LBB 2 Drives in Small Craft? (and LBB2 PP Fuel Rules)

[AnotherDilbert]

The power equation is P = k·a²

That has nothing to do with acceleration and everything to do with top speed.

Note that what you measure above is not acceleration, but the time it takes to travel 400 m, which is a velocity.

The above formula is just a way of writing E_k = mv², kinetic energy is proportional to the square of the velocity.

If you plot power versus acceleration times 0-100 km/h it will fall apart completely.



But it hints at that it takes more energy, and hence more power, to accelerate to higher velocities, whether using 1 G or 6 G. This is something Traveller studiously ignores, presumably for simplicity.

Example: For a 1000 kg car to accelerate from 0 to 100 m/s takes E_k = mv² = 1000 kg × (100 m/s)² = 10000000 J = 10 MJ. To accelerate from 100 m/s to 200 m/s takes E_k = 1000 kg × ( (200 m/s)² - 100 m/s)² ) = 1000 × ( 40000 - 10000 ) = 30000000 J = 30 MJ.

To do both in, say, 10 s each it would take a power of 10 MJ / 10 s = 1 MW ≈ 1360 hp for the first 100 m/s and 30 MJ / 10 s = 3 MW ≈ 4079 hp for the second. Note that this is with a constant acceleration (100 m/s in 10 s => 100 m/s / 10 s = 10 m/s² ≈ 1 G).


I'm not familiar with drag racing, but I think you will find that a car accelerates far faster in the first hundred meters than in the last hundred meters, due both wind resistance and the need for more power to accelerate at the higher speed. As expected from the above example.

 

[mike wightman]

Your kinetic energy formula is wrong...

 

[AnotherDilbert]

Your kinetic energy formula is wrong...

Ha, yes, obviously. Let's just say it is too early in the morning...

It should be E_k = mv²/2.


Changes the numbers, but not the argument...

 

[Straybow]

When you look at a drag race accelerometer there is indeed a drop-off at higher speeds, but air resistance is far smaller than the power required to accelerate the mass of the car. For example, my car may need only 50 HP to cruise at 60 mph, but use over 200 HP to accelerate 0-60 in about 11 seconds (half a G). There is also loss due to the energy in the rotation of the tires, especially with tire spin. This is the inefficiency of converting torque to linear force. This loss can be larger than the air drag at high speed, and almost all the power loss at low speed.

At a constant power level, acceleration will vary inversely with tire rotational rate in each gear. That means the acceleration drops as speed increases, and then jumps up again after a gear shift.

Nonetheless, the data matches P = k·a² very closely. 350 HP/ton at 1 G (~9 s), 1400 HP/ton at 2 G (~6½ s), with top fuel dragsters reaching 11k HP in a vehicle barely over a ton for ~5 G performance (a bit under 4 s).

From the article "Tech—Why the G-Meter is the Most Important Sensor for Going Fast"

Spoiler:

TLDR: The green line is accelerometer, the red line is drive shaft, and the white line is engine rpm.

F = m·a
E = m·a·d
P = m·a·d/t = m·a·v

But what is v? With a rocket the controlling v isn't the rocket's velocity, but the exhaust velocity of the reaction mass. But we don't have reaction mass, or exhaust velocity. What velocity do we use?

We can also rewrite the above as P = m·a·(a·t). Then we go to the thought experiment where the thrust isn't constant but occurs in short spurts. Between each spurt of acceleration there is an instantaneous inertial reference frame without acceleration, so that we don't accumulate velocity from an arbitrary external reference frame. We can use a 1 second increment, or sum up a number of tiny increments into a 1 second total to determine P = m·a·a·(1 s) for our instantaneous power measurement.

We have no idea what the mathematical and engineering constraints of gravity manipulation drive would be, but we can assign a number to k. I used 6.5, and I also throw in a space-and-mass factor via a 5/4 power applied to register tonnage over 1000 (actually the formula uses MW and kT, and applies the 5/4 power rule over 1.0 kT). I explain this in part as an overall efficiency of 0.40 in converting input electrical power to output grav power, and the rest being the effectiveness of the grav drive "grip" on the fabric of space. For deck gravity and i-comp, use 3.85 assuming plates above and below in push-pull giving a better "grip."

