RocketCat sez

Have you simply had it up to here with these impotent little momma's-boy rockets that take almost a year to crawl to Mars?

Then you want a herculean muscle-rocket, with rippling titanium washboard abs and huge geodesic truck-nuts! You want a Torchship! To heck with John's Law, who cares if the exhaust can evaporate Rhode Island? You wanna rocket with an obscenely high delta V, one that can crank out one g for days at a time. Say goodby to all that fussy Hohmann transfer nonsense, the only navigation you need is point-and-shoot.

It is a pity that torchships are currently science fiction. But they are unobtanium, not handwavium. Ain't no law of physics sez they are impossible, we just don't know how to make one. Yet.

And like all good unobtanium, even though we can't build it yet, we can calculate what it can do just fine.

The basic problem is that rocket engines seem to only come in two types:

Engine Types

Muscle engines have the thrust to perform Hohmann trajectories but the low specific impulse means they have to carry outrageous amounts of propellant. Which really cuts into the payload mass.

Fuel Economy engines need only modest amounts of propellant. Unfortunately the low thrust means they take forever to inject themselves into a Hohmann trajectory.

Some engines can "shift gears" from muscle to fuel-economy and back, but that still just trades one set of liabilities with another.

What rocket designers really want is an engine with both high thrust and high specific impulse. Such engines don't have to use weak Hohmanns, they can use fantastically expensive (but rapid) Brachistochrone trajectories.

Such engines are called "Torch Drives", a spacecraft with a torch drive is called a "Torchship."

The term "Torchship" was coined by Robert Heinlein in 1953 for his short story "Sky Lift", it is also featured in his stories Farmer in the Sky, Time for the Stars, and Double Star. Sometimes it is referred to as "Ortega's Torch".

Nowadays it is implied that a Torchship is some kind of high thrust fusion drive, but Heinlein meant it to mean a handwavium total-conversion mass-into-energy type drive.

The Commodore’s office was crowded. All present wore the torch, except a flight surgeon and Commodore Berrio himself, who wore the jets of a rocketship pilot.

He held a torcher’s contempt for the vast distance itself. Older pilots thought of interplanetary trips with a rocket-man’s bias, in terms of years — trips that a torch ship with steady acceleration covered in days. By the orbits that a rocketship must use the round trip to Jupiter takes over five years; Saturn is twice as far, Uranus twice again, Neptune still farther. No rocketship ever attempted Pluto; a round trip would take more than ninety years. But torch ships had won a foothold even there: Proserpina Station — cryology laboratory, cosmic radiation station, parallax observatory, physics laboratory, all in one quintuple dome against the unspeakable cold.

From SKY LIFT by Robert Heinlein (1953)

What is a Torchship?

What is the definition of a torchship? Well, it is kind of vague. It more or less boils down to "unreasonably powerful."

With most propulsion systems, there is an inverse relationship between thrust and specific-impulse/exhaust-velocity (if one is high the other is low). So Rick Robinson defines a torch drive as propulsion system with both high acceleration (from high thrust) and high exhaust velocity. Note that whether the drive is "high acceleration" or not depends upon the thrust of the drive and the total mass of the spacecraft, that is, it does not just depend upon the drive.

If you recall, a propulsion systems Thrust Power (Fp) is thrust times exhaust velocity, then divide by two. Which sort of combines Rick's two parameters defining a torch drive (but so does propellant mass flow). The Drive Table is helpfully sorted in order of increasing Thrust Power. Remember that each entry is for a single engine of that type, it is possible to have multiple engines. With multiple engines, the exhaust velocity stays the same, but the thrust is multiplied by the number of engines.

A spacecraft's Specific Power (Fsp) is its propulsion system's Thrust Power divided by the spacecraft's dry mass. Remember that dry mass is the mass of spacecraft fully loaded with cargo and everything, but no propellant.

Rick Robinson's rule of thumb is that a Torch Drive is a propulsion system with both both high acceleration and high thrust. His further rule of thumb is that a Torchship is a spacecraft with a Torch Drive and a specific power of one megawatt per kilogram or larger.

And as a side note, such a frightful amount of thrust power in your spacecraft exhaust could do severe damange to anything it hosed. One would almost say weapons-grade levels of damage. Clever readers have already mouthed the phrase the Kzinti Lesson. This might lead to torchships being reserved for military vessels only.

Another thing to keep in mind that as a rule of thumb, torchships are not subject to the Every gram counts rule, because they are unreasonably powerful.

When it comes to propulsion systems we might actually be able to build in the near future, the list includes Orion drives, Zubrin's nuclear salt water rocket, and maybe Medusa.

Another definition of a torchship is a spacecraft with more than 300 km/s total delta V and an acceleration greater than 0.01 g. Which may or may not fit with Rick's definition.

I am trying to triangulate on Rick's rule of thumb for Torch Drives. I'm toying with defining "high thrust" as 100,000 Newtons or higher, and "high exhaust velocity" as 100,000 m/s or higher.

And a thrust power of 100 gigawatts or higher. Which is a number I pulled out of the air by examining the drive table and picking a dividing line that pleased me.

And a propellant mass flow between 100 and 0.01 kg/sec.

I'm working on it, OK? Let me crunch some numbers and draw a few graphs and I'll get back to you.

Would Rick Robinson consider the good ship Polaris to be a torchship? Let's see.

Our design equips the Polaris with not one, not two, but three freaking Nuclear Salt Water rocket engines. 12,900,000 Newtons of thrust each for a total of 38,700,000 Newtons. And an exhaust velocity of 66,000 meters per second. So NSWRs are definitely torch drives. Is the Polaris a torchship?

  • Fp = ( (F * E) * Ve ) / 2
  • Fp = ( (12,900,000 * 3) * 66,000 ) / 2
  • Fp = ( 38,700,000 * 66,000 ) / 2
  • Fp = 2,554,200,000,000 / 2
  • Fp = 1,280,000,000,000 watts or 1.28 terawatts

The Polaris has a dry mass of 126,000 kilograms (126 metric tons).

  • Fsp = Fp / Me
  • Fsp = 1,280,000,000,000 / 126,000
  • Fsp = 10,200,000 watts per kilogram or 10.2 megawatts per kilogram

Oh, yes, Rick Robinson would say the Polaris is very much a torchship.

With ion engines, chemical engines, and nuclear torches we're facing a classic Newton's Third Law problem. Somehow the exhaust needs to have sufficient momentum for the opposite reaction to give the ship a good acceleration.

Chemical rockets solve the problem by expelling a ton of mass at a relatively low velocity. (high propellant mass flow but low exhaust velocity: SUV)

Ion drives expel a tiny amount of mass, so to get anywhere they get it moving FAST, but even at gigawatts of power they get a measly 0.0001g. (low propellant mass flow but high exhaust velocity: Economy)

Torch drives take a small-to-moderate amount of mass and use nuclear destruction to get it moving insanely fast. (medium propellant mass flow and high exhaust velocity: Torch) They're the only ones (insert disclaimer) with enough power per unit of reaction mass to get 0.3g constant acceleration conveniently. Even a perfect ion drive would need a phenomenal (read: impossible) amount of power input to match the performance of a nuclear explosion.

(A low propellant mass flow and low exhaust velocity engine would be utterly worthless)

From ON TORCHSHIPS comment by Eric (2010)

Torchship Performance

Fast Torchship
High GearLow Gear
Wet Mass1,000,000 kg
Dry Mass500,000 kg
Mass Ratio2
300,000 m/s50,000 m/s
ΔV200,000 m/s40,000 m/s
Thrust3,000,000 N14,700,000 N
3 m/s2
(0.3 g)
14.7 m/s2
(1.5 g)
Thrust Power450 GW370 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.) 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 Mass1,000,000 kg
Dry Mass500,000 kg
Mass Ratio2
300,000 m/s
ΔV200,000 m/s
Thrust290,000 N
0.29 m/s2
(30 mg)
Thrust Power45 GW
Really Fast Torchship
Wet Mass1,000,000 kg
Dry Mass500,000 kg
Mass Ratio2
3,000,000 m/s
ΔV2,000,000 m/s
Thrust3,000,000 N
3 m/s2
(0.3 g)
Thrust Power4.5 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.

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).

From ON TORCHSHIPS by Rick Robinson (2010)

Rick Robinson:

Those performance stats (for the Project Daedalus) are certainly torchlike, and in fact an exhaust velocity of 10,000 km/s is wasteful for nearly all Solar System travel — on most routes you just don't have time to reach more than a few hundred km/s.

Using STL starship technology on interplanetary routes is like using a jet plane to get around town.

Jean Remy:

There's no such thing as going somewhere "too fast". At least in terms of military strategy you'll want the ability to get somewhere faster than anyone else can, and damn the price at the pump. It is more costly to arrive at a battle late (and for want of a horse)

Rick Robinson:

Oh, I have nothing against speed! A better way to put it is that STL starships are geared all wrong for insystem travel, like driving city streets in 5th gear.

Luke Campbell:

Consider a 1,000 ton spacecraft with a 10,000 km/s exhaust velocity and an acceleration of 0.722 m/s/s. For a 1 AU trip at constant acceleration, flipping at the midpoint, it will take 10.5 days and consume 66 tons of propellant/fuel.

Now let's add extra mass into the exhaust stream, so that the spacecraft uses propellant at 16 times the rate but expells it at 1/4 the exhaust velocity (thus keeping the same power). This brings the acceleration up to 2.89 m/s/s. We will accelerate for 1/10 the distance, drift for 8/10 the distance, and then decelerate for 1/10 the distance. The trip now takes 7 days and uses 240 tons of propellant, of which only 7 tons is fuel.

Bulk inert (non-fuel) propellant is probably cheap (water or hydrogen). Fuel is probably expensive (He-3 and D). The second option gets you there faster and cheaper.

In Rick's analogy, high exhaust velocity, low thrust, low propellant flow corresponds to high gear. Low exhaust velocity, high thrust, high propellant flow is low gear. In this case, a lower gear than the default "interstellar" Daedelus thrust parameters is preferable.

