BB: Circumference of the earth at a given latitude
Erin
Given that the circumference of the earth at the equator is
*approximately* 24,000 miles, how does one figure the circumference of
the earth at 60'N Latitude (Anchorage)?
Erin.
John W. Vinson
wrote:
Interesting problem... not absolutely sure my logic is correct but... assuming
a spherical Earth (which isn't accurate due to tidal bulge):
The diameter of the small latitude circle would be the diameter at the equator
(7,927 miles) times the cosine of the latitude. At the equator it would be
7927 miles; at the pole it would be 0. Since cos(60) is 0.5, the diameter of
the small circle would be 3963.5 miles, so the circumference would be
pi*3963.5 = 12451 miles.
--
John the Wysard JVinson *at* Wysard Of Info *dot* com
Erin
<jvinson@STOP_SPAM.WysardOfInfo.com> wrote:
> On Sun, 20 Sep 2009 17:45:50 -0700 (PDT), Erin <karenan...@hushmail.com>
Okay, that makes sense. Think you can put that into 6th grader math
language?
Canth
Draw a circle. That represents a cross section of the Earth.
Draw a line across the middle of the circle. That represents the
Equator.
Draw a line vertically through the poles. The intersection of the two
lines is the centre of the Earth.
From your point 60 degrees North, draw a line to the centre of the
Earth. This is another Radius of the Earth.
Now draw a line vertically down from the 60 degree point. This line
will cut the Equator line & form a right angled triangle. The bottom
of that triangle is your small circle radius. The angle at the centre
of the Earth in this triangle is your latitude.
To calculate the length of the small circle radius, multiply the
cosine of the latitude by the radius of the Earth.
AS! ds++:+++ a++ c+++ p++ t+ f-- S+ p+ e++ h++ r++ n++ i+ P+ m++ M
Barry Gold
Not really. The problem is that no matter how you slice it, it
involves trigonometry, which is a 10th or 11th grade subject (in the
US).
Mind you, I think you _could_ explain trig to a 6th grader, but you
would be in effect explaining at least the first few weeks of an 11th
grade course, which means you also need to explain some of the
background for that course (probably including algebra).
That's if you insist on a purely numerical approach. Now, if you're
willing to use some aids...
Draw a horizontal line, representing the radius of the earth at the
equator (approx. 4,000 miles). Now take a protractor(*), and measure
off the latitude as an angle (in this case, 60 degrees). Draw another
line at that angle. Use a ruler (or compass) to make it the same
length as the first line.
Now draw a vertical line from the end of that line, down to the first
line (the earth's radius at the equator). Measure the distance from
the vertex of the angle (representing the center of the earth) to
where the vertical line crosses the horizontal line. Divide by the
length of the horizontal line. That gives you the ratio. Multiply
that by the circumference of the earth at the equator, and there's
your answer.
(This is, in a sense, analytic geometry -- another 11th grade class --
on the cheap. But I think it should be comprehensible to a reasonably
smart 6th grader.)
(*) Have 6th graders learned to use a protractor? If not, you'll have
to teach that, but it shouldn't take long (I hope).
--
Barry Gold, webmaster:
Conchord: http://www.conchord.org
Los Angeles Science Fantasy Society, Inc.: http://www.lasfsinc.org
Erin
THERE you go - thank you! I could get all of it into 6th grader
language except the cosign.
Joyce of Pendle
But ... but ...
The circumference of the earth is the same everywhere: it is a
great circle distance. Everyone else (so far) has talked about the
length of the latitude line (which probably has a name of its own).
This would be a minor nitpick except that you are going to be
teaching it to a class, so imho you need the info.
Joyce.
--
"The spear in the Other's heart is in your own: you are he." - Surak
Matthew Russotto
Consider a line drawn from the center of the earth to the surface at
60 degrees North latitude. This line's length is the radius of the
earth (R). If we use an earth-centered cartesian coordinate system where
the line is in the X-Y plane and the north pole is at (0,R), then the
X-coordinate of the line is equal to the radius of the circle at 60
degrees north latitude. By trignometry, that X coordinate is (R cos
60), or R/2. Since the circumference of a circle is directly
proportional to the radius, the circumference of the earth at 60
degrees North latitude is approximately 12,000 miles.
Which is all a long-winded way of saying multiply the circumference of
the earth by the cosine of the latitude. Assuming a spherical earth
and all that.
--
The problem with socialism is there's always
someone with less ability and more need.