 

[Straybow]

I misstated a bit. For G = 10 m/s² that 350 HP/ton would be 260 kW/ton, for a k factor of 2.6, and dividing by 0.4 I get 6.5, so I suppose the 0.4 efficiency is the combined input-output and grav drive volume-and-space interface "grip."

 

[whartung]

I misstated a bit. For G = 10 m/s² that 350 HP/ton would be 260 kW/ton, for a k factor of 2.6, and dividing by 0.4 I get 6.5, so I suppose the 0.4 efficiency is the combined input-output and grav drive volume-and-space interface "grip."

My physic-fu fell on the floor and the cat knocked it under the couch.

What is "k" here?

What does it represent?

 

[AnotherDilbert]

F = m·a

Yes, momentaneously.

E = m·a·d

Yes, but integrated over time, thereby baking in a velocity. Note that acceleration has gone from momentaneous acceleration to an average acceleration over time.

Basically, to cover the distance faster, you have to accelerate faster, achieving a higher top speed faster, so the power requirement goes up both acceleration and velocity.

It we instead take acceleration to a specified speed, instead of to a specified distance, the connection is different. For simplicity we assume constant acceleration, starting at 0 velocity and 0 distance.
E_k = mv²/2 (thanks Mike)
P = E_k/t = mv²/2t
Here v = a_avt so a_av = v/t where v is constant and t the variable. So
P = mv²/2t = mv/2 × a_av where mv/2 is a constant, so
P = k_somea_av or a_av = P/k_some.
Note that acceleration is not constant, but varies over time, hence a_av is average acceleration.


What we have found is that the connection between power and acceleration is dependant on the assumptions we start with.


In the generalised case:
E_k = mv²/2.
Differentiate over t to get the momentaneous power:
P = dE_k/dt = dmv²/2dt = mv × dv/dt = mva.
Power requirement is proportional to both velocity and acceleration.

We can simplify this a bit by assuming boundary conditions, e.g. accelerate 400 m or accelerate to fixed speed, but it gives us different simplifications, as we have already found. I.e. your formula is an artefact of your chosen assumptions, in this case acceleration from full stop to 400 m distance.


In your graph above we can see that, after the initial spin and rev up, momentaneous acceleration decreases with speed, as expected from the formula:
P = m×av, where P (power) and m (mass) are approximately constant, so
P / m / v = a = P/mv or a = k_other/v, where k_other = P/m.

We have no idea what the mathematical and engineering constraints of gravity manipulation drive would be, ...

Except that it produces constant acceleration, using constant power, regardless of velocity. It might just be a simplification, but that is the data we have, the same in all editions I have looked at, from CT to T5.

Obviously it does not just convert power from the power plant into kinetic energy (velocity), rendering the formulæ derived above non-applicable.



P.S. Please use a IMGW=xxx tag to include the graph. It's a PITA when the text don't wrap to the browser window width.

 

[AnotherDilbert]

My physic-fu fell on the floor and the cat knocked it under the couch.

What is "k" here?

What does it represent?

Just a constant, some value that does not vary.

P = ka⁴ is the same as saying P is proportional to a⁴.


Since it does not vary we don't have to keep track of what exact number it is until we have derived our formula and want to calculate a specific value.

 

[The Oz]

I did this some years ago. It may help.
http://www.travellerrpg.com/CotI/Discuss/showthread.php?p=581991

 

[Grav_Moped]

I did this some years ago. It may help.
http://www.travellerrpg.com/CotI/Discuss/showthread.php?p=581991

Looks good. I like the small/large bridge concept. I'll need to run the numbers to see exactly how hard using HG breaks the bridge sizes for canon LBB2 small craft, but it wouldn't surprise me if it's a big issue.

Some of the life support issues (restroom facilities) were probably meant to be addressed with small craft cabins.

The oddly high fuel burn rates of LBB2 Drive Table drives (especially with respect to HG, but also LBB2 Second Edition small craft drives, which may have been mostly HG) make a lot more sense when tankage is reduced to hours/days endurance rather than weeks/months. Which is something that isn't allowed by rules as written for ships 100Td and up, but would be necessary to use the Drive Table Drives (and interpolations thereof) in small craft.

