Gravity

Have any of you mathematicians come up with a good system for determining the
effects of gravity on a ship's path? I could calculate unpowered circular
orbits easily enough, but I'm not included to calculate anything else during a
game.

> Laserlight wrote:

I stumbled across Traveller Book 2 tonight serendipitously, thus I quote (page
28): "During the movement phase, lay out the vector of the ship to determine
where it will move. If the exact midpoint of the vector lies in a gravity
band, a gravity vector will be added....The gravity vector is parallel to a
line connecting the regular course midpoint to the planetary template center."

> Laserlight wrote:

1/4 gee increments and the equation was based on "Traveller
world sizes." I'll work out something more generic sometime today

[quoted original message omitted]

The formula, if I've done it correctly, is
r = 6.378 * sqrt (m/g)
r = radius in MU 6.378 = radius of Earth in MU (assumes 1 MU is 1000km) m =
mass of primary in Earth masses g = gravity quantity you want (in gees)

Note also that Traveller book 2 (my edition, anyway) has some errors in the
planetary stats, eg Jupiter is listed as 1318 Earth masses where I believe
it's actually 318; Saturn is 714 when IIRC it's more like 95. I suspect they
forget about density.

> Laserlight wrote:

> The formula, if I've done it correctly, is

Nyrath said:
> I don't know how well this will work with cinematic

I haven't had time to lay it out and see if it works. In theory your ship's
momentum is normal to the line between the ship and the planetary center.
However, you have to account for the gravity vector as well, so the ship
doesn't move along the circle, nor along a line tangent to the circle and
normal to the
line between ship and center--I'd think it would move on a chord
which continues after intersecting the circle, then gets pulled back to the
circle by the gravity vector.

Time to print up some graph paper with something like 1mm squares and see what
happens.

> > The formula, if I've done it correctly, is

"Earth masses" is easier to remember than 10^23 kg, or whatever it is.

> > The formula, if I've done it correctly, is

this formula works, but requires that you calculate the mass and radius of the
planet relative to earth. the equation to find g (gravitational
acceleration in m/s) is as follows, directly out of a physics textbook:

g=6.67*10^-11*M/r^2

^ is a symbol i used for "to the power of" M is the mass of the planet (or
star, etc.) in kg.
6.67*10^-11 is the gravitational constant
r is the distance from the centre of the primary mass to the point of
measurement

this would have to be changed to accept FT units (tons, MU, and whatever

turn length you have agreed apon)

if there is enough interest, i (or someone else) could do up a java program to
calculate this and post it somewhere.