If one has a beefy, muscular spacecraft propulsion system, you can ignore all that Hohmann nonsense. But it better be really powerful. It will have to be some science-fictional torchship, capable of accelerating at one g for days at a time, with obscenely high delta V ratings, far beyond foreseeable technology. A drive that will run full tilt into John's Law. What is the actual definition of a torchship? Well, it is kind of vague. It more or less boils down to "unreasonably powerful." Personally as a rule of thumb I'd say it was a propulsion system with a thrust power of 150 gigawatts or higher. Thrust power is thrust times exhaust velocity then divided by two (see equations). The entries in the Drive Table are sorted by thrust power for your convenience.
(The term "Torchship" was coined by Robert Heinlein, and is featured in his stories Farmer in the Sky, Time for the Stars, Double Star, and "Sky Lift". 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 total-conversion mass-into-energy drive.
But the presence of torchships doesn't mean mathematics no longer applies. You just need different equations.
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.
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.)
Just 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".
A Hohmann orbit is the maximum transit time / minimum deltaV mission. A "Brachistochrone" is a minimum transit time / maximum deltaV mission is where you accelerate constantly to the midpoint, flip over ("skew flip"), and decelerate to the destination (Weaker spacecraft will accelerate up to a certain velocity, coast for a while, then 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 down from a speed X to speed zero as it took to accelerate from speed zero to speed X. 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 turns to slow down to a stop, and end up either streaking off the edge of the map or smacking into Mars at a speed high enough to make a crater. 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.
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 x 30 x 93,000,000 x 5280. Eight gravities is: 8 x 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 x 30 x 93,000,000 x 5280 = 1/2 x 8 x 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 My Very Educated Mother Just Served Us Nine Pumpkins. 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:
In "Sky Lift" and Double Star, the crew spent the days of high thrust in acceleration couches that were like advanced waterbeds (called "cider presses"). In The Mote in God's Eye by Larry Niven and Jerry Pournelle, the captain's chair had a built-in "relief tube" (i.e., a rudimentary urinal) for use during prolonged periods of multi-g acceleration. There were also a few motorized acceleration couches used by damage control parties who had to move around during high gs. Such mobile couches also appeared in Joe Haldeman's The Forever War.