Optimizing Surface Pressure for Minimal Delta V

[Kerano]

The scenario: You have the ability to increase or decrease the surface atmospheric pressure on a given rocky body to any value of your choosing by adding or removing particles in bulk from the atmosphere. (Say it's several hundred years in the future, and the technology to terraform planets has been perfected. Or you're HarvesteR. )

The question: For a given rocky body of mass M and radius R, what is the optimum surface atmospheric pressure required to minimize the combined delta-v cost of landing and taking off to a stable orbit in a rocket? (No jet engines. Assume different sized bodies have similar density, temperature, and molecular composition. Assume a simple exponentially decaying atmosphere.)

Let me illustrate what I mean with a rough diagram representing a reasonably large rocky body. There are no numbers here, this is just a sketch to show the concept.

We're trying to find what quantity of atmosphere hits the "sweet spot" to minimize the total delta-v requirement to visit this large rocky body. We know that a rocky body with enough atmosphere is very nearly "free" to land on from a low orbit using parachutes. However, the same atmosphere which reduces the landing delta-v cost also makes takeoffs more expensive (remember, no jet engines in this scenario). Beyond a certain point, increasing the atmospheric pressure begins to increase the total delta-v cost of visiting the body.

Let's take a look at some rocky bodies in the KSP universe with the current (admittedly rather soupy) atmospheric model to see how they compare on this scale.

KSP Comparison #1:

- Kerbin costs about 4550 m/s of delta-v to take off to a 70 km orbit, but is nearly free to land on. If we approximate 50 m/s to lower the periapsis nicely into the atmosphere, that gives a total delta-v cost for a Kerbin visit of about 4600 m/s.

- Tylo costs around 2700-3000 m/s to land on (carefully) from a 20 km orbit, and about 2300 m/s to take off. Let's go with the lower end and say only 5000 m/s of delta-v is required for an ideal Tylo visit. Even with the optimistic estimate, that's still 400 m/s more than is required for a Kerbin visit, despite Tylo having an identical radius to Kerbin and only 80% of the gravity.

So Kerbin is somewhere in the the "sweet spot valley" in the diagram above, though not necessarily on the "sweet spot" itself - probably somewhere to the right of it, given the current soupy lower atmosphere which makes takeoffs more expensive than they need to be.

KSP Comparison #2:

- Duna costs about 1400 m/s to take off from to a 45 km orbit, but the landing is mostly free with a decent number of parachutes. Let's say 50 m/s to deorbit + 100 m/s to slow down just above the surface, for 150 m/s of landing delta-v. That means Duna's total delta-v cost is 1550 m/s.

- Moho and Vall are both smaller than Duna (250/300 km vs 320 km) and have lower surface gravity (0.275/0.235 g vs 0.300 g). Nonetheless both require around 1000 m/s delta-v for landing and 950 m/s for takeoff, for a total delta-v cost of 1950 m/s each. That is 400 m/s more than is required for a Duna visit, despite Duna's higher gravity and larger radius.

Duna is definitely in the "sweet spot valley" with its very low delta-v requirement. My suspicion is that Duna may actually be very close to the "sweet spot" itself - perhaps a fraction to the left. The atmosphere is just thick enough to make a landing almost free, but not too thick to make the takeoff more expensive.

Back to choosing the atmosphere for our arbitrary rocky planet. Now, I'm aware that the ideal atmosphere for minimizing total delta-v (at least with the current KSP 0.90.0 model) would probably look something like a step function: nice and thick next to the surface, then straight to vacuum a short way up (say above ~1 km). Real atmospheres decay exponentially though, so let's not cheat our way out of that requirement.

The scale height of an atmosphere determines the rate of exponential decay of pressure with altitude.

For the same surface pressure, the atmosphere will be thinner (and the vacuum of space closer) if there is:

- Higher surface gravity

- Lower average temperature

- Larger average molecule size

Alternatively, for the same surface pressure, the atmosphere will be thicker (and extend further into space) if there is:

- Lower surface gravity

- Higher average temperature

- Smaller average molecule size

So for very small bodies, having an atmosphere - even a weak one - could actually be a disadvantage. The low surface gravity would mean the thin atmosphere would be dispersed over a wide range of altitudes, so to satisfy the "stable orbit" requirement you'd need to descend from - and takeoff to - a higher orbit, increasing the delta-v requirement.

