700, wrote:
| >|
| >| >
| >| >W. Watson wrote:
| >| >> Suppose I take perfectly round pole about 2' long and 1/2" wide out
| >onto a
| >| >> very smooth surface, and let it cast a shadow from the Sun onto the
| >surface.
| >| >> Now I take out a perfectly straight metal ruler and lay it along the
| >shadow
| >| >> of the pole, and draw a line along one edge. Suppose I do this when
the
| >Sun
| >| >> crosses the local meridian and about 2 hours later. Will the angular
| >| >> distance agree with the computed azmithual angle as computed
accurately
| >from
| >| >> an ephemeris?
| >| >>
| >| >> I suspect not because the dark (umbra) shadow is elongated and wider
| >near
| >| >> the base of the pole, and narrows to the top of the shadow; thereby
| >| >> producing an incorrect azimuthal line. Even with one rod, I suspect
the
| >| >> drawn line will be a few degrees off the true azimuth line. It may
be
| >| >> possible that the line from each time is off by a dissimilar amount,
so
| >the
| >| >> difference is also incorrect. Perhaps my observation about the
shadow
| >is wrong.
| >| >>
| >| >> I've done this a few times, and get about a two degree difference
when
| >using
| >| >> the computed azimuth at, say, 2 pm or so. That is, one might expect
the
| >line
| >| >> at the meridian to be accurate on N-S, but when working back from
the 2
| >pm
| >| >> measurement, the angle to the N-S is off by 2 degrees.
| >| >>
| >| >> Any comments?
| >| >>
| >| >> Wayne T. Watson (Watson Adventures, Prop., Nevada City,
CA)
| >| >> (121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
| >| >> Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
| >| >> --
| >| >>
| >| >> "I have made a ceaseless effort not to ridicule, not to
bewail,
| >| >> not to scorn human actions, but to understand them."
| >| >> --Baruch Spinoza
| >| >>
| >| >> Web Page: <home.earthlink.net/~mtnviews>
| >| >
| >| >The main consideration in all this is that the top of the shadow
| >| >appears narrower because the shadow is farther from the pole there and
| >| >the 0.5 degree disk of the sun is making the edge indistinct. You
| >| >would have to find a way to mark this blurry edge at a point which
| >| >corresponds to the distinct bottom edge. That would be where it
| >| >becomes as dark as at the bottom. Also, the pole would need to be
| >| >perfectly vertical and the base perfectly flat and horizontal.
| >| Off the top of my head, the rate is really 15 degrees per hour
| >| multiplied by the cosine of your latitude. I havent worked it out but
| >| where did you get the idea that 2 hours equals 2 degrees?
| >| John Polasek
| >
| >Where did you get the idea anyone said it was?
| >"when working back from the 2 pm measurement, the angle to the N-S is
| >off by 2 degrees."
| >The 2:00 pm measurement is 30 +/- 2 degrees, subtract 30 and
| >the angle to N-S is +/- 2 degrees.
| >You've summarised it quite well with " I havent worked it out ".
| >Androcles.
| >
| >
| OK I didn't read it in detail. I saw
| "crosses the local meridian and about 2 hours later"
| and I saw "the angle to the N-S is off by 2 degrees" but did not
| notice he said it was an error of 2 degrees. (So that's the downside
| of speed-reading).
|
| So, actually his angle for two hours should be 30 degs x cos 39 = 23.3
| deg.
Huh?
The earth rotates 361 degrees per day relative to the sun,
a shadow at the equator shortens until noon and then lengthens,
a shadow at the N or S pole rotates 360 degrees relative to
the snow and remains the same length (or would if the Earth
were not tilted).
Since it is tilted, the shadow is formed not by the post in the snow
but by the Earth itself, which hides the shadow of the stick..
... we call that "night", but the shadow still goes through 15 degrees
an hour wherever you are.
There are longer "days" in summer with shorter nights, but that
kind of 'day' is not the same 'day' as 24 hours.
You will NOT see an angle to the *local* meridian of 23 degrees
at 2 PM local time.
| but he is not stating the obvious:
Why should he?
