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font face="Courier New, Courier, monospace"Off topic, but funny as
hell:br
br
==============================br
At New York's Kennedy airport today, an
individual laterbr
discovered to be a public school teacher was arrested
tryingbr
to board a flight while in possession of a ruler, a protractor,br
a setsquare, a slide rule, and a calculator. br
br
At a morning press conference, Attorney
general John Ashcroftbr
said he believes the man is a member of the
notorious al-gebrabr
movement. He is being charged by the FBI with
carrying weaponsbr
of math instruction. "Al-gebra is a fearsome cult,",
Ashcroftbr
said. "They desire average solutions by means and extremes,
andbr
sometimes go off on tangents in a search of absolute value. Theybr
use secret code names like "x" and "y" and refer to themselves asbr
"unknowns", but we have determined they belong to a commonbr
denominator
of the axis of medieval with coordinates in everybr
country. "As the
Greek philanderer Isosceles used to say, therebr
are 3 sides to every
>triangle," Ashcroft declared. br
br
When asked to comment on the arrest,
President Bush said, "If Godbr
had wanted us to have better weapons of
math instruction, He wouldbr
have given us more fingers and toes. "I am
gratified that ourbr
government has given us a sine that it is intent on
protracting usbr
from these math-dogs who are willing to disintegrate us
with calculusbr
disregard. Murky statisticians love to inflict plane on
every spherebr
of influence," the President said, adding: "Under the
circumferences,br
we must differentiate their root, make our point, and
draw the line."br
President Bush warned, "These weapons of math
instruction have thebr
potential to decimal everything in their math on a
scalene neverbr
before seen unless we become exponents of a Higher Power
and begin tobr
factor-in random facts of vertex." br
br
Attorney General Ashcroft said, "As our
Great Leader would say, readbr
my ellipse. Here is one principle he is
uncertainty of though theybr
continue to multiply, their days are
numbered as the hypotenuse tightensbr
around their necks." br
br
br
-- br
Jon Gauthierbr
br
Given the likely reaction to an increase inbr
terror-alert level to "severe threat imminent,"br
wouldn't a more appropriate alert color be brown?br
-Brad Simanek on a class="moz-txt-link-abbreviated" href="http://www.ruminate.com"www.ruminate.com/a/font
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Dear Lloyd,

I've picked this up rather late. If you still require the answer[s] then get
back to me including telling me whether you have a book of tables known as
"Norries Tables". If not I can give you an arithmetic method by calculator.
Basically you work out the D'Lat [difference in minutes of latitude] and
Dep(arture), then converting Dep into D'Long [differnce in minutes of
Longitude]. From which can be determined Dist(ance) and Co(arse) made good,
or of course the reverse having only course and distance find the rest.

Chris Spreckley



"Lloyd Sumpter" wrote in message
news
Hi,

I'm writing a program for Linux that displays position (from GPS) on a
scanned-in chart, and would like it to calculate distance from current
position to the cursor.

How do you calculate distance between two points using lat/long? If
they're due North/South, I can do it ( 1 minute of lat = 1 NM) but how do
you calculate distance from longitude? Perhaps some formula based on the
circumfrence of the Earth at the equator and the latitude?

Lloyd Sumpter

"H. Chris Spreckley" wrote in message ...
Dear Lloyd,

I've picked this up rather late. If you still require the answer[s] then get
back to me including telling me whether you have a book of tables known as
"Norries Tables". If not I can give you an arithmetic method by calculator.
Basically you work out the D'Lat [difference in minutes of latitude] and
Dep(arture), then converting Dep into D'Long [differnce in minutes of
Longitude]. From which can be determined Dist(ance) and Co(arse) made good,
or of course the reverse having only course and distance find the rest.

Chris Spreckley



"Lloyd Sumpter" wrote in message
news
Hi,

I'm writing a program for Linux that displays position (from GPS) on a
scanned-in chart, and would like it to calculate distance from current
position to the cursor.

