If there are 63,360 inches in a mile, we can do a quick calculation and find out how long the model's diameter is supposed to be. First, reduce the diameter of the earth by the scale factor (not via shrink-ray) and then convert to inches. diameter of model ≈ 5.02 in
From there, we can find everything we need to know about the scale model. The radius is half the diameter (2.51 in), which we can use to find the surface area. *SA* = 4π*r*^{2}
*SA* = 4π(2.51 in)^{2}
*SA* = 79.2 in^{2}
And the volume, too.
*V* = 66.2 in^{3}
We've found our measurements. To be completely, absolutely, positively, 100% certain that we have the right values, the best thing to do would be to check them using the surface area and volume ratios. The surface area of the earth would equal: *SA* = 4π*r*^{2}
*SA* = 4π(3956.5 mi)^{2}
*SA* ≈ 196,712,612 mi^{2}
If we want to find out the surface area of our scale model using that number, we have to make sure to square the scale factor *and* the miles-to-inches ratio. *SA* of model = 196,712,612 mi^{2} ×
*SA* of model ≈ 79 in^{2}
That's close enough. What about volume? *V* = ^{4}⁄_{3}π*r*^{3}
*V* = ^{4}⁄_{3}π(3956.5 mi)^{3}
*V* ≈ 259,431,149,227 mi^{3}
And now for some epic *cubic* unit conversion. *V* of model = 259,431,149,227 mi^{3} ×
*V* of model ≈ 66 in^{3}
Also pretty darn close. Looks like we did our work right, and your model of earth is going to be stellar. Actually it'll be planetary, but that's more than Pluto can say. |