Feet Per Minute Conversion Question

Gene Nygaard wrote in message . ..
On 26 Aug 2003 02:26:21 -0700, (Doug Hoffman)
wrote:

Nygaard is a troll. He is *not* a glider person. He gets his jollies
by trying to "show off" his supposedly superior knowledge of units of
measure. Ignore him.

"A" troll? Can't even get your metaphors right, can you?

http://w0rli.home.att.net/youare.swf

This is even better than _trolling_ (verb). This place has fools like
like you who just jump into my boat, you're so anxious to show off
your stupidity.

I rest my case.

On 27 Aug 2003 10:54:05 GMT, John Lee
wrote:

Navigational question for the day;
One minute of latitude = 1 nautical mile. At what point
on the earth does 1
minute of longatude = 1 nautical mile?
JJ Sinclair

Nowhere! However on the equator 1 second of longitude
equals 1 nautical mile

That isn't any more correct now than when Bill Dean
claimed that
earlier in this thread. See my reply to his message.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/

Still looking Gene...... the other stuff is interesting
though

(JJ Sinclair) wrote in message ...
Navigational question for the day;
One minute of latitude = 1 nautical mile. At what point on the earth does 1
minute of longatude = 1 nautical mile?
JJ Sinclair

8-9 degrees north or south LAT where EW radius of curvature = NS
radius of curvature.

With screwy "geodetic" latitude that gps uses, FAI 6371km sphere
errors are even worse than if using geocentric latitude on ellipsoid.
I think pythagorean theorum with scale factors below would match
ellipsoid better than sphere does for gliding distances.

Someone please check these equations. The instrument display makers
may want to use an ellipsoid some day:

dist = pi/180*[delta Lat or Long in deg]*radius of curvature

radius of curvature NS:
a^2*b^2/(a^2*cos^2(Lat)+b^2*sin^2(Lat))^(3/2)

radius of curvature EW:
a^2/sqrt(a^2+b^2*tan^2(Lat))

Jonathan

p.s. check my equations against tiny EW or NS distances at various
lats using the FAI java calculator.

(JJ Sinclair) wrote in message ...
Navigational question for the day;
One minute of latitude = 1 nautical mile. At what point on the earth does 1
minute of longatude = 1 nautical mile?
JJ Sinclair

(Didn't read the question the first time, and answered "When does a
minute of lat = a minute of long?" instead.)

Second guess:

One minute of latitude equals 1.000000000 nm only at a single latitude
somewhere between 44° and 45° N (or S) latitude. (Don't expect it to
fall on an even minute.)
One minute of longitude equals 1.000000000 nm only at a single
latitude somewhere between 3° and 4° (N or S) latitude. (Don't expect
it to fall on an even minute.)

Morals:
1) Don't count on those ticks on the chart - you'll fall short : )
2) Don't move to the equator to find square lat/long grids.

Jonathan Gere

Interesting reading all this cr@# about units,
for gliding varios [the oridinal question I think!] all you need
is 1 knot / 100 feet per minute
and 1 metre per sec / 2 knots.
IT REALLY IS THAT SIMPLE!!!!!!

And at what altitude would that be?

Just messin with you!

:-)

Ian

"Tom Claffey" wrote in message
om...
Interesting reading all this cr@# about units,
for gliding varios [the oridinal question I think!] all you need
is 1 knot / 100 feet per minute
and 1 metre per sec / 2 knots.
IT REALLY IS THAT SIMPLE!!!!!!

Pat Russell wrote:

My vario is calibrated in megaparsecs per millenium. Very
handy.

According to my calculations, this should be near 0.5 mm/s.
You need a very big scale.

On Thu, 28 Aug 2003 12:49:30 +0000, root
wrote:

Pat Russell wrote:

My vario is calibrated in megaparsecs per millenium. Very
handy.

According to my calculations, this should be near 0.5 mm/s.
You need a very big scale.

Try again. What do you think a millennium is? A parsec? The prefix
mega-?

