I Want to Hide
I want to hide in a hole, in a corner.
I've tried to imagine what shape would take up the least amount of space in a corner, have the least amount of surface area showing, be the least noticeable (mathematically anyway): spherical or triangular. I'm pretty sure I'd have to be spherical...
Supposing I had a volume of 2 and 7/9 cubic feet (a reasonable average and equal to 4800 cubic inches), I would have a surface area of 813.69 square inches if I assumed the shape of a triangular pyramid, and I would have a surface area of 688.04 square inches if I assumed the shape of an eighth of a sphere (I chose eighth of a sphere because I only want to occupy a corner; if I were any other fraction of a sphere, I would stand out from the corner more than necessary).
You may want to think that a one-dimensional line might be the shape with the least surface area, but, because you'd be taking three dimensions and forcing them into just one dimension, your line would, hypothetically, be infinite.
If you're interested, the math can be found below.
All this to say, I want to be an eighth of a sphere in a corner.
I need to find my radius.
Since 4800 in^3 will only be taking up a corner (one eighth of a sphere), I multiply by 8: 38400.
Volume of a sphere is (4/3)πr^3.
38400 = (4/3)πr^3
28800 = πr^3
9167.32 ≈ r^3
r ≈ 20.93 inches
Surface area of a sphere is 4πr^2.
4π(20.93)^2 ≈ 5504.34 in^2
However, I have to divide by 8 because I am just occupying a corner (and because I multiplied by 8 earlier in the calculation).
My surface area as an eighth of a sphere would be roughly 688.04 in^2.
I need to find the dimensions of the visible surface.
My volume is 4800 in^3.
Volume of a triangular pyramid is (1/3)AH where A is area of the base and H is the height of the pyramid.
Now, A is a triangle, so it is equal to (1/2)bh where b is the length of the base of the triangle and h is the height of the triangle.
Thus, 4800 = (1/3)((1/2)bh)H
However, since having an equilateral surface will yield the most conservative surface area, b = h = H.
Thus, 4800 = (1/3)((1/2)bb)b.
4800 = (1/6)b^3
28800 = b^3
b ≈ 30.65 in
This b, however, is only the distance from the corner in which I'm hiding to one of the corners of my visible surface. To find my surface area, I need the length of one of my corners to another of my corners; I also need to know my height. I'll draw a picture later, but, for now, just work with me.
This b can get me the length of one of my corners to another of my corners using the Pythagorean theorem. √(b^2+b^2) ≈ 43.35 in. I'll call this H. This, now, is the base for my surface area, but I still need my surface area height.
Using some trigonometry (sOHcAHtOA), I can determine that the height - which I'll call A since it's the adjacent - is B•cos 30. (Remember that my surface area is equilateral, which is how I know the angle is 30 degrees.)
A = H•cos 30 ≈ 37.54 in
Thus, my surface area is (1/2)HA ≈ 813.69 in^2.
Remember: You may want to think that a one-dimensional line might be the shape with the least surface area. However, if you took 4800 in^3 and tried to stretch it out line-thin, you'd be multiplying two dimension by 0 to reduce them, but you'd have to divide the remaining dimension by two numbers each approaching 0, which yields ∞.
I've tried to imagine what shape would take up the least amount of space in a corner, have the least amount of surface area showing, be the least noticeable (mathematically anyway): spherical or triangular. I'm pretty sure I'd have to be spherical...
Supposing I had a volume of 2 and 7/9 cubic feet (a reasonable average and equal to 4800 cubic inches), I would have a surface area of 813.69 square inches if I assumed the shape of a triangular pyramid, and I would have a surface area of 688.04 square inches if I assumed the shape of an eighth of a sphere (I chose eighth of a sphere because I only want to occupy a corner; if I were any other fraction of a sphere, I would stand out from the corner more than necessary).
You may want to think that a one-dimensional line might be the shape with the least surface area, but, because you'd be taking three dimensions and forcing them into just one dimension, your line would, hypothetically, be infinite.
If you're interested, the math can be found below.
All this to say, I want to be an eighth of a sphere in a corner.
Surface Area of an Eighth of a Sphere
I need to find my radius.
Since 4800 in^3 will only be taking up a corner (one eighth of a sphere), I multiply by 8: 38400.
Volume of a sphere is (4/3)πr^3.
38400 = (4/3)πr^3
28800 = πr^3
9167.32 ≈ r^3
r ≈ 20.93 inches
Surface area of a sphere is 4πr^2.
4π(20.93)^2 ≈ 5504.34 in^2
However, I have to divide by 8 because I am just occupying a corner (and because I multiplied by 8 earlier in the calculation).
My surface area as an eighth of a sphere would be roughly 688.04 in^2.
Surface Area of a Triangular Pyramid
I need to find the dimensions of the visible surface.
My volume is 4800 in^3.
Volume of a triangular pyramid is (1/3)AH where A is area of the base and H is the height of the pyramid.
Now, A is a triangle, so it is equal to (1/2)bh where b is the length of the base of the triangle and h is the height of the triangle.
Thus, 4800 = (1/3)((1/2)bh)H
However, since having an equilateral surface will yield the most conservative surface area, b = h = H.
Thus, 4800 = (1/3)((1/2)bb)b.
4800 = (1/6)b^3
28800 = b^3
b ≈ 30.65 in
This b, however, is only the distance from the corner in which I'm hiding to one of the corners of my visible surface. To find my surface area, I need the length of one of my corners to another of my corners; I also need to know my height. I'll draw a picture later, but, for now, just work with me.
This b can get me the length of one of my corners to another of my corners using the Pythagorean theorem. √(b^2+b^2) ≈ 43.35 in. I'll call this H. This, now, is the base for my surface area, but I still need my surface area height.
Using some trigonometry (sOHcAHtOA), I can determine that the height - which I'll call A since it's the adjacent - is B•cos 30. (Remember that my surface area is equilateral, which is how I know the angle is 30 degrees.)
A = H•cos 30 ≈ 37.54 in
Thus, my surface area is (1/2)HA ≈ 813.69 in^2.
Remember: You may want to think that a one-dimensional line might be the shape with the least surface area. However, if you took 4800 in^3 and tried to stretch it out line-thin, you'd be multiplying two dimension by 0 to reduce them, but you'd have to divide the remaining dimension by two numbers each approaching 0, which yields ∞.
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