1) The obvious is obvious or it would not be obvious
and it wasn't obvious to him.
2) he was asking a question, why does he get a two degree
error from what he expected to see.
I've given him part of the explanation and he said thank you,
but I left out part as well to see if he came back for more.
| the ephemeris value.
| Is it too much to ask what that angle was and what his reading was?
Yes, it is too much to ask (of him). His watch will not agree with a
sundial to within 15 degrees in the TIME ZONE in which he lives, and
that is because he sets his watch not to local time, but to a nomimal
value within a 15 degree spread. The time in Bristol is 10 minutes
behind the time in London (by sundial), and it was playing havoc with
railway timetables.
http://tinyurl.com/g8dk2
Androcles.
.
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
10 Sep 2006 09:26:24 AM |
|
|
On Sat, 09 Sep 2006 19:59:40 GMT, "Sorcerer"
<Headmaster@hogwarts.physics_b> wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:pl16g2d2glpn2ipcf36itvu1a8i97o35dl@4ax.com...
| On Sat, 09 Sep 2006 16:31:35 GMT, "Sorcerer"
| <Headmaster@hogwarts.physics_b> wrote:
|
| >
| >"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
| >news:dmm5g2ppq3rr9spkp3bh4g7en6aich15ig@4ax.com...
| >| On 8 Sep 2006 19:02:19 -0700, wrote:
| >|
| >| >
| >| >W. Watson wrote:
| >| >> Suppose I take perfectly round pole about 2' long and 1/2" wide out
| >onto a
| >| >> very smooth surface, and let it cast a shadow from the Sun onto the
| >surface.
| >| >> Now I take out a perfectly straight metal ruler and lay it along the
| >shadow
| >| >> of the pole, and draw a line along one edge. Suppose I do this when
the
| >Sun
| >| >> crosses the local meridian and about 2 hours later. Will the angular
| >| >> distance agree with the computed azmithual angle as computed
accurately
| >from
| >| >> an ephemeris?
| >| >>
| >| >> I suspect not because the dark (umbra) shadow is elongated and wider
| >near
| >| >> the base of the pole, and narrows to the top of the shadow; thereby
| >| >> producing an incorrect azimuthal line. Even with one rod, I suspect
the
| >| >> drawn line will be a few degrees off the true azimuth line. It may
be
| >| >> possible that the line from each time is off by a dissimilar amount,
so
| >the
| >| >> difference is also incorrect. Perhaps my observation about the
shadow
| >is wrong.
| >| >>
| >| >> I've done this a few times, and get about a two degree difference
when
| >using
| >| >> the computed azimuth at, say, 2 pm or so. That is, one might expect
the
| >line
| >| >> at the meridian to be accurate on N-S, but when working back from
the 2
| >pm
| >| >> measurement, the angle to the N-S is off by 2 degrees.
| >| >>
| >| >> Any comments?
| >| >>
| >| >> Wayne T. Watson (Watson Adventures, Prop., Nevada City,
CA)
| >| >> (121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
| >| >> Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
| >| >> --
| >| >>
| >| >> "I have made a ceaseless effort not to ridicule, not to
bewail,
| >| >> not to scorn human actions, but to understand them."
| >| >> --Baruch Spinoza
| >| >>
| >| >> Web Page: <home.earthlink.net/~mtnviews>
| >| >
| >| >The main consideration in all this is that the top of the shadow
| >| >appears narrower because the shadow is farther from the pole there and
| >| >the 0.5 degree disk of the sun is making the edge indistinct. You
| >| >would have to find a way to mark this blurry edge at a point which
| >| >corresponds to the distinct bottom edge. That would be where it
| >| >becomes as dark as at the bottom. Also, the pole would need to be
| >| >perfectly vertical and the base perfectly flat and horizontal.
| >| Off the top of my head, the rate is really 15 degrees per hour
| >| multiplied by the cosine of your latitude. I havent worked it out but
| >| where did you get the idea that 2 hours equals 2 degrees?
| >| John Polasek
| >
| >Where did you get the idea anyone said it was?
| >"when working back from the 2 pm measurement, the angle to the N-S is
| >off by 2 degrees."
| >The 2:00 pm measurement is 30 +/- 2 degrees, subtract 30 and
| >the angle to N-S is +/- 2 degrees.
| >You've summarised it quite well with " I havent worked it out ".
| >Androcles.
| >
| >
| OK I didn't read it in detail. I saw
| "crosses the local meridian and about 2 hours later"
| and I saw "the angle to the N-S is off by 2 degrees" but did not
| notice he said it was an error of 2 degrees. (So that's the downside
| of speed-reading).
|
| So, actually his angle for two hours should be 30 degs x cos 39 = 23.3
| deg.
