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What are the optimal burns for hopping between locations on a moon? How to plan?


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My Mun base lander has a mini lander/hopper that is powered by RCS thrust. Vernor engines, actually. I want to hop around and gather science in different biomes without wasting too much fuel. There are 4 biomes within a small distance of the initial landing site.

My intuition is that the most efficient way to get from point a to point b in the absence of atmosphere is a single launch burn, creating a suborbital trajectory, then a suicide burn. And... I guess you also need to figure out the optimal launch angle. I wouldn't be able to figure out any of that without knowing the distance between point and a and b... kerbalmaps gets me a rough estimate that I'm going about 15km each way.

I'm sure I'm overthinking this.

What I'm doing right now is pointing in the general direction I want to go, thrusting at 45 degrees until I have a little elevation and vertical speed, switch to map view, zoom in on my tiny orbit, and thrust toward the horizon until it looks like I'm headed to my landing zone, and get my horizontal velocity around 60m/s. From that point out, I point skyward and keep my vertical speed close to zero until it's time to land.

But it's pretty clear that the lower my horizontal speed is, the more dV I'm going to waste fighting gravity, but the faster it is, the more dV I spend starting and stopping. Even if this is somewhat inefficient, I have fuel to burn, literally. In theory, according to my last calculation, my vernor-hopper has about 2k dV. But after flying 15 kilometers, landing, flying back, landing again, way more than half of my dV is gone. So... my calculation is wrong or I'm wasting a ton of fuel. It still takes me quite a bit of fuel to properly dock the hopper on top of the base for refueling.

Any tips?

Side note: I would love the ability to plan a maneuver node on the ground. Like... ballistic maneuver node, lol.

Edited by MGM
Answered
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Technically, for a vacuum ballistic trajectory, the most efficient way would be to burn very quickly at takeoff only, at a 45° angle, until the point where you touch the ground again is on your target. That is the option that will take you the furthest for a set amount of dV. Then, to stop, suicide burn: wait until last moment to cancel all your velocity.

This is the most efficient way of doing suborbital hops, but might not be the easiest to realise.

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45° launch trajectory relies on the (incorrect) assumption of flat terrian. It can work for short hops, but longer traversals need to incorporate the curvature.

It might not be the optimal solution, but just establishing a suborbital hop followed by landing is a fairly good approach.

Remember, you will need about the same dV at either end since you are in the end just modifying the phase of your landed state.

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Thanks! Intuitively, I suspected this. Why 45 degrees? Maybe I should math that.

I've had multiple instances where I'd like to have the ability to target a spot on the ground from map view. But I think if I try your method, I can mostly do it with what I have.

Incidentally, I tested in orbit and found my mini-hopper does have 2300 delta-V. The calculations were correct. The fact that I spent all of it on two 15km trips, two landings, and one shabbily executed surface docking operation is alarming.

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45° launch trajectory relies on the (incorrect) assumption of flat terrian. It can work for short hops, but longer traversals need to incorporate the curvature.

Very good point. As OP mentioned distances of 15km, you might indeed want to consider curvature.

Then you'll have to draw a fictive line between your takeoff point and your target (so it will be underground), and make a 45° angle relative to this line (I think). I think that the fact that gravity is exerted on a single point should be negligible (radius of 200km, so angle of 4.3° according to my calculations)

EDIT: then again what I said relies on flat approximations, there should be an equation somewhere taking into account the curvature of the body.

Edited by Gaarst
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It so happens that, about a year or two ago, I wound up being interested in surface distance as a function of angle and velocity on the Mun, and wound up putting together a Desmos Graph to work that out.

huvowzrc9t.png

Vertical axis is distance traveled, measured along the spherical surface of the Mun, in kilometers. Horizontal axis is the launch pitch angle, measured in degrees above the horizontal. The furthest surface distance for a launch velocity of 500 m/s on the Mun turns out to be at an angle of about 25.72°, for an along-the-surface distance of about 269.3 km.The higher the launch velocity, the lower the launch pitch angle for maximum distance becomes, up until you reach orbital velocity, at which point a 0° launch will take you /effectively/ around the planet, and when you hit escape velocity, of course, you're not coming back down.

The actual graph is here, and allows the changing of the radius of the world being launched on (kilometers), the Standard Gravitational Parameter (in m^3/s^2), and launch velocity (in m/s)

https://www.desmos.com/calculator/eqvmftnmcg

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It so happens that, about a year or two ago, I wound up being interested in surface distance as a function of angle and velocity on the Mun, and wound up putting together a Desmos Graph to work that out.

