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 new path is in the xy plane and is given by

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} = ...

so we can write the integral as

$\int F(x'(s)

and we have ds = \frac{\sqrt{(x_i^2+(s \cos(\theta))^2}}{s \cos(\theta)} dx + \frac{1}{\sin(\theta)} dy

which under the substitution gives

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

Assume jet is oriented along y and that we are integrating along the direction (0, sin(\theta), cos(\theta))

where \theta is the angle of inclination.

We are interested in calculating the integral of B_p^2, B_x^2, and B_xB_p where B_p = cos(\theta)By + sin(\theta)Bz

If \theta = 0 this is just \int B_y^2 dz, \int B_x^2 dz, and \int B_x B_y dz

If \theta /= 0 then we have to tilt the integral and calculate B_p using the expression above.