EDG's thread made me want to fully work out the size of L1 jump points. Is it possible to guard a pirate point? What's the timescale of an attack? How many sensor platforms do you need to monitor? Are collisions between jumpers likely? The following is a calculation for the size for L1 jump points. Maybe it's useful to anyone who wants super-crunchy battletech? In any case, I'm happy to hear about any thoughts or corrections.

I'll use M

_{A} and M

_{C} to denote the mass of an ancillary body and M_P to denote the mass of the primary body, and we'll consider a 2-body system. This is obviously an approximation, although it should be reasonably good in many situations.

**The Center**The center of the L1 jump point (note: different from the L1 Lagrange point) is where the gravitational potentials of the two bodies cancel, thus:

M_{A}/d_{A}^{2} = M_{C}/d_{C}^{2}

where d

_{A} is the distance to A and d

_{C} is the distance to C. Pictorially, we have:

C-------------------------------------------------------------------------------------------------------------------------J--A

where J is the jump point.

This can be rewritten as:

M_{A}/d_{A}^{2} = M_{C}/(d_{CA}-d_{A})^{2}

where d

_{CA} is the distance between C and A. Using algebra we get:

(M_{C}-M_{A})d_{A}^{2}+2d_{A}d_{CA}M_{A}-d_{CA}^{2}M_{A}=0

Setting us up for the quadratic equation which has a solution at:

d_{A}=d_{CA}((M_{A}M_{C})^{0.5}-M_{A})/(M_{C}-M_{A})

Notably, in the limit where M

_{C} >> M

_{A}, this simplifies as:

d_{A}~=d_{CA}(M_{A}/M_{C})^{0.5}

**Radial Boundary**The radial boundary applies to the direction between the primary and ancillary bodies. Pictorially, it looks like:

C------------------------------------------------------------------------------------------------------------------------BJB-A

Where B is +/- radial boundary.

Calculating a boundary requires that we know the tolerance to non-flat spacetime for jumps. This is given by the jump limit which in Sol is set at 10.2AU (per SO page 133). Let's call this d

_{j}. Jump drives can tolerate

M_{S}/d_{j}^{2}

nonflatness where M

_{S} is the mass of Sol.

Using this, the equation for the radial boundary is:

M_{A}/d_{A}^{2} = M_{C}/(d_{CA}-d_{A})^{2} +/- M_{S}/d_{j}^{2}

An analytic solution is tricky here since this yields a quartic. This could be solved numerically, or we can make analytic progress after approximating M

_{C} >> M

_{A} and d

_{CA} >> d

_{A}. Under that approximation, we get:

M_{A}d_{CA}^{2} ~= d_{A}^{2}(M_{C} +/- M_{S}d_{CA}^{2}/d_{j}^{2})

Implying

d_{A}~=d_{CA}(M_{A}/(M_{C} +/- M_{S}d_{CA}^{2}/d_{j}^{2}))^{0.5}

Adding a further assumption that 1 >> M

_{S}d

_{CA}^{2}/(d

_{j}^{2}M

_{C}) this gives:

d_{A}~=d_{CA}(M_{A}/M_{C})^{0.5}(1 +/- M_{S}/M_{C} * d_{CA}^{2}/(d_{j}^{2}))^{0.5}

Which further approximates as:

d_{A}~=d_{CA}(M_{A}/M_{C})^{0.5}(1 +/- 1/2*M_{S}/M_{C} * d_{CA}^{2}/(d_{j}^{2}))

Implying the radial dimension of the jump point is about:

d_{r}~=d_{CA}(M_{A}/M_{C})^{0.5}M_{S}/M_{C} * d_{CA}^{2}/(d_{j}^{2})

**Perpendicular Boundary**The perpendicular boundary applies in all directions which are perpendicular to the radial direction. As you leave the radial axis at the center of the jump point a small part of the gravitational potential adds together rather than canceling. Pictorially, this looks like:

T

C------------------------------------------------------------------------------------------------------------------------BJB-A

T

In particular, the perpendicular distance to the boundary d

_{T} satisfies:

M_{A} Tan(d_{T}/d_{A})/d_{A}^{2} + M_{C} Tan(d_{T}/d_{C})/d_{C}^{2} = M_{S}/d_{j}^{2}

Solving this analytically is again tricky. However, when M

_{C} >> M

_{A} the second term on the left hand side can be neglected. Using this, we get:

d_{T} ~= d_{A} Tan^{-1}(M_{S} d_{A}^{2} /(d_{j}^{2}M_{A}))

**Examples**Since solutions involve mass ratios and distance ratios, it's relatively easy to do various calculations.

Planet | M_{A}/M_{C} | d_{CA} | d_{A} | radial width (approx) | perpendicular width (approx) | Notes |

Earth | 3.0*10^{-6} | 150*10^{6}km | ~260*10^{3}km | ~2497km=139 hexes | ~2497km = 139 hexes | Large enough that a single warship does not quite have fire control with direct fire weapons. Since the moon orbits at 384*10^{3} km it will substantially perturb the solution. |

Mars | 3.0*10^{-7} | 228*10^{6}km | ~125*10^{3}km | ~1200km = 67 hexes | ~2772km = 154 hexes |

Jupiter | 0.000955 | 778*10^{6}km | ~24.0*10^{6}km | ~231*10^{3}km = 12.8*10^{3}hexes | ~6.08*10^{6}km = 338*10^{3}hexes | Large enough to be beyond targeting range for a single large craft. |

Ceres | 4.8*10^{-10} | 414*10^{6}km | ~9070km | ~87km=5 hexes | ~663km=37 hexes | Small enough for a traffic jam. |

Looking at the trends, the L1 jump point gets larger the further from the center and the larger the ancillary. The perpendicular over radial width apparently also grows with distance from the primary.