Reentry and landing.

[Grav_Moped]

Fuel isn't an issue for the canonical Air/Raft.*

[An Air/Raft's] Range in time or distance on a world is effectively unlimited, requiring refueling from a ship's power plant every ten weeks or so.

The constraint on the top speed of an Air/Raft is aerodynamics. At higher altitudes with thinner (or no) air, there is little to stop it from accelerating to any arbitrary speed other than the potential to strike an obstacle.

Working backward from Striker "Design Sequence Tables" p. 5, an Air/Raft with its 120KPH top speed has 1.1G acceleration, of which 1G is committed to lift. So it accelerates upward at 0.1G, or ~1m/sec^2 until reaching its drag-limited velocity for its current altitude (which, as it's basically a 3m x 6m flat plate when ascending vertically, isn't going to be terribly high -- say, 30kph at low altitude). This drag-limited velocity will increase with altitude as atmospheric density decreases, becoming effectively unlimited as atmospheric density approaches vacuum.

At about 20km altitude, air density is about 0.1 of that at ground level; at 35km, it's 0.01, at 50km, it's 0.001...

So it basically takes one hour to get high enough that the drag-limited velocity is 300kph, and another 30 minutes to where it's 3000kph. It looks like drag is the limiting factor for ascent speed up to about 30km altitude, then it's basically constrained by the 0.1G acceleration available after accounting for lift. Getting up isn't going to be what takes the time, reaching orbital velocity will be. But even 1m/sec^2 adds up fast if there's no atmosphere to create drag... I don't see the problem.

I'm just eyeballing graphs here; someone with current math skills please let me know what the real numbers look like.

On the other hand, coming back down requires decelerating to a very low horizontal velocity before starting the descent, as you start with an orbital velocity of tens of thousands of kph and there's no mention of reentry shielding on an Air/Raft. That said, the idea of a "hot" aero-braked reentry in a heat-shielded Air/Raft has a certain visual and dramatic appeal...




*Which means you could use it to drive a perpetual motion machine. But then, that's true of all Traveller drives so we'll just pretend nobody notices.

 

[Ulsyus]

Second, you’re applying the rocket equation, a mathematical constant of how hard it is to get X amount of payload into Y orbit and the energy needed to fight gravity on the way up. That’s all well and good - just not with a vehicle whose entire premise is flying by modifying gravity.

All current (contemporary) planning has to be done around the concept of lifting a certain weight out of the Earth's gravity well. The mass of the rocket plus payload gives it a 1:1 weight at the bottom of the gravity well.

What happens when there's no weight to the vehicle as a result of it's CG engine? How does that change the energy required to achieve escape velocity? Wouldn't that change also work in reverse when determining the picture of what a Free Trader would look like when entering an atmosphere in order to land at a downport?

 

[Straybow]

With respects to air rafts making it to orbit, I simply do not allow it. According to Maxwell Hunter in Thrust Into Space, it takes about 4 kilowatts to put a pound of mass into orbit. So, to put an air raft at 4 tons into orbit would take 4 kilowatts times 8820 pounds/4000 kilograms means that it will take 35.28 Megawatts used with 100% efficiency to put the air raft into orbit. To do that in 8 hours means your power plant is putting out a minimum of 4.41 Megawatts per hour. A kilowatt is equivalent to 1.34 horsepower. A Megawatt is equivalent to 1,340 horsepower. Multiply that by 4.41 Megawatts gets you a minimum power output of the air raft power plant of 5,909.4 horsepower per hour. The air raft has a top speed of 100 kilometers per hour in a standard atmosphere. A World War 2 B-17 had a top speed of around 300 miles per hour/482 kilometers per hour on an output of 4,800 horsepower per hour. Compare the B-17 size and drag to the air raft.

Now, tell me again how an air raft with a top speed of 100 kilometers per hour in the atmosphere can get into orbit.

Well, it's about 32 MJ/kilo at 100% efficiency. Of that only about 1.6 MJ is required to raise 1 kg to 160 km under 10m/s² gravity (from good old m·a·d). If you're doing that over 8 hours, your vertical velocity is a pokey 20 km/h, and you don't really need to account for drag. You're using 0.11 MW for a 2 ton vehicle, befitting the anemic 100 km/h top speed.



Getting up to orbit is cheap and easy, albeit slow. You have antigrav to keep you there, so it is also safe. You then need to accelerate the vehicle to 7.8 km/s to match speed with anything in ballistic orbit, which takes the remaining 30.4 MJ/kg. If your vehicle is 2 mT with a 0.11 MW drive, that will take about 160 hours. It is far quicker for the ship to match your speed (about 13 minutes at 1 G) to take you aboard.