What I'd really love to see is a "heat map" indicating the optimal surface pressure required to minimize the total delta-v for a given rocky body radius, assuming a simple atmospheric model. (For KSP and for the real world!) My own speculation is that it might look something like this:

The purple line indicates the minimum total delta-v requirement for that particular body radius - along the vertical on the graph. Note that this is not a line of constant delta-v, but rather a line connecting minimum values for total delta-v for different body sizes. (Remember we're assuming similar values for density, temperature, etc, between differently sized bodies.)

It's possible I may be wrong with this speculative heat map - I haven't done any actual calculations, just a few quick thought experiments. In particular I haven't considered whether the purple line flattens out or continues to increase at the higher values of radius. And of course there aren't any real numbers yet. If you reckon you can improve on any of this stuff, please be my guest.

To finish off, I thought it'd be interesting to look at a few bodies in the real solar system to see how they compare. I used this map for delta-v values.

Real Comparison #1:

- Earth requires 9400 m/s delta-v to take off to a low orbit, and is close to free to land on. Say 9500 m/s total delta-v including the deorbit burn.

- Venus is also nearly free to land on (ignoring the damaging atmosphere), but requires about 27000 m/s to reach orbit. Total delta-v around 27100 m/s.

Venus requires nearly three times as much delta-v for takeoff plus landing as Earth, despite the fact that Venus is 5% smaller, 20% less massive, and has a lower surface gravity (~0.9 g). It appears that it is more favorable in terms of total delta-v to have a surface pressure of 1 atm than 90 atm for a body of ~Earth size. (But would Earth be closer to the "sweet spot" if it had a thicker atmosphere or a thinner one? My hunch is thinner.)

Real Comparison #2:

- Mars requires about 3800 m/s delta-v to take off from. According to this paper, the Curiosity descent required about 300-350 m/s of delta-v from RCS and the skycrane. Let's round up to a total landing plus takeoff delta-v requirement of 4200 m/s.

- Mercury requires around 3100 m/s delta-v to land on from orbit, and takeoff would require a similar amount. Total delta-v required is thus something like 6200 m/s.

Mars is about 2000 m/s cheaper to visit from orbit than Mercury in terms of total delta-v, despite being 1.4x larger, 2x more massive, and having slightly more surface gravity. It appears that it is more favorable in terms of total delta-v to have a surface pressure of 0.006 atm than 10^-14 atm for a body of ~Mars size. (Is Mars close to the "sweet spot" for minimal total delta-v like Duna is in the KSP universe? It seems possible.)

Real Comparison #3:

- Titan is nearly free to land on, but requires 7600 m/s to reach a high orbit due to its thick atmosphere. Let's estimate the total delta-v including deorbit at 7700 m/s.

- Ganymede requires around 2000 m/s to land on and another 2000 m/s to take off, for a total delta-v budget of 4000 m/s.

Titan has very similar mass, radius and surface gravity to Ganymede - about 5-10% less for Titan in each case. Yet Titan has an atmosphere (surface pressure ~1.5 atm) while Ganymede has (almost) none. It appears that a near-vacuum is more favorable than a surface pressure of 1.5 atm in terms of total delta-v for a body of ~Titan size. (Much more favorable in fact, indicating that the surface pressure should probably be much lower than 1 atm to minimize total delta-v... maybe closer to Mars surface pressure?)

Edited January 12, 2015 by Kerano

[vexx32]

I think this has a much better place, and is likely to achieve its purpose far faster, in the Science Labs.

[Guest]

Looks very interesting, I will definitely read through this again.

But I would leave away the 100m/s landing Delta-v for slowing down on Duna descend - I never do this on my Duna landing, I just pack LOTS of chutes.

[Bill Phil]

Ummm...

No atmosphere...

[Ralathon]

Ummm...

No atmosphere...