Vertical axis is distance traveled, measured along the spherical surface of the Mun, in kilometers. Horizontal axis is the launch pitch angle, measured in degrees above the horizontal. The furthest surface distance for a launch velocity of 500 m/s on the Mun turns out to be at an angle of about 25.72°, for an along-the-surface distance of about 269.3 km.The higher the launch velocity, the lower the launch pitch angle for maximum distance becomes, up until you reach orbital velocity, at which point a 0° launch will take you /effectively/ around the planet, and when you hit escape velocity, of course, you're not coming back down.

The actual graph is here, and allows the changing of the radius of the world being launched on (kilometers), the Standard Gravitational Parameter (in m^3/s^2), and launch velocity (in m/s)

https://www.desmos.com/calculator/eqvmftnmcg

According to your graph, optimal angle for a 15 km hop is 44°, I was close enough with my 45° flat surface approximation :sticktongue:

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Whoa! Terrific answers. Awesome information!

Kudos to mhoram for the link to the perfect answer with all the info I needed that happens to cite "page 65 of the 1993 edition of Prussing and Conway's Orbital Mechanics textbook." I would love to have that on my shelf!

Bigger kudos to maltesh for the plot demonstrating that the furthest distance traveled is actually at 25 degrees. Math win!

Biggest kudos to Gaarst for accidentally guessing the correct answer, lol. Jeb would be proud!

Thanks guys!

I'm probably about to mark this as answered, but I want to play with that desmos plot a bit. It gives you the optimum angle for greatest distance given a velocity. I'm looking for angle with the lowest launch velocity given a travel distance. Thanks for the science/math, all!

[EDIT]

I did play with the graph a bit. The velocity I'm looking for is about 154m/s at about 44 degrees.

iqRfNfx.png?1

Cool tool, maltesh!

Edited by MGM
Added a graph, fixed formatting
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  • 2 years later...

The solutions presented here are great! However, they have a couple more foundational assumptions than what has been mentioned. One is atmosphere, though since this question was primarily about the Mun, that's a moot one. Still, it's worth noting that on a body with atmosphere, optimal launch angle will be considerably higher for all but the very shortest hops, up to a pretty typical gravity turn for hops more than about (gut feel) 1/4 of the way around the body.

The other assumption, though, is very relevant on the Mun, and that is TWR. From what I can tell at a glance, the solutions here only involve an instantaneous kick (i.e. assumption of infinite impulse), as if you were launching a ship via catapult. This is fine for a one or two biome direct ascent ship with TWR of 2:1 (Kerbin) or higher (12:1+ for Mun), but a high-efficiency lunar lander designed for multiple biome hops is typically not going to have that kind of TWR. For a Munar TWR of 1.5:1 to 2:1, burns are going to be, in my experience, on order of 20 to 30% of total flight time, which means some calculus needs to get involved for a truly accurate flight plan.

I could -probably- work that out if I felt like braining that hard at this time of night, but what I think more players are interested in is a more "Kerbal" approach, less akin to plotting a gravity assist trajectory by hand, and more akin to "put the planets at 12:00 and 2:30, then make a maneuver node on the light/dark side of the planet and then fiddle until you get an encounter... then fiddle more until the after-encounter orbit is about what you want".

I suspect a true optimal approach based on non-instantaneous thrust would be something more akin to finding the minimum of a function of time that balances vertical time including both burns with horizontal velocity to cover the distance in that amount of time, at least for a brute force numerical/mathematical approach, with a more nuanced approach directly balancing gravity drag losses into the mix.

My gut feel for a more "kerbal" approach is that the simpler methods presented above (which pretty much boil down to "launch at 45° for short hops, scaling down to 25° for hemispheric hops") gives a pretty close answer, but only if your vector at the -end- of the launch burn is at the angle specified. With a low TWR, this could mean your actual NavBall heading could be a -lot- higher during the burn. It should work out to something like half the angle to verticle for a TWR of 2:1, if my instinct is correct. If you have KER installed (Simple Orbit, or any other mod that splits velocity into horizontal and vertical components), you can get a decent feel for this angle by looking at the relative ratio of vertical and horizontal velocity. So for a short hop, the "Kerbal Way (tm)" would likely be to keep Vertical and Horizontal speed equal during the burn, whatever angle that actually requires, with the throttle pinned to the floor. For a long hop, "close enough" would probably put horizontal about double vertical during the burn.