 

[mike wightman]

An air raft is supposed to be able to be made at Tech Level 8, and is supposed to be able to carry 4 tons of cargo. To carry 4 tons of cargo, which I assume is actually mass, not volume, you need a reasonably substantial vehicle. I will stick with the 4 tons of mass rate, and simply assume that it might be carrying something.

TL8 in CT includes man portable laser carbines, fusion power plants and the air/raft. Despite what is written by lots of people in CT terms our real world TL has yet to hit TL8, I usually give a TL of 7.7 when people ask in my games.

The energy requirement has absolutely nothing to do with the rocket equation, but simply indicates how much energy is required to put one pound of mass in orbit around Earth. That energy can come from a black powder rocket, Space Shuttle launch, Cavorite, Dean Drive, or the air raft lift and drive propulsion mechanism, but it has to be supplied. That does assume 100 per cent efficiency, so the actual energy required is going to be considerably more.

You haven't mentioned energy in your post, you used power which is where the confusion lies and why the maths is wrong.
In terms of energy in order to lift a 1kg mass to a height of 200km against earth's gravity you will need to supply 1x200000x9.8=1.96MJ. It will then immediately fall back to earth unless you also give it orbital velocity which is approximately 7.5km/s. This would require kinetic energy so 0.5x1x7500x7500=28MJ.

So you have to supply 30MJ of energy per kg. 4ton air/raft = 120GJ

Now the air/raft takes 8 hours so we can calculate the rate at which energy is supplied or power 120000000000/8x60x60=4300000J/s or 4.3MW. So through the magic of grav technology the air/raft can convert 4.3MW

How much energy does manipulating gravity take? I do not know, but I do know that it will take 550 foot-pounds of energy to lift one pound to 550 feet in one second in Earth's gravity. That is what one horsepower can do. As for ignoring the immediate conditions, the following comes from LBB Book 3, page 23, 1981 edition, my emphasis.

Traveller does things in metric to make the maths easier.

An air raft has a cruise speed of 100 kilometers per hour in a standard atmosphere, with a top speed of maybe 120 kilometers per hour. That gives it range of between 62 and 75 miles per hour, or about the speed on an Interstate Highway. That is it. No faster. That gives it a power plant capable of forward motion comparable to the average car, so around 60 horsepower, or about 45 kilowatts. Then it has to add the energy cost of countering gravity based on the weight of the raft, passengers, and cargo. Now, you are telling me that this air raft, which cannot keep up with the average traffic speed on a U.S. Interstate, can generate enough forward motion to go into a stable orbit around Earth, requiring a velocity, however delivered, of around 26,000 feet per second. I fear that my "willing suspension of disbelief" becomes very unwilling at this point.

A 45kW power plant is nowhere near powerful enough, you need it to be one hundred times more powerful

Not the above is just approximate, you would have to take into account energy lost due to air resistance, efficiency of the power plant and grav modules etc.

 

[mike wightman]

What happens when there's no weight to the vehicle as a result of it's CG engine? How does that change the energy required to achieve escape velocity? Wouldn't that change also work in reverse when determining the picture of what a Free Trader would look like when entering an atmosphere in order to land at a downport?

It still has mass which is what is used in the potential energy and kinetic energy equations.

Now if grav modules reduce mass...

 

[Straybow]

In terms of energy in order to lift a 1kg mass to a height of 200km against earth's gravity you will need to supply 1x200000x9.8=1.96MJ. It will then immediately fall back to earth unless you also give it orbital velocity which is approximately 7.5km/s. This would require kinetic energy so 0.5x1x7500x7500=28MJ.

Hah, beat you to it!
You neglected that the grav drive will keep you at orbital altitude if it can push you up there. You only need to add velocity to match ballistic orbit. You'll still have whatever angular momentum you started with, maybe around 300-500 m/s depending on lattitude for an Earth-like planet (size and rotational period).

 

[kilemall]

This brings up a whole point for me re: grav drives that makes me itchy.


Fine on anti-gravity being a neutralization AND a push against fields.


Fine on gravitic manipulation being able to create artificial gravity.


What I don't get is the getting velocity to go towards a planet as though you have thrust towards it, outside the gravity of the planet itself.


Seems like in orbit you have to use your thrust part for lateral push against velocity so you would naturally deorbit, but that's not the same as pushing the engines 'down'.


