# *** SPOILER *** — Engineering – Space, Time, Orbits, Tools, Ships

Distance of Saka from Saloa KM:  2,250,000,000
Average speed to do it in 9 months KM/h: 347,222 = 2250000000 / (24*30*9)
Time to accelerate to 347,222 KM/h at 1g: 2h 45mins ( @ 1/2 g = 5h 30mins)
———————————————–

Average speed to do it in 1 month KM/h: 3,125,000 = 2250000000 / (24*30*1)
Time to accelerate to 3,125,000 KM/h at 1g: 24h  ( @ 1/2 g = 49h)
———————————————–

Calculates the time required and travel distance from the acceleration and velocity. https://keisan.casio.com/exec/system/1224829579

Kaiper Belt

Oort Cloud

The last two diagrams show the caravan in thrust and coast mode. In coast mode the thin lines are supposed to be wires holding the rotating sections together while they spin.

A bit of background:
The ’Torch’ is a Banksian ‘knife missile’ type object. They have very powerful AI to control the magnets etc to keep the fusion burn going, e.g. to control a flexing plasma… which means they just happen to be also be smart and chatty…

The way I work is to think of a narrative element and then do my best to come up with plausible science to deliver that.
I had read about laser ignited fusion, and magnetic confinement fusion, and I thought the incredible energy density of this (still very theoretical) hafnium isomer technology could deliver the spark to ignite fusion.

The torches already exist in my other books, so I pretty much need to stay with that. They are also sub components of larger fusion drives, providing the igniter. The general idea is that the fusion reaction generates electricity as well as direct thrust, the electricity is used to power superconducting magnets, which confines the fusion, which in turn generates more power, and therefore more confinement… A bit like the way a jet engine uses thrust to drive the compressor which generates more thrust…

At one point I have said that the torches on their own are able to accelerate at at least 8gs.

I have not given any performance envelope for the skiffs which use redundant torches to ignite a larger version of the same engine design. This time using the Torch as the igniter, just as the torch uses the hafnium gamma emission as its igniter.

To your question:
However let’s get the basic data first – how far and how fast?
Currently the plot calls for:
1) The large caravan ships with several hundred people pushed by several (4-10) of the skiff shuttles to fly in from the Kuiper belt, and then back out again. This would be a long journey, and about a year would be fine, but could be shorter. The acceleration phase could use additional smaller ships which would then return to their base. But the ships that accompanied the caravan would need to be sufficient to decelerate it and then re-accelerated for the journey back out. It would be okay to refuel in the in the solar system. The skiffs are also offensive and defensive assets as the solar system is not entirely friendly.
2) one of the dagger skiffs making the journey from the Kuiper belt to earth alone. I have this taking about nine months with two crew. I would like these smaller ships to be capable of high acceleration, for example 3-5gs, when necessary. Longer, more efficient burns would be used when getting up to high speeds. I imagine that the amount of fuel they can carry would be the limiting factor on long journeys… although they are not really designed for flying such long distances. For this type of mission they would usually be coupled with a larger vessel which stores fuel and provides more comfortable habitation.

Ideally, I would like to have a bunch of equations that I could plug numbers into to give me things like max acceleration, journey time, fuel burnt etc. I might make a JavaScript calculator on a page so that I can use it as a tool to quickly give me numbers so I don’t make any glaring mistakes.