Rick Robinson:

'Gearing' is highly desirable even if the drive won't produce surface lift thrust from any significant body. Each deep space mission also has its own optimum balance of acceleration and delta v, favoring an adjustable drive.

(ed note: given the mission delta V, the optimal exhaust velocity is Δv * 0.72 . See here.)

From ON TORCHSHIPS comments (2010)

Calculating the performance of a spaceship can be complicated. But if the ship is powerful enough, we can ignore gravity fields. It is then fairly easy. The ship will accelerate to a maximum speed and then turn around and slow down at its destination. Fusion or annihilation-drive ships will probably do this. They will apply power all the time, speeding up and slowing down.(ed note: a "brachistochrone" trajectory)

In this simple case, all the important performance parameters can be expressed on a single graph. This one is drawn for the case when 90% of the starting mass is propellant. (ed note: a mass ratio of 10) Jet velocity (exhaust velocity) and starting acceleration are the graph scales. Distance for several bodies are shown. Mars varies greatly; I used 150 million kilometers. Trip times and specific power levels are also shown. "Specific power" expresses how much power the ship generates for each kilogram of its mass, that is, its total power divided by its mass. The propellant the ship will carry is not included in the mass value. (ed note: Specific Power (Fsp) is its propulsion system's Thrust Power divided by the spacecraft's dry mass. On the chart specific power is the diagonal dotted lines.)

An example: Suppose your ship can produce 100 kW/kg of jet power (specific power). You wish to fly to Jupiter. Where the 100 kW/kg and Jupiter lines cross on the graph, read a jet velocity of 300,000 m/s (Isp = 30,000) and an initial acceleration of nearly 0.01g. Your trip will take about two months.

The upper area of the graph shows that high performance is needed to reach the nearest stars. Even generation ships will need, in addition to very high jet velocities, power on the order of 100 kW/kg. The space shuttle orbiter produces about 100 kW/kg with its three engines. The high power needed for starflight precludes its attainment with means such as electric propulsion.

From TO THE STARS by Gordon Woodcock (1983). Collected in Islands In The Sky, edited by Stanley Schmidt and Robert Zubrin (1996).

Torch Drive Heat

The main problem with torch drives is getting rid of the waste heat generated by such a monster.

Remember that Thrust Power (Fp) is thrust times exhaust velocity, then divide by two. Thrust power is rated in watts, and if you look at the ratings for torch drives, well, that's a lotta watts.

Unavoidably some of those watts are going to become waste heat.

If you have a Resistojet with a thrust power of 700 watts, and 10% becomes waste heat, the heat will be 70 watts. This is only about as much as emitted by an old style incandescent bulb, not a problem.

But if you have Rick's Fast Torchship with a thrust power of 450 gigawatts, the heat will be 45,000 megawatts. Since your average drive assembly can survive no more than 5 megawatts, it will glow blue-white for a fraction of a second before it vaporizes. Along with most of the spacecraft.

A spacecraft with a chemical engine can have a torchship like specific power of 1MW/kg, but the chemical engine is not a torch drive because of its pathetic exhaust velocity. Chemical engines have no waste heat problem. Yes they have huge amounts of it, but they can easily use their huge propellant mass flow to get rid of it. Basically the exhaust plume acts like a heat radiator. The technical term is "open-cycle cooling".

Unfortunately torch drives cannot use that trick. High propellant flow equals pathetic specific impulse. Torch dives have large specific impulse ratings, which means low propellant flows, which means they will have to rely upon something else to keep themselves from melting. There isn't enough propellant in the exhaust plume to carry away the heat.

mDot = F / Ve


  • F = Thrust (Newtons)
  • mDot = Propellant Mass Flow (kg/s)
  • Ve = Exhaust Velocity (m/s)

As you can see from the above equation, if both thrust and specific impulse is torche-drive high, the propellant mass flow will be small.

Reaction Chamber Size

So open-cycle cooling is out.

If one has no science-fictional force fields, as a rule of thumb the maximum heat load allowed on the drive assembly is around 5 MW/m2. This is the theoretical ultimate, for an actual propulsion system it will probably be quite a bit less. For a back of the envelope calculation:

Af = sqrt[(1/El) * (1 / (4 * π))]

Rc = sqrt[H] * Af


Af = Attunation factor. Anthony Jackson says 0.126, Luke Campbell says 0.133
El = Maximum heat load (MW/m2). Anthony Jackson says 5.0, Luke Campbell says 4.5
π = pi = 3.141592...
H = reaction chamber waste heat (megawatts)
Rc = reaction chamber radius (meters)
sqrt[x] = square root of x

As a first approximation, for most propulsion systems one can get away with using the thrust power for H. But see magnetic nozzle waste heat below.

Science-fictional technologies can cut the value of H to a percentage of thrust power by somehow preventing the waste heat from getting to the chamber walls (e.g., Larry Niven's technobabble crystal-zinc tubes lined with magic force fields).

Only use this equation if H is above 4,000 MW (4 GW) or so, and if the propulsion system is a thermal type (i.e., fission, fusion, or antimatter). It does not work on electrostatic or electromagnetic propulsion systems.

(this equation courtesy of Anthony Jackson and Luke Campbell)


Say your propulsion system has an exhaust velocity of 5.4e6 m/s and a thrust of 2.5e6 N. Now Fp=(F*Ve)/2 so the thrust power is 6.7e12 W. So, 6.7e12 watts divided by 1.0e6 watts per megawatt gives us 6.7e6 megawatts.

Assuming Anthony Jackson's more liberal 5.0 MW/m2, this means Af = 0.126

Plugging this into the equation results in sqrt[6.7e6 MW] * 0.126 = drive chamber radius of 326 meters or a diameter of almost half a mile. Ouch.

Equation Derivation

Here is how the above equation was derived. If you could care less, skip over this box.

It is based on the good old Inverse-square law.

The reaction chamber is assumed to be spherical. Obviously the larger the radius of the chamber, the more surface area it has, and the given amount of waste heat has to be spread thinner in order to cover the entire area. If you only have one pat of butter, the more slices of toast means the lesser amount of butter each slice gets.

El is the Maximum heat load, or how many megawatts per square meter the engine can take before the blasted thing starts melting. Anthony Jackson says 5.0 MW/m2.

The idea is to expand the radius of the reaction chamber such that the inverse-square law attenuates the waste heat to the point where it is below the maximum heat load. Then we are golden.

The attenuation due to the inverse square law is:

ISLA = (4 * π * Rc2)


ISLA = attenuation due to the inverse square law
π = pi = 3.141592...
Rc = reaction chamber radius (meters)

The heat load on the reaction chamber walls is:

Cl = H / ISLA


H = waste heat (megawatts)
Cl = heat load on chamber wall (MW/m2)

Merging the equations:

Cl = H / (4 * π * Rc2)

Solve for Rc:

Cl = H / (4 * π * Rc2)
4 * π * Rc2 = H / Cl
4 * π * Rc2 = H * (1/Cl)
Rc2 = (H * (1/Cl)) / (4 * π)
Rc2 = H * (1/Cl) * (1 / (4 * π))
Rc = sqrt[H] * sqrt[(1/Cl) * (1 / (4 * π))]

Looking at the last equation, take the right half and swap Cl for El to get:

Af = sqrt[(1/El) * (1 / (4 * π))]

and the entire equation is where we get:

Rc = sqrt[H] * Af

which is what we were trying to derive. QED.

Playing with these figures will show that enclosing a thermal torch drive inside a reaction chamber made of matter is impractical. Unless you think a drive chamber a half mile in diameter is reasonable.

Therefore, the main strategy is to try and direct the drive energy with magnetic fields instead of metal walls. See "Magnetic Nozzle" below.

With these propulsion systems, H is not equal to thrust power. It is instead equal to the fraction of thrust power that is being wasted. In other words the reaction energy that cannot be contained and directed by the magnetic nozzle. Which usually boils down to neutrons, x-rays, and any other reaction products that are not charged particles.

For instance, D-T (deuterium-tritium) fusion produces 80% of its energy in the form of uncharged neutrons and 20% in the form of charged particles. The charged particles are directed as thrust by the magnetic nozzle, so they are not counted as wasted energy. The pesky neutrons cannot be so directed, so they do count as wasted energy. Therefore in this case H is equal to 0.8 * thrust power.

Magnetic Nozzles

So enclosing a thermal torch drive inside a reaction chamber made out of matter appears to be a dead end.

Therefore, the main strategy is to try and direct the star-core hot exhaust with energy instead of matter: using magnetic fields instead of metal walls. In science fiction terms, this would be a force-field rocket nozzle.

Rick Robinson says that even subtorch high-end drives will have an open latticework to support a magnetic containment nozzle. He likes to call it a "lantern", because it will glow brilliantly. Which is putting it mildly. Heinlein calls them "torches" and John Lumpkin calls them "candles."


High specific impulse thermodynamic rockets benefit from a nozzle that is not limited by the melting point of its material.

Magnetic nozzles direct the exhausted flow of ions or a conductive plasma by use of magnetic fields instead of walls made of solid materials (see de Laval nozzle). If superconducting coils are used, these must be thermally shielded to remain in the superconducting range.

The design illustrated operates at a throat magnetic field strength of 25 Teslas and a nozzle efficiency of 86%.

From HIGH FRONTIER by Philip Eklund

All advanced concepts have the same problem when they try to produce high thrust at high specific impulse (greater than 1500 s) in a compact engine.

The problem is independent of the type of engine, whether it involves high-power electromagnetic thrusters; atomic hydrogen, metallic hydrogen, or metastable atom fuels; solar, laser, or microwave-heated thrusters; and fission-, fusion-, or antimatter-powered reaction chambers. The high-energy exhaust from any of these processes produces a blazing plasma that melts or evaporates any reaction chamber and nozzle made of ordinary materials.

One solution is to make or shield the reaction chamber and nozzle with magnetic fields. Research in this field is still in its preliminary phases [Gerwin, 1990] and we are not sure a good design is possible.