 

[Grav_Moped]

What if the Drive Table formulae can be projected downward, with "Small Craft Drives" being a combination of the maneuver drive and power plant, even if this means that small "drives" get shrunk a bit by the negative maneuver drive tonnage?

This corresponds to the First Edition concept that a power plant was only needed by the maneuver drive.

Small Craft Drives are then 2.5Td per 100 G-tons thrust and cost MCr6 per 100G-tons thrust. Minimum size is 1Td. (Not positive about this, maybe 2Td?) Reducing output below 40 G-tons thrust (1Td) (alt: 80 tons thrust/2Td) reduces cost and fuel consumption rates, but not size.

Small Craft Drives are self-powered. They have agility=Gs. They cannot accept external power to increase Gs, and they do not generate energy points. They do provide enough power to operate small craft components that do not require energy points (bridge, life support, computers up to Mod/2bis, missile launchers and sandcasters). Components requiring energy points must be supported by a separate power plant extrapolated from the LBB2 drive table with 1EP= a rating of 1 in a 100Td hull. This includes jump drives in any rule set later than LBB2:'77.

I'll work out fuel consumption later; it'll be scaled down to a per-day basis from the 20Td/month of the Size A drive in a 100Td hull, with the assumption that it can be reduced to 1/6 of that while at idle. Separate power plants have their own fuel consumption rates in addition to the drive fuel consumption, and can be powered down entirely when not needed to support components that use energy points.

Examples:

Spoiler:

100Tons (2G in 50Td)
2.5 tons
MCr 6

50Tons (1G in 50Td)
1.25 tons
MCr 3

40 tons (2G in 20Td)
1Td
MCr 2.4

20 tons (1G in 20Td)
1Td (would be 0.5Td)
MCr 1.2

This becomes inefficient in starship sizes (100Td and up).

 

[Straybow]

[snip lots of stuff based on conservation of energy]

I'm not using conservation of energy. A gravity slingshot in orbital mechanics uses conservation of angular momentum. We believe in conservation of energy, therefore we assume that the energy gained in the slingshot comes from the orbital energy of the planet, but we can't measure planetary velocities down to attometers per second to test that assumption.

We can simplify this a bit by assuming boundary conditions, e.g. accelerate 400 m or accelerate to fixed speed, but it gives us different simplifications, as we have already found. I.e. your formula is an artefact of your chosen assumptions, in this case acceleration from full stop to 400 m distance.

Or we can look at empirical data of HP/ton vs acceleration. QED.

In your graph above we can see that, after the initial spin and rev up, momentaneous acceleration decreases with speed, as expected from the formula:
P = m×av, where P (power) and m (mass) are approximately constant, so
P / m / v = a = P/mv or a = k_other/v, where k_other = P/m.

Largely explained as the need to translate the energy input through the use of a rolling wheel, where the rotation rate of the wheel effects the energy translated to linear velocity. Again, a conservation of energy calculation. We're saying that grav drive does not directly impart energy to the vehicle, but rather uses the gravitational energy of space and of nearby masses to impart energy to the vehicle.

P.S. Please use a IMGW=xxx tag to include the graph. It's a PITA when the text don't wrap to the browser window width.

Yeah, the way to do that dropped out of my head, and there is no longer a BB code cheat link in the edit window. I tried editing it, it didn't work. I spoilered it instead.

 

[AnotherDilbert]

I'm not using conservation of energy.

Doesn't matter if you use it, it works anyway...

A gravity slingshot in orbital mechanics uses conservation of angular momentum. We believe in conservation of energy, therefore we assume that the energy gained in the slingshot comes from the orbital energy of the planet, but we can't measure planetary velocities down to attometers per second to test that assumption.

Yes, we assume conservation of energy works, for a good reason. It is a very well tested theory, and so far no physicist have caught the universe cheating (much).

Or we can look at empirical data of HP/ton vs acceleration.

Yes, your formula presumably holds, on a drag strip but not in general.

Check what your formula becomes if you change the distance to say 1 km. The 400 m distance is baked into your proposed k above.

Check what your formula becomes if you look at data for 0-100 km/h acceleration times. I guess it becomes something like P = k_0-100a, as derived in my last post.