Ship’s characteristics:
• Powerplant : fusion
• Size : 35 m (length) x ~ 17.5 m (width) x ~ 10m (height)
• Fuel : offboard
• Desired Performance : deep interplanetary (kuiper belt)
Power per reaction:
• Deuterium (22H) + Tritium (33H) = Helium (44He) + 3.5 MeV (source)
• Helium to Carbon (42+42→12624He+24He→612C) = 7.275 MeV (net; by way of Beryllium) (source)
• Helium + Carbon to Oxygen = 7.162 MeV
Events per volume of material:
Thinking that “100 years from now” technology could give you the best available 10−2110−21 reactions per cubic meter, per second.
Let’s say most of the ship (35 meters long, 17.5 meters wide, 10 meters high) is engine, and about nearly all of that is reaction vessel (which is a big assumption compared to today’s technology).
Reaction vessel is 35×17.5×1035×17.5×10 = 6,125 cubic meters Power generated is 3.5×10−21×6,1253.5MeV×10−21×6,125 = 2.14×10−162.14×10−16 MeV
That’s not gonna work.
Assuming “100 years from now” technology can also give you better reaction rates. Since we’re going bold, let’s say 25% of ˙m˙ is reacting.
So, each gram of 50-50 Deuterium/Tritium mix (assuming hydrogen fuel) (avg. molar mass 2.5) contains 12.5×6.02×102312.5×6.02×1023 reagents, 25% of which react, and produce 3.5 MeV on each reaction (and assuming the energy imparted the fast neutrons is unrecoverable). 2.107 ×1023×1023 MeV ≈≈ 3.3 ×1010×1010 Joules of energy (33 gigajoules) per second. Or 33 GW. Scaling up to kilograms of reactants would give 33 TW.
Let’s say the deductions of power to run accessories is negligible (but maybe it isn’t). What’s the ˙m˙ and ve out the back? 3.3×1013=12×1×2→6.6×1013=2→=3.3×1013=12×1×v2→6.6×1013=v2→ve= 8,124,038 meters per second (2.7% c)
˙m˙ is 1 kilogram per second. The rocket equation (excluding nozzle effect) is =˙F=m˙ve
Thrust, therefore, is 8,124,038 Newtons (8.1 MN) per kilogram of fuel mix. That’s in the same region as the 764 kN produced by space shuttle main engines.
What’s the peak amount of fuel that can flow through the ship?
Say the fuel is stored in liquid form, and the fuel line can be no bigger than the ships’ width (17.5 meters). And let’s say it’s circular. =2≈240A=πr2≈240 square meters. The volume is equal to that area times the rate in which fuel is being brought aboard by the pumps. The density of liquid hydrogen is 70 kg per cubic meter.
Let’s say 1 ms for now, giving ˙=70×240=m˙fuel=70×240= 16,800 kilograms per second.
Applying that to the thrust per engine : 16,800 ×× 8,124,038 Newtons ≈≈ 136 giganewtons.
You can go higher or lower on fuel flow rate. And select a different fuel. Or re-adjust down the huge “future technology” boost to reaction rate.

## Performance

The mass-energy efficiency of this set-up is the ˙2Eextractedm˙c2. You only get 33 TJ per kilogram. That’s a mass-energy efficiency of 3.6×10−73.6×10−7 or 0.000036%
This is important for evaluation long-range performance (like interplanetary travel). To accelerate your load to a cruise velocity of 0.01c requires a mass-energy of (0.01)2(0.01c)2, times the mass of your ship (or 0.0001 m in this scenario).
How much fuel mass your motor requires to confer this energy is taken by dividing the mass-energy requirement (0.0001 m) by your mass-energy efficiency (3.6 \times 10^{-7}) getting 277 kilograms of fuel mass required for every kg of payload (and does not include deceleration).
That’s clearly not going to work. So, let’s try a lower top speed: 0.001c ~ 2.7 kilograms per kg of payload.
Considering both acceleration and deceleration, then, for every kg of payload, you’ll need 2.7 / (2.7 + 1) = 72% of your mass to be fuel. And another 72% of the residual to be fuel for deceleration, giving 72% + (72% x 0.28) = 92% of your payload mass will need to be fuel.
Maybe an even slower top speed. Let’s try 0.0001c ~ 0.027 fuel/pay = … most of your cargo capacity can be payload, instead of fuel.
What’s that do to performance? 0.0001c is a cruise velocity of ~ 30,000 meters per second / 100,000 kilometers per hour. To travel a distance of 1 AU at this cruise speed (approx. 8 light-minutes) takes about 55 days.

## Thrust Performance

To push a 150 ton load (150,000 kg) with 4 ships each providing 8.1 MN of thrust up to a cruising velocity of 0.001c, you could get accelerations up to 216 2ms2 (or ~21 gees). Not sure you wouldn’t want to limit to only a few gees. At that acceleration, you’d reach cruise velocity of 300,000 ms in 23.1 minutes.
If 150 tons is not inclusive of fuel, you’ll need 1,725 tons of fuel. Total weight = 1,875 tons. Acceleration in this case would be 17.28 2ms2, and it would take 4 hours, 49 minutes (approx) to reach cruising speed.
Acceleration will actually vary over the course of the burn (as the weight being pushed starts to drop), being lower than the average value at the beginning of the maneuver and higher than the average at the end.

