In January 1912, Scott reached the South Pole and discovered Amundsen's flag, which was aided by Amundsen's team placing multiple flags in the area to mark the location. Without GPS, Scott relied on precise navigation techniques, particularly latitude measurements, as longitude readings are less reliable near the poles. The discussion highlights that one minute of latitude is approximately 1.85 kilometers, while the distance for longitude varies significantly depending on proximity to the poles. The conversation also touches on the challenges of accurately determining position in polar regions due to the convergence of longitude lines. Overall, the combination of careful navigation and the strategic placement of flags facilitated Scott's discovery of Amundsen's flag.
#1
Stephanus
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Dear PF Forum,
In January 1912, Scott reached South Pole and found Amundsen flag.
So, in millions of km square white place, Scott reached the very spot that Amundsen posted his (Amundsen) flag.
How was the method? Considering there were no GPS back then. Could Scott navigate very precisely? Or in the white ice very far away, somehow they could spot the black flag?
Just curious
FWIW - for navigation the error in longitude becomes pretty bad near either pole. Latitude much less so. At the equator ~590m is one minute of longitude and
at 89 deg 59, one minute is less than 1 meter. Because if you walk in a circle around the pole at that latitude, a full circle is about 3.7km in diameter.
Or, 1 minute is approximtely 3.7 / (360 * 60) . So longitude is often reported as indeterminate at very high latitudes.
#6
Stephanus
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jim mcnamara said:
FWIW - for navigation the error in longitude becomes pretty bad near either pole. Latitude much less so. ...
Stephanus said:
But, that 1 minute latitude (the same on every spot on earth) would be...
Amundsen didn't just place a single flag. His team placed several flags around the approximate location of the pole, so even with some measurement uncertainty the pole was within the marked area. It is not so hard to spot a flag if everything else is white.
#8
Stephanus
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mfb said:
Amundsen didn't just place a single flag. His team placed several flags around the approximate location of the pole, so even with some measurement uncertainty the pole was within the marked area. It is not so hard to spot a flag if everything else is white.
Several? Thanks FYI. I didn't know that.
#9
Stephanus
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mfb said:
Amundsen didn't just place a single flag. His team placed several flags around the approximate location of the pole, so even with some measurement uncertainty the pole was within the marked area. It is not so hard to spot a flag if everything else is white.
Wait...
What about the letter that Scott found which had instruction if I'm not mistaken "to send the letter to the king"? There are several, too?
I don't think he wrote multiple letters. A tent and a flag at the central position, and several flags in the surrounding area to increase the chance to (a) cover the pole and (b) let Scott find the area with the central tent.
For Scott's web page, scroll down a little until you see "download the KML file here"
#12
Devon Fletcher
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jim mcnamara said:
FWIW - for navigation the error in longitude becomes pretty bad near either pole. Latitude much less so. At the equator ~590m is one minute of longitude and
at 89 deg 59, one minute is less than 1 meter. Because if you walk in a circle around the pole at that latitude, a full circle is about 3.7km in diameter.
Or, 1 minute is approximtely 3.7 / (360 * 60) . So longitude is often reported as indeterminate at very high latitudes.
At the poles, of course, longitude is irrelevant, all lines of longitude come to a point. So latitude is the relevant measurement, and this had long been practised, and could be determined to pretty good resolution, from shots on any of a large number of bodies.
#13
Stephanus
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Devon Fletcher said:
At the poles, of course, longitude is irrelevant, all lines of longitude come to a point...
Stephanus said:
But, that 1 minute latitude (the same on every spot on earth) would be...
At the equator ~590m is one minute of longitude...
Why isn't this roughly the same as one minute of latitude?
#18
Stephanus
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jim mcnamara said:
FWIW - ... Latitude much less so. At the equator ~590m is one minute of longitude and
at 89 deg 59, one minute is less...
insightful said:
Why isn't this roughly the same as one minute of latitude?
If it's in the equator it should be the same.
But I think 1 minute longitude (in the equator) or 1 minute lattitude is about 40 thousands KM / 360 degrees = 111.1111 KM
And if 1 degree is 60 minutes then should it be 111/60 = 1.85 KM per minute?
I know this thread is not about cartography, but since we are in the heat of it, is 1 minute latitude = 1.85 or 1 minute latitude = 590m?
#19
insightful
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Stephanus said:
If it's in the equator it should be the same.
Agreed.