That's only when going up. But OP is asking about a combined launch and landing. In a vacuum you'd need the launch dV again to get back down while with an atmosphere it is free (provided you have chutes)

First of all, you need to realize that atmospheric drag IRL works differently than in KSP. The air at sea level isn't made of soup in the real world. So losses to drag are much smaller than you'd expect. Usually in the range of just barely more than a 100 m/s. And most of that is thanks to reduced ISP on the engines. The main problem with an atmosphere is the dynamic pressure breaking things during ascent.

Gut feeling tells me that the sweet spot is somewhere around 0.25 atmosphere and 0.5 atmosphere if we take everything into account. Sure, it'd be cheaper during launch to go with a near vacuum like Mars, but the terminal velocity on a chute is just too high there. Just look at the kind of nonsense engineers needed to put Curiosity on the surface.

Meanwhile the earth has too high an atmospheric pressure. If we had a lower atmospheric pressure we wouldn't need to worry about deviating from a zero lift trajectory and our rockets could have more efficient shapes (Like a sphere).

So yea, gut feeling is telling me somewhere between 0.25 and 0.5 so we get a bit more leeway in rocket design yet can still use chutes.

[K^2]

No atmosphere. Build a giant magrail on the ground. Use it for launches and landings. We should do this on the Moon, by the way. Soon-ish.

Edit: But just for reference, a good estimate of atmospheric losses during launch is 4gH/v_t, where g is surface gravity, H is scale height, and v_t is terminal velocity at the surface. The derivation's a bit insane, and there are a lot of assumptions there, but it will get you within a ballpark for estimates.

Edited January 12, 2015 by K^2

[shynung]

I struggle to understand how would one attempt to land on an airless body using a giant magrail. It's like trying to shoot a bullet at another gun and having said bullet fly into the barrel of said gun through the muzzle.

[Ralathon]

Edit: But just for reference, a good estimate of atmospheric losses during launch is 4gH/v_t, where g is surface gravity, H is scale height, and v_t is terminal velocity at the surface. The derivation's a bit insane, and there are a lot of assumptions there, but it will get you within a ballpark for estimates.

That's for KSP I presume? Because tossing some ballpark values for earth in there (g=10m/s^2, H=8.5e3 and v=200) gives me 1.7km/s of dV lost to drag. Which is way higher than it actually is for a RL rocket, where it is usually less than 100m/s.

[Kerano]

That's for KSP I presume? Because tossing some ballpark values for earth in there (g=10m/s^2, H=8.5e3 and v=200) gives me 1.7km/s of dV lost to drag. Which is way higher than it actually is for a RL rocket, where it is usually less than 100m/s.

I believe the formula is supposed to take into account both gravity and atmospheric drag losses. At least, that's what some of the folks quoting it in the past have said (most of whom are on this forum, based on a quick googling).

I'd like to get to the bottom of this 4gH/v_t approximation though, because I can't seem to find any proof of it - whether rigorous or hand-wavy. The earliest reference from a quick googling seems to be by K^2 himself here, and it sounds like the guy who made the solar system delta-v map used the same equation without checking it or doing any sort of derivation. So I'd really like to know more of the details of how this equation was arrived at, and whether it is appropriate to apply to real solar system bodies or only KSP ones. If the equation doesn't hold up well in the real world, the solar system delta-v map on reddit probably needs a lot of fixing.

In particular, on a body with a very thin to non-existent atmosphere (such as Mercury), v_t tends towards infinity, and the 4gH/v_t approximation tends towards 0. This does not make sense if the equation is supposed to take account of both gravity and atmospheric drag, since you still have losses due to gravity when taking off in a vacuum.

So yeah, please post (or link to) the derivation. I can take the math.

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No atmosphere. Build a giant magrail on the ground. Use it for launches and landings. We should do this on the Moon, by the way. Soon-ish.

I can see how takeoff would work, but I can't imagine that you could do safe landings reliably with that sort of setup. The slightest miscalculation or gravitational anomaly and you're splattered across the surface at near-orbital velocity. I'd much prefer to take the space elevator.

The question I asked was about using basic rocket propulsion though. So we're assuming no fancy structures pre-existing on the body, nor any method of refuelling once on the surface. We can use parachutes for the descent if the body has an adequate atmosphere, but we have to bring all the propellant required for ascent down with us from orbit. That's the scenario I'd like to solve for.

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Looks very interesting, I will definitely read through this again.

