Loaded vs. Unloaded M-ratings

[phavoc]

Does anyone recalculate their drive speeds based on the ships loadout, especially when the power/displacement ratio is big.

For example, a 10,000 ton tanker that has set aside 9,500 tons for fuel. To get 1G fully loaded the Mdrive is 100 tons and the PP is 150 tons. But if it's running empty then it's quite over-powered.

So would you treat it as capable of 6g's empty, or would you say it needed a 6G drive and power plant to do so?

In High Guard they have a 30ton tug with a 14G-capable drive mounted. But I can't find anything that talks about how they calculate how much it could shift by itself.

 

[GypsyComet]

It is all done by displacement, not mass.

In the smaller range, a 600 ton ship with Maneuver J has 3gs of thrust. Strap 100 tons of tanks on the outside and it is now a 700 ton ship with 2gs of thrust, because that's what J gets you in a 700 ton ship.

Using High Guard percentage drives, you add the displacement of the tug (in your example) and the ship it is pushing together and recalc the performance of the drive based on the new volume.

 

[locarno24]

Because the drive's creating a gravity well, not directly applying a force of X newtons, your mass is supposedly irrelevant to the acceleration you get.

A 1 tonne object accelerates 'down' a gravity well at the exact same rate as a 1,000 tonne object of the same volume.

 

[Wil Mireu]

Because the drive's creating a gravity well, not directly applying a force of X newtons, your mass is supposedly irrelevant to the acceleration you get.

Which would be fine if that's actually what the drive is doing... but it isn't. No gravity wells are being created. It is applying a force of X newtons by pushing against the thruster plates.

We've been through this already here: http://forum.mongoosepublishing.com/viewtopic.php?p=763792#p763792

 

[Sigtrygg]

Which is pretty much why the reactionless drive thruster plate explanation is less than satisfactory.

It wasn't very good when DGP came up with it and it remains a pretty useless explanation for how reaction less thrusters tie in with canon.

The artificial gravity well is actually a better explanation, as is making the manoeuvre drive a variant of the Alcubierre warp drive.

http://www.extremetech.com/extreme/164326-nasa-discusses-its-warp-drive-research-prepares-to-create-a-warp-bubble-in-the-lab

 

[Wil Mireu]

I think it actually adds too many complications (actually raised by locarno). Such as - if there's a gravity well outside the ship, does it attract other stuff to it?

It's a lot easier to say that it's a self-contained thing that has no effect outside the ship, as described in TNE.

 

[phavoc]

I'm personally working from the assumption that if your drive rating can go down by changing the mass, then it would also go up by decreasing it.

But I see it more of an 'artsy' estimation, because I don't want to have to recalc the speed of a free trader running on empty. In the example I cited the mass was decreased by about 90%, so that's a massive change. Though I admit I don't have a cut-off assumption of when and where it should start.

From the responses it seems to be a solid 'maybe'...?

 

[spirochete]

It is all done by displacement, not mass.

In the smaller range, a 600 ton ship with Maneuver J has 3gs of thrust. Strap 100 tons of tanks on the outside and it is now a 700 ton ship with 2gs of thrust, because that's what J gets you in a 700 ton ship.

What if the payload is entirely internal? If acceleration is determined by displacement (non-newtonian), then I can load the cargo bay with 700 metric tons of lead and still get 3G acceleration. The ship occupies the same volume regardless of its internal cargo.

 

[phavoc]

It is all done by displacement, not mass.

In the smaller range, a 600 ton ship with Maneuver J has 3gs of thrust. Strap 100 tons of tanks on the outside and it is now a 700 ton ship with 2gs of thrust, because that's what J gets you in a 700 ton ship.

What if the payload is entirely internal? If acceleration is determined by displacement (non-newtonian), then I can load the cargo bay with 700 metric tons of lead and still get 3G acceleration.

Your example doesn't quite fit with GypsyComet's. Traveller doesn't take into account mass, just displacement. You could be carrying 500Dtons of Lead, or 500Dtons of aerogel and the performances are the same, using the rules. So internal/external would blend together.

 

[Sigtrygg]

Yup, you can carry 100tons (that's 1400m^3) of lead, feathers, hydrogen - it does't matter.

Only a couple of versions of Traveller over the years have worried about the actual mass of the ship in tonnes (1000 of kg) affecting drive performance.

In nearly every version the ship displacement in tons is a measure of volume only and only the volume counts when figuring drive performance.

In '77 CT it was implied in the trade rules that the maximum mass in a cargo displacement ton would be 1000kg. So 10 tonnes of lead would take up 10 tons of cargo hold. But this disappeared in later iterations of Traveller.

 

[Sigtrygg]

I'm personally working from the assumption that if your drive rating can go down by changing the mass, then it would also go up by decreasing it.

But I see it more of an 'artsy' estimation, because I don't want to have to recalc the speed of a free trader running on empty. In the example I cited the mass was decreased by about 90%, so that's a massive change. Though I admit I don't have a cut-off assumption of when and where it should start.

From the responses it seems to be a solid 'maybe'...?

You are confusing the displacement ton with the actual measure of mass - 1000kg or tonnes abbreviated to ton.

A 200ton free trader will have an actual mass closer to 2,000,000kg - 2000tonnes.