[Gerwin, 1990] Gerwin, R. A., G. J. Marklin, A. G. Sgro, A. H. Glasser. 1990. Characterization of Plasma Flow Through Magnetic Nozzles. Los Alamos National Laboratory. Final Technical Report AL-TR-89-092, February 1990, on Contract RPL: 69018 with Air Force Astronautics Laboratory, Edwards AFB, CA 93523,1 May 1986 to 30 April 1987.

Simply stated, a magnetic nozzle converts thermal energy of a plasma into directed kinetic energy. This conversion is achieved using a magnetic field contoured similarly to the solid walls of a conventional nozzle (see, for example, Fig. 1). The applied magnetic field in most cases possesses cylindrical symmetry and is formed using permanent magnets or electromagnetic coils, which confines the plasma and acts as an effective "magnetic wall" through which the thermal plasma expands into vacuum. Applications include laboratory simulations of space plasmas, surface processing, and plasma propulsion for spaceflight.

Magnetic nozzle research at the EPPDyL began in 2008. The goal of this research is to understand the fundamental physical processes of the plasma flow through the nozzle and their impact on the nozzle performance. Specifically, we are working towards answering the following important questions:

  • How do the dynamics of the plasma flow through the magnetic nozzle influence the exhaust plume structure?
  • How is the exhausted plasma affected by the relationship between the physical and magnetic goemetries?
  • How does the magnetized plasma detach from the applied magnetic field?
  • These questions will be answered through a combination of theoretical, experimental, and computational research. Ultimately, knowledge of these processes will yield fundamental scaling laws for the performance of magnetic nozzles for plasma propulsion.

    From MAGNETIC NOZZLE PLASMA DYNAMICS AND DETACHMENT by Electric Propulsion and Plasma Dynamics Lab at Princeton University

    And as for torch ships — a bigger problem than creating the hellish inferno of nuclear fire needed to propel your spacecraft is keeping your spacecraft from evaporating under the intense x-ray, gamma-ray, and neutron irradiation.

    My best guess for accomplishing this is keeping the torch flame outside the spacecraft, and couple the plasma to the spacecraft using magnetic fields to give you thrust. The latter trick probably means several conductive — or, more likely, superconductive — loops of cable surrounding the torch flame but at a good healthy radius.

    To protect your field generating cable, you will need to shield them, probably with blade-like tungsten structures edge-on to the torch. The large surface area of the blades gives you a lot of radiator surface while only intercepting a small amount of radiation (only what is needed to shield the cables), tungsten does a good job of scattering neutrons away without heating up much, at shallow angles of incidence x-rays (but not neutrons or gammas) are reflected away, and tungsten can heat up to yellow-white hot without evaporating too much.

    An unusual consequence of the latter, and the relatively small emissions from your optically thin thrust plasma, is that visually the tungsten shields will be the brightest part of your spacecraft when under thrust — the intensity will be the same as that of an incandescent filament from a light bulb, but you will have a lot more area to radiate from.

    These resulting torch craft don't end up looking much like conventional rockets. You get loopy filigree and from the support and field cables, and graceful glowing sheets from the heat shields, enclosing a volume that is probably much larger than the passenger/payload section.

    A recent thread on got me thinking about how to protect your lantern structure from the radiation of a Magic Fusion Torch (MFT). here are my current thoughts I figured I'd throw out for the group mind to pick over.

    My design goal was to get a shield that did not require active cooling in a moderately compact package. For illustration, I will use a 10 TW D-3He laser inertial confinement fusion (ICF) torch.

    Materials: You need something that can withstand high temperatures. Briefly, this comes down to graphite, tungsten, and silcon carbide.

    • Graphite has sublimation vapor pressure of 1 atmosphere at about 3900 K, and a vapor pressure of 10 Pa at 2840 K so we would probably want to keep the temperature below 2500 K. Additional properties that will become important later are that high energy neutrons in graphite have a mean free path of about 6 cm before scattering, and each scattering will transfer on average about 14% of the neutron's energy to the graphite.
    • Tungsten melts at 3695 K, but experience with incandescent bulbs tells us it is best to keep the temperature below 3000 K to prevent sublimation (putting the tungsten in a quartz container and flooding it with halogen gas can help re-deposit the evaporated tungsten back on the tungsten, but you need to keep the quartz cool enough that it does not melt). The mean free path of high energy neutrons in tungsten is about 2.5 cm, and each collision between a neutron and a tungsten nucleus imparts about 1% of the neutron's energy to the tungsten. Further, at angles of incidence of 0.5 degrees or less, you get total external reflection of 8 keV x-rays off of a tungsten surface. This critical angle increases as the x-ray energy decreases, so for soft x-rays tungsten at low angles will reflect most of the incident x-ray flux.
    • Silicon carbide sublimates at 3003 K, presumably this is where the vapor pressure is 1 atmosphere. We would thus want to keep the temperature of SiC components at a maximum of somewhere around 2000 to 2500 K. The benefit of SiC is that it is an excellent thermal conductor.

    All data on neutrons was taken for the 2.5 MeV neutrons produced by D-D fusion.

    (ed note: Deuterium-Helium 3 fusion does not produce neutrons. Unfortunately you cannot rely upon the deuterium nuclei to obligingly react only with the Helium-3 nuclei. Some will insist upon reacting with other deuterium nuclei in the vicinity, and that reaction does produce those nasty neutrons)

    My idea, therefore, is to point a very narrow wedge of tungsten at the point of the fusion torch flame. A 0.3 degree angle is a slope of 200:1, so we will take two 2 mm thick sheets of tungsten and join them at an opening angle of 0.6 degrees. Pointed directly at the fusion source, we get a 0.3 angle of incidence. This will specularly reflect most x-rays. Plots I have seen for low angle x-ray reflectance indicates at least 99% of the x-rays will be reflected.

    A neutron encountering the tungsten shield will pass through 40 cm of tungsten for a 2 mm thick sheet at a 200:1 slope. This is on the order of 16 interaction lengths, so essentially all incident neutrons will be scattered. Because the thickness of the sheets is much less than the mean free path of the neutrons, essentially all neutrons will only scatter once. Thus, about 1% of the intercepted neutron energy will be deposited in the tungsten, the rest will be scattered away.

    (ed note: MW is megawatt {millions of watts or 106 watts}, GW is gigawatts {billions of watts or 109 watts}, TW is terawatt {trillions of watts or 1012 watts} )

    We now have a structure that will absorb at most only 1% of the fusion radiation it intercepts. It also has a radiating area of 200 times the cross section it presents to the radiation source. If instead of a wedge, we used a single inclined sheet, we could double this radiating area at the expense of a shield that is twice as long. If the tungsten can withstand a maximum temperature of 3000 K before sublimation becomes problematic, it will be radiating 4.6 MW/m2 of surface area. This corresponds to 200 * 4.6 MW/m2 = 920 MW/m2 of cross section presented to the fusion source. Since only 1% of the incident radiation is absorbed, this structure can withstand radiant intensities of 92 GW/m2 before it suffers from excessive sublimation. If we limit the temperature to 2500 K, the allowable radiant flux is still 44 GW/m2.

    For the 10 TW D-3He torch, we need to deal with about 2 TW of x-rays and about 0.5 TW of neutrons from the D-D side reaction. This gives a total radiant intensity of 200 GW/r2, where r is the distance from the fusion source. This gives us a stand-off distance of 1.5 meters for 3000 K, or 2.1 meters for 2500 K.

    There is the additional consideration of the size of the lantern structures to be protected. If you need cables and support beams 10 cm across, then the wedge will need to stick out 10 meters from the lantern structures. This gives a lantern radius of around 12 meters. If you can get by with 2 cm wide structures, you can build your lantern as compact as 3.5 to 4 meters in radius.

    This analysis neglects radiant heat from one shield being intercepted by another shield. It also ignores how you would protect the optical elements used to focus the laser on the fuel pellet. You would also probably want additional neutron shielding for superconducting cables that need to be cooled (steel support structures wouldn't need this), since some neutrons will be scattered at low angles, and other parts of the lantern structure will be scattering neutrons all over the place. You will probably need active cooling for the superconductors, but at least you will not need to pressurize the coolant at hundreds of atmospheres to contain it at 3000 K.

    Isaac Kuo

    This is assuming the fusion source is a point source of radiation, of course.

    You can also double the radiating area with a wedge by doubling the shield length and halving the sheet thickness.

    Luke Campbell

    Good call. The inverse process, using a thicker sheet at a larger angle, would probably be useful for keeping the length of the wedge down. In particular, you could have the wedge be very sharp near the fusion source, and then have a wider angle as you keep the tungsten at a constant temperature. Otherwise, you end up with a wedge that is 3000 K at 1.5 meters from the fusion source, 2500 K at 2.1 meters, 2000 K at 3.3 meters, 1500 K at 6 meters, and so on.

    Anthony Jackson

    Luke Campbell: We now have a structure that will absorb at most only 1% of the fusion radiation it intercepts. It also has a radiating area of 200 times the cross section it presents to the radiation source.

    Sadly, the tip does not, so what will happen is that the tip will melt/sublimate, becoming blunt, and it will continue sublimating backwards.

    Luke Campbell

    This should depend on the sharpness of the tip and the thermal conductivity of the material. If the thermal conductivity is high enough and the amount of energy absorbed by the tip is low enough due to the tip being very sharp, you should be able to rapidly conduct that heat away to adjacent highly inclined bits and keep the temperature down.

    I think. I'll have to do the analysis later.

    Anthony Jackson

    That may argue for diamond or graphite, for superior thermal conductivity. Also, if you can get the tip to be translucent, its absorption goes down.

    In general, though, it's important to ask what the purpose of the lantern shield is. Usually, it's envisioned as a support structure for the magnetic nozzle and/or igniter mechanism.

    Luke Campbell

    Anthony Jackson: That may argue for diamond or graphite, for superior thermal conductivity. Also, if you can get the tip to be translucent, its absorption goes down.

