What is the formula for calculating in-system traveling time in Mongoose Traveller 2? On page 153 of the MGT2 core rulebook there is a table, but I'm looking for the formula behind the table to insert into an excel sheet I'm making.

As far as I can tell this is what is said about the calculation in the book:

assume the ship is undertaking a journey from rest, that it accelerates continuously to midpoint of the trip, then decelerates to rest again. 1G is equal to approximately 10 metres per second per second

## Best Answer

This table is, indeed, calculated by assuming that a ship can produce constant acceleration away from its origin, instantaneously pivot \$180^\circ\$ at the midpoint of its journey, and constantly decelerate for the second half of its journey. The greatest veloticy thus attained (w.r.t. origin) in the table above is roughly 1 million m/s--approximately 1/3 of one percent of

c--so we're comfortable sticking with classical calculations.The formula that produces the above times is:

$$ t = 2 \times \sqrt{\dfrac d a} $$

where \$t\$ is measured in seconds, \$d\$ in meters, and \$a\$ in meters/seconds

^{2}. (G is rounded to 10 m/s^{2}for ease of calculation. Note that the table gives distances in km, so you've got to tack on three zeroes to end up in meters.)Those who would like a refresher on their classical kinematics, read on:

Recall that the distance travelled when starting at rest and undertaking constant acceleration is given by

$$ d= \frac 1 2 at^2 $$

Solving for t gives us

$$ t=\sqrt{2 \times \frac d a} $$

In our situation we consider the time (t

_{1/2}) to accelerate to the midpoint of the journey (d_{1/2}):$$ t_{\frac 1 2}=\sqrt{\dfrac{2 \times d_{\frac 1 2}}{a}} = \sqrt{\dfrac d a} $$

Doubling this gives a total trip time of

$$ t=2 \times \sqrt{\frac d a} $$