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
This is the table that I'm looking for the formula for:
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/seconds2. (G is rounded to 10 m/s2 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 (t1/2) to accelerate to the midpoint of the journey (d1/2):
$$ t_{\frac 1 2}=\sqrt{\dfrac{2 \times d_{\frac 1 2}}{a}} = \sqrt{\dfrac d a} $$
Doubling this gives a total trip time of
$$ t=2 \times \sqrt{\frac d a} $$