Erin
Actually *I'm* NOT teaching it to a class -- it was part of a math
question that my daughter brought home from HER class. You can see my
problem! The wording I used was *exactly* the wording that came home
-- so if *I* was confused by the question and needed reinforcement
that I wasn't going nuts, I can only imagine how other parents were
looking at it.
I e-mailed the teacher this morning (Monday) and got clarification.
Turns out, there WAS missing information. Fortunately, the teacher is
open to realizing he made a mistake and has sent out a correction :-)
Erin
Joyce of Pendle
> Actually *I'm* NOT teaching it to a class -- it was part of a
> math question that my daughter brought home from HER class. You
> can see my problem! The wording I used was *exactly* the wording
> that came home -- so if *I* was confused by the question and
> needed reinforcement that I wasn't going nuts, I can only imagine
> how other parents were looking at it.
I recognize it! I had my children come home with homework that was
to say the least open to misunderstanding, and I didn't handle it
well. I'm glad you were able to ask for clarification. And get it!
> I e-mailed the teacher this morning (Monday) and got
> clarification. Turns out, there WAS missing information.
> Fortunately, the teacher is open to realizing he made a mistake
> and has sent out a correction :-)
Joyce.
Wes Struebing
So what was the missing information?
Erin
The ratio of the length of the line at the equator and the length of
the line at 60'N (which is about 2:1). That would have made ALL the
difference :-).
Erin
MajorOz
> Erin wrote:
> > Yeah, okay, I should already know this, but I'm having a "duh"
> > moment.
> > Given that the circumference of the earth at the equator is
> > *approximately* 24,000 miles, how does one figure the
> > circumference of the earth at 60'N Latitude (Anchorage)?
>
> But ... but ...
>
> The circumference of the earth is the same everywhere: it is a
> great circle distance. Everyone else (so far) has talked about the
> length of the latitude line (which probably has a name of its own).
You are, of course, correct. It is a common question over beer.
If it was used as a question for sixth graders, however, the teacher
should be tarred and feathered, as it is antithetical to educational
purpose, as are most "trick" questions, designed principally to show
how teddibly clever the teacher is.
cheers
oz, been there but never done that
Margo
You are both right of course, but if t the question had been stated
as, " What is the circumference of a chord through the earth at 60 deg
lattitude?" In that case, it is dependent on the radius of the chord
at 60 deg...and we've already had that answered. {:>)
Margo
Erin
Well, as I said, the teacher was very open to admitting and correcting
his error (which, I believe, negates the "tarred and feathered"
penalty). I'd be hard pressed to believe anyone who said they'd
never made a mistake in doing their job.
Erin
Margo
Oh, dear. *vexed* I did not intend to agree with the "tarred and
feathered" penalty, certainly not for an honest mistake.
Margo
Wes Struebing
Yeah - that *would* make it a tad easier...<G>
Matthew Russotto
MajorOz <Maj...@centurytel.net> wrote:
>On Sep 21, 6:29=A0am, Joyce of Pendle <pen...@invalid.invalid> wrote:
>> Erin wrote:
>> > Yeah, okay, I should already know this, but I'm having a "duh"
>> > moment.
>> > Given that the circumference of the earth at the equator is
>> > *approximately* 24,000 miles, how does one figure the
>> > circumference of the earth at 60'N Latitude (Anchorage)?
>>
>> But ... but ...
>>
>> The circumference of the earth is the same everywhere: it is a
>> great circle distance. Everyone else (so far) has talked about the
>> length of the latitude line (which probably has a name of its own).
>
>You are, of course, correct. It is a common question over beer.
The length of a great circle through Anchorage will be less than that
of the equator, however, as the Earth is not spherical. Calculations
(done in WGS84, please) are left as an exercise for the student.
Andrewd...@outlook.com
we learn 1degree of longitude = 60nm at the equator so for a rough calculation multiple choice we would do \/
So for this question we would do 60x360xCos(Lat)
360x60xcos(60)=10,800Nm
Which is the answer they are looking for in exams, I know mathematically it is wrong but hey, we’ve only got 300 lives In our hands, who needs accuracy. (to clarify I would prefer to calculate the proper method although it is not what the CAA request)
Barry Gold
oblate spheroid, so one really should use the equation for the cross
section of an ellipse with a small eccentricity at the given latitude,
then multiply that local "diameter" by pi.
But... the error is quite small as a percentage of the distance you have
to traverse.
Besides that, you don't fly from point A to point B by following a
circle of latitude. You follow a Great Circle route, or as close to it
as ATC and weather will allow.