    True, although carbon cannot withstand as high of temperatures as tungsten in vacuum (C has 10 Pa vapor pressure at 2840 K, W has 10 Pa vapor pressure at 3770 K). Carbon also has the disadvantage that it absorbs an order of magnitude more energy when struck by a neutron. However, since the tip is so small, you probably only need a very thin layer of carbon. Neutrons are so penetrating that most of them will just pass through the thin carbon layer.

    At these temperatures, diamond is not available. It decomposes into graphite at 1700 K. For the same reasons, any delicate nanostructure of carbon will probably decompose into graphite at some temperature less than 2000 K.

    One option would be a tungsten/carbon multilayer mirror. These are currently used for soft x-ray reflection, and can achieve normal incidence soft x-ray reflectivities of around 0.4 to 0.6, with theory predicting normal incidence reflectivities of up to 0.8. This is nowhere near as good as the .99+ reflectivity for grazing incidence, of course, but can still cut down the amount absorbed. It would be a very thin surface structure located only at the tip, to cut down on neutron absorption. If the outer layer were tungsten, it might be able to contain the carbon vapor at temperatures close to 3000 K.

    Anthony Jackson: In general, though, it's important to ask what the purpose of the lantern shield is. Usually, it's envisioned as a support structure for the magnetic nozzle and/or igniter mechanism.

    I was thinking of shielding the superconducting cables (which may need to be at cryogentic temperatures, and even in settings with room temperature superconductors still probably couldn't exceed a few hundred K) from the x-rays and neutrons. In addition, I want to shield the support structures that transmit the thrust from the magloops to the spacecraft. Presumably, these are made out of steel (possibly reactor steel, to handle scattered neutrons), and would weaken around 1000 K or a bit higher and melt at 1800 K. Cryogenic coolant pipes can be run behind the structural members, but these also need to be protected from the intense radiation.

    The igniter mechanism could possibly be protected from the radiation if the ICF used heavy ions. In this case, the beams would be steered with magnetic fields and the steering magnets could be protected with the tungsten shields. For laser ICF, you could use a highly inclined thin sheet of silicon carbide (SiC) in front of the mirror. The SiC would reflect the x-rays and scatter the neutrons like the tungsten, but it would be a transparent window to the laser light. Unfortunately, the transparency means it will not radiate its heat efficiently, and neutrons will transfer more of their energy to the lower mass silicon and carbon atoms that they would to heavy tungsten. Perhaps it could be actively cooled by a transparent molten salt that pumps the heat away to a black body radiator. With a set-up like this you could probably get the mirrors within a few tens of meters of that 10 GW fusion source.

    Anthony Jackson

    Luke Campbell: The igniter mechanism could possibly be protected from the radiation if the ICF used heavy ions.

    Of course, unless you're using self-ignited targets (Mini-Mag Orion) or antimatter, dealing with incidental heat from the method by which you generate the beam may be a bigger problem than shielding the torch. You can probably use your mag-loop to steal energy from the drive (the Mini-Mag Orion does that), probably without a lot of waste heat, but you'll have some, and you'll have some more from your ion beams, so it's still hard to see waste heat less than about 0.1% of drive power, all of it at fairly low temperatures (operating temperature of generator and drive beam).

    Luke Campbell

    That is a good point. I wonder if it is worthwhile to use a heat pump to increase the temperature of the radiator coolant, in order to increase the rate at which entropy is radiated away?

    Anthony Jackson

    Depends on assumptions. For perfect heat pumps, a net efficiency of 25% minimizes radiator area, but the weight penalty elsewhere may be more important than the penalty for radiator area. For example, in Attack Vector: Tactical, going from 25% efficient reactors with 1,500K radiators to 50% efficient reactors with 1,000K radiators would increase radiator size by 69%, but it would reduce reactor fuel consumption by 50% and, while the storage capacity of heat sinks would be reduced (perhaps by as much as 50%), the quantity of heat generated would be reduced by even more (by a factor of 3), and the plumbing would be a lot easier to work with.

    Isaac Kuo

    Luke Campbell: This should depend on the sharpness of the tip and the thermal conductivity of the material.

    You can greatly reduce the area of the tips by using an edge shaped like a sharp sawtooth. This more or less transitions a wedge shape into a series of sharp knife-points. You have grazing angles everywhere except for the tips of the peaks and the saddle-points of the valleys.

    Anthony Jackson: That may argue for diamond or graphite, for superior thermal conductivity. Also, if you can get the tip to be translucent, its absorption goes down.

    Another possibility is to actively cool the edge. Instead of joining the two sides together, the two sides of the wedge simply fit together by spring forces. Between the two sides is a "coolant plate" that slides in/out. When it slides out, the wedge sides are forced apart and thermal conduction transfers heat from the wedge edges to the coolant plate. When it slides in, the wedge sides join together.

    Anthony Jackson: In general, though, it's important to ask what the purpose of the lantern shield is. Usually, it's envisioned as a support structure for the magnetic nozzle and/or igniter mechanism.

    It's meant to shield the superconducting magnetic nozzle loop(s) from neutral radiation.

    But you also need a shield for the bulk of your ship...the ship itself may be protected by a very large wedge shape; the ship's hull being a flat wedge shape. This provides more volume than a cone.

    Think of it like the design of a house where the roof needs to be steeply angled to let snow fall off. A seemingly compact square or circle floorplan is very bad because it requires a very tall roof. In contrast, a long rectangle floorplan can be covered with a shorter roof.

    So, the ship's overall shape could look like a thin slice of cantaloupe, along with a superconducting loop ring. This loop ring has a conical fringe for its shield.

    Luke Campbell

    Isaac Kuo: You can greatly reduce the area of the tips by using an edge shaped like a sharp sawtooth. This more or less transitions a wedge shape into a series of sharp knife-points. You have grazing angles everywhere except for the tips of the peaks and the saddle-points of the valleys.

    I don't think you want any valleys — the sides of the valleys would be radiating their heat on to each other rather than into space.

    I expect you could get the knife-edge of the tungsten wedge to be only a few atoms wide — very sharp steel knives are only a few atoms wide, and some ceramics can get a mono-atomic edge.

    Isaac Kuo: Another possibility is to actively cool the edge. Instead of joining the two sides together, the two sides of the wedge simply fit together by spring forces. Between the two sides is a "coolant plate" that slides in/out. When it slides out, the wedge sides are forced apart and thermal conduction transfers heat from the wedge edges to the coolant plate. When it slides in, the wedge sides join together.

    This adds a considerable amount of complexity over passive cooling. If necessary, it would be best to design it so that the coolant plate were retracted and protected by the tungsten wedge during each fusion pulse. Then it could perform its cooling in between the pulses. It would be best to make it out of heavy elements, to minimize energy transfer from neutrons, since that close to the tip you are likely to get plenty of scattered neutrons hitting it.

    Isaac Kuo

    Luke Campbell: I don't think you want any valleys — the sides of the valleys would be radiating their heat on to each other rather than into space.

    Not these valleys. These valleys are on a sawtooth knife-edge, so the sides still radiate almost perpendicular to the overall plane.

    Geometrically, consider the wedge to be the product of sweeping the straight edge at a 1:200 slope. The alternate sawtooth would be created by sweeping a sharp sawtooth shape at the same slope.

    Rick Robinson


    But I agree with whoever it was upthread who said that the problem here is the knife edge, or even if you sawtooth it the points. No matter how fine you make it there will be a blunt surface facing the full brunt of the fire. As it sublimates away the flat surface will widen, leading to progressive and accelerating failure.

    This does not invalidate the concept, but I think the radius will have to be much more conservative to keep the knife edge within the limits of some form of active cooling.

    The tougher problem in torch design seems to be the one Anthony brought up. If you suppose that about 1 percent of torch output will be in the form of relatively low-temperature heat generated by the various gizmos that keep the torch running, for a 1 TW torch that will be 10 GW of low-grade heat to dump from secondary radiators, probably at much less than 1 MW/m2.

    I tend to handwave a 2-stage control function, where the torch cycle is sustained by pulsations in a plasma (or some such), and the onboard gizmos are required only to stabilize and control that secondary plasma, nominally permitting about another hundredfold reduction in onboard low-grade waste heat, to about 100 MW.

    Luke Campbell

    Torch drive describes a class of thrusters for spacecraft propulsion in which a high energy yield detonation or pulse is initiated at a high rate external to the spacecraft. Magnetic fields are used to deflect the plasma produced by the pulse to generate thrust. By detonating the pulse outside the main structure of the spacecraft, the neutral radiation by-products of the pulse (neutrons, bremmstrahlung x-rays, gamma rays, and thermal radiation) can mostly escape into space, without creating a large thermal load aboard the spacecraft. This allows very high energy thrusters, which can combine both high thrust and high delta-V simultaneously.

    A torch drive requires one or more current carrying loops surrounding the reaction region to produce the magnetic field that deflects the plasma and charged radiation from the drive pulses. Typically, these loops are made of superconductors, since any normal conductor would quickly melt or vaporize under the high currents needed to produce the field. The field coils are backed by a high tensile strength support to withstand the magnet-current back reaction from bursting the superconductor.

    These field coils must be protected from the intense neutral radiation that the drive pulses produce. This is because a superconductor that becomes too hot will cease to superconduct. A common shielding design is a sheet of tungsten with a "V"-shaped cross section at a very narrow opening angle, resembling a knife-edge. The point of the V faces the radiation source, the field coil runs along the open top of the V. At very small angles of incidence, tungsten makes a good reflector of x-rays so that most x-rays are simply reflected at low angles away from the field coils and into space. Since tungsten atoms are much heavier than neutrons, a collision between a tungsten nucleus and a neutron results in the neutron rebounding with most of its original energy, delivering only 1% of its energy on average to the tungsten shield. This scatters the neutrons away from the field coils. The narrow opening angle means that the tungsten knife edge is essentially a sheet perpendicular to the incoming radiation, allowing a large radiator area compared to the cross section exposed to the radiation.

