Suborbital Lunar Hops

[Hop_David]

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.

This is one of the things I was interested in looking at.

Spirit and Opportunity landed on Mars using airbags. But after aerobraking and firing of retrorockets, the impact was only 24 meters/sec. That's about 86 km/hour or 54 miles per hour.

A suborbital hop from the lunar pole to equator would impact at about 1.53 km/s which is about 3400 mph.

Impact wouldn't fall to 54 miles per hour until the distance between departure and destination was shrunk to ~tenth of degree, about 400 yards.

But inert supplies would be much less fragile than a rover. For buik commodities a higher impact speed might be acceptable. If 250 mph crash is okay, the departure and destination could be 7.5 km apart. This might be a way to get stuff from the rim of Whipple Crater to the crater floor.

[Proponent]

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

That'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.

EDIT: Replaced attachment with corrected version.

[sdsds]

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 trajectory
of 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?

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!

[Hop_David]

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 trajectory
of 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?

Yes.

Specific orbit energy is v^2/2 - mu/r. First term is kinetic, 2nd potential. As you say, mu/r is fixed.
Mu/r = G * moon mass / 1738 km

Another way to see it is looking at the vis viva equation: v^2 = mu (2/r - 1/a). When a gets smaller, the term subtracted from 2/r gets bigger, which makes the whole thing smaller.

[Hop_David]

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

That'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.

I've found a geometrical way to demonstrate Alan SE's pretty result:



Using a well known reflective property of the ellipse it can be seen that
2 alpha + (pi - theta)/2 = pi.
A little straight forward algebra gives
alpha = (pi + theta)/4.

Angle between position vector and velocity vector is alpha + (pi - pheta)/2 which comes out to (3 pi - theta)/4

A horizontal velocity vector is pi/2 from the position vector. When pi/2 is subtracted from (3 pi - theta)/4 you get Alan's pleasing result.

I am still looking at my drawings searching for a visual demonstration of Proponent's nice find. Reading the first part of Proponent's pdf I can see Kepler style drawings but it becomes opaque to me when he starts using differential calculus.

[KelvinZero]

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 v²/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

[AlanSE]

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!

I don't think anyone has posted this explicitly.  However, the information within this thread is sufficient to obtain it.  This is what I have:

1/2 v^2  = (mu/r) sin(theta/2)/(sin(theta/2)+1)
1/2 v^2  = (mu/r) x/(x+r)

I'm giving this in terms of the specific kinetic energy.  Here, x is the distance of the launch point from the y-axis in some of the illustrations used in this thread.  Suffice it to say that x=r sin(theta/2).  Also, mu=GM.  The radius of the moon is just r.  The following is a plot of fraction of mu/r that the kinetic energy at launch is.  To be more articulate, this is the ratio of the kinetic energy to the (negative of) gravitational energy at launch.



Or put into Wolfram alpha: plot sin(theta/2)/(sin(theta/2)+1)  for theta=0..Pi
Or Google

This is, as far as I'm concerned, the same plot as what was shown in the text I quoted.  The final 0.5 value is what's expected for a fully circular orbit, in the sense that v=sqrt(GM/r).  The requisite pieces of information are the following:

the distance f1->p->f2 is 2a.
2a = x + r
a = -mu / (2 eng)
eng = v^2 / 2 - mu / r

[Proponent]

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!

See equation (12) in the PDF I wrote (and just revised to correct errors).  Just drop the factor GM/R to get speeds in terms of circular speed, of course.  There's something funny with that plot, by the way -- take-off speeds should always be less than circular.

EDIT:  Added missing closing bracket to quote block.

[Proponent]

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 v²/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

True, if we're making the journey by rocket, then it's more complicated.  The answer will depend on specific impulse.  I suspect that, as a rule of thumb, the minimum propellant consumption occurs about when the delta-V for a hop (take-off delta-V plus landing delta-V) is about equal to effective exhaust velocity.  If the required per-hop delta-V is higher, then the exponential factor in the rocket equation starts to bite.

With circular speed at the lunar surface being about 1600 m/s and typical exhaust velocities twice that or better, I think it's likely that single hops are best.

[Hop_David]

Also of interest is the time of flight for the suborbital hop.

To look at fraction of area swept out, Kepler would vertically scale an ellipse by a factor of 1/sqrt(1-e²) to make it a circle.



I do the same. Without loss of generality, we can choose our units so a = 1. Things are simpler when working with a unit circle.

It's well known semi-latus rectum has length a(1-e²). When (1-e²) is scaled by a factor of 1/sqrt(1-e²) you get sqrt(1-e²).

We have a blue circular section with central angle acos(e). It's area is acos(e)/2.

We have a pink triangle of base e and height sqrt(1-e²). It's area is e * sqrt(1-e²)/2.

The circle section and triangle are mirrored below the major axis, so we can lose the denominator of 2. Total colored area is acos(e) + e * sqrt(1-e²).

The suborbital's fraction of the total period is (acos(e) + e * sqrt(1-e²))/pi

Orbital period is 2 pi * (a³/mu)^1/2. So time of flight is
(acos(e) + e * sqrt(1-e²)) *  (a³/mu)^1/2.