From Adam:

One thing to keep in mind, which might be of relevance, is the Minimum Jet-Power required for any mission. Basically the engine has to work 2/3 (1/3 acceleration, 1/3 coasting, 1/3 braking) of the total mission time to operate at the Minimum Power. That’s important because typically total power is proportional to the engine mass. If the engine has waste heat – and all engines do – then it needs to dispose of it too. Fusion drives run super hot – not just the plasma, but stuff like x-rays and neutrons. So minimum power is important. If your desired cruising speed is V, then the total mission delta-V is 3 times V – your maximum speed is 1.5 x V and of course twice that as you have to brake to stop. For a Minimum Power mission the required Specific Jet-Power is 4.8 x (S^2/T^3) W/kg – when S (displacement/distance) is in AU and T is in years. For a 1 year mission to 50 AU, that’s 12 kW/kg specific jet-power. If your vehicle masses 1,000 tons, then its jet-power is 12 GW. The fusion power needed will be significantly higher because of the inefficiencies in any rocket system. If it’s burning 3He+D (Helium 3 plus Deuterium) then it gets about 600 GJ per gram of fuel. Say the efficiency is 40 percent – so 240 GJ per gram fuel becomes jet-power . 1 gram supplies 20 seconds of jet-power.

Another useful conversion factor is that 1 AU in 1 year is 4.74 km/s. So 50 AU/year is 237 km/s, so the top-speed is 355.5 km/s and the delta-vee is 711 km/s. Max mass-ratio for minimum power is 4.42, so the exhaust velocity is (711/LN(4.42)) = 478.4 km/s. The acceleration is 0.0338 m/s^2 or 3.45 milligees. Short duration high-acceleration boosters might be needed if you want multi-gee maneuvers.

Hopefully that’ll give you some more ideas to chew on.

Adam

So lets say a cylinder shuttle is 120tons dry mass ans 120 tons hydrogen fuel.

ROCKETPUNK MANIFESTO: ON

TORCHSHIPS

Fast Torchship
High Gear Low Gear
Wet Mass 1,000,000 kg
Dry Mass 500,000 kg
Mass Ratio 2
Exhaust
Velocity 300,000 m/s 50,000 m/s
ΔV 200,000 m/s 40,000 m/s
Thrust 3,000,000 N 14,700,000 N
Starting
Acceleration 3 m/s2
(0.3 g) 14.7 m/s2
(1.5 g)
Thrust Power 450 GW 370 GW

Let us say that our ship has a mass of 1000 tons, and a modest exhaust velocity of 300 km/s, a mere 0.001 c. In the name of further modesty we will set our acceleration at less than one third g, 3 meters per second squared. This ship is very much on the low end of torchships; Father Heinlein would hardly recognize her. If our torch has VASIMR style capability, we can trade specific impulse for thrust; dial exhaust velocity down to 50 km/s to develop 1.5 g, enough to lift from Earth. (but see below)

(ed note: trading specific impulse for thrust is shifting from high gear to low gear)

Setting surface lift aside, let’s look at travel performance from low Earth orbit. We will reach escape velocity about 15 minutes after lighting up, a good Oberth boot, but with this ship it hardly matters.

(ed note: 15 minutes of 1.5 g is 900 seconds of 14.7 m/s2 which is a ΔV cost of about 13,000 m/s)

Suppose half our departure mass is propellant, giving us about 180 km/s of delta v in the tanks

(ed note: 200,000 m/s total – 13,000 m/s liftoff = 187,000 m/s.)

Since we must make departure and arrival burns, our transfer speed (relative to Earth) is about 90 km/s.

(ed note: 180,000 / 2 = 90,000 m/s.)

Holding acceleration to 0.3 g as we burn off mass, we’ll reach 90 km/s in less than eight hours,

(ed note: 90 km/s divided by 0.3 g is 90,000 m/s / 3 m/s2 = 30,000 seconds. 30,000 / 3,600 = 8.3 hours.)

…at about three times lunar distance from Earth. (If we were going to the Moon, we’d need to swing around three hours out for our deceleration burn.)

For any deep space mission we’ll end up coasting most of the way, and 90 km/s is some pretty fast coasting. Even this low end torchship lets you take retrograde hyperboloid orbits, which is Isaac Newton’s way of saying you can pretty much point & scoot.

You will truck along at 1 AU every three weeks, reaching Mars in little more than a week at opposition, though up to two months if you travel off season. The near side of the asteroid belt is a month from Earth, Jupiter three months in season. For Saturn and beyond things do get prolonged.

Still, 90 days for the inner system out to Jupiter is nothing shabby, and there’s no apparent High Magic involved.

How much power output does out modest torch drive have? The answer, skipping simple but tedious math, is not quite half a terawatt, 450 GW.

(ed note: (3,000,000 N thrust * 300,000 m/s exhaust velocity) / 2 = 450,000,000,000 watts = 450 GW.)

This baby has 600 million horses under the hood. That’s effective thrust power. Hotel power for the hab is extra, and so is getting rid of the waste heat.

Putting it another way, this drive puts out a tenth of a kiloton per second, plus losses.

(ed note: 450,000,000,000 watts / 4.184 × 1012 = 0.108 kilotons per second.)

And yes, Orion is the one currently semi plausible drive that can deliver this level of power and performance. I say semi plausible because, broadly political issues aside, I suspect that Orion enthusiasts gloss over the engineering details of deliberately nuking yourself thousands of times. Like banging your head on the table, it only feels good when you stop.
Reasonably Fast Torchship
Wet Mass 1,000,000 kg
Dry Mass 500,000 kg
Mass Ratio 2
Exhaust
Velocity 300,000 m/s
ΔV 200,000 m/s
Thrust 290,000 N
Starting
Acceleration 0.29 m/s2
(30 mg)
Thrust Power 45 GW
Really Fast Torchship
Wet Mass 1,000,000 kg
Dry Mass 500,000 kg
Mass Ratio 2
Exhaust
Velocity 3,000,000 m/s
ΔV 2,000,000 m/s
Thrust 3,000,000 N
Starting
Acceleration 3 m/s2
(0.3 g)
Thrust Power 4.5 TW

The good news, for practical, Reasonably Fast space travel, is that we don’t need those near-terawatt burns. They don’t really save much time on interplanetary missions — we could reduce drive power by 90 percent, and acceleration to 30 milligees — the acceleration of a freight train — and only add three days to travel time.

(Our reduced ‘sub-torch’ fusion drive is still putting out a non trivial 45 gigawatts of thrust power, which happens to be close to the effective thrust power of the Saturn V first stage.)

(ed note: (290,000 N thrust * 300,000 m/s exhaust velocity) / 2 = 45,000,000,000 watts = 45 GW.)

If you want Really Fast space travel, however, you will need more than this. Go back to our torchship and increase her drive exhaust velocity by tenfold, to 3000 km/s, and mission delta v to 1800 km/s, while keeping the same comfortable 0.3 g acceleration.

A classic brachistochrone orbit, under power using our full delta v, takes a week and carries us 270 million km, 1.8 AU.

transitDeltaV = 2 * sqrt[ D * A ]
((transitDeltaV / 2)^2) / A = D
((1,800,000 / 2)^2) / 3 = D
270,000,000,000 m = D
270,000,000 km = D
270 million km = D

Add a week of coasting in the middle and you’re at Jupiter. Saturn is about three weeks’ travel, and you can reach distant Eris in 6 months.

Drive power output of our upgraded torchship is now 4.5 TW, about a third the current power output of the human race. Which in itself is no argument against it. Controlling the reaction and getting rid of the waste heat are more immediate concerns.

(ed note: (3,000,000 N thrust * 3,000,000 m/s exhaust velocity) / 2 = 4,500,000,000,000 watts = 4.5 TW.)

As for a true, Heinleinian torchship? Heinlein’s torch is a mass conversion torch. He sensibly avoids any details of the physics, but apparently the backwash is a mix of radiation, AKA photon drive, and neutrons, probably relativistic.

Torchship Lewis and Clark, pictured above, is about 60 meters in diameter, masses in on the order of 50,000 tons, and in Time for the Stars she begins her relativistic interstellar mission by launching from the Pacific Ocean at 3 g. I don’t know how to adjust the rocket equations for relativity, but the naive, relativity-ignoring calculation gives a power output of 225 petawatts, AKA 225,000 TW, AKA 53 megatons per second.

Do not try this trick at your homeworld. I don’t know whether Heinlein never checked this calculation, did it and ignored the results, or did it and decided that a few dozen gigatons — 12 torchships are launched, one after another — was no big deal since it was off in the middle of the Pacific somewhere.

Maybe he only did the calculation later, because in other stories the torchships sensibly remain in orbit (served by NERVA style nuke thermal shuttles).

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