    Rates of tungsten sublimation become problematic at temperatures above 3000 K, so the shield is typically placed far enough away from the drive pulses to keep its temperature at or below this value. At all times, the shield must be kept below 3695 K, the melting point of tungsten. At these temperatures, the tungsten knife-edge sheets are radiating at a blazing yellow-white color, with the intensity of an M-class star or an old style incandescent bulb filament. A torch drive in operation appears as a brilliant flare primarily from the thermal radiation of the tungsten shields — the drive pulses themselves emit relatively little visible light in comparison. With a 200:1 aspect ratio for the length of the knife blade to its width, a heat shield that absorbs 1% of the incident radiation and scatters the rest can withstand an incident intensity of 90 GW/m2.

    A plasma in a magnetic field will expand against the field while the energy density of the plasma is greater than the energy density of the field. In SI units, the energy density of the field in J/m3 is given by


    where B is the magnitude of the magnetic field in tesla and mu_0 is the magnetic constant (mu_0 = 4 pi * 10-7 N/A2). If the field coils produce a uniform field, then a drive pulse with energy E in its plasma will expand against the field until it reaches a volume V such that the following relation approximately holds

    V = 2 mu_0 E/B2

    Assuming the radius is approximately spherical, the pulse's blast will expand into a fireball with a diameter d approximately the cube root of this volume. The time t it will take for the pulse to expand to a stop before it is deflected is approximately

    t = d/V_ex

    where V_ex is the velocity of the torch drive exhaust. If a second pulse is detonated before the first pulse has been fully deflected, it will add its energy to that of the first and require a larger volume to hold the combined fireball.

    The size of the field coils is set by the requirement that the tungsten shield remain cool enough not to sublimate, and for the drive pulse fireballs not to contact the tungsten shield. Since the tungsten shield extends a considerable distance toward the detonation point (the distance from the field coil to the tip of the shield is typically 200 times or more the width of the field coil), the field coil must be set far enough back that the tip of the tungsten shield does not evaporate.

    If there are multiple field coils producing the magnetic field, then the knife-blade heat shield of one coil will scatter neutrons and radiate thermal heat onto the heat shields of the other field coils, compromising their ability to shed heat. Consequently, many designs use only a single field coil despite the loss of efficiency. Those torch drives that use multiple field coils typically space them at distances significantly larger than the length of the knife-blade shield.

    As previously mentioned, the tungsten knife-blade heat shield that protects the field coils will glow very brightly. At 3000 K, it will radiate 4.6 MW/m2 of heat as thermal radiation. As an example, consider a torch drive with 2 cm wide field coils, 4 meter long knife-blade heat shields that are exposed to 90 GW/m2, and 10 MW of neutral radiation produced by the drive pulses. At this rated intensity, the tip of the heat shield can be as close as 9 meters to the detonation point. This produces a radiating disk with an outer radius of 13 meters and an inner radius of 9 meters, which will therefore radiate 2.6 GW of heat and 76 Glm of luminous power combined from its front and back. The apparent brightness will depend on the angle of the disk with respect to the observer, but unless the disk is edge-on, the unaided dark-adapted eye of a baseline human could detect the disk at a distance of approximately 1 Gm (gigameter) and would appear as an apparent magnitude 0 point of light at approximately 0.05 Gm. For comparison, the distance between Sol and old Earth is approximately 150 Gm. A dedicated 1 meter aperture early alarm scanning scope could detect the disk at approximately 600 Tm (terameter) with 1 kilosecond exposure time. For comparison, this is over 1% of the distance from Sol to its nearest stellar neighbor, Alpha Centauri.


    Constant Acceleration Equations

    Conventional rocket engines are so weak that their burn times are generally measured in minutes. They spend lots of time coasting. Torchships on the other hand can burn for hours or days, up to burning for the entire duration of the trip.

    Constant Acceleration Or Death

    As a side note, since these equations assume a constant acceleration, the rocket's thrust will have to be continually reduced as the expended propellant continually reduces the spacecraft mass.

    Remember that acceleration is equal to rocket thrust divided by spacecraft mass. If the mass goes down the acceleration will go up. To keep the acceleration the same you'll have to manually reduce the thrust. You'd want to do this anyway or the rising acceleration will do a Solomon Epstein on the crew and squish them like bugs.

    Constant-acceleration Delta-V

    To calculate the deltaV resulting from accelerating at a fixed rate for a certain period of time:

    plainVanillaDeltaV = A * t


    • plainVanillaDeltaV = deltaV result (m/s)
    • A = acceleration (m/s2)
    • t = duration of acceleration (seconds)

    Obviously if you want to calculate duration or acceleration level:

    t = plainVanillaDeltaV / A

    A = plainVanillaDeltaV / t


    (ed note: Gabriel GAB Fonseca derived equations to calculate the total distance traveled by a constantly accelerating rocket taking into account the change in mass as the propellant is expended. Understand that the Distance is how far from the starting point the spacecraft will be when the tanks go dry or the burn ends. After that the spacecraft will continue to coast at whatever its plainVanillaDeltaV is, gradually increasing the distance traveled.)

    (ed note: What follows is Mr. Fonseca's explanation)

    To calculate the distance travelled by a rocket, one needs to start somewhere. That somewhere should, under ideal circumstances, be an equation that our aspiring rocketeer actually knows. In my case, that equation would be the formula for the acceleration of a rocket as a function of time, as seen bellow:

    Eq.1: a(t) = (mDot * Ve) / (M - mDot * t)


    a(t) = acceleration of a rocket as a function of time
    mDot = propellant mass flow
    Ve = engine exhaust velocity
    M = rocket mass when full
    t = burn duration

    Fun fact to those that don't speak Calculeese: Velocity is nothing more than the derivative of distance by time. The version we all know and love from high school, V = d/t, is actually just the formula for the average speed of an object. If you want to know the instantaneous speed of an object, that is, it's speed at an exact given moment, you have to do it Newton style, so we go and take that derivative and find V. (Calculus looks harder than it actually is, but we don't really need to go into all the nitty-gritty details — for now just take my word for it.)

    But wait, there's more! Do you know what happens if we take the derivate of V as a function of time? Why, we get the object's acceleration! I'll be darned, so it turns out that the acceleration is the double derivative of distance by time!

    You might be asking, "sure, that's all well and good, but how does this help us?". Why, we now have a relation that connects distance and acceleration: d'' = a (Those apostrophes mean we're taking the double derivative of d or distance). So now we've got ourselves a happy little equation we can use to find out the formula for distance as a function of time! We just have solve the equation we just found backwards, using the inverse operation of the derivative.

    "But what is the inverse operation of the derivate?", you ask? Well, that would be the integral, that... No, wait, don't run! Don't worry, you won't be doing any of this heavy duty calculus yourself, I promise you, ok?

    So, where were we? Oh, yes: To find out distance as a function of time, we'll have to integrate the acceleration as a function of time, twice. To aid me in demonstrating these equations, we'll have with us a little test rocket with known parameters — our rocket will have:

    Mass when Full (M) = 5000 Kg
    Mass when Empty (Me) = 1000 Kg
    Engine's Mass Flow (mDot) = 40 Kg/s
    Exhaust Velocity (Ve) = 9320 m/s

    With these, we can fill in the equation for Acceleration as a function of time — after one last thing, that is: We need to know how long our rocket takes to burn completely dry. This is rather simple; we just subtract from the rocket's full mass it's mass when completely empty, to find out how much fuel it has (5000Kg - 1000Kg = 4000 Kg). Knowing that every second 40 Kilograms are drained from our rocket, we can find out how many second it takes to drain all our fuel and propellant (4000/40 = 100). We now know that the total burn time of our test rocket is 100 seconds.

    So, plugging our values into the formula, we have:

    Eq.1a: a(t) = (40 * 9320)/(5000 - 40t)

    To find out what the rocket's acceleration was during a specific moment of it's burn, just plonk it's value in seconds after burn start on the little t to find out. If we graph this equation, with the x axis being time (in seconds) and the y axis being acceleration (in m/s2):

    All good so far, but now we need to integrate. Before we do that however, first I'll tell you a little secret: There are actually two ways of integrating something. There's the definite integral, where you have an upper and a lower limit to your function, and you have the indefinite integral, where the function is not limited and it just keeps going (There’s actually more to it than this, but the exact definition won’t be that important to our purposes). So, if we take the definite integral with our lower limit being the start of our burn (0 seconds) and the upper limit as the burn's end (100 seconds), after all the numbers have been properly crunched, we have the following result:

    Result 1a: 9320 * ln(5) = approximately 15000

    If you don't know how to integrate, but want to follow along, don't worry — you can integrate using Wolfram, the free computational engine. Just type in

    definite integral of (372800/(5000-40x)) lower limit 0 and upper limit 100

    and presto! Out comes our solution.

    "Wait a second… this looks oddly familiar". Well, it should! That's none other than Konstantin Tsiolkovsky's Rocket Equation, and if you're interested in spacecraft at all, you know this formula by heart! Our outputted formula has an exhaust velocity (9320) multiplied by the natural logarithm of a rocket's mass ratio (5), just like the rocket equation! It turns out that the math we just did is exactly what Tsiolkovsky did on the early days of the 20th century. We have calculated our test rocket's ΔV the hard way.

    If we are stumbling into rocket equations, we must surely be walking in the right direction! So now, if we take the indefinite integral of the formula without our test values inputed, we should get…

    Result 1b: -Ve * ln(M - mDot * t)

    If you want to do this on Wolfram, we will have to do a bit of mathematical hocus-pocus. Wolfy can't understand the notation (variable names) we are using, so we will have to

    write M as a
    write mDot as b
    write Ve as c
    write t as x

    Putting it all together, the Wolfram unreadable (mDot * Ve)/(M - mDot * t) becomes the Wolfram readable:

    indefinite integral of (c*b)/(a - bx)

    "Wait, this isn't the rocket equation"!

    (ed note: because ΔV = -Ve * ln(M - mDot * t) does not look quite like ΔV = Ve * ln(M / Me) )

    Well, actually, it is the rocket equation — it's just an alternate form of writing it! No, seriously, I'm telling you it is. How, you ask? Well, uhm… Look, just take my word for it, ok?

