Surface area of a world

[Refplace]

The Chalice World is about 15,000 miles radius so 7.07*10^8 or (I think) 707,000,000 miles, is that sq miles?
Based on A=πr2
Earth is 196.9 million miles squared according to google.
Since my diameter is not much bigger than earths at the equator that seems too large a difference, leading me to think my math is wrong.

Can I get a confirmation from the hive mind?

[RyanW]

Originally Posted by Refplace

The Chalice World is about 15,000 miles radius so 7.07*10^8 or (I think) 707,000,000 miles, is that sq miles?
Based on A=πr2
Earth is 196.9 million miles squared according to google.
Since my diameter is not much bigger than earths at the equator that seems too large a difference, leading me to think my math is wrong.

Can I get a confirmation from the hive mind?

Pi*r^2 would give you the area of a circle, not the surface area of a sphere. Surface area is 4*pi*r^2. The actual surface area would be about 2,830,000,000.

I think, though, you might be confusing radius and diameter. 15,000 miles is a lot higher than the radius of the Earth.

(Assuming Chalice World is a sphere, which may not be true.)

[Refplace]

Originally Posted by RyanW

Pi*r^2 would give you the area of a circle, not the surface area of a sphere. Surface area is 4*pi*r^2. The actual surface area would be about 2,830,000,000.

I think, though, you might be confusing radius and diameter. 15,000 miles is a lot higher than the radius of the Earth.

(Assuming Chalice World is a sphere, which may not be true.)

The earth area I picked up from google, not calculated. However I might have read KM as miles or looked at radius on my phone because now I see it as 7,917.5 miles diameter.
Chalice World is a disc (at least the part I'm dealing with here) with a radius of about 15,000 miles (50 hexes (300 mile) from center to edge.
So did I do it right?

[whswhs]

Originally Posted by Refplace

Chalice World is a disc (at least the part I'm dealing with here) with a radius of about 15,000 miles (50 hexes (300 mile) from center to edge.
So did I do it right?

Is it flat? Or the inner surface of a shallow cone? Or some other shape? "Chalice" suggests some sort of cup shape, which will be bigger than pi r squared.

[Refplace]

Originally Posted by whswhs

Is it flat? Or the inner surface of a shallow cone? Or some other shape? "Chalice" suggests some sort of cup shape, which will be bigger than pi r squared.

Yeah, its not flat but earth isn't either. Most of the usable area is a gradual slope from the walls to the center. Think of the surface as the water level in a half filled chalice.
But we dont need an exact surface area, I just was curious about the general ballpark and assuming flat is "close enough".

[Daigoro]

Would it be fair to describe the shape of the chalice as a hemisphere at the base with a cylinder extending upward from the rim?

[Refplace]

Originally Posted by Daigoro

Would it be fair to describe the shape of the chalice as a hemisphere at the base with a cylinder extending upward from the rim?

Yes, though for practical purposes that really doesnt matter for surface area, though it affects total volume.

Think of a wine glass half filled with wine or water. The effective surface area (at least all the inhabitants would care about) is just the surface area of the water.

[whswhs]

If we envision it as a wine glass, its surface is that of a cone. The area of a cone is pi times r times square root (r squared plus h squared), where r is the horizontal distance from the axis to the outer rim, and h is the vertical distance from the bottom to the highest level. You have r, it seems, so you just need h.

If h is negligible, then r squared plus h squared is close to r squared, its square root is close to r, and the total surface is close to pi r squared. As the depth increases, so does the discrepancy.

[Refplace]

Originally Posted by whswhs

If we envision it as a wine glass, its surface is that of a cone. The area of a cone is pi times r times square root (r squared plus h squared), where r is the horizontal distance from the axis to the outer rim, and h is the vertical distance from the bottom to the highest level. You have r, it seems, so you just need h.

If h is negligible, then r squared plus h squared is close to r squared, its square root is close to r, and the total surface is close to pi r squared. As the depth increases, so does the discrepancy.

So since I have specified h is negligible and r is 15,000 miles is the result 707,000,000 sq miles?
I think thats right, but since this forum has much better math minds than mine I thought I would ask here for confirmation or correction.

[Daigoro]

Originally Posted by Refplace

Yes, though for practical purposes that really doesnt matter for surface area, though it affects total volume.

Think of a wine glass half filled with wine or water. The effective surface area (at least all the inhabitants would care about) is just the surface area of the water.

Well, the surface area of a hemisphere (2piR^2) is half that of a sphere (4piR^2), and therefore double that of a disc of the same radius.