I'm going to do a blog entry Transportation on Airless Worlds. It will link to a spreadsheet that will include Luna, Mercury, Ceres, and a few other round, airless worlds. I will toss in Mars because it's nearly airless.

Proponent and AlanSE would you mind if I mentioned you? Proponent, I'd like to upload your pdf and link to it.

[Hop_David]

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 you have propellents with exhaust velocities on the order of 4 km/s, there's not a whole lot of incentive to break delta V budgets of 2*1.6 km/s (or less) into smaller hops. This is even more true if using rail guns or trebuchets.

There may be other reasons to break a big suborbital hop into smaller hops, though. If your destination is close to pi radians (180 degrees) away, flight path angle is close to zero. So crashing into mountains becomes a problem. Getting the flight path angle big enough to avoid obstacles might be a reason to break it smaller hops.

On Mars, atmosphere might become a hassle for longer suborbital hops that require greater velocity. And with lower flight path angle, the payload spends more time in Mars' lower (and denser) atmosphere.

[Proponent]

Please feel free to use the PDF anyway you like.

I suspect the most efficient way to clear obstacles on long hops would be to use a slightly off-optimal trajectory with a larger apoapsis.  If that doesn't do the trick and we're using rockets rather than mass drivers or something like that, then use a two-impulse trajectory.  For example, if a 10-km obstacle is very close to the origin (worst case), boost vertically at 180 m/s, coast up to 10 km, then boost horizontally.

I could see multi-hop trajectories being useful if some sort of ISRU propellant of low specific impulse were being used.

[AlanSE]

The question of energy efficiency is also intriguing.  As others have pointed out, the optimal number of hops is trivial for this problem.  Traveling the entire distance in one hop requires the least amount of energy.  But this is jumping the gun a little bit, because the optimal solution for the limit-case of small hops isn't known to begin with!  Here is my math:

Small Hops:
h(t) = v0 t - 1/2 g t^2   = { 0 = t ( v0 - g t / 2 ) }
t = 2 v0 / g   (the useful solution)
v0 = sqrt(2) v
distance = v0 t = 2 v0^2 / g = 4 v^2 / g
(mass-distance/energy) = distance / (1/2 v^2) = 8 / g

for 1,000 kg:
(distance/energy) = 8 / ((1.622 m/s^2)*(1,000 kg)) = 3.065e-6 mile/J
 x (1.3e8 J/gallon) = 398.4 mpg
with 20% conversion efficiency --> 80 mpg

In the first block of math, the 8/g result contains nothing about the distance of the hop.  That means that any number of hops is equally energy efficient.  I won't do the general case which takes into account the planet's curvature, but I'll compromise by entertaining the largest possible hop.

Full 180 degree movement:
1/2 v^2 = 1/2 GM/R
(mass-distance/energy) = Pi R / (1/2 GM/R) = 2 Pi R^2 / (GM)

for 1,000 kg:
(distance/energy) = Pi*(radius of the moon)^2*2/(G*(mass of the moon)*(1000 kg)) = 2.4e-6 mile/J
 x (1.3e8 J/gallon) = 312 mpg
with 20% conversion efficiency --> 62.4 mpg

We see that this is clearly more energy efficient, which suggests that the efficiency of the hop improves monotonically as the angle increases.  However, this isn't predicted by the flat-surface Newtonian mechanics.  The improvements are only due to the benefit of centripetal acceleration.

Of course, the miles-per-gallon figure is artificial.  Obviously nothing would be running on gas on the moon.  However, this still gives a reference to compare the method's efficiency to the drivetrain efficiency of common vehicles.  Actually, the above numbers are extremely generous.  Even the 2012 Toyota Camry weighs close to 1,500 kg.  That puts it into surprisingly close parity with the above numbers.

Actually, since the Camry performs that well on Earth, we would suspect a moon-car to be even more efficient.  The wheel and drivetrain friction are mostly constant energy losses, whereas the drag increases with speed.  We discussed this a great deal on Physics Stack Exchange.  For normal automobiles, their efficiency suffers at low speeds due to reduced engine efficiency, which is why optimal speed is somewhere around 55 mph.  That means that the Earth mpg is a very poor predictor of efficiency of a moon vehicle, because air resistance is still relatively high at 55 mph.

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.

[Hop_David]

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.

Yes, I am hoping there would eventually be roads. Or, better yet, railroads.

But initially there will be no transportation infrastructure on other bodies. It's my belief that transportation between the first mines and manufacturing centers would need to be done by suborbital hops.

I like to imagine very advanced infrastructure where underground tunnel rails connect points in straight line chords. Here's an exercise: If there were a mohole linking Ceres' north and south poles, how much energy would it take to travel from one pole to the other?

[IslandPlaya]

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.)

[Proponent]

I was wondering if, with very good guidance you could use electromagnetic mass drivers for both launch and landing.  With electromagnetic braking on landing, it would in principle be possible to recover much of the energy.

[IslandPlaya]

Good point. All true.
Formidable engineering though!

[Hop_David]

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

That is my opinion. Here is a model. If memory serves I got it from Dr. John Stockton:



I believe the round trip time would be the same as orbital period at surface. In the case of Ceres, that'd be about 3.4 hours.

[Hop_David]

I did a blog post on this. My thanks to AlanSE and Proponent.

[Proponent]

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.)

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.