Also seems to me that if there is no 'pull' option like the 'thrust/push', that your gravtank isn't anywhere as maneuverable as a conventionally thrusting rocket or jet in that your down speed is dependent on planetary gravity.


And if you DO have an 'attract towards gravity field' option for lifters/M-drives, why aren't there tractor beams?

 

[Adam Dray]

Yeah, Main Sequence has antigrav (at least, for now) but it's partial or full gravity negation, with no "push" or "pull" effect. You still rely on thrusters for translation, but you need only enough to overcome wind resistance, if you're not in a hurry to reach orbit.

To land, you use partial antigrav to negate most of a planet's natural pull, and then use thrusters to control descent velocity.

Why antigrav is low-energy, whereas it ought to require all the mass-energy of the entire planet you're drifting toward, is handwavium.

 

[shaunhilburn]

Given that gas giant atmospheric refueling is a controlled re-entry, I'm curious about how starships enter ordinary rocky world atmospheres. I'm thinking it's ordinary powered flight without all the reentry theatrics and special effects, but I've never seen it addressed in the rules.

Grav vehicles are capable of slowly reaching orbit and making slow and gentle descents back to the surface. Starships are equipped with even more capable "maneuver" drives. Even unstreamlined starships should be capable of landing in any atmosphere. It's not canon, but the game designers weren't propulsion engineers and didn;t contemplete such things.

The days of hypersonic aerobraking re-entry are from the early days of spaceflight; TL 6 to mid-TL 8.

 

[Grav_Moped]

The constraint on the top speed of an Air/Raft is aerodynamics. At higher altitudes with thinner (or no) air, there is little to stop it from accelerating to any arbitrary speed other than the potential to strike an obstacle.

Working backward from Striker "Design Sequence Tables" p. 5, an Air/Raft with its 120KPH top speed has 1.1G acceleration, of which 1G is committed to lift. So it accelerates upward at 0.1G, or ~1m/sec^2 until reaching its drag-limited velocity for its current altitude (which, as it's basically a 3m x 6m flat plate when ascending vertically, isn't going to be terribly high -- say, 30kph [ETA: =500m/sec] at low altitude). This drag-limited velocity will increase with altitude as atmospheric density decreases, becoming effectively unlimited as atmospheric density approaches vacuum.

At about 20km altitude, air density is about 0.1 of that at ground level; at 35km, it's 0.01, at 50km, it's 0.001...

Did some math. For an Air/Raft accelerating vertically at 1m/sec^2, air density (and thus drag) falls off faster than its ability to accelerate. Therefore, aerodynamic drag is NOT a limiting factor going up, so the time to orbital altitude (but not velocity) is T=2 * [sqrt(D/A)] (as seen in LBB2). This formula works because the Air/Raft needs to have a zero vertical vector at orbital altitude (if the orbit is to be circular).

The flight path is such that halfway to orbital altitude, it starts accelerating laterally at 0.2G while committing only 0.9G to lift, leaving 0.1G uncompensated to kill the vertical component of its vector. Once that's neutralized (at orbital altitude), it then continues to accelerate laterally at 0.1G (with 1G lift), gradually increasing the lateral component of its thrust as it approaches orbital velocity (since at orbital velocity and altitude, you're in orbit by definition and don't need vertical thrust).

 

[Blue Ghost]

Grav vehicles are capable of slowly reaching orbit and making slow and gentle descents back to the surface. Starships are equipped with even more capable "maneuver" drives. Even unstreamlined starships should be capable of landing in any atmosphere. It's not canon, but the game designers weren't propulsion engineers and didn;t contemplete such things.

The days of hypersonic aerobraking re-entry are from the early days of spaceflight; TL 6 to mid-TL 8.

Yeah, it is implicit, but I've never seen the rules explicitly say "this is how it's done". I was just curious.

 

[Straybow]

[FONT=arial,helvetica]Classic Type A render-

{Reasonably sized picture here}

Classic Type R render-

{Giant format-blowing picture here}[/FONT]

Regarding the scale of the ship, passenger volume for a 747-400 is 876 m³ and cargo is 159 m³, plus about 30 m³ for fuel, totaling 1065 m³. That converts to 76 dT. The volume includes the inner part of the wings holding fuel tanks but ignores the rest of the flight surfaces. The 400 dT subbie needs to be five times as big.



