# Thread: Arctic ice

1. ## Arctic ice

Hi mathematicicans!

Can anyone solve this problem for me?

I have an image from space that looks down on the north pole and I have estimated the area of ice cover of a flat earth in this image as a percentage of the area of the flat earth.

I need to estimate what the same area of coverage would be VIEWED FROM THE EQUATOR.

Ive tried to work it out, I think it will be some sore of integration based on the sin of the latitude? Im not sure

Any takers??

2. Originally Posted by Tpod
Hi mathematicicans!

Can anyone solve this problem for me?

I have an image from space that looks down on the north pole and I have estimated the area of ice cover of a flat earth in this image as a percentage of the area of the flat earth.

I need to estimate what the same area of coverage would be VIEWED FROM THE EQUATOR.

Ive tried to work it out, I think it will be some sore of integration based on the sin of the latitude? Im not sure

Any takers??
Try rewording this more clearly. It is unclear what you are referering to when you talk about flat earth areas etc.

CB

3. I have a tennis ball and I draw a big spot (circle) on it. I pretend the ball is the earth and the large spot is the north pole. I take a picture of the tennis ball from the equator (90 degrees from the centre of the spot).
What percentage of the earth's area IN MY PICTURE is covered by the spot, if:
12% of the surface is covered when the picture is taken fron directly above the spot.

Imagine you take a photograph of the north pole from above. You will have a 2d image of the earth. The area of the polar ice (a crude approximation of a circle) can be expressed as a percentage of the area of the earth.

I have this value as 12%.

I need to know what the approximate value would be of the same ice if viewed from the equator.

4. please see attached.

I have a tennis ball and I draw a big spot (circle) on it. I pretend the ball is the earth and the large spot is the north pole. I take a picture of the tennis ball from the equator (90 degrees from the centre of the spot).
What percentage of the earth's area IN MY PICTURE is covered by the spot, if:
12% of the surface is covered when the picture is taken fron directly above the spot.

Imagine you take a photograph of the north pole from above. You will have a 2d image of the earth. The area of the polar ice (a crude approximation of a circle) can be expressed as a percentage of the area of the earth.

I have this value as 12%.

I need to know what the approximate value would be of the same ice if viewed from the equator.

5. Originally Posted by Tpod
please see attached.

I have a tennis ball and I draw a big spot (circle) on it. I pretend the ball is the earth and the large spot is the north pole. I take a picture of the tennis ball from the equator (90 degrees from the centre of the spot).
What percentage of the earth's area IN MY PICTURE is covered by the spot, if:
12% of the surface is covered when the picture is taken fron directly above the spot.

Imagine you take a photograph of the north pole from above. You will have a 2d image of the earth. The area of the polar ice (a crude approximation of a circle) can be expressed as a percentage of the area of the earth.

I have this value as 12%.

I need to know what the approximate value would be of the same ice if viewed from the equator.

12% = 0.12 $= \dfrac{\pi r_i^2}{\pi r_e^2}$

Thus the radius of the polar ice $r_i = 0.3464$ of the earth radius.

Arcsin(0.3464)= 0.35374 radians

Area of the segment viewed from a point where the latitude lines are parallel:

$
0.35374 - \cos(0.35374 \text{radians})\times 0.3464 = 0.0288
$

$\dfrac{0.0288}{\pi}=0.0092$ or 0.92%

Approximately 1%.

NOTE:
Your image indicates a side view of the earth at a point far from the earth and NORTH of latitude 69.73 degrees north.
If you know the distance from the surface of the earth and the distance north or south of the equator a better approximation is possible.
My calculation is for a flat latitude line.
If you were at some point directly above the equator (some high altitude) the "shadow" of the polar ice would appear as crescent shaped.

6. Originally Posted by aidan
12% = 0.12 $= \dfrac{\pi r_i^2}{\pi r_e^2}$

Thus the radius of the polar ice $r_i = 0.3464$ of the earth radius.

Arcsin(0.3464)= 0.35374 radians

Area of the segment viewed from a point where the latitude lines are parallel:

$
0.35374 - \cos(0.35374 \text{radians})\times 0.3464 = 0.0288
$

Approximately 3%.

NOTE:
Your image indicates a side view of the earth at a point far from the earth and NORTH of latitude 69.73 degrees north.
If you know the distance from the surface of the earth and the distance north or south of the equator a better approximation is possible.
My calculation is for a flat latitude line.
If you were at some point directly above the equator (some high altitude) the "shadow" of the polar ice would appear as crescent shaped.
You need to divide the area by the area of the unit circle to get the required percentage, which becomes ~1%

CB

7. sorry CB not sure what you mean

8. Originally Posted by Tpod
sorry CB not sure what you mean
CB means (in a very polite way) that I slipped on the ice and ruptured my final statement.
The 0.0288 is the area, NOT the ratio.

You need one final calculation:

$\dfrac{0.0288}{\pi} = 0.00917$ equals 0.92%

Sorry for the mistake.