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 unobtainium, 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 unobtainium, even though we can't build it yet, we can calculate what it can do just fine.

[radio static noises followed by blaring heavy metal]

I can't thrive on ion drive.
I need that rocket to burn.
Kick the ass, give it the gas.
I've got that need... for I... s... p.

[guitar riff]

Solar sail is an epic fail.
I need that rocket to burn.
Full on thrust is what I lust.
I've got that need... for I... s... p.

(Can you say) Constant acceleration?

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.

• 

• 

(ed note: The liner Pegasus is a conventional ion-drive spacecraft using finely divided dust as propellant. Like most non-torchships it has barely enough propellant to perform the mission.

The Federal cruiser Acheron is an honest-to-Heinlen torchship, using a hand-waving reactionless drive.)

"That's the background, but now it's only the less important part of the story. There are weapons here as well. Some have been completed and tested, others are waiting for the final adjustments. I'm bringing the key components for what may be the decisive one. That's why Earth may owe you a greater debt than it can ever pay. Don't interrupt—we're nearly there and this is what I really want to tell you. The radio was not telling the truth about that twenty hours of safety. That's what the Federation wants us to believe, and we hope they go on thinking we've been fooled. But we've spotted their ships, and they're approaching ten times as fast as anything that's ever moved through space before. I'm afraid they've got a fundamental new method of propulsion—I only hope it hasn't given them new weapons as well. We've not much more than three hours before they get here, assuming they don't step up their speed still further. You could stay, but for your own safety I advise you to turn around and drive like hell back to the Observatory. If anything starts to happen while you're still out in the open, get under cover as quickly as possible. Go down into a crevasse—anywhere you can find shelter—and stay there until it's all over. Now good-by and good luck. I hope we have a chance of meeting again, when this business is finished."

---

THE LINER Pegasus, with three hundred passengers arid a crew of sixty, was only four days out from Earth when the war began and ended. For some hours there had been a great confusion and alarm on board, as the radio messages from Earth and Federation were intercepted. Captain Halstead had been forced to take firm measures with some of the passengers, who wished to turn back rather than go on to Mars and an uncertain future as prisoners of war. It was not easy to blame them; Earth was still so close that it was a beautiful silver crescent, with the Moon a fainter and smaller echo beside it. Even from here, more than a million kilometers away, the energies that had just flamed across the face of the Moon had been clearly visible, and had done little to restore the morale of the passengers.

They could not understand that the law of celestial mechanics admit of no appeal. The Pegasus was barely clear of Earth, and still weeks from her intended goal. But she had reached her orbiting speed, and had launched herself like a giant projectile on the path that would lead inevitably to Mars, under the guidance of the sun's all-pervading gravity. There could be no turning back: that would be a maneuver involving an impossible amount of propellant. The Pegasus carried enough dust in her tanks to match velocity with Mars at the end of her orbit, and to allow for reasonable course corrections en route. Her nuclear reactors could provide energy for a dozen voyages—but sheer energy was useless if there was no propellant mass to eject. Whether she wanted to or not, the Pegasus was headed for Mars with the inevitability of a runaway streetcar. Captain Halstead did not anticipate a pleasant trip.

The words MAYDAY, MAYDAY came crashing out of the radio and banished all other preoccupations of the Pegasus and her crew. For three hundred years, in air and sea and space, these words had alerted rescue organizations, had made captains change their course and race to the aid of stricken comrades. But there was so little that the commander of a spaceship could do; in the whole history of astronautics, there have been only three cases of a successful rescue operation in space.

There are two main reasons for this, only one of which is widely advertised by the shipping lines. Any serious disaster in space is extremely rare; almost all accidents occur during planetfall or departure. Once a ship has reached space, and has swung into the orbit that will lead it effortlessly to its destination, it is safe from all hazards except internal, mechanical troubles. Such troubles occur more often than the passengers ever know, but are usually trivial and are quietly dealt with by the crew. All spaceships, by law, are built in several independent sections, any one of which can serve as a refuge in an emergency. So the worst that ever happens is that some uncomfortable hours are spent by all while an irate captain breathes heavily down the neck of his engineering officer.

