[Spaceships] Drive economics

[AlexanderHowl]

A Hoffman trajectory is just a more general name for the types of manuevers that take advantage of the differencs in orbital velocities between two objects (a Hoffman transfer orbit is merely the most time efficient such trajectory). By the way, a SM+11 barge is only three times as massive as an SM+10 tug. As for the delta-v, the normal delta-v requirements to go from lunar orbit to Mars orbit is only 3.6 km/s (1.3 km/s if you are brave enough to use aerobreaking), so you can have a much lower delta-v requirement than a Hoffman transfer orbit.

[the-red-scare]

There really isn’t such a thing as a Hoffman anything, it’s Hohmann, as DaltonS pointed out more than once. But that’s really neither here or there.

[DaltonS]

Originally Posted by AlexanderHowl

A Hoffman trajectory is just a more general name for the types of manuevers that take advantage of the differencs in orbital velocities between two objects (a Hoffman transfer orbit is merely the most time efficient such trajectory). By the way, a SM+11 barge is only three times as massive as an SM+10 tug. As for the delta-v, the normal delta-v requirements to go from lunar orbit to Mars orbit is only 3.6 km/s (1.3 km/s if you are brave enough to use aerobreaking), so you can have a much lower delta-v requirement than a Hoffman transfer orbit.

A Quarterhorse-Class Deep Space Tug (TL9) is SM+9. Tugs have a history of being a lot smaller than the boats they push around. I just wanted to show that yes, it could be done with a published vehicle but it would take a bit of effort to do it realistically (which meant allowing the crew to go home).

I'm confused about your mention of Lunar orbit launches. I'm working from the numbers I've found in the Spaceships books and some Pyramid articles and I don't remember lunar orbit launches being mentioned at all there. I wouldn't be surprised if you are right (it does make sense in my head anyway) but I would like to know your sources for these numbers. Parking in lunar orbit or L5 feels right to me (Spaceships assumes LEO is the starting point) and knowing where your numbers come from would make me much more comfortable with them. (And yes, shaving a bit off my required ΔV would be nice too.)

Dalton “who doesn't mind learning new things” Spence

[AlexanderHowl]

I blame autocorrect. Anyway, there are delta-v maps on Wikipedia when you look up delta-v budget. Now, they are parts of the maps that work best with high thrust drive, but I do not think that the Lunar Orbit to Earth C3 spur would be one of them (you may want to add another 0.7 km/s just in case). In any case, you are talking about a range of trajectories that primarily close the distance through the difference in orbital velocities, meaning that the trip would take anywhere from 0.65 years to 1.3 years, depending on the relative positions of Earth and Mars at launch. I imagine that the shorter trips would probably cost more than the longer trips, though the average charge would be enough for reasonable profit.

In the case of the tug that I suggested, it would probably earn an average of $20M per month. Of that amount, it would spend $9M on standard expenses and $8M on hydrogen reaction mass, leaving $3M of profit. Assuming eight 'pushes' per month, it would charge $2.5M per push on average, meaning that a SM+11 barge would pay $105/ton of cargo capacity (assuming 16 cargo holds, 3 steel armor, and 1 external clamp). The barge would then be caught by another barge on the other end, who would likely charge another $105/ton, because no one would not want an unguided barge to attempt aerobreaking.

[Michael Thayne]

Originally Posted by Agemegos

What you need to use here are the economic concepts of "generalised cost", and "behavioural value of time" for passengers or "value of time in transit" for cargo.

I won't go into the derivation. Basically, passengers will choose the travel option that minimises the generalised cost of travel, where the generalised cost is the fare (inclusive of e.g. insurance, taxes, and incidental expenses) plus the travel time times the individual passenger's BVOT. Behavioural values of time ought to be similar to wage rates (after taxes and fringe benefits etc.), but there are complicating factors e.g. when there are deadlines, when working hours are not freely adjustable, etc..

Customers despatching cargo will seek to minimise a generalised cost that is basically the sum of freight (plus insurance, loading, port fees, duties etc.) plus their opportunity cost of capital times the time in transit, plus depreciation (in the case of a perishable cargo). The opportunity cost of capital ought to look a bit like the interest rate, but there are complications in the cases e.g. of restricted access to capital markets, tax deductibility of interest, premiums on interest for e.g. risk….

