Goal is to integrate various functions along lines of sight which is constrained to be in the y-z plane and has components B_x and B_\perp

I = \int f(B_\perp(s), B_x(s)) ds

\theta is the tilt angle of the jet towards the observer

so B_\perp(s) = B_y(s) \cos(\theta) + B_z(s) \sin(\theta)

Each pixel in the resulting image will be given by the integral of the function along the path

S = x_i, y_i+s \sin(\theta), s \cos(\theta)

where s=0 corresponds to crossing of the xy plane

and we have x_i-x_{i-1} = \delta x and y_i-y_{i-1} = \delta y

and \delta y = \cos(\theta) \delta x

For 3D we just need to create the expressions in visit, create the lineouts, and then integrate the resulting query for each x_i and y_i

For 2.5D, we need to transform the integral along s (normally in the yz plane) into an integral in the xy plane.

Since everything is axi-symmetric, the value at x_i,y_i+s \sin(\theta), s \cos(\theta) can be inferred by rotating the corresponding solution at

\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]

by an angle

\phi = \cos^{-1} \frac{x_i}{\sqrt{x_i^2+\left ( s \cos (\theta) \right)^2}} .

So our integral is now

\int f( \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 ) ds

Now, if \theta = 0, this simplifies to

\int f( \sqrt{x_i^2+s ^2}, y_i, 0 ) ds

and we can undergo a chance of variables

x' = \sqrt{x_i^2+s^2}

and

dx' = \frac{s}{\sqrt{x_i^2+s^2}} = \frac{\sqrt{x'^2-x_i^2}}{x'} ds

so our integral becomes

2 \int_{x_p}^{\infty} f( x', y_i, 0 ) \frac{x'}{\sqrt{x'^2-x_i^2} } dx'

and B_\perp is unchanged, while B_x = B_\phi \sin(\theta) = B_\phi \frac{\sqrt{x'^2-x_i^2}}{x'}

So our new path is in the xy plane and is given by

S'(s)=\left [ \sqrt{x_i^2+\left(s \cos(\theta) \right ) ^2}, y_i+s \sin(\theta), 0 \right ]

And we have \frac{dS'}{ds}=\sqrt{ \left ( \frac{s \cos(\theta)^2}{\sqrt{x_i^2+\left ( s \cos(\theta) \right)^2} }\right ) ^2 + \sin(\theta)^2}

So our path integral becomes

\int_s f(B(x_i,y_i+s \sin(\theta), s \cos(\theta))) \frac{dS}{ds} ds = \int_{s} { f(B'(\sqrt{x_i^2+\left ( s \cos(\theta) \right )^2}, y_i+s \sin(\theta),0)) \frac{dS'}{ds} ds}

Where B' is the rotated vector B through the angle \phi.

Now if \theta=0, this simplifies to

\int_{s} { f(B'(\sqrt{x_i^2+s^2}, y_i,0)) \frac{\sqrt{x_i^2+s^2}}{s} ds}

Now B_\perp is unchanged under rotations about the jet axis, however B_x will be given by B_\phi \cos \phi or B_x = B_\phi \frac{x_i}{\sqrt{x_i^2+(s \cos(\theta))^2}}

And finally we can change the path integral to an x integral under the substitution

x=\sqrt{x_i^2+s^2}

or

s = \sqrt{x^2-x_i^2}

and

ds/dx = \frac{x}{\sqrt{x^2-x_i^2}}

This gives

\int_{x} { f(B'(x, y_i,0)) \frac{x^2}{x^2-x_i^2} dx}

and B_x'(x,y_i,0)=B_\phi(x,y_i,0) \frac{x_i}{x}