The second reason why space rescues are so rare is that they are almost impossible, from the nature of things. Spaceships travel at enormous velocities on exactly calculated paths, which do not permit of major alterations—as the passengers of the Pegasus were now beginning to appreciate. The orbit any ship follows from one planet to another is unique; no other vessel will ever follow the same path again, among the changing patterns of the planets. There are no "shipping lanes" in space and it is rare indeed for one ship to pass within a million kilometers of another. Even when this does happen, the difference of speed is almost always so great that contact is impossible.

All these thoughts flashed through Captain Halstead's mind when the message came down to him from Signals. He read the position and course of the distressed ship—the velocity figure must have been garbled in transmission, it was so ridiculously high. Almost certainly, there was nothing he could do—they were too far away, and it would take days to reach them.

Then he noticed the name at the end of the message. He thought he was familiar with every ship in space, but this was a new one to him. He stared in bewilderment for a moment before he suddenly realized just who was calling for his assistance. Enmity vanishes when men are in peril on sea or in space. Captain Halstead leaned over his control desk said: "Signals! Get me their captain."

"He's on circuit, sir. You can go ahead." Captain Halstead cleared his throat. This was a novel experience, and not a pleasant one. It gave him no sort of satisfaction to tell even an enemy that he could do nothing to save him.

"Captain Halstead, Pegasus, speaking," he began. "You're too far away for contact. Our operational reserve is less than ten kilometers a second. I've no need to compute, I can see it's impossible. Have you any suggestions? Please confirm your velocity; we were given an incorrect figure."

The reply, after a four-second time-lag that seemed doubly maddening in these circumstances, was unexpected and astonishing. (distance of 2 light-seconds, about 599,585 kilometers or twice the Terra-Luna distance)

"Commodore Brennan, Federal cruiser Acheron. I can confirm our velocity figure. We can contact you in two hours, and make all course corrections ourselves. (I calculate average speed of 83 km/s) We still have power, but must abandon ship in less than three hours. Our radiation shielding has gone, and the main reactor is becoming unstable. We've got manual control on it, and it will be safe for at least an hour after we reach you. But we can't guarantee it beyond then." Captain Halstead felt the scalp crawl at the back of his neck. He did not know how a reactor could became unstable, but he knew what would happen if one did. There were a good many things about the Acheron he did not understand—her speed, above all—but there was one point that emerged very clearly upon which Commodore Brennan must be left in no doubt.

"Pegasus to Acheron," he replied. "I have three hundred passengers aboard. I cannot hazard my ship if there is danger of an explosion."

"There is no danger, I can guarantee that. We will have at least five minutes' warning, which will give us ample time to get clear of you."

---

Then the ship's speakers announced, in an almost quietly conversational tone: Pegasus to Acheron. We've got all your men out of the locks. No casualties. A few hemorrhages. Give us five minutes to get ready for the next batch."

They lost one man on the last transfer. He panicked and they had to slam the lock shut without him, rather than risk the lives of all the others. It seemed a pity that they could not all have made it, but for the moment everyone was too thankful to worry about that.

There was only one thing still to be done. Commodore Brennan, the last man aboard the Acheron, adjusted the timing circuit that would start the drive in thirty seconds. That would give him long enough; even in his clumsy spacesuit he could get out of the open hatch in half that time. It was cutting it fine, but only he and his engineering officer knew how narrow the margin was.

He threw the switch and dived for the hatch. He had already reached the Pegasus when the ship he had commanded, still loaded with millions of kilowatt-centuries of energy, came to life for the last time and dwindled silently toward the stars of the Milky Way (more than 6.3×10¹⁸ joules, equivalent to 1.5 gigaton nuclear weapon).

The explosion was easily visible among all the inner planets. It blew to nothingness the last ambitions of the Federation, and the last fears of Earth.

(ed note: the article uses the term "starcruiser", which in context more or less means "torchship")

Wars in space will never be like Star Wars. “Starfighters” will not engage in dogfights with unlimited maneuverability and range. An actual conflict in space would be slow and deliberate, requiring prepositioning of weapons and meticulous planning.