How do you calculate distance between two points using lat/long? If
they're due North/South, I can do it ( 1 minute of lat = 1 NM) but how do
you calculate distance from longitude? Perhaps some formula based on the
circumfrence of the Earth at the equator and the latitude?

Lloyd Sumpter


How come nobody has suggesed the simple midlatitude estimate? If all
you want is the distance from a point (lat0, lon0) on your chart and
the cursor, and you're only planning to use this over short distances
(say, less than 100 NM) AND lat0 is between 60°N and 60°S or so, a
good estimate of distance, which can even be done on a hand
calculator, is

D = K * SQRT((lat - lat0)^2 + cos(lat0)*(lon - lon0)^2)

where

D = desired distance
K = 60 (approx) = conversion factor from degrees of latitude to
nautical miles
(lat,lon) = cursor position (deg)
(lat0, lon0) = point on chart from which distance is desired (deg)

This easy estimate uses a tangent-plane approximation to a round-earth
model and relies on the fact that lines of longitude (meridians) are
approximately parallel in the mid latitudes. Under the indicated
conditions it's accurate to dozens of feet, and for short distances
(under a mile, say), to within a foot or two.

"H. Chris Spreckley" wrote in message ...
Dear Lloyd,

I've picked this up rather late. If you still require the answer[s] then get
back to me including telling me whether you have a book of tables known as
"Norries Tables". If not I can give you an arithmetic method by calculator.
Basically you work out the D'Lat [difference in minutes of latitude] and
Dep(arture), then converting Dep into D'Long [differnce in minutes of
Longitude]. From which can be determined Dist(ance) and Co(arse) made good,
or of course the reverse having only course and distance find the rest.

Chris Spreckley



"Lloyd Sumpter" wrote in message
news
Hi,

I'm writing a program for Linux that displays position (from GPS) on a
scanned-in chart, and would like it to calculate distance from current
position to the cursor.

How do you calculate distance between two points using lat/long? If
they're due North/South, I can do it ( 1 minute of lat = 1 NM) but how do
you calculate distance from longitude? Perhaps some formula based on the
circumfrence of the Earth at the equator and the latitude?

Lloyd Sumpter


How come nobody has suggesed the simple midlatitude estimate? If all
you want is the distance from a point (lat0, lon0) on your chart and
the cursor, and you're only planning to use this over short distances
(say, less than 100 NM) AND lat0 is between 60°N and 60°S or so, a
good estimate of distance, which can even be done on a hand
calculator, is

D = K * SQRT((lat - lat0)^2 + cos(lat0)*(lon - lon0)^2)

where

D = desired distance
K = 60 (approx) = conversion factor from degrees of latitude to
nautical miles
(lat,lon) = cursor position (deg)
(lat0, lon0) = point on chart from which distance is desired (deg)

This easy estimate uses a tangent-plane approximation to a round-earth
model and relies on the fact that lines of longitude (meridians) are
approximately parallel in the mid latitudes. Under the indicated
conditions it's accurate to dozens of feet, and for short distances
(under a mile, say), to within a foot or two.

Dear Lloyd,

I've picked this up rather late. If you still require the answer[s] then get
back to me including telling me whether you have a book of tables known as
"Norries Tables". If not I can give you an arithmetic method by calculator.
Basically you work out the D'Lat [difference in minutes of latitude] and
Dep(arture), then converting Dep into D'Long [differnce in minutes of
Longitude]. From which can be determined Dist(ance) and Co(arse) made good,
or of course the reverse having only course and distance find the rest.

Chris Spreckley



"Lloyd Sumpter" wrote in message
news
Hi,

I'm writing a program for Linux that displays position (from GPS) on a
scanned-in chart, and would like it to calculate distance from current
position to the cursor.

How do you calculate distance between two points using lat/long? If
they're due North/South, I can do it ( 1 minute of lat = 1 NM) but how do
you calculate distance from longitude? Perhaps some formula based on the
circumfrence of the Earth at the equator and the latitude?

Lloyd Sumpter