Largely explained as the need to translate the energy input through the use of a rolling wheel, where the rotation rate of the wheel effects the energy translated to linear velocity. Again, a conservation of energy calculation.

I don't understand what you mean. Are you saying the car looses grip as the wheels spins faster, or are you just saying it takes energy hence power to spin up the wheels?

As far as I know neither would explain almost halving the acceleration, as in your example graph.

Physics does not stop working just because you don't believe in it. It obviously takes more power to accelerate at higher speed, or with constant power you accelerate less at higher speed. Just as your graph shows.

We're saying that grav drive does not directly impart energy to the vehicle, but rather uses the gravitational energy of space and of nearby masses to impart energy to the vehicle.

Yes? So your proposed formula does not apply?

 

[Condottiere]

I thought we had a clean acceleration due to lack of friction and moving parts.

 

[Proneutron]

I thought we had a clean acceleration due to lack of friction and moving parts.

In space, using Trav M-Drives (thrust plate or grav) we do.

 

[kilemall]

Small Craft Drives are self-powered. They have agility=Gs. They cannot accept external power to increase Gs, and they do not generate energy points. They do provide enough power to operate small craft components that do not require energy points (bridge, life support, computers up to Mod/2bis, missile launchers and sandcasters). Components requiring energy points must be supported by a separate power plant extrapolated from the LBB2 drive table with 1EP= a rating of 1 in a 100Td hull. This includes jump drives in any rule set later than LBB2:'77.

I'll work out fuel consumption later; it'll be scaled down to a per-day basis from the 20Td/month of the Size A drive in a 100Td hull, with the assumption that it can be reduced to 1/6 of that while at idle. Separate power plants have their own fuel consumption rates in addition to the drive fuel consumption, and can be powered down entirely when not needed to support components that use energy points.

Problem is several of those canon CT small craft DO have a limited capacity to mount energy weapons, so there is clearly some power generation capability beyond 'just' M-drive and basics.

 

[AnotherDilbert]

Problem is several of those canon CT small craft DO have a limited capacity to mount energy weapons, so there is clearly some power generation capability beyond 'just' M-drive and basics.

Not beyond M-drive requirement, instead of M-drive.

LBB2'81 small craft have exactly as much power as required by LBB5'80, and can use it for either weapons, computers, or propulsion.


Example: A Pinnace has 5 G in a 40 Dt hull, so 5 × 40 / 100 = 2 EP, so can power two lasers (or one fusion gun). If it uses it's 2 EP for weapons, it has no power left for propulsion, so Agility = 0.


Since LBB2 does not care about power, you can use both the weapons and the M-drive at the same time in the LBB2 combat system.

 

[kilemall]

Good summation, just saying the LBB2 canon versions apparently have more juice or at least more efficiency to allow both maneuver and weapons, if one compares HG EP use as a baseline.



Speaking of which, a small craft power issue dodge I have messed around with before is using capacitors as ersatz short term batteries to charge up ahead of time either on the host system or during flight and then enable full power use at need.



You could have a PP1 power and then a couple tons of capacitor for high-G/combat, for instance.

 

[Grav_Moped]

Good summation, just saying the LBB2 canon versions apparently have more juice or at least more efficiency to allow both maneuver and weapons, if one compares HG EP use as a baseline.

That's true of everything in LBB2. What is this strange "energy point" thing of which you speak?

Speaking of which, a small craft power issue dodge I have messed around with before is using capacitors as ersatz short term batteries to charge up ahead of time either on the host system or during flight and then enable full power use at need.



You could have a PP1 power and then a couple tons of capacitor for high-G/combat, for instance.

Maybe ok for small craft, but problematic if this gets extended to ships.

 

[AnotherDilbert]

... just saying the LBB2 canon versions apparently have more juice or at least more efficiency to allow both maneuver and weapons, if one compares HG EP use as a baseline.

It's the same edition (CT) and we can use ships designed with either system in either combat system, so it's presumably not a difference in design.

Since both systems are equally valid, they must be different views of the same reality. Perhaps LBB2 is lower intensity combat with alternating manoeuvre and fire and LBB5 is all out combat. What LBB2 calls double fire is what LBB5 calls normal fire, or something like that...