    (If you really want a proof, derive this formula and the traditional form of the rocket equation by time, they will both yield the same answer — the formula we just integrated)

    So now we have the formula for the speed of a rocket as a function of time! We are halfway there! Our new equation now looks like this:

    Eq.2: V(t) = Ev * ln(M/(M - mDot * t))

    If we plug in our test rocket's values and plot it's graph, we have the following, with the x axis as time (in seconds) and the y axis as speed (in m/s):

    Now, to integrate this formula! Before we proceed to the last step however, I want to confess something to you: I had to do this bit twice. The first time, the humble potato that now writes to you tried to integrate the formula with the values for the test rocket plonked in, yielding a monstrous equation which actually didn't really work at all. I guess that what I'm trying to say is, basically: Kids, always check your math. It's only through our mistakes that we learn how to do it right.

    Anyway, now that that is out of the way, we can finally be about integrating our second equation. I warn you though, even if you do know calculus, this operation is not for the faint hearted — I have no shame in admitting that I had to rely on the computer to crunch the numbers for me on this one. If we input our equation into Wolfram:

    indefinite integral of c*ln(a/(a - bx))

    we finally get our result; what we've all been waiting for!

    Eq.3: d(t) = Ve * ((t - M/mDot) * ln(M/(M - mDot * t)) + t)

    Plugging our little test rocket's numbers in it, we get:

    9320 * ((100 - 5000/40) * ln(5000/(5000 - 40* 100)) + 100) =
    9320 * (-25 * ln(5) + 100) =
    9320 * (-40.236 + 100) =
    9320 * 59.764 =
    557,000.966 meters

    So the distance travelled by our test rocket in a full burn is ~ 557,001 meters! (557 Kilometres, or ~ 348 Miles, or 0.1858% of a light-second)

    If we graph this equation, with the x axis as time (in seconds) and the y axis as distance from the starting point (in meters):

    The magic is that this equation will work for any burn, as long as we don't start goofing around and plugging "t"s higher than the time to drain our tanks fully dry.

    So now, a worked example: I want to calculate the distance travelled by my test rocket when it burns to half it's ΔV reserve — so we need to know how much mass our rocket has after expending half of it's ΔV. Knowing our test rocket's full ΔV is ~15,000m/s:

    Eq.4a: 7499.98 = 9320 * ln(Mc/1000)

    Where Mc the current mass, that is, what we want to find. So, isolating Mc:

    Eq.4b: Mc = e(7499.98 / 9320) * 1000 = 2236.068 Kg

    Now that we know our current mass, we know that we have expended (5000 - 2236.098) 2763.932 Kg of fuel and propellant. If we divide this by our engine's mass flow, we get the time it took to expend this amount, which is (2763.932/40) 69.0983 seconds. If we plonk down this as our t, we get:

    9320 * ((69.0983 - 5000/40) * ln(5000/(5000 - 40 * 69.0983) + 69.0983) =
    9320 * ((-55.9017) * ln(2.236) + 69.0983) =
    9320 * 24.113 =
    224,734.5 meters from the starting point

    If we want to know the distance travelled when our test rocket, starting from rest, has half of it's ΔV remaining up to when it has a quarter of it left, however, what we do is:

    Eq.5a: 3749.99 = 9320 * ln(Md/1000)

    Where Md is the mass when we have 1/4 of our ΔV. Isolating Md:

    Eq.5b: Md = e(3749.99 / 9320) *1000 = 1495.349 Kg

    Which means that the rocket expended 5000 - 1495.349 = 3504.651 Kg of fuel and propellant, which takes 3504.651/40 = 87.6162 seconds. Knowing that to have only half of it's ΔV left our rocket must have already burned for 69.0983 seconds, we can find out the burn time to go from half to a quarter ΔV subtracting: (87.6162 - 69.0983 = 18.5179). So, plugging our mass at the start of the burn (when we have half our ΔV left in the tanks) and our burn time to the formula:

    9320 * ((18.5179 - 2236.068/40) * ln(2236.068/(2236.068 - 40 * 18.5179)) + 18.5179) =
    9320 * ((-37.3838) * ln(1.4953) + 18.5179) =
    32409.3 meters from the starting point

    But that result is for if our rocket was originally at rest at the start of the burn — so what if we were already moving at those 7500 m/s our half-empty ΔV reserve seems to imply? Simple, we just add in our starting speed times our burn's time!

    Eq.3d: 7500 * 18.5179 + 32409.3 = 171,293.5 meters travelled from the starting point

    As such, if we want to write the formula so that it can handle any situation, we have:

    Eq.6: Vo * t + Ve * ((t - Mc/mDot) * ln(Mc/(Mc - mDot) + t) = total distance travelled during prograde burn

    Well, for now that would be all; still, I'm sure that it is possible to re-write this formula to give you time as a function of distance — but that is something for another day!

    Brachistochrone Equations

    What exactly is a Brachistochrone anyway?

    A Hohmann orbit is the maximum transit time / minimum deltaV mission. Weak spacecraft use this because they do not have a lot of deltaV. All current space probes use Hohmann because currently there ain't no such thing as a strong propulsion system.

    A "Brachistochrone" is a minimum transit time / maximum deltaV mission. Torchships use this because they have lots of deltaV to spare.

    With a constant-acceleration mission, you aim your rocket at the destination and burn at a constant acceleration. The trouble is that almost everybody wants to come to a stop at their destination, nobody want to go streaking past it at twenty kilometers per second.

    Rocket Engine 101: you point your rocket exhaust in the exact opposite direction your ship is traveling in order to speed up (accelerate), you point the exhaust in the exact same direction in order to slow down (decelerate). So for a brachistochrone you accelerate constantly to the midpoint, flip over (Heinlein calls this a "skew flip", The Expanse calls it a "flip-and-burn"), and decelerate to the destination. You come to a halt at the destination and everybody is happy.

    Weaker torchships will accelerate up to a certain velocity, coast for a while, then flip and decelerate to rest.

    Brachistochrone missions are not only of shorter mission time, but they also are not constrained by launch windows the way Hohmann are. You can launch any time you like.

    You Can't Stop-on-a-Dime

    It is very important to note that it takes exactly the same amount of time to slow from a speed X to speed zero as it took to accelerate from speed zero to speed X. There is no way to jam on the brakes for a stop-on-a-dime halt. Other than "lithobraking" aka "crashing into a planet".

    People who played the ancient boardgame Triplanetary or the new game Voidstriker discovered this the hard way. They would spend five turns accelerating to a blinding speed, find out to their horror that it would take five more turns to slow down to a stop. A lot of novice players had their spaceships go streaking off the edge of the map or smacking into Mars fast enough to make a crater before they learned this.

    This is why a Brachistochrone accelerates to the mid-way point then decelerates the rest of the trip. The idea is to come to a complete stop at your destination, which means taking as much time to brake as you took to accelerate.

    In the first episode of The Expanse, the good ship Canterbury has been doing a leisurely burn accelerating from Saturn's rings to Ceres. Suddenly they have to come to a screeching halt in order to help a ship in distress called the Scopuli. In order to shed all their delta V they have to do a "flip-and-burn", flipping the ship over so the thrust opposes their vector. The "burn" part is where they put everybody into accelerations couches and pump them full of anti-acceleration drugs so they don't die under the multiple-gee thrust. Again if they decelerate with the same leisurly burn they accelerated with, it will take exactly the same time, ending up with them overshooting the Scopuli. By doing a violent deceleration they do not overshoot, but they do injury to the crew and severe damage to the ship.


    As RocketCat pointed out, torchships are unobtanium. Therefore, it doesn't mean the math no longer applies. You just need different equations.

    Please note these are nice simple equations, but only hold true if the spacecraft is not traveling at significant fractions of the speed of light (below 0.14 c). If it is, then you have to use the huge ugly relativistic equations with big nasty pointed teeth and covered in hyperbolic trigometric functions.

    Travel Distance

    First figure the distance between the two planets, say Mars and Terra.

    The "superior" planet is the one farthest from the Sun, and the "inferior" planet is nearest. The distance from the Sun and the superior planet is Ds and the distance between the Sun and the inferior is Di. No "church lady" jokes please.

    Obviously the maximum distance between the planets is when they are on the opposite sides of the Sun, the distance being Ds + Di. And of course the minimum is when they are on the same side, distance being Ds - Di. Upon reflection you will discover that the average distance between the planets is Ds. (when averaging, Di cancels out.)

    (maximumDistance + minimumDistance) / 2 = average distance
    ((Ds + Di) + (Ds - Di)) / 2 = average distance
    ((Ds) + (Ds)) / 2 = average distance
    2 * Ds / 2 = average distance
    Ds = average distance

    So either just use Ds or randomly choose a distance between the max and min.

    If you want to actually calculate the distance between two planets on a given date, be my guest but I'm not qualified to explain how. Do a web search for a software "orrery".

    TRANSIT TIME given distance and desired acceleration

    If you know the desired acceleration of your spacecraft (generally one g or 9.81 m/s2) and wish to calculate the transit time, the Brachistochrone equation is

    T = 2 * sqrt[ D/A ]


    • T = transit time (seconds)
    • D = distance (meters)
    • A = acceleration (m/s2)
    • sqrt[x] = square root of x

    Remember that

    • AU * 1.49e11 = meters
    • 1 g of acceleration = 9.81 m/s2
    • one-tenth g of acceleration = 0.981 m/s2
    • one one-hundredth g of acceleration = 0.0981 m/s2

    Divide time in seconds by

    • 3600 for hours
    • 86400 for days
    • 2592000 for (30 day) months
    • 31536000 for years

    Timothy Charters worked out the following equation. It is the above transit time equation for weaker spacecraft that have to coast during the midpoint

    T = ((D - (A * t^2)) / (A * t)) + (2*t)


    • T = transit time (seconds)
    • D = distance (meters)
    • A = acceleration (m/s2)
    • t = duration of acceleration phase (seconds), just the acceleration phase only, NOT the acceleration+deceleration phase.