Does anyone feel like doing more math to see if my gut feel for the "Kerbalized Sub-Orbital Hop Flight Plan" is correct? I.e. keep vertical and horizontal speed equal during the launch burn for short hops, and 1:2 for long trips?

Edited by FirroSeranel
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On 9/3/2015 at 2:38 PM, MGM said:

Thanks! Intuitively, I suspected this. Why 45 degrees? Maybe I should math that.

Math in spoiler.  ;)

Spoiler

You're on a flat surface in a uniform gravity field.  You fire a projectile at some velocity v, at some angle θ above the horizontal.  How far downrange does it travel?

Well, the distance x will equal the horizontal velocity, times the amount of time it spends until impact:

  • x = vx * t

So, let's figure out vx and t, shall we?  :)

vx is pretty straightforward; it's just the horizontal component of v, thus:

  • vx = v * cosθ

So, how much time does it spend until impact?

Well, it initially is heading upward at some speed vy.  So, the time it takes to reach the peak of its trajectory (i.e. for the vertical component of its velocity to reach zero) will be vy / g, where g is the acceleration of gravity.  So... that's how much time it spends going up.  It will spend an equal amount of time coming back down, so to find the total time, just multiply by 2, and we get:

  • t = 2 * vy / g

Well, vy = v * sinθ, so this gives us:

  • t = 2v * sinθ / g

Great!  So now we have both vx and t expressed in terms of v, θ, and g, so we can multiply them together to get the distance traveled, x.

  • x  =  vx * t
  • x  =  (v * cosθ) * (2v * sinθ / g)
  • x  =  (v2/g) * 2sinθcosθ

Dusting off ye olde high school trigonometry textbook, we find that 2sinθcosθ is one of the well-known trigonometric identities, and turns out to equal sin2θ.

That means we can rewrite our equation very simply as:

x  =  (v2/g) * sin2θ

So... if v and g are fixed, and we're trying to find a way to maximize x by picking the optimum angle... it means we want to pick whatever value of θ is such that sin2θ has the maximum value.

The sine function reaches its greatest value, 1, when an angle is 90 degrees.  So, maximum distance happens when  is 90 degrees, i.e. when θ = 45 degrees.

Q.E.D.  ;)

 

...Also, if you want to know "what's the farthest I can go" ... well, at the optimum θ = 45 degrees, sin2θ is just 1, and it becomes simply

xmax  =  v2 / g

Or, if you'd like to turn it around as "what velocity do I need to travel some distance X", it's just

v = sqrt(g * xmax)

So, for example, if you're on the Mun where g = 1.63 m/s2, and you want to go somewhere 30 km away, then v = sqrt(1.63 * 30000) = 221 m/s.  You'd need to launch with 221 m/s at 45 degrees to cover that distance.

Note that all of this, of course, is assuming a flat surface with uniform gravity, so for a spherical surface and gravity field such as a planet or moon, it's an approximation that only works well if the distance traveled is fairly small compared to the radius of the celestial body in question.

 

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14 minutes ago, Gargamel said:

Only took ya 3 years to respond there Snark.

  • Check how old the thread is before you post --> reasonable to expect.
  • Check how old every single post is when someone necros something and doesn't say "This is a necro, but" --> not so much.  :mad:

It's all good, but... @FirroSeranel, there's nothing wrong with reviving a long-dead thread, where appropriate-- but if you're doing so, one way to be kind to your fellow forum-goers is to explicitly mention "This is a necro, but..." or something along those lines, so that the people who see your post-- and who don't happen to notice that you're responding to an ancient thread-- will be aware of that fact and can make a point of not responding to years-dead questions unless they are specifically so inclined.  ;)

The forum makes it very easy, in a variety of ways (not least that big loud "this is a very old thread! are you sure you want to do this?" banner) to avoid accidentally necroing a thread without realizing it... but it's not so easy to notice after the revival.

Just a suggestion. ;)

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Ah, heh. Well it seemed like a good thread to revive, as it has open questions that I don't feel have ever been answered truly sufficiently.

I actually made a suggestion that Scott Manley might take this one on; it seems right up his alley, both to do the real physics first, and then to come up with the "Kerbal Way" after.

But yes, this was a necro. :P

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This thread is quite old. Please consider starting a new thread rather than reviving this one.

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