And there's the larger error: your flying time (and fuel use) will
depend not only on the length of the geodesic from A to B, but also on:
* local wind conditions, which may speed you up (tailwind) or slow you
down (headwind)
* variations from the "ideal" route required by the need to avoid other
planes.
* choosing a non-geodesic route in order to get the maximum wind boost
or minimum wind drag, as the case may be.
* and a couple of other small things I can't remember off the top of my
head.
I suspect those changes are much larger than the error due to assuming
the Earth is spherical.
And anyway, this isn't hyperspace, where you need to consult the
Ven-Tura tables -- or better yet, the Caylon updates to the tables -- or
you never arrive anywhere at all.
The ground is visible (most of the time), so you can (if necessary) fly
the way a recreational pilot does, navigating by landmarks. And you
probably have a GPS receiver (these days), so you know where you are
(latitude, longitude, and AGL altitude) to within about 10 feet.
TL;DR: Don't sweat the small stuff.
--
I do so have a memory. It's backed up on DVD... somewhere...
JimP
wrote:
>> I understand this may not be strictly correct, but for the purposes of aviation (which I assume a number of people will visit this post from)
>> we learn 1degree of longitude = 60nm at the equator so for a rough calculation multiple choice we would do \/
>> So for this question we would do 60x360xCos(Lat)
>> 360x60xcos(60)=10,800Nm
>Right. That assumes that the Earth is a sphere, and as we know it's an
>oblate spheroid, so one really should use the equation for the cross
>section of an ellipse with a small eccentricity at the given latitude,
>then multiply that local "diameter" by pi.
>
>But... the error is quite small as a percentage of the distance you have
>to traverse.
--
Jim
Barry Gold
> On Tue, 14 Jan 2020 06:45:26 -0800, Barry Gold<Barry...@ca.rr.com>
> wrote:
>>> I understand this may not be strictly correct, but for the purposes of aviation (which I assume a number of people will visit this post from)
>>> we learn 1degree of longitude = 60nm at the equator so for a rough calculation multiple choice we would do \/
>>> So for this question we would do 60x360xCos(Lat)
>>> 360x60xcos(60)=10,800Nm
>> oblate spheroid, so one really should use the equation for the cross
>> section of an ellipse with a small eccentricity at the given latitude,
>> then multiply that local "diameter" by pi.
>>
>> But... the error is quite small as a percentage of the distance you have
>> to traverse.
> Thanks for the explanation Barry. Calculations like that confuses me.
as if Earth were a true sphere.
Tim Merrigan
>head.
--
I pledge allegiance to the Constitution of the United States of America,
and to the republic which it established, one nation, from many peoples,
promising liberty and justice for all.
Feel free to use the above variant pledge in your own postings.
Tim Merrigan
JimP
wrote:
being easier if we presumed a spherical cow, but since they don't
exist we would use a different method.
We just sort of sat there, and then he said that was his one big joke
for the year... but if we became math majors we would hear more things
like that, including actual calculations.
I didn't, I became a computer repair tech with my BS in Computer
Science.
--
Jim
Basil D
> we learn 1degree of longitude = 60nm
> at the equator so for a rough calculation multiple choice we would do \/
> So for this question we would do 60x360xCos(Lat)
> 360x60xcos(60)=10,800Nm
At 60° it's smarter to use the polar diameter, which is 12,713,505 meters, plus or
minus a couple. That's a circumference (assuming a sphere) of 39,940,654 meters
(rounded to the nearest meter). The cosine of 60° is 0.5, giving an answer of
13,970,327 m. How you get from that to 10,800 ???'s is your affair.
~~Basil
Barry Gold
Miles.
Basil D
~~Basil
Dave Bowman
> Yeah, okay, I should already know this, but I'm having a "duh" moment.
> Given that the circumference of the earth at the equator is
> *approximately* 24,000 miles, how does one figure the circumference of
> the earth at 60'N Latitude (Anchorage)?
The circumference of the 60° latitude line is 12450.73 miles.
Go to ==> https://www.vcalc.com/wiki/MichaelBartmess/Rotational+Speed+at+Latitude
Enter your latitude to get the rotational speed in miles per hour.
Multiply the rotational speed (in mph) at your latitude by 23.934472 (as in, number of hours per day in reference to distant stars, not the sun).
That will give you the approximate circumference value of 60° latitude.
D.J.
<arct...@gmail.com> wrote:
that.
Welcome to the Show.
--
Jim