Even defense analysts working on the matter understate the physical constraints on warfare in space. In a recent War on the Rocks article, Jeff Becker claims that the era of “starcruisers” — spacecraft whose maneuvering is not principally dictated by orbital mechanics — is closer than people think. While Becker examines some meaningful technological developments, his analysis — like other work on the topic — does not recognize the challenges and physics that would be involved in fighting a conflict in space.

Policymakers and defense planners need to have a realistic understanding of what is physically possible and practical. As we explain in a recent research paper and accompanying video, in any space war physical limitations will constrain both the movement of assets and overall strategy. Space is big and satellites maneuver slowly while following predictable trajectories. Major limitations such as these mean that starcruisers like the Millennium Falcon or Galactica will never exist.

Physical Constraints in Space

Spacecraft and space weapons cannot defy physics. As they develop their ideas about spacepower, strategists should internalize five major points and use them to attain advantages, rather than working against physical realities.

Satellites move quickly. Those in commonly used circular orbits move at speeds of 6,700 to 18,000 miles per hour (three to eight kilometers per second), depending on their altitude. In comparison, an average bullet only travels about 1,700 miles per hour (0.75 kilometers per second). To an Earth-bound observer, satellite speeds might suggest that science fiction is the right way to imagine space war, but speed is not actually the biggest driver of fighting in space.

Satellites move predictably. Due to orbital dynamics, satellites exhibit a strict relationship between altitude, speed, and orbit shape. Gravity dictates that satellites at lower altitudes always move more quickly than those at higher altitudes. This makes their path predictable — and deviating from an orbit is costly and slow — which means that engagements, whether peaceful or aggressive, can be planned long ahead of time regardless of the speed at which satellites are traveling.

Space is big. The volume of space between low Earth orbit and geostationary orbit is about 50 trillion cubic miles (200 trillion cubic kilometers). That is 190 times bigger than the volume of Earth. Even if satellites and spacecraft are designed to have more energy for maneuverability, distances in space are so big that extensive maneuvering will remain painstakingly slow.

Timing is everything. Within the confines of the Earth’s atmosphere, airplanes, tanks, and ships can move in multiple directions. By comparison, satellites are always moving in a circular or elliptical path due to the gravitational pull of Earth. The nature of conflict often requires two competing weapon systems to get close to each other, in this case “on-orbit.” Therefore, timing is everything.

Satellites maneuver slowly. While satellites move quickly along their orbit, purposeful maneuvers to a different orbit are relatively slow. That is because space is big. Since gravity dictates the motion of a satellite, deviating from the prescribed orbit requires the use of an engine to maneuver. However, those deliberate maneuvers can only happen at certain points in an orbit. Adopting a mindset and a strategy of slow and deliberate movements is needed, rather than thinking about flashy “light speed” jumps.

The X-37B: Not a Forerunner to Starcruisers

A starcruiser will not exist because it would have to defy these physical constraints, especially the realities that satellites move predictably and maneuver slowly. Becker argues that two current programs — the X-37B and SpaceX’s Starship — show that starcruisers will be here soon enough that analysts need to begin thinking through the implications for spacepower. But, in reality, there’s no prospect of a technological breakthrough that will allow spacecraft to effectively “beat” physics. Neither the X-37B nor the Starship show that the fundamentals of spacepower will change anytime soon. Those programs should not give false hope to those who dream big about space.

The X-37B spacecraft offers real benefits because of its overall reusability and its ability to conduct experiments and then return them to Earth. It is unique in that it operates in low Earth orbit and leverages the atmosphere to maneuver. That allows it to move in unpredictable ways, but it is not a precursor to starcruisers. To understand why, consider the concept of “delta-v,” which is the change in velocity required to alter a spacecraft’s trajectory. Since there is a strict relationship between a satellite’s velocity and its orbit, a change in velocity will necessarily change the orbit. To deviate from the path determined by orbital dynamics, delta-v is required. The amount of delta-v carried by a satellite is known as its “delta-v budget.” This metric is independent of the satellite’s size and propellant type. Thus, contrary to what Becker suggests, there is no concept of delta-v efficiency. What matters is the overall delta-v capacity of a satellite and how quickly it can be applied to accelerate the satellite.

