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| Dr. Distance |
 Posted: Sat Jul 02, 2005 1:48 pm |
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What is the formula for loxodromic ("Rhumb") distance.
There is a PDF paper on the web but it seems confusing and the examples he uses++("table 1") aren't even right!
-------------- ÀÎÅÍ³Ý Ä«¸®½º¸¶ KORNET ------------- |
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| Proginoskes |
| Posted: Sat Jul 02, 2005 6:22 pm |
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Dr. Distance wrote:
[quote:4e87ed0014]What is the formula for loxodromic ("Rhumb") distance.
There is a PDF paper on the web but it seems confusing and the examples he uses++("table 1") aren't even right!
[/quote:4e87ed0014]
Are you talking about the James Alexander page (at
http://www.cwru.edu/artsci/math/alexander/mathmag349-356.pdf )?
--- Christopher Heckman |
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| Randy Poe |
| Posted: Sat Jul 02, 2005 6:25 pm |
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Dr. Distance wrote:
[quote:0f2bd981df]What is the formula for loxodromic ("Rhumb") distance.
There is a PDF paper on the web
[/quote:0f2bd981df]
Was it this one?
http://www.cwru.edu/artsci/math/alexander/mathmag349-356.pdf
Equation (4) seems to be the easiest to use:
D = R*|L_2 - L_1|*|sec(theta)|
where R = radius of earth, L_1, L_2 are the latitudes of the
points of interest, and theta is the bearing, which I believe
is defined clockwise from North (theta = 0 for due north
90 deg for east, 180 deg for south, and 270 for west).
The latitudes should be in radians for the formula above.
[quote:0f2bd981df]but it seems confusing and the examples he uses++("table 1")
aren't even right!
[/quote:0f2bd981df]
Hard to comment when you don't tell us what you're looking
at. However, if it's the paper I cited above, perhaps you
could say what you mean by "aren't even right".
- Randy |
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| Timothy Little |
| Posted: Mon Jul 04, 2005 2:19 am |
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Nomen Nescio wrote:
[quote:16b63bdf9c]Using R = 6371, these are the |results| I get for table 1 (including
theta (in radians), in parenthesis).
5567.817 5809.135 (1.3626575822)
4031.637 4031.637 ( .0029212129)
10996.670 16215.879 (1.5650755933)
16254.490 24544.476 (1.2175972203)
Since D_gc distances are way off, I have no idea if the D_rh
distances are right or not!
[/quote:16b63bdf9c]
I presume this is the data from the table in question, presented in
similar manner (without theta, which was not explicitly shown in the
original table):
5564 5802
4024 4030
11019 14380
16230 16408
The D_gc distances appear to be the same to within 0.2%, which hardly
seems "way off". Two of your loxodromic distances do appear to differ
significantly. Looking at the largest discrepancy, for Canberra:
New York: +0.71122 radians north, -1.29096 radians east
Canberra: -0.61988 radians north, +2.60345 radians east
Using equation (6) from the article, the values of Sigma are
New York: Sigma = +0.78009
Canberra: Sigma = -0.66388
From equation (5), D_rh = 16393 if I choose the shortest loxodrome,
which heads west, or D_rh = 24393 if I choose the shortest eastward
loxodrome. My first figure agrees very closely with the article, my
second agrees quite closely with yours.
Hence I suspect that you calculated for the eastward path, rather than
the shorter westward one.
- Tim |
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| Timothy Little |
| Posted: Mon Jul 04, 2005 1:11 pm |
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Guest
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Anonymous wrote:
[quote:dcd8cc0472]Well considering that they are simple great cirle distances and the
radius is a fixed 6371 and he goes to the km, then rounding should
be within +/-1km, certainly not more than 2, regardless of the
percentage.
[/quote:dcd8cc0472]
The Earth is not quite a sphere. The polar radius is about 6357 km,
while the equatorial radius is about 6378 km. The figure of 6371 km
you used is just the mean radius.
Since the original article gave formulas for the ellipsoidal model, it
is quite likely that the table was calculated using that more accurate
model rather than the simpler spherical one.
- Tim |
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| Kaimbridge |
| Posted: Thu Aug 04, 2005 1:01 pm |
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<pre>
Proginoskes wrote in thread
http://groups.google.ca/group/sci.math/index/browse_frm/thread/db5ead1af6d53c19
[quote:6a9d733c64]Anonymous via the Cypherpunks Tonga Remailer wrote:
Okay like I asked the other fellow, for that "shortest loxodrome"
as measured on a globe, don't you just multiply the longitude
difference by the average latitude cosine to find the spherically
adjusted longitude difference, then calculate the hypotenuse using
Pythagorean?