    Note that the coast duration time is of course = T - (2*t)

    ACCELERATION given distance and desired transit time

    If you know the desired transit time and wish to calculate the required acceleration, the equation is

    A = (4 * D) / T2

    Keep in mind that prolonged periods of acceleration a greater than one g is very bad for the crew's health.

    Yes, it is supposed to be 2 * sqrt[ D/A ], NOT sqrt[ 2 * D/A ]

    Don't be confused. You might think that the Brachistochrone equation should be T = sqrt[ 2 * D/A ] instead of T = 2 * sqrt[ D/A ], since your physics textbook states that D = 0.5 * A * T^2. The confusion is because the D in the physics book refers to the mid-way distance, not the total distance.

    This changes the physics book equation from

    D = 0.5 * A * t^2


    D * 0.5 = 0.5 * A * t^2

    Solving for t gives us t = sqrt(D/A) where t is the time to the mid-way distance. Since it takes an equal amount of time to slow down, the total trip time T is twice that or T = 2 * sqrt( D/A ). Which is the Brachistochrone equation given above.



    Basic Brachistochrone Delta-V

    Now, just how brawny a rocket are we talking about? Take the distance and acceleration from above and plug it into the following equation:

    transitDeltaV = 2 * sqrt[ D * A ]


    • transitDeltaV = transit deltaV required (m/s)

    Detailed Delta-V

    The rocket on the Brachistochrone trip will also have to match orbital velocity with the target planet. In Hohmann orbits, this was included in the total.

    orbitalVelocity = sqrt[ (G * M) / R ]


    • orbitalVelocity = planet's orbital velocity (m/s)
    • G = 0.00000000006673 (Gravitational constant)
    • M = mass of primary (kg), for the Sun: 1.989e30
    • R = distance between planet and primary (meters) (semi-major axis or orbital radius)

    If you are talking about missions between planets in the solar system, the equation becomes

    orbitalVelocity = sqrt[1.33e20 / R ]

    Figure the orbital velocity of the start planet and destination planet, subtract the smaller from the larger, and the result is the matchOrbitDeltaV

    matchOrbitDeltaV = sqrt[1.33e20 / Di ] - sqrt[1.33e20 / Ds ]

    If the rocket lifts off and/or lands, that takes deltaV as well.

    liftoffDeltaV = sqrt[ (G * Pm) / Pr ]


    • liftoffDeltaV = deltaV to lift off or land on a planet (m/s)
    • G = 0.00000000006673
    • Pm = planet's mass (kg)
    • Pr = planet's radius (m)

    The total mission deltaV is therefore:

    totalDeltaV = sqrt(liftoffDeltaV2 + transitDeltaV2) + sqrt(matchOrbitDeltaV2 + landDeltaV2)

    Do a bit of calculation and you will see how such performance is outrageously beyond the capability of any drive system in the table I gave you.

    If you want to cheat, you can look up some of the missions in Jon Roger's Mission Table.


    For some ballpark estimates, you can use my handy-dandy Transit Time Nomogram. A nomogram is an obsolete mathematical calculation device related to a slide rule. It is a set of scales printed on a sheet of paper, and read with the help of a ruler or straight-edge. While obsolete, it does have some advantages when trying to visualize a range of solutions. Print out the nomogram, grab a ruler, and follow my example. You can also purchase an 11" x 17" poster of this nomogram at . Standard disclaimer: I constructed this nomogram but I am not a rocket scientist. There may be errors. Use at your own risk.

    Let's say that our spacecraft is 1.5 ktons (1.5 kilo-tons or 1500 metric tons). It has a single Gas-Core Nuclear Thermal Rocket engine (NTR-GAS MAX) and has a (totally ridiculous) mass ratio of 20. The equation for figuring a spacecraft's total DeltaV is Δv = Ve * ln[R]. On your pocket calculator, 98,000 * ln[20] = 98,000 * 2.9957 = 300,000 m/s = 300 km/s. Ideally this should be on the transit nomogram, but the blasted thing was getting crowded enough as it is. This calculation is on a separate nomogram found here.

    The mission is to travel a distance of 0.4 AU (about the distance between the Sun and the planet Mercury). Using a constant boost brachistochrone trajectory, how long will the ship take to travel that distance?

    Examine the nomogram. On the Ship Mass scale, locate the 1.5 kton tick mark. On the Engine Type scale, locate the NTR-GAS MAX tick mark. Lay a straight-edge on the 1.5 kton and NTR-GAS MAX tick marks and examine where the edge crosses the Acceleration scale. Congratulations, you've just calculated the ship's maximum acceleration:2 meters per second per second (m/s2).

    For your convenience, the acceleration scale is also labeled with the minimum lift off values for various planets.

    So we know our ship has a maximum acceleration of 2 m/s2 and a maximum DeltaV of 300 km/s. As long as we stay under both of those limits we will be fine.

    On the Acceleration scale, locate the 2 m/s2 tick mark. On the Destination Distance scale, locate the 0.4 AU tick mark. Lay a straight-edge on the two tick marks and examine where it intersects the Transit time scale. It says that the trip will take just a bit under four days.

    But wait! Check where the edge crosses the Total DeltaV scale. Uh oh, it says almost 750 km/s, and our ship can only do 300 km/s before its propellant tanks run dry. Our ship cannot do this trajectory.

    The key is to remember that 2 m/s2 is the ship's maximum acceleration, nothing is preventing us from throttling down the engines a bit to lower the DeltaV cost. This is where a nomogram is superior to a calculator, in that you can visualize a range of solutions.

    Pivot the straight-edge on the 0.4 AU tick mark. Pivot it until it crosses the 300 km/s tick on the Total DeltaV scale. Now you can read the other mission values: 0.4 m/s2 acceleration and a trip time of a bit over a week. Since this mission has parameters that are under both the DeltaV and Acceleration limits of our ship, the ship can perform this mission (we will assume that the ship has enough life-support to keep the crew alive for a week or so).

    Of course, if you want to have some spare DeltaV left in your propellant tanks at the mission destination, you don't have to use it all just getting there. For instance, you can pivot around the 250 km/s DeltaV tick mark to find a good mission. You will arrive at the destination with 300 - 250 = 50 km/s still in your tanks.


    Which reminded me that I had not worked out how long it would take to get home on a one-gee boost, if it turned out that I could not arrange automatic piloting at eight gees. I was stymied on getting out of the cell, I hadn't even nibbled at what I would do if I did get out (correction: when I got out), but I could work ballistics.

    I didn't need books. I've met people, even in this day and age, who can't tell a star from a planet and who think of astronomical distances simply as "big." They remind me of those primitives who have just four numbers: one, two, three, and "many." But any tenderfoot Scout knows the basic facts and a fellow bitten by the space bug (such as myself) usually knows a number of figures.

    "Mother very thoughtfully made a jelly sandwich under no protest." Could you forget that after saying it a few times? Okay, lay it out so:

    AASTEROIDS(assorted prices, unimportant)

    The "prices" are distances from the Sun in astronomical units. An A.U. is the mean distance of Earth from Sun, 93,000,000 miles. It is easier to remember one figure that everybody knows and some little figures than it is to remember figures in millions and billions. I use dollar signs because a figure has more flavor if I think of it as money — which Dad considers deplorable. Some way you must remember them, or you don't know your own neighborhood.

    Now we come to a joker. The list says that Pluto's distance is thirty-nine and a half times Earth's distance. But Pluto and Mercury have very eccentric orbits and Pluto's is a dilly; its distance varies almost two billion miles, more than the distance from the Sun to Uranus. Pluto creeps to the orbit of Neptune and a hair inside, then swings way out and stays there a couple of centuries — it makes only four round trips in a thousand years.

    But I had seen that article about how Pluto was coming into its "summer." So I knew it was close to the orbit of Neptune now, and would be for the rest of my life-my life expectancy in Centerville; I didn't look like a preferred risk here. That gave an easy figure — 30 astronomical units.

    Acceleration problems are simple s=1/2 at2; distance equals half the acceleration times the square of elapsed time. If astrogation were that simple any sophomore could pilot a rocket ship — the complications come from gravitational fields and the fact that everything moves fourteen directions at once. But I could disregard gravitational fields and planetary motions; at the speeds a wormface ship makes neither factor matters until you are very close. I wanted a rough answer.

    I missed my slipstick. Dad says that anyone who can't use a slide rule is a cultural illiterate and should not be allowed to vote. Mine is a beauty — a K&E 20" Log-log Duplex Decitrig. Dad surprised me with it after I mastered a ten-inch polyphase. We ate potato soup that week — but Dad says you should always budget luxuries first. I knew where it was. Home on my desk.

    No matter. I had figures, formula, pencil and paper.

    First a check problem. Fats had said "Pluto," "five days," and "eight gravities."

    It's a two-piece problem; accelerate for half time (and half distance); do a skew-flip and decelerate the other half time (and distance). You can't use the whole distance in the equation, as "time" appears as a square — it's a parabolic. Was Pluto in opposition? Or quadrature? Or conjunction? Nobody looks at Pluto — so why remember where it is on the ecliptic? Oh, well, the average distance was 30 A.U.s — that would give a close-enough answer. Half that distance, in feet, is: 1/2 × 30 × 93,000,000 × 5280. Eight gravities is: 8 × 32.2 ft./sec./sec. — speed increases by 258 feet per second every second up to skew-flip and decreases just as fast thereafter.

    So — 1/2 × 30 × 93,000,000 × 5280 = 1/2 × 8 × 32.2 x t2 — and you wind up with the time for half the trip, in seconds. Double that for full trip. Divide by 3600 to get hours; divide by 24 and you have days. On a slide rule such a problem takes forty seconds, most of it to get your decimal point correct. It's as easy as computing sales tax.

    It took me at least an hour and almost as long to prove it, using a different sequence — and a third time, because the answers didn't match (I had forgotten to multiply by 5280, and had "miles" on one side and "feet" on the other — a no-good way to do arithmetic) — then a fourth time because my confidence was shaken. I tell you, the slide rule is the greatest invention since girls.