Here's a picture of a Lufthansa 747 (the first one, if I get the gist of the page) with a crowd of people. Its web address implies it to be a 747-8, but it also says the date was 1968, which would have to be a 100 or 200 model demonstrator since first delivery was 1969. The short upper deck would also indicate 100 or 200, rather than 300, 400, or 8. The cars also support a very early date.

 

[Grav_Moped]

Now, to do the rest of the math for the Air/Raft. We'll assume Earth (Size 8) as a reference.

Low Earth Orbit (LEO) is about 2000 km (2,000,000 m) altitude. Orbital velocity at that altitude is about 28,000 km/h.

Time T to altitude (in seconds) is
T=2 * (sqrt(2,000,000/0.1))
T= 8944 seconds, or 149 minutes (about 2.5 hours) to the altitude of LEO (2000 km) with no lateral velocity.

Time to 1000 km altitude (the point at which the antigravity is dialed back to 0.9G to start killing the upward velocity component) is 4472 seconds (74 minutes).
After that point, lateral acceleration at 0.1G commences. (I'm going with a pessimistic assumption that an Air/Raft cannot apply its lifting force of 1G in any direction other than vertically, so it has only the 0.1G available for lateral acceleration).

Time T in seconds, Acceleration A in m/sec^2, Velocity V in m/sec
V=AT^2
V=1*T^2
Desired V=28,000,000 m/sec (orbital velocity)
28,000,000=1*T^2
28,000,000=T^2
sqrt(28,000,000) =T
T=5300 (approximately)
5300 seconds is about 90 minutes.

So, with lateral acceleration starting at 74 minutes after takeoff, LEO altitude is reached in 2.5 hours and orbital velocity at that altitude achieved 15 minutes later.

Descent reverses this process: Decelerate for 15 minutes at 0.1G with no gravity compensation (de-orbit), then decelerate at 0.1G for 75 minutes with 0.9G gravity compensation (accelerating descent), then descend vertically with 1.1G compensation to landing (retarded descent). No aerobraking involved. With a heat shield and other mods for coming in hot, the descent time could be cut in half -- it'd look like an Apollo capsule re-entry, and subject the vehicle to about 3G for a brief period.

Is this a fair approximation, and is my math correct?
This only describes the time needed to put an Air/Raft into its own orbit. Additional time (possibly a lot of time) will be needed if attempting to rendezvous with a craft or station in a specific orbit that does not pass directly over the Air/Raft's takeoff site, or whose orbital position will be ahead of or behind the point at which the Air/Raft achieves steady-state orbit, or both.

 

[mike wightman]

The flight path is such that halfway to orbital altitude, it starts accelerating laterally at 0.2G while committing only 0.9G to lift, leaving 0.1G uncompensated to kill the vertical component of its vector. Once that's neutralized (at orbital altitude), it then continues to accelerate laterally at 0.1G (with 1G lift), gradually increasing the lateral component of its thrust as it approaches orbital velocity (since at orbital velocity and altitude, you're in orbit by definition and don't need vertical thrust).

You are describing an air/raft at an orbital altitude that is gravitationally 'buoyant' but it is not in orbit. It is not in orbit until it has built up 7-8km/s orbital velocity, ie it can switch off its grav modules and remain in orbit. If it does not have this velocity then it will fall back to the planet when it switches its grav modules off.

LEO is between 200 and 2000km.

 

[kilemall]

Neat, take it up with the render artist. My objective was to clarify the classic rendering of the form factors in question between his description of what sounded like a Type R vs. a Type A.



The specific artist' mistake could be understandable given that the classic William Keith renditions had similar relative dimensions.

Caswell's version is a little longer-

One thing to keep in mind is the 747 cabin is 6 meters wide, the S7 Type R is 15 meters wide and looks like 12 meters on the floor plan, so it is considerably wider then the plane.



Dividing up the floor space, looks like the ceiling of the cargo area shoud at least be 4 to 4.5 meters- I'd prefer the latter just for making it another multiple of 1.5.



The small craft is 20 tons and so should be 5% of the volume of the Type R. By the Carswell rendition looks to me like the Type R should stretch one launch longer, the Keith and one I posted version should go two launches longer minimum.


The air/raft in the picture of course is crazy big if the launch is 20 tons, or everything about the ship is miniaturized and no perspective can explain it.


Whew, don't like those 3.85 ton containers. Pretty whacky sizing, go 4 dtons, 5/10 dtons, or go home.


Did come up with an interesting idea while looking at those pics- the 5-dton containers I specced before should be same length or half height of the 10-dton containers, so they take up the same floor space and each 5-dton is just bolted onto the top of another 5-dton. That would then allow high ceiling ships to have a third half-layer on top of the other containers, and bolt-ons for 1-dton containers.