No. Pythagorean's Theorem only is true in a flat plane. Geometry on a
sphere is a bit harder. But you will get a close answer if the
differences are small.
[/quote:6a9d733c64]
Of course the Pythagorean Theorem is at work here....It just hasn't
been presented right! P=)
*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*
Right Triangle Properties
======================== HLeg
VLeg = cos{Az} * HyP = cot{Az} * HLeg; |¯¯¯¯¯¯¯¯¯/
| /
HLeg = sin{Az} * HyP = tan{Az} * VLeg; | /
V | / H
HLeg HLeg VLeg L | / y
Az = atan{----} = asin{----} = acos{----}; e | / P
VLeg HyP HyP g | /
|Az/
HyP = [VLeg^2 + HLeg^2]^.5 | /
|/
HLeg VLeg
= VLeg * [1 + (----)^2]^.5 = HLeg * [(----)^2 + 1]^.5,
VLeg HLeg
= VLeg * [1 + tan{Az}^2]^.5 = HLeg * [cot{Az}^2 + 1]^.5,
= VLeg * sec{Az} = HLeg * csc{Az};
*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*^-^*
In the case of a globe's common geographical graticule, the "legs"
are created as so:
Where iG{Lat} = acosh{sec{Lat}} = atanh{sin{Lat}},
= ln{sec{Lat}*[1+sin{Lat}]};
(The "Inverse Gudermannian Function")
iG'{Lat} = sec{Lat};
UT=oo iG'{Lat_tn} UT=oo sec{Lat_tn}
iG[BLat,DLat] = SUM ----------- = SUM -----------;
TN=1 UT TN=1 UT
(Lat_1 = BLat, Lat_ut = DLat)
iG{DLat(^)BLat} = iG{DLat} - iG{BLat} = iG[BLat,DLat] * LD;
then
VLeg = |LD| = |DLat - BLat| = cos{Azg} * HyP = cot{Azg} * HLeg;
|MD| = |DLong - BLong| = tan{Azg} * |iG{DLat(^)BLat}|,
= sin{Azg} * iG[BLat,DLat] * HyP,
= iG[BLat,DLat] * HLeg;
|MD|
HLeg = ------------- = sin{Azg} * HyP = tan{Azg} * VLeg;
iG[BLat,DLat]
MD
HyP = [VLeg^2 + HLeg^2]^.5 = [LD^2 + (-------------)^2]^.5;
iG[BLat,DLat]
HLeg | MD |
Azg = atan{----} = atan{| ------------------ |};
VLeg | iG[BLat,DLat] * LD |
Intuitively, yes, it would seem that integration would be of HLeg,
via an averaging of cos{Lat} adjusted values of MD:
HLeg{DLat} HLeg{BLat<Lat<DLat}
/¯¯¯¯¯¯¯¯¯¯\
|¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯|
V / \ V V | | V
L / \ L ----> L | | L
e / \ e e | | e
g/ \g g | | g
/____________________\ |_______________|
HLeg{BLat} HLeg{BLat<Lat<DLat}
HLeg{Lat} = cos{Lat} * |MD|;
HLeg = HLeg{BLat<Lat<DLat} = cos{BLat<Lat<DLat} * |MD|,
= sin'{BLat<Lat<DLat} * |MD|,
= sin[BLat,DLat] * |MD|;
thus the OP's suggestion--which I would have agreed with, up until
now--to "multiply the longitude difference by the average latitude
cosine to find the spherically adjusted longitude difference".
However, as I finally realized, the integration ISN'T
HLeg{BLat<Lat<DLat} = cos{BLat<Lat<DLat} * |MD|,
= sin[BLat,DLat] * |MD|,
but rather
|MD| = sec{BLat<Lat<DLat} * HLeg = iG[BLat,DLat] * HLeg;
Or, since loxodromic means "along a constant angle",
= tan{Azg} * |iG{DLat(^)BLat}|;
thus
|MD| |MD|
HLeg = ------------------ = -------------;
sec{BLat<Lat<DLat} iG[BLat,DLat]
The spherical loxodromic distance and directed azimuths are now
attainable:
LoxDxS = Radius * HyP;
BAzg = Azg:If LD < 0 then BAzg = 180° - Azg
If MD < 0 then BAzg = 360° - BAzg;
DAzg = Azg:If LD > 0 then DAzg = 180° - Azg
If MD > 0 then DAzg = 360° - DAzg;
This is for the spherical.