The X-37B has a total delta-v budget of 3,100 meters per second for a mission. That is not radically different to the delta-v budget of a typical low-thrust geostationary orbit communications satellite. That type of satellite spends 1,800 meters per second of delta-v just to get to its orbit and then requires about 55 meters per second per year to maintain its orbit. If we also include what is needed for re-positioning, disposal, and other maneuvers, a typical geostationary orbit satellite’s delta-v budget approaches 4,000 meters per second. Even more strikingly, the X-37B’s delta-v budget is small compared to the commercially available Boeing 702SP satellite bus, which first flew in 2012 and carries up to 6,800 meters per second of delta-v. Nobody thinks these satellites are forerunners to starcruisers. Neither is the X-37B.

The Starship: Evolutionary Technology

SpaceX’s Starship is very exciting, but for the purposes of space warfare it is evolutionary not revolutionary. Starship is a huge, reusable rocket and spacecraft. It is able to refuel on-orbit, which provides it with a delta-v budget of 6,900 meters per second and the maneuverability that comes with it. If Starship was fully topped off outside of the Earth’s atmosphere, this uniquely large spacecraft would be able to perform more maneuvers and enjoy a longer lifetime. While these gains are real, they do not eliminate the constraints on maneuvering due to orbital mechanics. Making decisions about the number and location of fueling depots would also be a complex challenge, precisely because space is so big.

We are all for gaining refueling capabilities on-orbit, but for different reasons than one might think. Once the Space Force can have hundreds of millions of pounds of propellant in space waiting to refuel spacecraft, it will have the means to greatly improve freedom of movement and prolong the lifetime of satellites. But that will not make space operations analogous to air operations. Even Starship’s large delta-v budget is too small to enable it to match the orbit of many satellites. We calculated that if a fully loaded Starship is in an equatorial orbit (zero-degree inclination) at an altitude of 250 miles (400 kilometers), it would barely have enough delta-v to match the International Space Station’s orbit, at 51 degrees inclination. It definitely could not match satellites in sun-synchronous orbits, which have an inclination of about 98 degrees and are a common orbit for Earth observation satellites.

Among its major capabilities, the Space Force needs the ability to quickly launch payloads into their desired orbits and Starship could play an important role in performing that mission. But that need can also be met better by other systems. Recently, the Space Force demonstrated its ability to rapidly put satellites into orbit by using the air-launched Pegasus rocket. That rocket has a critical advantage over Starship: While Starship can currently only launch from two locations — Boca Chica, Texas, and Cape Canaveral, Florida — Pegasus can already be launched from a number of locations worldwide.

Science-fiction movies have conditioned people to believe that advanced propulsion systems enable travel in a straight line through space, unconstrained by orbital mechanics. However, our calculations show that in order to travel in a straight line from an altitude of 400 kilometers to 1,000 kilometers, a spacecraft would need at least 40,000 meters per second of delta-v split into two equal, instantaneous bursts about 30 seconds apart — one to start the maneuver and one to stop it. To go from 400 kilometers to geostationary orbit — an altitude of 35,786 kilometers — a spacecraft would need at least 2,400,000 meters per second of delta-v split into two equal, instantaneous bursts about 30 seconds apart. The spacecraft would also need to survive the immense forces generated by accelerating that fast. This is not happening in our lifetime. Nor is it clear why this would be desirable. At the very least, the Starship, just like the X-37B, is nowhere close to achieving this type of maneuver. Even if they were, supporting infrastructure would be required to fight in space. Currently, the Space Force does not have the communications, navigation, and other capabilities necessary for the effective use of a starcruiser or other advanced spacecraft. That should not alarm American policymakers, precisely because starcruisers will not exist.

Starcruisers: The Stuff of Science Fiction

Without having delta-v budgets orders of magnitude greater than that currently being assumed for Starship, orbital mechanics bound what spacecraft can do. As we have written elsewhere, large applications of delta-v by a spacecraft in a low Earth orbit can move it into orbits that either hit Earth or that enter the Van Allen belts, a region of charged particles that can reduce the spacecraft’s lifetime or permanently damage it. By trying to be fast, a starcruiser would be forced to make many large corrective burns to avoid these fates, destroying any “efficiency” provided by the large delta-v load.