Stephanus said:
But I think 1 minute longitude (in the equator) or 1 minute lattitude is about 40 thousands KM / 360 degrees = 111.1111 KM
I think you mean degree, not minute.
Stephanus said:
And if 1 degree is 60 minutes then should it be 111/60 = 1.85 KM per minute?
Agreed.
Stephanus said:
is 1 minute latitude = 1.85 or 1 minute latitude = 590m?
I'd say both 1 minute latitude and 1 minute longitude at the equator should be about 1.85km.
#20
Stephanus
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insightful said:
I'd say both 1 minute latitude and 1 minute longitude at the equator should be about 1.85km.
So, the honourable Jim McNamara made a mistake?
jim mcnamara said:
FWIW... latitude much less so. At the equator ~590m is one minute of longitude and
at 89 deg 59, one minute is less...
Now, a new question:
At what latitude that 1 minute is 590 m?
Condition: Earth is sphere not oblates
Circumference: Exactly 40 thousands KM
Consider a slice of cone.
Bottom (the real circumference) diameter is ##\frac{40000}{\pi} = 12732.4##
Upper (smaller) diameter is proportional to 590m/1.85
While 1.85 is 1 minute = 40000/360/60
##\frac{590}{\frac{40000}{360*60}} * \frac{40000}{pi}##
Where 1 minute degree of longitude is 590 m
This is the diameter: ##\frac{590*360*60}{\pi}## in metres
This is the circumference ##590*360*60## in metres
This is the angle: ##acos(0.59*360*60/40000) = 55.6## north/south?
No matter what, I think Jim's statement
At equator 1 minute longitude is not 590 m.
If it's in the equator it should be the same.
But I think 1 degree longitude (in the equator) or 1 degree latitude is about 40 thousands KM / 360 degrees = 111.1111 KM
And if 1 degree is 60 minutes then should it be 111/60 = 1.85 KM per minute?
I know this thread is not about cartography, but since we are in the heat of it, is 1 minute latitude = 1.85 or 1 minute latitude = 590m?
Thanks for the corrections. I was just lazy doing back of the napkin calculations and flaatening a sphere into 2D. My bad. But the point I was trying to make was and still is: It is hard, or very problematic(depending on your point of view), to get accurate longitude readings, when you are with 1 minute latitude of either pole. Amundsen and Scott did not have GPS, either.
#23
Stephanus
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jim mcnamara said:
Thanks for the corrections. I was just lazy doing back of the napkin calculations and flaatening a sphere into 2D. My bad. But the point I was trying to make was and still is: It is hard, or very problematic(depending on your point of view), to get accurate longitude readings, when you are with 1 minute latitude of either pole. Amundsen and Scott did not have GPS, either.
Thank you very much for both of your replies Jim. I'm no scientist, and I try to make sense out of your answer 590 vs 1.85 km.
Perhaps the Earth is oblate?
Perhaps the equator does not really divide the south/north hemisphere equally?
But even so, the answer is far from 590 m. Much less oblate, it will be more then 1.85 KM
Thanks again
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
// one_min.c distance between two longitudinal points on a "circle" drawn 1 minute latitude from South
// pole of Earth, each point on the drawn circle, one minute of longitude apart.
int main( void)
{
double polar_rad=6356; // polar radius of Earth km
double polar_rad_m=polar_rad * 1000; // km-> meters
double diam_e= 2 * polar_rad_m; // polar diameter in meters
double polar_circum=diam_e * M_PI; // diameter of Earth * pi = circumference in meters
double lat_1_min= polar_circum / (360 * 60); // in meters
// Use 2D euclidean map of earth
// a circle drawn at 89deg 59min 0sec with constant radius from the S pole
double circle= (lat_1_min + lat_1_min ) * M_PI; // circumference of 1 minute circle
printf("Distance between 2 longitudinal points \n");
printf(" at latitude 89deg 59min and 1 minute longitude apart: %.2f meters\n",
circle/( 360 * 80) );
return 0;
}
Commands to get output from code:
Code:
$ gcc -lm -Wall one_min.c -o one_min
$ ./one_min
Distance between 2 longitudinal points
at latitude 89deg 59min and 1 minute longitude apart: 0.40 meters
The correct answer is less than one meter. Which is not what I posted before. Given that most GPS receivers have an accuracy of +/- 3m, finding an accurate longitude for any latitude very close to the pole would be difficult.