Now let's make it elliptically palpable.
Referring here to an oblate ellipsoid (a > b):
a,b = equatorial, polar radii;
Oz = acos{b/a} (sin{Oz} = eccentricity);
M{Lat} = a * E0'{0:Lat},
= (b^2/a^4) * N(Lat}^3;
(Meridional arcradius/radius of curvature)
N{Lat} = a * E1'{0:Lat},
= a^2/[(a * cos{Lat})^2 + (b * sin{mlat})^2]^.5;
(Perpendicular/"Normal" arcradius)
In the paper that the OP was referencing,
http://www.cwru.edu/artsci/math/alexander/mathmag349-356.pdf
(extending the notation from above)
1 + sin{Lat} * sin{Oz}
iGe{Lat} = iG{Lat} - .5 * sin{Oz} * ln{----------------------}
1 - sin{Lat} * sin{Oz}
or
= iG{Lat} - sin{Oz} * iG{LatZ}
(LatZ = asin{sin{Lat}*sin{Oz}})
M{Lat} E0'{0:Lat}
iGe'{Lat} = sec{Lat} * ------ = sec{Lat} * ----------;
N{Lat} E1'{0:Lat}
| MD |
Aze = atan{| ---------------- |};
| iGe{DLat(^)BLat} |
Though not explicitly stated, the paper suggests that the elliptical
(loxodromic) distance = a * VLeg * sec{Aze}.
This is not possible, however, as the north-south case shows:
a * VLeg * sec{Aze} = a * |LD| * sec{0} = a * |LD|!
Instead, the ellipticity is integrated into HyP.
As M{Lat}/E0'{0:Lat} defines the vertical arcradius and
N{Lat}/E1'{0:Lat} the horizontal:
a * HyP{Lat} = a * [(VLeg * E0'{0:Lat})^2 +
(HLeg * E1'{0:Lat})^2]^.5,
= [(VLeg * M{Lat})^2 + (HLeg * N{Lat})^2]^.5,
= [(cos{Azg} * HyP * M{Lat})^2 +
(sin{Azg} * HyP * N{Lat})^2]^.5,
= [(cos{Azg} * M{Lat})^2 +
(sin{Azg} * N{Lat})^2]^.5 * HyP,
= O{Azg:Lat} * HyP,
= a * ZG'{Azg:Lat} * HyP;
Therefore,
LoxDxE = O{Azg:BLat<Lat<DLat} * HyP,
UT=oo O{Azg:Lat_tn}
= SUM ------------- * HyP;
TN=1 UT
The elliptical azimuth has its own confliction:
sin{Azg} * N{Lat} = [O{Azg:Lat}^2 - (cos{Azg} * M{Lat})^2]^.5,
M{Lat}
= O{Azg:Lat} * [1 - (cos{Azg} * ----------)^2]^.5,
O{Azg:Lat}
= O{Azg:Lat} * sin{Aze{Lat}};
Likewise,
cos{Azg} * M{Lat} = O{Azg:Lat} * cos{Aze{Lat}};
Hence,
N{Lat}
sin{Aze{Lat}} = sin{Azg} * ----------;
O{Azg:Lat}
M{Lat}
cos{Aze{Lat}} = cos{Azg} * ----------;
O{Azg:Lat}
N{Lat} HLeg * N{Lat}
tan{Aze{Lat}} = tan{Azg} * ------ = -------------;
M{Lat} VLeg * M{Lat}
| MD | N{Lat}
= atan{| ------------------ | * ------};
| iG[BLat,DLat] * LD | M{Lat}
Just as with the distance, the elliptical azimuth prescribed in the
paper referenced by the OP also appears to be flawed.
The path of the loxodromic is "along a constant angle".
On a sphere, this is unambiguous: Azg equals one value throughout the
loxodromic path.
On an ellipsoid, as well, the graticular Azg is still constant
throughout.
But, excluding north-south and east-west cases, the elliptical azimuth
varies at each point along the way.