Discussions focused on delta-v overshadow another key component of conflict in space: time. Since space is big, maneuvering in space is slow, even with large delta-v budgets. Maneuvering to rendezvous with another satellite, whether to refuel or attack, requires the approaching satellite to match orbital planes and phasing with the other one. Both satellites will need to end up in the same spot, at the same time, with no velocity difference. Larger delta-v budgets certainly enable faster transfers to different orbits, but it may still take days for a satellite to reach its target. As the Space Force’s capstone doctrine emphasizes:

Constrained freedom of maneuver is a defining attribute of space operations. … Due to extreme velocities, the amount of energy required to reach a different orbit may be significant enough to render the option unfeasible or impractical.

By embracing the constrained nature of space movement, the Space Force and other space operators can use Starship-like capabilities to enhance their warfighting abilities, rather than fight needless battles against natural forces that will inevitably win.

Looking Ahead

Emerging space technologies are exciting. They promise to create more responsive and reusable launch vehicles and to provide refueling and repair opportunities on-orbit. However, contrary to the hopes of some defense analysts, they will not allow the Space Force to defy physics. Policy planning and strategizing should be based on reality and built upon an understanding of astrodynamics. Having spacecraft with massive amounts of fuel will certainly be beneficial to space operations, but that will not take space warfare into the realm of science fiction. Any conflict in space would be a slow and deliberate affair. An effective spacepower strategy should be developed with physics-based constraints in mind, not based on dreams of starcruisers.

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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

And yes, Orion is the one currently semi plausible drive that can deliver this level of power and performance. I say semi plausible because, broadly political issues aside, I suspect that Orion enthusiasts gloss over the engineering details of deliberately nuking yourself thousands of times. Like banging your head on the table, it only feels good when you stop.

---

Reasonably Fast Torchship
Wet Mass	1,000,000 kg
Dry Mass	500,000 kg
Mass Ratio	2
Exhaust Velocity	300,000 m/s
ΔV	200,000 m/s
Thrust	290,000 N
Starting Acceleration	0.29 m/s2 (30 mg)
Thrust Power	45 GW
Really Fast Torchship
Wet Mass	1,000,000 kg
Dry Mass	500,000 kg
Mass Ratio	2
Exhaust Velocity	3,000,000 m/s
ΔV	2,000,000 m/s
Thrust	3,000,000 N
Starting Acceleration	3 m/s2 (0.3 g)
Thrust Power	4.5 TW

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

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

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

---

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

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

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

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

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

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

---

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

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

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

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

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 Δ_v * 0.72 . See here.)

• 

• 

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 (F_sp) 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.

Say your propulsion system has an exhaust velocity of 5.4e6 m/s and a thrust of 2.5e6 N. Now F_p=(F*V_e)/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 A_f = 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.

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 a spherical hollow ball, with a hole for the exhaust nozzle. 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 you spread it over means the lesser amount of butter each slice gets.

E_l 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 the engine won't melt.

The attenuation due to the inverse square law is:

ISLA = (4 * π * R_c²)

where:

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

The heat load on the reaction chamber walls is:

C_l = H / ISLA

where:

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

Merging the equations:

C_l = H / (4 * π * R_c²)

Solve for R_c:

C_l = H / (4 * π * R_c²)
4 * π * R_c² = H / C_l
4 * π * R_c² = H * (1/C_l)
R_c² = (H * (1/C_l)) / (4 * π)
R_c² = H * (1/C_l) * (1 / (4 * π))
R_c = sqrt[H] * sqrt[(1/C_l) * (1 / (4 * π))]

Looking at the last equation, take the right half and swap C_l for E_l to get:

A_f = sqrt[(1/E_l) * (1 / (4 * π))]

and the entire equation is where we get:

R_c = sqrt[H] * A_f

which is what we were trying to derive. QED.

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(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)

where:

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):

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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

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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.

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"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)

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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)

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"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):

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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!

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

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

Eq.3a:
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):

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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:

Eq.3b:
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:

Eq.3c:
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: V^o * 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!