Approximating drag coefficient

I’ve been trying to approximate the drag coefficient based on a frontal shape of an object. here’s what I’ve come up with. We can represent the shape of an object as a function f(x) (x is length, f is height, a 2D problem), the object is moving along the x-axis. So I assume that the drag force in x direction is the projection of the force lying on the normal line to f(x) and equal in magnitude to the flow force at a point to which the flow force is applied, as shown on the sketch.

We can find the angle between the x-axis and the object – that’s the arctan of the derivative. Then the angle between the normal line and the drag force vector is $frac {pi}{2}-arctan(f'(x))=text{arccot}(f'(x))$. The drag coefficient C at a single point will be the sine of it. So we need to sum it along the entire length of an object, let it be L. Then we get the following equation: $$C=vertfrac 1Lint_0^L sin(text{arccot}(f'(x)))dxvert=vertfrac 1Lint_0^L frac {f'(x)}{sqrt {1+f'(x)^2}}dxvert.$$ I assume that it’s pretty much right, because it yields 0 for a horizontal line, 1 for a vertical line and 0.5 for a sphere. But the drag is affected by many other factors. The question is can the total drag coefficient be split into several ones that depend on different factors? For example, $C=C_fs+C_ss+C_v…$, where $C_fs$ is the coefficient based on the frontal shape (the one we’ve found), $C_ss$ is affected by the side shape, $C_v$ depends on viscosity and so on. And can each of them be calculated (or approximated)?