    But I got a proved answer. Five and a half days. I was on Pluto.

    (Ed note: I learned it as


    In Slide Rule terminology: K&E is Keuffel & Esser, noted manufacturer of quality slide rules. 20 inches is twice the size and accuracy of a standard slide rule. Log-log means the rule possesses expanded logarithmic scales. Duplex means there are scales on both sides of the rule and the cursor is double sided. Decitrig means the rule possesses decimal trigometric scales.)

    From HAVE SPACE SUIT - WILL TRAVEL by Robert A. Heinlein, 1958

    Thanks to Charles Martin for this analysis:

    In Heinlein's short story "Sky Lift", the torchship on an emergency run to Pluto colony does 3.5 g for nine days and 15 hours. 3.5 g is approximately 35 m/s2 and 9d15h is 831,600 seconds. 35 m/s2 * 831,600 s = 29,100,000 m/s total deltaV.

    Assume a mass ratio of 4. Most of Heinlein's ships had a mass ratio of 3, 4 is reasonable for an emergency trip.

    Ve = Δv / ln[R] so 29,100,000 / 1.39 = 21,000,000 m/s exhaust velocity or seven percent of the speed of light.

    A glance at the engine table show that this is way up there, second only to the maximum possible Antimatter Beam-Core propulsion, and twice the maximum of Inertial Confinement Fusion. If Heinlein's torchship can manage a Ve of ten percent lightspeed it can get away with a mass ratio of 3.

    Charles Martin

         “How high, sir?”
         Berrio hesitated. “Three and one-half gravities.”
         Three and a half g’s! That wasn’t a boost — that was a pullout. Joe heard the surgeon protest, “I’m sorry, sir, but three gravities is all I can approve.”
         Berrio frowned. “Legally, it’s up to the captain. But three hundred lives depend on it.”

         Kleuger said, “Doctor, let’s see that curve.” The surgeon slid a paper across the desk; Kleuger moved it so that Joe could see it. “Here’s the scoop, Appleby—”
         A curve started high, dropped very slowly, made a sudden “knee” and dropped rapidly. The surgeon put his finger on the “knee.” “Here,” he said soberly, “is where the donors are suffering from loss of blood as much as the patients. After that it’s hopeless, without a new source of blood.”
         “How did you get this curve?” Joe asked.
         “It’s the empirical equation of Larkin’s disease applied to two hundred eighty-nine people.”
         Appleby noted vertical lines each marked with an acceleration and a time. Far to the right was one marked: “1 g—18 days” That was the standard trip; it would arrive after the epidemic had burned out. Two gravities cut it to twelve days seventeen hours; even so, half the colony would be dead. Three g’s was better but still bad. He could see why the Commodore wanted them to risk three-and-a-half kicks; that line touched the “knee,” at nine days fifteen hours. That way they could save almost everybody, but, oh, brother!
         The time advantage dropped off by inverse squares. Eighteen days required one gravity, so nine days took four, while four-and-a-half days required a fantastic sixteen gravities. But someone had drawn a line at “16 g—4.5 days.” “Hey! This plot must be for a robot-torch — that’s the ticket! Is there one available?”
         Berrio said gently, “Yes. But what are its chances?”

         Joe shut up. Even between the inner planets robots often went astray. In four-billion-odd miles the chance that one could hit close enough to be caught by radio control was slim. “We’ll try,” Berrio promised. “If it succeeds, I’ll call you at once.” He looked at Kleuger. “Captain, time is short. I must have your decision.”
         Kleuger turned to the surgeon. “Doctor, why not another half gravity? I recall a report on a chimpanzee who was centrifuged at high g for an amazingly long time.”
         “A chimpanzee is not a man.”
         Joe blurted out, “How much did this chimp stand, Surgeon?”
         “Three and a quarter gravities for twenty-seven days.”
         “He did? What shape was he in when the test ended?”
         “He wasn’t,” the doctor grunted.

         The ship was built for high boost; controls were over the pilots’ tanks, where they could be fingered without lifting a hand. The flight surgeon and an assistant fitted Kleuger into one tank while two medical technicians arranged Joe in his. One of them asked, “Underwear smooth? No wrinkles?”
         “I guess.”
         “I’ll check.” He did so, then arranged fittings necessary to a man who must remain in one position for days. “The nipple left of your mouth is water; the two on your right are glucose and bouillon.”
         “No solids?”
         The surgeon turned in the air and answered, “You don’t need any, you won’t want any, and you mustn’t have any. And be careful in swallowing.”
         “I’ve boosted before.”
         “Sure, sure. But be careful.”
         Each tank was like an oversized bathtub filled with a liquid denser than water. The top was covered by a rubbery sheet, gasketed at the edges; during boost each man would float with the sheet conforming to his body. The Salamander being still in free orbit, everything was weightless and the sheet now served to keep the fluid from floating out. The attendants centered Appleby against the sheet and fastened him with sticky tape, then placed his own acceleration collar, tailored to him, behind his head.

         The room had no ports and needed none. The area in front of Joe’s face was filled with screens, instruments, radar, and data displays; near his forehead was his eyepiece for the coelostat. A light blinked green as the passenger tube broke its anchors; Kleuger caught Joe’s eye in a mirror mounted opposite them. “Report, Mister.”
         “Minus seven’ minutes oh four. Tracking. Torch warm and idle. Green for light-off.”
         “Stand by while I check orientation.” Kleuger’s eyes disappeared into his coelostat eyepiece.

         When the counter flashed the last thirty seconds he forgot his foregone leave. The lust to travel possessed him. To go, no matter where, anywhere go! He smiled as the torch lit off.
         Then weight hit him.
         At three and one-half gravities he weighed six hundred and thirty pounds. It felt as if a load of sand had landed on him, squeezing his chest, making him helpless, forcing his head against his collar. He strove to relax, to let the supporting liquid hold him together. It was all right to tighten up for a pullout, but for a long boost one must relax. He breathed shallowly and slowly; the air was pure oxygen, little lung action was needed. But he labored just to breathe. He could feel his heart struggling to pump blood grown heavy through squeezed vessels. This is awful! he admitted. I’m not sure I can take it. He had once had four g for nine minutes but he had forgotten how bad it was.

         Joe then found that he had forgotten, while working, his unbearable weight. It felt worse than ever. His neck ached and he suspected that there was a wrinkle under his left calf. He wiggled in the tank to smooth it, but it made it worse.

         He tried to rest — as if a man could when buried under sandbags.
         His bones ached and the wrinkle became a nagging nuisance. The pain in his neck got worse; apparently he had wrenched it at light-off. He turned his head, but there were just two positions — bad and worse. Closing his eyes, he attempted to sleep. Ten minutes later he was wider awake than ever, his mind on three things, the lump in his neck, the irritation under his leg, and the squeezing weight.
         Look, bud, he told himself, this is a long boost. Take it easy, or adrenalin exhaustion will get you. As the book says, “The ideal pilot is relaxed and unworried. Sanguine in temperament, he never borrows trouble.” Why, you chair-warming so-and-so! Were you at three and a half g’s when you wrote that twaddle?

         The integrating accelerograph displayed elapsed time, velocity, and distance, in dead-reckoning for empty space. Under these windows were three more which showed the same by the precomputed tape controlling the torch; by comparing, Joe could tell how results matched predictions. The torch had been lit off for less than seven hours, speed was nearly two million miles per hour and they were over six million miles out. A third display corrected these figures for the Sun’s field, but Joe ignored this; near Earth’s orbit the Sun pulls only one two-thousandth of a gravity — a gnat’s whisker, allowed for in precomputation. Joe merely noted that tape and D.R. agreed; he wanted an outside check.

         His ribs hurt, each breath carried the stab of pleurisy. His hands and feet felt “pins-and-needles” from scanty circulation. He wiggled them, which produced crawling sensations and wearied him. So he held still and watched the speed soar. It increased seventy-seven miles per hour every second, more than a quarter million miles per hour every hour. For once he envied rocketship pilots; they took forever to get anywhere but they got there in comfort.
         Without the torch, men would never have ventured much past Mars. e = mc2, mass is energy, and a pound of sand equals fifteen billion horsepower-hours. An atomic rocketship uses but a fraction of one percent of that energy, whereas the new torchers used better than eighty percent. The conversion chamber of a torch was a tiny sun; particles expelled from it approached the speed of light.

         “Oh, there’s one thing I don’t understand, uh, what I don’t understand is, uh, this: why do I have to go, uh, to the geriatrics clinic at Luna City? That’s for old people, uh? That’s what I’ve always understood — the way I understand it. Sir?”
         The surgeon cut in, “I told you, Joe. They have the very best physiotherapy. We got special permission for you.”
         Joe looked perplexed. “Is that right, sir? I feel funny, going to an old folks’, uh, hospital?”
         “That’s right, son.”
         Joe grinned sheepishly. “Okay, sir, uh, if you say so.”
         They started to leave. “Doctor — stay a moment. Messenger, help Mr. Appleby.”
         “Joe, can you make it?”
         “Uh, sure! My legs are lots better — see?” He went out, leaning on the messenger.

         Berrio said, “Doctor, tell me straight: will Joe get well?”
         “No, sir.”
         “Will he get better?’
         “Some, perhaps. Lunar gravity makes it easy to get the most out of what a man has left.”
         “But will his mind clear up?”
         The doctor hesitated. “It’s this way, sir. Heavy acceleration is a speeded-up aging process. Tissues break down, capillaries rupture, the heart does many times its proper work. And there is hypoxia, from failure to deliver enough oxygen to the brain.”
         The Commodore struck his desk an angry blow. The surgeon said gently, “Don’t take it so hard, sir.”
         “Damn it, man — think of the way he was. Just a kid, all bounce and vinegar — now look at him! He’s an old man — senile.”
         “Look at it this way,” urged the surgeon, “you expended one man, but you saved two hundred and seventy.”

    From SKY LIFT by Robert Heinlein (1953)

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