Thanks
#25
sandersr
3
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Stephanus said:
View attachment 94838
Good, at least there are US, UK, France, Denmark? Russia? Australia?
double polar_rad=6356; // polar radius of Earth km
double polar_rad_m=polar_rad * 1000; // km-> meters
double diam_e= 2 * polar_rad_m; // polar diameter in meters
double polar_circum=diam_e * M_PI; // diameter of Earth * pi = circumference in meters
This will give polar_circum=39,935,925.8124
double lat_1_min= polar_circum / (360 * 60); // in meters
This will give lat_1_min = 1,848.8855
double circle= (lat_1_min + lat_1_min ) * M_PI; // circumference of 1 minute circle
Now this I don't understand. It's like you're drawing, as you said, 2D circle from 3D arc.
I think, assuming the Earth is sphere not oblate, the circle (longitude) at 89059 is...
First I'll find n from Cos(@)
Perhaps I should start again
R = polar_rad_m
n = Cos(@) * R
Let Cos(89,59 min) = 0.00029089.
R = 6,356,000
So n = 0.00029089 * 6,356,000 = 1,848.8854
This gives slightly different from yours: 1,848.8855
Perhaps it has something to do with Lim sine (x) /x, x ≈ 0.
printf("Distance between 2 longitudinal points \n");
printf(" at latitude 89deg 59min and 1 minute longitude apart: %.2f meters\n",
circle/( 360 * 80) );
Did you mean circle/(360*60)?
May I rewrite your source code?
Code:
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
// one_min.c distance between two longitudinal points on a "circle" drawn 1 minute latitude from South
// pole of Earth, each point on the drawn circle, one minute of longitude apart.
int main( void)
{
double polar_rad=6356; // polar radius of Earth km
double polar_rad_m=polar_rad * 1000; // km-> meters
double diam_e= 2 * polar_rad_m; // polar diameter in meters
double polar_circum=diam_e * M_PI; // diameter of Earth * pi = circumference in meters
double lat_1_min= polar_circum / (360 * 60); // in meters
// We don't need that line above.
// Use 2D euclidean map of earth
// a circle drawn at 89deg 59min 0sec with constant radius from the S pole
double nDegree = (89+59/60)/180 * M_PI; // add by me
// double circle = (lat_1_min + lat_1_min ) * M_PI; // circumference of 1 minute circle
double n = Cos(nDegree) * polar_rad_m;
double circle = n * M_PI * 2; // circumference of 1 minute circle
printf("Distance between 2 longitudinal points \n");
printf(" at latitude 89deg 59min and 1 minute longitude apart: %.2f meters\n",
circle/( 360 * 60) ); // changed to circle/(360 * 60);
return 0;
}
I change double lat_1_min= polar_circum / (360 * 60); // in meters
double circle = (lat_1_min + lat_1_min ) * M_PI; // circumference of 1 minute circle
to
double nDegree = (89+59/60)/180 * M_PI; // add by me
double n = Cos(nDegree) * polar_rad_m;
double circle = n * M_PI * 2; // circumference of 1 minute circle
I hope you can give correction to my source code.
It's been a good discussion though.
The air above the South Pole is cold, dry and clear. Close to vertical star sights suffer very low diffraction. In summer at the pole there is very little day-night variation in sunlight, temperature or seeing.
Observation of the changing declination from the vertical of a star, four times over 18 hours, will give a very accurate direction and angular distance to the pole.
Optical observations will be very accurate as the telescope can be rotated to the reciprocal bearing to cancel errors. The vertical reference will be a bigger problem as the gravity field may not be exactly parallel with the pole due to nearby ice depth irregularities.
Well, it was daylight, the sun was by far the easiest star to observe. You can compare its position with the direction of gravitational acceleration - if the angle does not change over the course of a day, you are at the pole.
#29
Stephanus
1,316
104
mfb said:
Well, it was daylight, the sun was by far the easiest star to observe. You can compare its position with the direction of gravitational acceleration - if the angle does not change over the course of a day, you are at the pole.
Yes and no. The Earth axis tils 23.50 in 91 days. So, every day it's about 0.30 variants. But I get your point.
Sure, the yearly variations are quite significant.
Scott arrived at the pole January 17, less than a month after (southern hemisphere's) summer solstice. I get 0.21 degrees per day. This has to be compared to 0.018 degrees daily variation from being 1 km away from the pole. Not trivial, but of course both teams knew about those effects.
Amundsen was there just a few days before the summer solstice, where the daily variation was just ~0.04 degrees.