Wanted to second this. If the behavioral value of time thing is confusing to you, an example might help. I grew up with my entire extended family in every state but the one my parents settled down in. We visited most of those aunts, uncles, and grandparents somewhat frequently. In the case of relatives a few states over, we would drive, but for very far away relatives we would fly. I asked my parents about this, and they explained that it was actually cheaper to fly, because driving cross-country would have required my dad (a dentist) to take a lot of time off work and forgo all the money he could have made doing dentistry instead of sitting in a car.

So yes, for purposes of figuring out which drive is "better" for a passenger ship, I would treat all the passengers as if they were crew needing to be paid. Not that that actually ups the ticket price, since the passengers are paying in their time rather than actual cash.

Also note that if passengers find being in space really unpleasant (or if health effects can't be fully mitigated), this will push things towards faster methods of travel. The opposite happens if they regard space travel as a fun vacation, or if they can do some sort of work (office work?) while in transit.

[Agemegos]

Originally Posted by the-red-scare

There really isn’t such a thing as a Hoffman anything, it’s Hohmann, as DaltonS pointed out more than once. But that’s really neither here or there.

Further, the Hohmann transfer orbit minimises the delta-vee requirement, it is not time-efficient but rather slow.

[Anthony]

The basic way any simple transfer orbit works is that it's an elliptical orbit that intersects the orbits of both the source and the destination.

Any given shape of orbit has a minimum time it takes to do that once, and a number of degree it traverses doing so (it is possible to use a path that takes more than the minimum time, since your transfer orbit crosses each orbit multiple times, but doing so is not generally useful).

For any transfer orbit, you must time your departure such that you arrive at the destination orbit at the same time as the planet you're trying to reach. The appropriate timing occurs once per synodic period (2.15 years for Earth-Mars).

The Hohmann transfer orbit is the least-energy solution for this equation; it's an elliptical orbit with perihelion at the inner orbit and aphelion at the outer orbit, and a semi-major orbit that is the average of the two. In the case of Earth to Mars, it takes 0.71 years and progresses by 180 degrees. As the Earth shifts by 360 degrees per year, and Mars shifts by 191 degrees per year, during this time Earth progresses by 256 degrees, and Mars progresses by 136 degrees. Since we will arrive at 180 degrees from where we started, we need Mars to have started out 44 degrees ahead, and in the reverse direction, we need Earth to have started out 76 degrees behind.

Thus, on our outgoing path, we start with Mars 44 degrees ahead. When we arrive, Earth will be 76 degrees ahead. We need earth to be 76 degrees behind (208 degrees), and Earth overtakes Mars at 169 degrees per year , so we need to wait at Mars for (208/169)=1.23 years. When we arrive at Earth, Mars is 44 degrees behind and we need it to be 44 degrees ahead, so we need to wait at Earth for (272/169)=1.69 years. Total time is 1.42 years in transit, 2.92 years waiting, total 4.36 years (which probably means I made a rounding error, it should be 4.3 years).

If you want to go every period instead of every other period, you need a different path. I believe the Mars cycler model is just an orbital path with a total period equal to the synodic period (2.15 years), which works but has the flaw that one leg is very long. The other option is a significantly faster path, which I haven't done the math for at the moment.

[Jinumon]

Originally Posted by AlexanderHowl

No, that is the simplified assumption for costs (Spaceships 2, p. 31). The maintenance costs for cheap and very cheap ships are in addition to the 1.5% of the cost per month. With cheap spacecraft, this ends up being 2.5% of its cost per month. With very cheap spacecraft, this ends up being 5.5% of its cost per month.

But don't you only have to pay for those maintenance costs if you choose to ignore everything else, including fuel consumption, bank payments, crew salaries, etc? Or is the 1.5% just something you tack on because you don't think it's realistic to have crew maintain the ship with no cost for replacement parts/materials?

Jinumon

[AlexanderHowl]

The 1.5% includes the normal maintenance costs of good quality spaceships. It is talked about in Spaceships 2.

[Jinumon]

Originally Posted by AlexanderHowl

The 1.5% includes the normal maintenance costs of good quality spaceships. It is talked about in Spaceships 2.

I get that, but that's only if you use the "Fudging It" rule. Otherwise it just says, "It's assumed that the crew performs general maintenance." Spaceships 2, p. 31.

Jinumon