Given BLat, DLat and MLat (equals .5 * [BLat + DLat]), let
Azg = 73°, Aze{BLat} = 73.12°, Aze{MLat} = 73.35°
and Aze{DLat} = 73.51°,
then, in this hypothetical example,
Aze{BLat<Lat<DLat} = 73.29°,
but
Aze{BLat<Lat<MLat} = 73.22° and Aze{MLat<Lat<DLat} = 73.41°,
meaning, standing at BLat in the direction Azg, a fixed, averaged
elliptical azimuth--as the cited paper prescribes--will change as the
loxodromic arc grows: But then one is not dealing with a constant
angle and, therefore, not a loxodrome!
One *could* argue that the loxodromic angle is elliptical, therefore
Aze is constant and the graticular/spherical Azg is what varies, in
the same way that the orthodromic geodesic is calculated on an
"auxiliary sphere"--but this would defeat the whole purpose and
advantage of using loxodromic calculations, namely, navigational ease
and cartographical uniformity!
Thence,
N{Lat} HLeg*N{Lat}
Aze{Lat} = atan{tan{Azg}*------} = atan{-----------}
M{Lat} VLeg*M{Lat}
BAze = Aze{BLat}:If LD < 0 then BAze = 180° - Aze{BLat}
If MD < 0 then BAze = 360° - BAze;
DAze = Aze{DLat}:If LD > 0 then DAze = 180° - Aze{DLat}
If MD > 0 then DAze = 360° - DAze;
As this post should show, the Pythagorean Theorem IS exclusively
applicable in finding loxodromic distance and azimuth.
~Kaimbridge~
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</pre> |
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| Kaimbridge |
| Posted: Fri Aug 05, 2005 10:40 am |
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<pre>
Kaimbridge wrote:
[quote:6841d0ac7b]
Proginoskes wrote in thread http://groups.google.ca/group/sci.math/index/browse_frm/thread/db5ead1af6d53c19
Anonymous via the Cypherpunks Tonga Remailer wrote:
Okay like I asked the other fellow, for that "shortest loxodrome"
as measured on a globe, don't you just multiply the longitude
difference by the average latitude cosine to find the spherically
adjusted longitude difference, then calculate the hypotenuse using
Pythagorean?
No. Pythagorean's Theorem only is true in a flat plane. Geometry on a
sphere is a bit harder. But you will get a close answer if the
differences are small.
Of course the Pythagorean Theorem is at work here....It just hasn't
been presented right! P=)
[/quote:6841d0ac7b]
If the posting came out garbled, try one of these links:
http://usenet.mail2web.com/cgi-bin/dnewsweb.exe?cmd=article&group=sci.math&item=850790
http://mathforum.org/kb/plaintext.jspa?messageID=3875787
[quote:6841d0ac7b]Intuitively, yes, it would seem that integration would be of HLeg,
via an averaging of cos{Lat} adjusted values of MD:
HLeg{DLat} HLeg{BLat<Lat<DLat}
/¯¯¯¯¯¯¯¯¯¯\
|¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯|
V / \ V V | | V
L / \ L ----> L | | L
e / \ e e | | e
g/ \g g | | g
/____________________\ |_______________|
HLeg{BLat} HLeg{BLat<Lat<DLat}
[/quote:6841d0ac7b]
That should be:
HLeg{DLat} HLeg{BLat<Lat<DLat}
/¯¯¯¯¯¯¯¯¯¯\ |¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯|
V / \ V V | | V
L / \ L ----> L | | L
e / \ e e | | e
g/ \g g | | g
/____________________\ |_______________|
HLeg{BLat} HLeg{BLat<Lat<DLat}
[quote:6841d0ac7b]HLeg{Lat} = cos{Lat} * |MD|;
HLeg = HLeg{BLat<Lat<DLat} = cos{BLat<Lat<DLat} * |MD|,
= sin'{BLat<Lat<DLat} * |MD|,
= sin[BLat,DLat] * |MD|;
[/quote:6841d0ac7b]
~Kaimbridge~
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Wanted—Kaimbridge (w/mugshot!):
http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html
----------
Digitology—The Grand Theory Of The Universe:
http://www.angelfire.com/ma2/digitology/index.html
***** Void Where Permitted; Limit 0 Per Customer. *****
</pre> |
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| Peter Duniho |
| Posted: Fri Aug 05, 2005 12:07 pm |
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Guest
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"Kaimbridge" <Kaimbridge@Gmail.com> wrote in message
news:42F39673.5000001@Gmail.com...
[quote:baa1edcb41][snipped]
[/quote:baa1edcb41]
This is entirely off-topic for rec.aviation.piloting. Please do not
cross-post inappropriately.
Thank you. |
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