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Hello,

Wind resistance against a Sphere.

Only half of the sphere's surface area is used, the half facing the wind.

However, due to the curvature of the surface, is it true that the wind is really only pressing wholey or directly against 2/3 of that surface ?

Thanks,

Chris

No, the wind is pressing on half the surface - the half facing the wind.

However, the pressure applied to the sphere is lessened due to the slope of the sphere. This is true for most surfaces facing wind or air resistance. It's why cars and planes are sloped, and while it is a very complex subject to grapple with aerodynamics, some vector calculus can explain why a sphere - in ideal conditions - feels less force than, say, a flat disk.

(If all you wanted was a qualitative answer, you can stop here, but if you want an explanation with technical details, read on.)

Assume the wind is blowing so as to apply 1 Newton of force per square meter (N/m^2 is a unit known as the Pascal, or Pa, which measures pressure). Imagine you have a disk of radius 1, and it is flat, like a paddle of some sort. You hold it up with the wind facing directly at it and you feel π Pa of force due to the area of the disk being π.

Turn it 90 degrees and now the wind passes through and gives no pressure, of course.

But what about a sphere? A sphere will feel some of the pressure but not all of it - somewhere between the two extremes in the case above. And of course, if you rotate a sphere, nothing happens to the pressure (it has rotational symmetry).

The surface of the sphere can be parametrized by x^2 + y^2 + z^2 = 1.

Then, the unit normal vector at any point (x,y,z) is, literally, (x,y,z). It's very convenient.

Assuming the wind blows in the x direction, the force at any point is (x,y,z).(1,0,0), where that is the dot product of the two vectors. Of course, this is really an infinitesimal quantity, so we need to compute an integral of this over the surface of the sphere (or rather, the half facing the wind).

∫ (x,y,z).(1,0,0) dV = ∫ x dV

This is not as easy as it looks since the integral is over the surface of the half-sphere, x^2+y^2+z^2=1 with x>0.

You wind up with a triple-integral where:

x goes from 0 to 1

y goes from -√(1-x^2) to +√(1-x^2)

z goes from -√(1-x^2-y^2) to +√(1-x^2-y^2)

of the quantity x dV.

You can use some symmetry to reduce it to the integral:

x goes from 0 to 1

y goes from 0 to +√(1-x^2)

z goes from 0 to +√(1-x^2-y^2)

of the quantity 4x dV.

You could also use different coordinates, which is a somewhat more clever approach but yields a very similar integral in the end. The final answer is π/4, which means that actually, the factor by which the pressure diminishes is not 2/3, but rather, 1/4 !

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