Huh?
The earth rotates 361 degrees per day relative to the sun,
a shadow at the equator shortens until noon and then lengthens,
a shadow at the N or S pole rotates 360 degrees relative to
the snow and remains the same length (or would if the Earth
were not tilted).
Since it is tilted, the shadow is formed not by the post in the snow
but by the Earth itself, which hides the shadow of the stick..
.. we call that "night", but the shadow still goes through 15 degrees
an hour wherever you are.
There are longer "days" in summer with shorter nights, but that
kind of 'day' is not the same 'day' as 24 hours.
You will NOT see an angle to the *local* meridian of 23 degrees
at 2 PM local time.
| but he is not stating the obvious:
Why should he?
1) The obvious is obvious or it would not be obvious
and it wasn't obvious to him.
2) he was asking a question, why does he get a two degree
error from what he expected to see.
I've given him part of the explanation and he said thank you,
but I left out part as well to see if he came back for more.
| the ephemeris value.
| Is it too much to ask what that angle was and what his reading was?
Yes, it is too much to ask (of him). His watch will not agree with a
sundial to within 15 degrees in the TIME ZONE in which he lives, and
that is because he sets his watch not to local time, but to a nomimal
value within a 15 degree spread. The time in Bristol is 10 minutes
behind the time in London (by sundial), and it was playing havoc with
railway timetables.
http://tinyurl.com/g8dk2
Androcles.
The effective angular rate at his latitude is 15 deg/hr times the
cosine of his latitude which he stated was 39 degrees, so 11.66
deg/hour.
He's talking of time *difference* of two hours so he's talking about
*difference* in angle (and then a 2 degree +-? error). Absolute time
it seems to me might only affect how he uses the ephemeris, of which
anyone using an ephemeris should be aware.
Yes it would be nice to post the expected and actual numbers. But
since we find ourselves arguing alone, you would think the OP would
chime in otherwise I declare him to be a WOPMUG*.
John Polasek
*Why OP's make us guess
.
|
|
|
| User: "Sorcerer" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
10 Sep 2006 09:45:31 AM |
|
|
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:qt78g25rjgbcfir1i3m3vbh8ldhv8qognv@4ax.com...
| On Sat, 09 Sep 2006 19:59:40 GMT, "Sorcerer"
| <Headmaster@hogwarts.physics_b> wrote:
|
| >
| >"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
| >news:pl16g2d2glpn2ipcf36itvu1a8i97o35dl@4ax.com...
| >| On Sat, 09 Sep 2006 16:31:35 GMT, "Sorcerer"
| >| <Headmaster@hogwarts.physics_b> wrote:
| >|
| >| >
| >| >"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
| >| >news:dmm5g2ppq3rr9spkp3bh4g7en6aich15ig@4ax.com...
| >| >| On 8 Sep 2006 19:02:19 -0700, wrote:
| >| >|
| >| >| >
| >| >| >W. Watson wrote:
| >| >| >> Suppose I take perfectly round pole about 2' long and 1/2" wide
out
| >| >onto a
| >| >| >> very smooth surface, and let it cast a shadow from the Sun onto
the
| >| >surface.
| >| >| >> Now I take out a perfectly straight metal ruler and lay it along
the
| >| >shadow
| >| >| >> of the pole, and draw a line along one edge. Suppose I do this
when
| >the
| >| >Sun
| >| >| >> crosses the local meridian and about 2 hours later. Will the
angular
| >| >| >> distance agree with the computed azmithual angle as computed
| >accurately
| >| >from
| >| >| >> an ephemeris?
