| | 22 | |
| | 23 | For mass, the solution for the kernel constants are trivial. For half of the cone (say the top), we have: |
| | 24 | |
| | 25 | $\Sigma \Sigma \frac{m_{out}}{2} k_r k_\theta R \Theta = \frac{M_{out}}{2} $ |
| | 26 | |
| | 27 | which gives: |
| | 28 | |
| | 29 | $k_r k_\theta = \frac{1}{\Sigma R(r,r_{out})}\frac{1}{\Sigma\Theta(\theta, \theta_{out})}$ |
| | 30 | |
| | 31 | Thus to fill the outflow kernel with mass, the algorithm would proceed as follows: |
| | 32 | |
| | 33 | 1. Solve for $k_r$ and $k_\theta$ |
| | 34 | 2. For all cells within $r<r_{out}$ for which the jet axis > 0 apply $\rho_{i,j,k}=k_r k_\theta R \theta \frac{M_{out}}{2}$ |
| | 35 | 3. Repeat for the bottom half |