 

[spirochete]

I'm personally working from the assumption that if your drive rating can go down by changing the mass, then it would also go up by decreasing it.

Ditto. It' basic Newtonian mechanics. F = ma, therefore a = F/m.

This way, a type A free trader with 1G nominal acceleration can liftoff from a 1.2 G planet by reducing its load (payload AND fuel). Less burden = more acceleration.

 

[DickTurpin]

I'm personally working from the assumption that if your drive rating can go down by changing the mass, then it would also go up by decreasing it.

Ditto. It' basic Newtonian mechanics. F = ma, therefore a = F/m.

This way, a type A free trader with 1G nominal acceleration can liftoff from a 1.2 G planet by reducing its load (payload AND fuel). Less burden = more acceleration.

No, not in Traveller it doesn't. With the exception of character encumbrance (the most often ignored part of any RPG), mass does not exist in the Traveller universe. Even when using reaction thrusters, where mass would be critical in any sane physical universe, the mass of the ship does not figure into the equation at all.

Diving into the atmosphere of a gas giant to skim fuel or landing in any planet with more than 1G would be suicide in a thrust-1 ship if mass, and gravity, mattered. The idea is to transport characters to places and put them into situations that allow role playing, not to simulate the real world. Whether the game system uses anti-grav drives or a teleport spell, it is the destination, not the technical details of the journey that matters.

 

[phavoc]

You are confusing the displacement ton with the actual measure of mass - 1000kg or tonnes abbreviated to ton.

A 200ton free trader will have an actual mass closer to 2,000,000kg - 2000tonnes.

No, I wasn't. MG Traveller doesn't take mass into account, only displacement (DTons). GURPS Traveller did take mass AND displacement into account. From a gaming perspective it does make sense, as otherwise you'd have to calculate mass AND displacement for every item your ship carried to ensure it wasn't overloaded. That doesn't really add to the gaming pleasure from my perspective.

 

[dragoner]

That doesn't really add to the gaming pleasure from my perspective.

True, as the calculations would be structural as well to make sure the ship didn't tear itself apart during acceleration and maneuvering.

 

[Sigtrygg]

I'm personally working from the assumption that if your drive rating can go down by changing the mass, then it would also go up by decreasing it.

Ditto. It' basic Newtonian mechanics. F = ma, therefore a = F/m.

This way, a type A free trader with 1G nominal acceleration can liftoff from a 1.2 G planet by reducing its load (payload AND fuel). Less burden = more acceleration.

Except you are using magic reactionless thrusters that don't obey newtonian rules.

And you do not need an acceleration equal to or greater than the planet's to take off an eventually reach orbit.

 

[F33D]

Because the drive's creating a gravity well, not directly applying a force of X newtons, your mass is supposedly irrelevant to the acceleration you get.

A 1 tonne object accelerates 'down' a gravity well at the exact same rate as a 1,000 tonne object of the same volume.

Correct. MgT, unlike MT, uses Grav creation (Gravity Maneuver Drives) for deep space drive. MT (Mega Trav) didn't use Gravity drives for deep space and thus DID take load out into account somewhat. (see agility rating calculation for that game system)

 

[Wil Mireu]

Except you are using magic reactionless thrusters that don't obey newtonian rules.

Actually they do generally obey Newton's laws. The only part they don't obey is the bit about throwing something out of the back to move the ship forwards. But otherwise, F=ma.

And you do not need an acceleration equal to or greater than the planet's to take off an eventually reach orbit.

No, but it's harder. You have to slowly increase your altitude and make many orbits around the planet (which takes longer) instead of just going straight up or down.

 

[Sigtrygg]

Actually they do generally obey Newton's laws. The only part they don't obey is the bit about throwing something out of the back to move the ship forwards. But otherwise, F=ma.

Since they use hull displacement volume instead of mass they actually obey:

F = (hull displacement).a

Somehow the energy produced by the power plant is turned into an acceleration of the hull volume, mass is not a factor.

And being reactionless they disobey the third law too

The only one they obey is that objects maintain their velocity unless acted upon by a force.

 

[Wil Mireu]

Since they use hull displacement volume instead of mass they actually obey:

F = (hull displacement).a

Somehow the energy produced by the power plant is turned into an acceleration of the hull volume, mass is not a factor.

Sorry, but that's just wrong.

The drives moves the mass of the ship. The mass may not be explictly stated, but to claim that means it doesn't act on the mass is just silly. What you need to do is convert the volume into a mass.

A displacement ton is 14 m³ of volume. That volume contains stuff that has mass. If it's just 1dt of air at 20°C (density 1.2 kg/m³), it'd have a mass of about 16kg. If it's 1dt of steel (density of 8000 kg/m³), it'd have a mass of 112,000 kg (112 tons). So it's best to assume a 'bulk density' for the whole ship since we don't know what each cubic metre is made of.

Let's say it's the density of water (1000 kg/m³) for argument's sake, since it's not solid steel and there's a lot of air in there. So 1dt of ship would mass 14,000 kg (14 tons). The mass of a ship with a volume of 100dt (and carrying no cargo) would therefore be 1400 tons. That's how much mass (for an unloaded ship) that the drive would have to move, and that's what you put into the F=ma equation.