| >| >| >>
| >| >| >> I suspect not because the dark (umbra) shadow is elongated and
wider
| >| >near
| >| >| >> the base of the pole, and narrows to the top of the shadow;
thereby
| >| >| >> producing an incorrect azimuthal line. Even with one rod, I
suspect
| >the
| >| >| >> drawn line will be a few degrees off the true azimuth line. It
may
| >be
| >| >| >> possible that the line from each time is off by a dissimilar
amount,
| >so
| >| >the
| >| >| >> difference is also incorrect. Perhaps my observation about the
| >shadow
| >| >is wrong.
| >| >| >>
| >| >| >> I've done this a few times, and get about a two degree difference
| >when
| >| >using
| >| >| >> the computed azimuth at, say, 2 pm or so. That is, one might
expect
| >the
| >| >line
| >| >| >> at the meridian to be accurate on N-S, but when working back from
| >the 2
| >| >pm
| >| >| >> measurement, the angle to the N-S is off by 2 degrees.
| >| >| >>
| >| >| >> Any comments?
| >| >| >>
| >| >| >> Wayne T. Watson (Watson Adventures, Prop., Nevada City,
| >CA)
| >| >| >> (121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
| >| >| >> Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
| >| >| >> --
| >| >| >>
| >| >| >> "I have made a ceaseless effort not to ridicule, not to
| >bewail,
| >| >| >> not to scorn human actions, but to understand them."
| >| >| >> --Baruch Spinoza
| >| >| >>
| >| >| >> Web Page: <home.earthlink.net/~mtnviews>
| >| >| >
| >| >| >The main consideration in all this is that the top of the shadow
| >| >| >appears narrower because the shadow is farther from the pole there
and
| >| >| >the 0.5 degree disk of the sun is making the edge indistinct. You
| >| >| >would have to find a way to mark this blurry edge at a point which
| >| >| >corresponds to the distinct bottom edge. That would be where it
| >| >| >becomes as dark as at the bottom. Also, the pole would need to be
| >| >| >perfectly vertical and the base perfectly flat and horizontal.
| >| >| Off the top of my head, the rate is really 15 degrees per hour
| >| >| multiplied by the cosine of your latitude. I havent worked it out
but
| >| >| where did you get the idea that 2 hours equals 2 degrees?
| >| >| John Polasek
| >| >
| >| >Where did you get the idea anyone said it was?
| >| >"when working back from the 2 pm measurement, the angle to the N-S is
| >| >off by 2 degrees."
| >| >The 2:00 pm measurement is 30 +/- 2 degrees, subtract 30 and
| >| >the angle to N-S is +/- 2 degrees.
| >| >You've summarised it quite well with " I havent worked it out ".
| >| >Androcles.
| >| >
| >| >
| >| OK I didn't read it in detail. I saw
| >| "crosses the local meridian and about 2 hours later"
| >| and I saw "the angle to the N-S is off by 2 degrees" but did not
| >| notice he said it was an error of 2 degrees. (So that's the downside
| >| of speed-reading).
| >|
| >| So, actually his angle for two hours should be 30 degs x cos 39 = 23.3
| >| deg.
| >
| >Huh?
| >The earth rotates 361 degrees per day relative to the sun,
| >a shadow at the equator shortens until noon and then lengthens,
| >a shadow at the N or S pole rotates 360 degrees relative to
| >the snow and remains the same length (or would if the Earth
| >were not tilted).
| >Since it is tilted, the shadow is formed not by the post in the snow
| > but by the Earth itself, which hides the shadow of the stick..
| >.. we call that "night", but the shadow still goes through 15 degrees
| >an hour wherever you are.
| >There are longer "days" in summer with shorter nights, but that
| >kind of 'day' is not the same 'day' as 24 hours.
| >You will NOT see an angle to the *local* meridian of 23 degrees
| >at 2 PM local time.
| >
| >| but he is not stating the obvious:
| >
| >Why should he?
| >1) The obvious is obvious or it would not be obvious
| > and it wasn't obvious to him.
| >2) he was asking a question, why does he get a two degree
| >error from what he expected to see.
