# Scott found Amundsen Flag

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1. Jan 16, 2016

### Stephanus

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

2. Jan 16, 2016

### Bystander

"Shooting" one minute of latitude isn't trivial, but it's not all that unusual either.

3. Jan 16, 2016

### Stephanus

But, that 1 minute latitude (the same on every spot on earth) would be... 10 thousands km / 90 / 60 = 1.85 km?

4. Jan 16, 2016

### Bystander

Well within eyeshot.

5. Jan 20, 2016

### Staff: Mentor

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. Jan 20, 2016

### Stephanus

D'accord

7. Jan 21, 2016

### Staff: Mentor

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. Jan 22, 2016

### Stephanus

Several? Thanks FYI. I didn't know that.

9. Jan 22, 2016

### Stephanus

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?

10. Jan 22, 2016

### Staff: Mentor

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.

11. Jan 22, 2016

### Stephanus

Scott route to South Pole
Amundsen route to South Pole

For Scott's web page, scroll down a little until you see "download the KML file here"

12. Jan 24, 2016

### Devon Fletcher

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. Jan 25, 2016

### Stephanus

D'accord

14. Jan 26, 2016

### Stephanus

15. Jan 26, 2016

### Stephanus

Good, at least there are US, UK, France, Denmark? Russia? Australia?

16. Jan 26, 2016

### Staff: Mentor

17. Jan 26, 2016

### insightful

Why isn't this roughly the same as one minute of latitude?

18. Jan 26, 2016

### Stephanus

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. Jan 26, 2016

### insightful

Agreed.
I think you mean degree, not minute.
Agreed.
I'd say both 1 minute latitude and 1 minute longitude at the equator should be about 1.85km.

20. Jan 26, 2016

### Stephanus

So, the honourable Jim McNamara made a mistake?
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 statment
At equator 1 minute longitude is not 590 m.