Then it occurred to me that departure and destination points on the Lunar surface could be corners of a Lambert space triangle. Of course, the third point would be the moon's center.

So basically it's like trying a to throw a baseball fast enough and at the right trajectory to reach the other side of the Moon. To put it in layman's terms. I suppose a mass driver could be used to effect a propellantless journey from on part of the moon to another in this way.

The second focus of the ellipse would lie in the middle of the chord connecting departure and destination points. Major axis of the ellipse would be r(1+sin(θ/2)). e would be cos(θ/2) / (1+sin(θ/2)). Attached is a pic that might make this clear.Also attached is a spreadsheet giving velocity and angle of suborbital launch from surface. In this spreadsheet I put the angular separation between the pole and an equatorial point - 90 degrees. This suborbital launch would be 1.53 km/s at an angle of 22.5 degrees. Typing into the pink cells user can change angle between departure and destination. The user can also type in radius and mass for other bodies.

Also, I would like to better understand the physics that supports the conclusion that the second focus is the center of the chord connecting the launch and landing sites!

Why is this in Avanced Concepts?

I don't want to offend anyone, but honestly any good student in a first year university physics class should be able to solve this problem.

In the image I found it useful to shade in the Moon. Regarding what this is saying, it might be a way to find the angle and velocity at which to launch from point A so that you land at point B, using minimum energy. But I think it minimizes the launch energy, with the expectation that doing so also minimizes the sum of launch and landing energies, and that both launch and landing are instantaneous impulses.

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

Quote from: ChrisWilson68 on 04/29/2014 05:53 amWhy is this in Avanced Concepts?It's about transportation between points A and B on a roughly spherical body like the Moon. It could also be applied to other airless, spherical bodies like Ceres, Mercury, Enceladus, etc.Taking off from one points A to B on the lunar surface assumes a railgun, sling, or perhaps ISRU propellant. Else, why not just go directly from earth to point B?I'd like to see suborbital hops on the moon become routine. But right now it's Advanced Concepts, in my opinion.

And yes, freshmen aerospace gives the vis viva equation.Simple, yes. Obvious? Well, then you are being insulting as it took a few years for this to occur to me.

Quote from: sdsds on 04/29/2014 06:45 amAlso, I would like to better understand the physics that supports the conclusion that the second focus is the center of the chord connecting the launch and landing sites!Recall that the energy of an elliptical orbit of semi-major axis a is -0.5GM/a. Therefore, minimizing the energy of the transfer orbit is equivalent to minimizing a. Let be T the angle between the major axis and a point on the ellipse as seen from the moon's center, where the other focus is at T=0 (in other words, T is theta/2 in sdsds's diagram, above). Then the radius of a point on the ellipse is r = (1 - e^{2}) a / (1 - e cos T), where e is the eccentricity (usually the denominator is 1 + e cos T, but the other focus is taken to be at T=180^{o}).Now set r to the moon's radius. In fact, it's convenient to measure distances in units of the moon's radius, so r=1. Solve the radius equation for a as a function of e and minimize. The distance from one focus to another is 2ea, which, at minimum a, turns out to be cosT. QED.

Thinking of various possibilities, the shortest possible colored leg seems to be going to the focus on the center of the chord connecting the two points.

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.

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.)

Quote from: IslandPlaya on 05/06/2014 03:54 pmAssuming 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.)And, in fact, if the body is spherically symmetric and of uniform density, then travel along a straight line connecting any two points on the surface takes no energy, if friction can be avoided. And the trip time is always the same, regardless of the distance between the two points on the surface.

And, in fact, if the body is spherically symmetric and of uniform density, then travel along a straight line connecting any two points on the surface takes no energy, if friction can be avoided. And the trip time is always the same, regardless of the distance between the two points on the surface.

Wendy Kreiger also told me trip time is always the same. I'm willing to take your word as well as Wendy's. But I would be happier if I knew why.

Quote from: Proponent on 06/17/2014 06:46 pmAnd, in fact, if the body is spherically symmetric and of uniform density, then travel along a straight line connecting any two points on the surface takes no energy, if friction can be avoided. And the trip time is always the same, regardless of the distance between the two points on the surface.It's a free energy ride if we're willing to let gravity do all the work. But then we're stuck with trip time on order of half an hour to an hour for most rocks.But we can shorten trip time by investing energy to accelerate at the beginning of the trip. But then the payload would need to be slowed before it reaches its destination. The deceleration at the end of the trip could put energy back into the system. So it seems to me faster trips could also be done for very little energy.

See equation (9) of the analysis attached to this post.

Quote from: Hop_David on 06/17/2014 10:20 pmWendy Kreiger also told me trip time is always the same. I'm willing to take your word as well as Wendy's. But I would be happier if I knew why.Without using calculus, we can easily show how the problem reduces to simple harmonic motion.Consider a particle constrained to travel along chord parallel to the y-axis. The acceleration of a particle along the chord is -(GM/R^{3})y. (I'm sure you can figure this out yourself: just calculate the acceleration of gravity at a given x and y, and then determine its y-component.) The form of the acceleration is the same as for a mass on a spring or for a small-amplitude pendulum (restoring force is proportional to displacement). In both of those cases, the period of oscillation is independent of amplitude. The acceleration is also independent of x, so it doesn't matter which chord (i.e., between which two points on the surface) we are travelling.Does that help?

A quick geometrical demonstartion that second focus lies on center of chord connecting two points....

NASA Developing drones/hoppers for Mars, asteriods and Moon. http://www.parabolicarc.com/2015/08/09/nasa-mars-drones/For the Moon a small H2O2 hopper could operate off one the Xprize landers eg MoonExpress MX-1 which uses H2O2.One big plus of hopper using lander fuel is that all the lander's reserve fuel can be used for exploration. A 10kg hopper + 2kg of H202 (12Kg total) with 150 ISP thrusters would have 10Km radius.4kg fuel would give 30km radius, ideal for exploring around Shackleton crater (21km diameter). http://space.stackexchange.com/questions/4413/lunar-sub-orbital-trajectoriesNB 1degree is 30Km.

This might be of interest:"Developing an Aerial Transport Infrastructure for Lunar Exploration" by David L. AkinAbstract http://www.lpi.usra.edu/meetings/leagilewg2008/pdf/4096.pdfPresentation http://www.lpi.usra.edu/meetings/leagilewg2008/presentations/oct29pmSalonIII/Akin4096.pdf