 

[Blue Ghost]

Strangely enough I started a story with a Type-R. I think Ian Stead has the bridge windows set further back, and inset into the superstructure.

 

[Timerover51]

Now, to do the rest of the math for the Air/Raft. We'll assume Earth (Size 8) as a reference.

Low Earth Orbit (LEO) is about 2000 km (2,000,000 m) altitude. Orbital velocity at that altitude is about 28,000 km/h.

Time T to altitude (in seconds) is
T=2 * (sqrt(2,000,000/0.1))
T= 8944 seconds, or 149 minutes (about 2.5 hours) to the altitude of LEO (2000 km) with no lateral velocity.

Time to 1000 km altitude (the point at which the antigravity is dialed back to 0.9G to start killing the upward velocity component) is 4472 seconds (74 minutes).
After that point, lateral acceleration at 0.1G commences. (I'm going with a pessimistic assumption that an Air/Raft cannot apply its lifting force of 1G in any direction other than vertically, so it has only the 0.1G available for lateral acceleration).

Time T in seconds, Acceleration A in m/sec^2, Velocity V in m/sec
V=AT^2
V=1*T^2
Desired V=28,000,000 m/sec (orbital velocity)
28,000,000=1*T^2
28,000,000=T^2
sqrt(28,000,000) =T
T=5300 (approximately)
5300 seconds is about 90 minutes.

So, with lateral acceleration starting at 74 minutes after takeoff, LEO altitude is reached in 2.5 hours and orbital velocity at that altitude achieved 15 minutes later.

Descent reverses this process: Decelerate for 15 minutes at 0.1G with no gravity compensation (de-orbit), then decelerate at 0.1G for 75 minutes with 0.9G gravity compensation (accelerating descent), then descend vertically with 1.1G compensation to landing (retarded descent). No aerobraking involved. With a heat shield and other mods for coming in hot, the descent time could be cut in half -- it'd look like an Apollo capsule re-entry, and subject the vehicle to about 3G for a brief period.

Is this a fair approximation, and is my math correct?
This only describes the time needed to put an Air/Raft into its own orbit. Additional time (possibly a lot of time) will be needed if attempting to rendezvous with a craft or station in a specific orbit that does not pass directly over the Air/Raft's takeoff site, or whose orbital position will be ahead of or behind the point at which the Air/Raft achieves steady-state orbit, or both.

I am not sure if you actually believe that or are just looking for an argument. As I stated previously, there is a point where my "willing suspension of disbelief" ends, and the air raft getting into orbit is well beyond that point. It is your version of the Traveller Universe. It is not mine. I am out of the discussion.

 

[whartung]

I am not sure if you actually believe that or are just looking for an argument. As I stated previously, there is a point where my "willing suspension of disbelief" ends, and the air raft getting into orbit is well beyond that point. It is your version of the Traveller Universe. It is not mine. I am out of the discussion.

It's simply a problem of thrust. The reasons that aircraft can't reach orbit is simply that they run out of air, lose lift, and lose power (with their air breathing engines).

The question is what constrains an air raft?

Do the grav modules lose oomph as the acceleration due to gravity falls away? Do the air rafts have an air breathing motor for thrust? Do the grav modules lose the capability of maintaining buoyancy at the higher altitudes? (And, boy, thats curious since there's less gravity to repel against, arguably the higher you go the easier it gets.)

So, the question is what keeps you air rafts from attaining orbit (beside that they're open topped, and the beverages in the cup holders tend to boil off as you climb).

 

[JimMarn]

Planet gravity goes up a long distance. I don't remember the exact distance Earth's and Luna's gravity balances, but I think its about two-thirds the way to the moon. After that, Luna's gravity takes charge.

And aren't there air rafts that are enclosed/air tight ?

 

[Xerxeskingofking]

Planet gravity goes up a long distance. I don't remember the exact distance Earth's and Luna's gravity balances, but I think its about two-thirds the way to the moon. After that, Luna's gravity takes charge.

And aren't there air rafts that are enclosed/air tight ?

Wiki places the earth-moon L1 point (the Lagrange point were the gravity of earth and moon are equal, that lies between the two bodies), as about 84% of the moons orbit.


at 300-400Km (roughly the height the ISS orbits, so a viable low earth orbit), the pull of earths gravity is about 0.9g, so you'd still need most of your thrust aimed downward to "hover" at this altitude.