1 Mpc/ka = 978 Gm/s, or about 3262 times the speed of light. He's got
bigger problems than you thought.

In other words, even one parsec (without the prefix) per millenium is
nearly 1 Mm/s (978 km/s). Thus 1 microparsec per millennium is equal
to a meter per second, within the precision many of you accept with
knots and the like.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/

On 27 Aug 2003 20:30:05 GMT, (JJ Sinclair) wrote:

John Lee wrote.
Nowhere! However on the equator 1 second of longitude
equals 1 nautical mile


Hmmmm? When I went to navigation school, 1 degree of latitude anywhere in the
world and 1 degree of longitude (on the equator) was made up of 60 minutes and
each minute was 1 nautical mile. A second would be 6080 devided by 60 or 101
feet. Please don't tell me they have changed all that. I'm not talking about
the nit-pickers that want to say there is an + or - a foot or two in a nautical
mile, but I would like to believe we still have 360 degrees around the equator
and 90 degrees north and 90 degrees south latitude.
JJ Sinclair

Maybe Eratosthenes still thought the Earth was a near-perfect sphere,
when he made a fairly reasonable calculation of its diameter Maybe
Columbus wasn't even aware of the true shape of the poles.

But we've known about the flattening at the poles for about four
centuries at least, probably longer than nautical miles have existed.
Certainly since long before the French scientists in the 1790s
designed the meter to be 1/10000000 of the distance from the equator
to the North Pole.

We do, of course, have 360 degrees around the equator. When we agree
on a starting point (e.g., the point where it crosses the meridian
through Greenwich, England, the one most often used now), any
particular place will always be the same number of degrees from it.

We also have 90 degrees between the equator and either pole. The
equator is always 0 degrees and the poles 90 degrees. But in between,
there are at least three different ways of measuring latitude:
geocentric latitude (the angle formed at the center of the Earth),
geodetic latitude (the one normally used, the angle formed between the
line normal to the tangent of the ellipsoid and the axis of rotation),
and the angles used in the parametric formulas representing an
ellipse. These don't agree with each other at any place not on the
equator or the poles.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/

Gene Nygaard wrote:

On Thu, 28 Aug 2003 12:49:30 +0000, root
wrote:

Pat Russell wrote:

My vario is calibrated in megaparsecs per millenium. Very
handy.

According to my calculations, this should be near 0.5 mm/s.
You need a very big scale.

Try again. What do you think a millennium is? A parsec? The prefix
mega-?

1 Mpc/ka = 978 Gm/s, or about 3262 times the speed of light. He's got
bigger problems than you thought.

In other words, even one parsec (without the prefix) per millenium is
nearly 1 Mm/s (978 km/s). Thus 1 microparsec per millennium is equal
to a meter per second, within the precision many of you accept with
knots and the like.

Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/

I just forgot the mega, I computed a parsec/millenium. Here is what I
did:

1 parsec is the distance of something having a parallax of 1 second, i.e.
the distance (from the sun) where the diameter of the orbit of the earth
is seen with an angle of 1 second. I don't remember this diameter but I
remember that the light coming from the sun needs 8 minutes to do that,
so the diameter should be 8*2*60*3*10^8 (diameter = 2*radius, 60 seconds
in a minute, speed of light 3*10^8 m/s) = 4800000000 m. 1 second is PI/(180*60*60)
radians, so 1 parsec is 4800000000/(PI/(180*60*60)), this is roughly 10^15 m.
A year is nearly 365.25 days, i.e. 365.25*24*60*60 seconds = 1980281535681600 s.
A millenium is 1980281535681600*1000 = 1980281535681600000 seconds or roughly
2*10^18 s. So a parsec/millenium is 10^15/(2*10^18) = 0.5*10^-3 m/s. A megaparsec
per millenium should be ~ 500 m/s. A huge unit for a vario in a glider, although
well under the speed of the light.

Well I was never good in calculations, so if there is a error, please point where.