| >I've given him part of the explanation and he said thank you,
| >but I left out part as well to see if he came back for more.
| >
| >
| >| the ephemeris value.
| >| Is it too much to ask what that angle was and what his reading was?
| >
| >Yes, it is too much to ask (of him). His watch will not agree with a
| >sundial to within 15 degrees in the TIME ZONE in which he lives, and
| >that is because he sets his watch not to local time, but to a nomimal
| >value within a 15 degree spread. The time in Bristol is 10 minutes
| >behind the time in London (by sundial), and it was playing havoc with
| >railway timetables.
| >http://tinyurl.com/g8dk2
| >
| >Androcles.
| >
| The effective angular rate at his latitude is 15 deg/hr times the
| cosine of his latitude which he stated was 39 degrees, so 11.66
| deg/hour.
You are speed reading again, not taking in what I said, FUCKWIT!
The effective angular rate at his latitude is 15 deg/hr times ONE.
Androcles.
.
|
|
|
| User: "John C. Polasek" |
|
| Title: Re: The Shadow of a Pole and Azimuth |
11 Sep 2006 04:45:26 PM |
|
|
On Sun, 10 Sep 2006 14:45:31 GMT, "Sorcerer"
<Headmaster@hogwarts.physics_b> wrote:
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:qt78g25rjgbcfir1i3m3vbh8ldhv8qognv@4ax.com...
| On Sat, 09 Sep 2006 19:59:40 GMT, "Sorcerer"
| <Headmaster@hogwarts.physics_b> wrote:
|
| >
| >"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
| >news:pl16g2d2glpn2ipcf36itvu1a8i97o35dl@4ax.com...
| >| On Sat, 09 Sep 2006 16:31:35 GMT, "Sorcerer"
| >| <Headmaster@hogwarts.physics_b> wrote:
| >|
| >| >
| >| >"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
| >| >news:dmm5g2ppq3rr9spkp3bh4g7en6aich15ig@4ax.com...
| >| >| On 8 Sep 2006 19:02:19 -0700, wrote:
| >| >|
| >| >| >
| >| >| >W. Watson wrote:
| >| >| >> Suppose I take perfectly round pole about 2' long and 1/2" wide
out
| >| >onto a
| >| >| >> very smooth surface, and let it cast a shadow from the Sun onto
the
| >| >surface.
| >| >| >> Now I take out a perfectly straight metal ruler and lay it along
the
| >| >shadow
| >| >| >> of the pole, and draw a line along one edge. Suppose I do this
when
| >the
| >| >Sun
| >| >| >> crosses the local meridian and about 2 hours later. Will the
angular
| >| >| >> distance agree with the computed azmithual angle as computed
| >accurately
| >| >from
| >| >| >> an ephemeris?
| >| >| >>
| >| >| >> I suspect not because the dark (umbra) shadow is elongated and
wider
| >| >near
| >| >| >> the base of the pole, and narrows to the top of the shadow;
thereby
| >| >| >> producing an incorrect azimuthal line. Even with one rod, I
suspect
| >the
| >| >| >> drawn line will be a few degrees off the true azimuth line. It
may
| >be
| >| >| >> possible that the line from each time is off by a dissimilar
amount,
| >so
| >| >the
| >| >| >> difference is also incorrect. Perhaps my observation about the
| >shadow
| >| >is wrong.
| >| >| >>
| >| >| >> I've done this a few times, and get about a two degree difference
| >when
| >| >using
| >| >| >> the computed azimuth at, say, 2 pm or so. That is, one might
expect
| >the
| >| >line
| >| >| >> at the meridian to be accurate on N-S, but when working back from
| >the 2
| >| >pm
| >| >| >> measurement, the angle to the N-S is off by 2 degrees.
| >| >| >>
| >| >| >> Any comments?
| >| >| >>
| >| >| >> Wayne T. Watson (Watson Adventures, Prop., Nevada City,
| >CA)
| >| >| >> (121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
| >| >| >> Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
| >| >| >> --
| >| >| >>
| >| >| >> "I have made a ceaseless effort not to ridicule, not to
| >bewail,
| >| >| >> not to scorn human actions, but to understand them."
