Flowlines of blobs I have the following formula for blobs/metaballs, which is said to be the same as the one used for electromagnetism:
\begin{gather}
f(x,y,z) = \frac{d(A,B)}{\sqrt{(x-xA)^2 + (y-yA)^2 + (z-zA)^2}} \\
g(x,y,z) = \frac{d(C,D)}{\sqrt{(x-xC)^2 + (y-yC)^2 + (z-zC)^2}} \\
h(x,y,z) = f(x,y,z) + g(x,y,z) - \text{Threshold}_1 \\
i(x,y,z) = \min(\text{Threshold}_2, \max(0, h(x,y,z)))
\end{gather}
This produces the equipotential lines that orbit the charges. However, what is the formula for the perpendicular flowlines?
 A: If $h(x,y,z)=C$ (that is, a level set of $h(x,y,z)$) gives you the equipotential surface, then $\nabla h(x,y,z)=\langle\frac{\partial h}{\partial x},\frac{\partial h}{\partial y},\frac{\partial h}{\partial z}\rangle$ is a vector that gives you the vectorfield perpendicular to the level sets (or, keeping with your electromagnetism analogy, the resulting electric field).
To find the flow curve $\gamma(t)$ from a given point, we solve the initial value problem
$$
\gamma(0)=\langle x_0,y_0,z_0\rangle\\
\gamma'(t)= \nabla h(\gamma(t))
$$
Let me know if that's along the lines of what you're looking for, and if you want clarification on that.

My attempt to clarify:
For the sake of making this as understandable as possible, I will stick to the electromagnetism analogy.  The "flow curve" we want is the trajectory of a point charge under the approximation that it exactly follows the direction of the electric field.
The first step is to find the direction of the electric field (a vector field), given the voltage field
$$
V(x,y,z) = f(x,y,z)+g(x,y,z)
$$
The components of the electric field are given by
$$
E_x=-\frac{\partial V(x,y,z)}{\partial x}\qquad
E_y=-\frac{\partial V(x,y,z)}{\partial y}\qquad
E_z=-\frac{\partial V(x,y,z)}{\partial z}
$$
These are the partial derivatives of $V(x,y,z)$.  We find, for example, that
$$
\begin{align}
E_x &= -\frac{\partial V}{\partial x}\\
&= -\left(\frac{\partial f}{\partial x} + \frac{\partial g}{\partial x}\right)\\
&= -\frac{\partial}{\partial x}\frac{k_1}{\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}} 
-\frac{\partial}{\partial x}\frac{k_2}{\sqrt{(x-x_2)^2+(y-y_2)^2+(z-z_2)^2}}\\
&=
\end{align}
$$
editing...
