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".
Weak propulsion systems have limited thrust and delta V. They have to use weak trajectories, most commonly Hohmann trajectories. Torchships, on the other hand, have large thrust and delta V. They can use expensive trajectories such as brachistochrone.
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.
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.
Fast Torchship High Gear Low Gear Wet Mass 1,000,000 kg Dry Mass 500,000 kg Mass Ratio 2 Exhaust
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
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
300,000 m/s ΔV 200,000 m/s Thrust 290,000 N Starting
Thrust Power 45 GW Really Fast Torchship Wet Mass 1,000,000 kg Dry Mass 500,000 kg Mass Ratio 2 Exhaust
3,000,000 m/s ΔV 2,000,000 m/s Thrust 3,000,000 N Starting
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).
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)
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.
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."
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.
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"), and decelerate to the destination. 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.
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.
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".
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
- 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)
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.
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.
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)
The rocket 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 = 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:
Mother MERCURY $.39 Very VENUS $.72 Thoughtfully TERRA $1.00 Made MARS $1.50 A ASTEROIDS (assorted prices, unimportant) Jelly JUPITER $5.20 Sandwich SATURN $9.50 Under URANUS $19.00 No NEPTUNE $30.00 Protest PLUTO $39.50
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.)
Thanks to Charles Martin for this analysis: