Well if your only sending bulk materials then you could place them in an oblong shaped travel container like a rhombus or caltrop shape and place airbags or a woven TPS over the exterior to protect the container from crash damage.

Quote from: AlanSE on 04/29/2014 04:09 pmAllow me to propose an alternative closed-form version of this. This follows directly from the calculations in your spreadsheet. In fact, it is just a reduced form of all the formulas there.optimal launch angle = (Pi - theta) / 4 = (180 - 90) / 4 = 22.5That's pretty. Thanks!It's late and I'm tired. At the moment don't have energy to reduce the equations, but plugging different quantities for theta into my spreadsheet does indeed consistently give (pi - theta) / 4.

Allow me to propose an alternative closed-form version of this. This follows directly from the calculations in your spreadsheet. In fact, it is just a reduced form of all the formulas there.optimal launch angle = (Pi - theta) / 4 = (180 - 90) / 4 = 22.5

Wow, the geometric insights, confirmed by the calculus, are quite stunning.Can I ask for confirmation of one underlying bit of the reasoning? Proponent writes: "The trajectory requiring the least initial speed is the trajectoryof least energy." That's because for all possible trajectories, the potential energy at the launch point is the same, so the only thing left to minimize is the kinetic energy, and minimizing kinetic energy is equivalent to minimizing speed. Yes?

Quote from: Hop_David on 05/02/2014 07:36 amQuote from: AlanSE on 04/29/2014 04:09 pmAllow me to propose an alternative closed-form version of this. This follows directly from the calculations in your spreadsheet. In fact, it is just a reduced form of all the formulas there.optimal launch angle = (Pi - theta) / 4 = (180 - 90) / 4 = 22.5That's pretty. Thanks!It's late and I'm tired. At the moment don't have energy to reduce the equations, but plugging different quantities for theta into my spreadsheet does indeed consistently give (pi - theta) / 4.I didn't believe this at first, but I was wrong. The details are worked out in the attachment (corrections welcome). It also turns out that the tangent of the optimal launch angle is equal to the eccentricity of the optimal trajectory.

It's definitely more efficient to do it in a single hop.Consider, for example, the case where the distance to be travelled is much smaller than the moon's radius, so we can just pretend the moon is flat. Then, by Tartaglia's formula, the range of a hop is v^{2}/g, where v is the take-off speed and g is the local acceleration of gravity. A take-off speed of 100 m/s will get you a certain distance. To cover the same distance in two equal hops, each take-off would have to be at 70.7 m/s, giving a total delta-V of 141 m/s. And that's ignoring the braking needed at the end of each hop.

Also, did I miss it? Is there a simple equation giving the initial velocity v0 as a function of theta, normalized with the circular orbital velocity at r=R (a constant, vC)? Obviously when theta=0 then v0=0, and as mentioned when theta=pi then v0=vC. And the function must be monotonically increasing on the interval. Plotting a few points it sure looks simple!

Is there a simple equation giving the initial velocity v0 as a function of theta, normalized with the circular orbital velocity at r=R (a constant, vC)? Obviously when theta=0 then v0=0, and as mentioned when theta=pi then v0=vC. And the function must be monotonically increasing on the interval. Plotting a few points it sure looks simple!

Quote from: Proponent on 05/03/2014 02:03 pmIt's definitely more efficient to do it in a single hop.Consider, for example, the case where the distance to be travelled is much smaller than the moon's radius, so we can just pretend the moon is flat. Then, by Tartaglia's formula, the range of a hop is v^{2}/g, where v is the take-off speed and g is the local acceleration of gravity. A take-off speed of 100 m/s will get you a certain distance. To cover the same distance in two equal hops, each take-off would have to be at 70.7 m/s, giving a total delta-V of 141 m/s. And that's ignoring the braking needed at the end of each hop.Isn't that comparing the case without refueling? I know fuel goes up exponentially with delta-V, so couldn't a 70% delta-v easily work out to be about half the effort, and two 70% hops of equivalent effort?When I saw that V^2 component my first guess was that distance travelled was proportional to kinetic energy no matter how you split it.. but such a simplistic approach would probably only work for travel by trebuchet

I know fuel goes up exponentially with delta-V, so couldn't a 70% delta-v easily work out to be about half the effort, and two 70% hops of equivalent effort?

Put together, it looks like a normal vehicle would use less energy than the orbital hops, although they would obviously take longer. The delivery of energy is an issue, but could be done by an electrified road. If you just built a (mostly) conventional railroad, this would demolish the efficiency of the alternatives. Train tracks seem impractical. A hardened surface would seem to be the most practical road. Even if you had a space cannon for suborbital hops, you would still send the bulk materials along this road if I understand the physics well enough.

... a round trip would take no energy at all. (Neglect losses in the system of course.)

Assuming the distances between the COM are different between the North and South poles and you have some way of converting linear motion into a storeable form then a round trip would take no energy at all. (Neglect losses in the system of course.)