| >| >| >> --Baruch Spinoza
| >| >| >>
| >| >| >> Web Page: <home.earthlink.net/~mtnviews>
| >| >| >
| >| >| >The main consideration in all this is that the top of the shadow
| >| >| >appears narrower because the shadow is farther from the pole there
and
| >| >| >the 0.5 degree disk of the sun is making the edge indistinct. You
| >| >| >would have to find a way to mark this blurry edge at a point which
| >| >| >corresponds to the distinct bottom edge. That would be where it
| >| >| >becomes as dark as at the bottom. Also, the pole would need to be
| >| >| >perfectly vertical and the base perfectly flat and horizontal.
| >| >| Off the top of my head, the rate is really 15 degrees per hour
| >| >| multiplied by the cosine of your latitude. I havent worked it out
but
| >| >| where did you get the idea that 2 hours equals 2 degrees?
| >| >| John Polasek
| >| >
| >| >Where did you get the idea anyone said it was?
| >| >"when working back from the 2 pm measurement, the angle to the N-S is
| >| >off by 2 degrees."
| >| >The 2:00 pm measurement is 30 +/- 2 degrees, subtract 30 and
| >| >the angle to N-S is +/- 2 degrees.
| >| >You've summarised it quite well with " I havent worked it out ".
| >| >Androcles.
| >| >
| >| >
| >| OK I didn't read it in detail. I saw
| >| "crosses the local meridian and about 2 hours later"
| >| and I saw "the angle to the N-S is off by 2 degrees" but did not
| >| notice he said it was an error of 2 degrees. (So that's the downside
| >| of speed-reading).
| >|
| >| So, actually his angle for two hours should be 30 degs x cos 39 = 23.3
| >| deg.
| >
| >Huh?
| >The earth rotates 361 degrees per day relative to the sun,
| >a shadow at the equator shortens until noon and then lengthens,
| >a shadow at the N or S pole rotates 360 degrees relative to
| >the snow and remains the same length (or would if the Earth
| >were not tilted).
| >Since it is tilted, the shadow is formed not by the post in the snow
| > but by the Earth itself, which hides the shadow of the stick..
| >.. we call that "night", but the shadow still goes through 15 degrees
| >an hour wherever you are.
| >There are longer "days" in summer with shorter nights, but that
| >kind of 'day' is not the same 'day' as 24 hours.
| >You will NOT see an angle to the *local* meridian of 23 degrees
| >at 2 PM local time.
| >
| >| but he is not stating the obvious:
| >
| >Why should he?
| >1) The obvious is obvious or it would not be obvious
| > and it wasn't obvious to him.
| >2) he was asking a question, why does he get a two degree
| >error from what he expected to see.
| >I've given him part of the explanation and he said thank you,
| >but I left out part as well to see if he came back for more.
| >
| >
| >| the ephemeris value.
| >| Is it too much to ask what that angle was and what his reading was?
| >
| >Yes, it is too much to ask (of him). His watch will not agree with a
| >sundial to within 15 degrees in the TIME ZONE in which he lives, and
| >that is because he sets his watch not to local time, but to a nomimal
| >value within a 15 degree spread. The time in Bristol is 10 minutes
| >behind the time in London (by sundial), and it was playing havoc with
| >railway timetables.
| >http://tinyurl.com/g8dk2
| >
| >Androcles.
| >
| The effective angular rate at his latitude is 15 deg/hr times the
| cosine of his latitude which he stated was 39 degrees, so 11.66
| deg/hour.
You are speed reading again, not taking in what I said, FUCKWIT!
The effective angular rate at his latitude is 15 deg/hr times ONE.
Androcles.
Until we hear some input from our client Watson I guess I shouldnt say
anything but I will say
At north pole his multiplier is 1, at the equator it's 0 and at the
south pole it's -1. So I guess it's the sine of the latitude or cosine
of the co-latitude, but who cares, Watson doesn't.
(I had a previous response that failed to get transmitted).
.
|
|
|
|
|
|
|
|
|
|
|
|
Related Articles |
|
|
