Calculating transit times

[Question]

Starting from page 68 in strat ops, the table on page 69 is confusing.

The table states that the time needed to travel 5k km in atmospheric row 1 is 1.6 hours, and page 68 says that assumes 3 hexes per turn (or velocity 3). I'm not sure if this includes spheroid craft or just aerodyne.

However 3 hexes a turn is 1,440 meters per turn or per 60 seconds (3 hexes * 16 hexes on the ground map * 30 meters per hex). That equals to 57.9 minutes. There is a time difference between ground row and atmospheric row 1, but the page does not appear to explain the calculations that were used to arrive at that time difference.

Page 70 has a 1G transit time, but how fast is 1G?

The numbers also seem to conflict with the numbers given in TW page 80 where it states that ground level to the atmospheric interface is 0-107 KM. If those numbers are true, and assuming a craft is travelling at the maximum velocity allowed, it should take under an hour for a dropship to land after entering the interface layer. This seems suspiciously low...

[Akjosch]

Page 70 has a 1G transit time, but how fast is 1G?

"1G" is not speed (or velocity), it's acceleration - specifically, about 9.807 m/s² (alternative forms: 127,100 km/h² or 79,000 mi/h²). It's the standard transit acceleration for interplanetary travel, simply because it's the standard force of gravity on Terra's surface.

The further you travel at 1g, the faster you are in the middle of it, where you need to turn and start breaking. For any sort of non-trivial distance, that means your travel times don't scale linearly with distance, as they would with constant speed.

The formulas are explained in more detail on page 259 of SO. If you're interested in the maximum speed (the one at the midpoint of the transit), that's just the time spent so far - half of the total travel time - times the acceleration. So if you're accelerating for 10 days (=240 hour) straight, your speed is ...

240 h * 127,100 km/h² = 30,504,000 km/h

... or ...

240 h * 79,000 mi/h² = 18,960,000 mi/h

... until you hit a significant portion of light speed, but that's highly unlikely given the typical fuel tank size in the BattleTech universe.

[Question]

The thing is, if i have a 3/5 dropship, howdo i know when i have reached 1G and should stop accelerating?

[noisenerd]

Think of taking off in a car: you only get the sense of being pushed back in your seat while the car is accelerating, it doesn't matter if you're doing 25mph or 125mph when you stop accelerating, the G-force that pushes you back stops too. Same thing here.

Where the book gives the transit time for traveling at 1G, it's talking about traveling under 1G of acceleration the whole time, so the ship would never stop accelerating on that trip. Like Akjosch said, you will have to accelerate in the other direction in order to stop at the destination, but that's still acceleration.

[idea weenie]

The thing is, if i have a 3/5 dropship, howdo i know when i have reached 1G and should stop accelerating?

At a thrust of 2, you have reached 1 G of acceleration.  Acceleration and velocity are two different things.  Velocity is a measure of how far you have moved over an amount of time.  Acceleration is a measure of how much the rate of movement changes over time.

For example, you have two cars.  One has a top speed of 100 mph, the other has a top speed of 60 mph.  The first car takes 20 seconds to reach its top speed, for an acceleration of 5 mph/s.  However, the second car has a higher acceleration, and can reach its top speed in 3 seconds, for an acceleration of 20 mph/s.

The fun part is that in space, there is no top speed, just how efficient your engine is, and how much fuel you have on board (and oxygen and food).

A burn-day means the ship spent 1 day accelerating at 1G.  Assuming a 10 day transit trip from the planet to the Jump point (zenith or nadir), that means the Dropship spent 5 days thrusting at 1 G towards the waiting Jumpship, spent a few minutes rotating 180 (so its transit drive is nw pointed at the Jumpship), and spending the next 5 days thrusting at 1G to come to a stop relative to the Jumpship.  Really good pilots and navigators can bring themselves to zero-zero relative to the Jumpship in only a few space hexes.  Really bad navigators tell the Captain that when they said zenith, they meant nadir.

[theagent]

Ok, there might be a scale issue for the travel times in-atmosphere.  In Atmophere Row 1, it takes 18.5 minutes (1,110 seconds) to travel 1,000 km, for a speed of about 3,243 km/h; that matches up to Mach 3 for higher altitudes (1,081 km/h calculated; the X-43 flew at an altitude of 29km, where Mach 1 was 1,083 km/h).  That works out to about 300 m/sec.

So, if 1 hex per turn is Mach 1, then the scale per hex depends on the length of the turn:
 -- at 10 seconds per turn, each hex would be 3,000 m (300 x 10).  However, the lower atmospheric hexes are only 500m each (equivalent to 1 map sheet).
 -- at 1 minute per turn (same as space combat rounds), each hex would be 18,000 m (300 x 60), or 18 km. This is the actual size of space hexes.

I think the issue is that it isn't clear that the high-altitude map uses the space scale.