But I would leave away the 100m/s landing Delta-v for slowing down on Duna descend - I never do this on my Duna landing, I just pack LOTS of chutes.

I was being conservative with the Duna landing - it's possible to land only on chutes, though for large payloads (50+ tonnes) it seems more efficient to do a short burn towards the end. Otherwise you're adding quite a high mass of chutes.

In any event, the point is that Duna is much cheaper to land on than Vall or Moho, despite both being smaller with weaker gravity. In fact the total delta-v budget for Duna is only slightly more than for the Mun or Eeloo - both of which require around 650 m/s for descent and 600 m/s for ascent. That's a total of 1250 m/s, which is only 200-300 m/s less than the requirement for Duna despite the fact that it has a 1.5x larger radius, 2x the gravity, and is 4.5x more massive.

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First of all, you need to realize that atmospheric drag IRL works differently than in KSP. The air at sea level isn't made of soup in the real world. So losses to drag are much smaller than you'd expect. Usually in the range of just barely more than a 100 m/s. And most of that is thanks to reduced ISP on the engines. The main problem with an atmosphere is the dynamic pressure breaking things during ascent.

Gut feeling tells me that the sweet spot is somewhere around 0.25 atmosphere and 0.5 atmosphere if we take everything into account. Sure, it'd be cheaper during launch to go with a near vacuum like Mars, but the terminal velocity on a chute is just too high there. Just look at the kind of nonsense engineers needed to put Curiosity on the surface.

Meanwhile the earth has too high an atmospheric pressure. If we had a lower atmospheric pressure we wouldn't need to worry about deviating from a zero lift trajectory and our rockets could have more efficient shapes (Like a sphere).

So yea, gut feeling is telling me somewhere between 0.25 and 0.5 so we get a bit more leeway in rocket design yet can still use chutes.

Very interesting. I agree that Mars probably is a bit to the left of the sweet spot - the atmosphere could probably be thicker to reduce that landing delta-v without increasing the takeoff delta-v too much. My hunch is that the sweet spot is closer to Mars atmosphere levels than Earth levels though - so perhaps somewhere around 0.1 atm of surface pressure would be optimal? Certainly somewhere between the lower bound of 0.01 atm and the upper bound of 1.00 atm anyway, at least for a rocky body of that size.

What about for much smaller or larger bodies - how might the optimal surface pressure change, if at all? For supersized Earths? Moons like Titan, Ganymede, or our own? Asteroids like Ceres? Comets like 67P? My speculation is that smaller bodies will be better off with a slightly lower surface pressure (or possibly even a vacuum), while larger bodies would be cheaper to visit with a slightly higher surface pressure. But how big or small is that "slightly"?

Edited January 13, 2015 by Kerano

[cantab]

If you're optimising for delta-V, a thin atmosphere and massive amounts of parachute in most cases. All that parachute will be heavy, increasing the amount of fuel required. Even if you dump them before takeoff, you still had to transfer it to the planet in the first place.

Which is why delta-V is generally a silly thing to optimise for. Cost is ultimately what counts but is complicated to determine. Payload fraction is easier to determine and more closely relates to cost, though you would need to assume an engine performance.

[K^2]

I'd like to get to the bottom of this 4gH/v_t approximation though, because I can't seem to find any proof of it - whether rigorous or hand-wavy.

So yeah, please post (or link to) the derivation. I can take the math.

It's entirely possible that somebody either derived it or arrived at it experimentally before me, but I've never seen such use or derivation. Derivation does include some hand-waving as far as gravity turn goes, but I'll get to that in a moment. As far as I know, the only reason it's a poor approximation for real planets is because atmosphere isn't exactly exponential, rocket's drag/mass ratio changes as fuel as consumed, and drag isn't strictly quadratic at hypersonic speeds. Nonetheless, I'm not aware of any simple formula that does better, so it's a good starting point before you refine computations with a proper model or testing.

The formula works for airless or nearly so bodies so long as you assume that TWR is not your limiting factor. With infinite TWR you escape from an empty world by accelerating to orbital velocity instantly. That way, there are no losses to gravity. Of course, that's not how real world works. So if you want to consider it a practical limitation, the approximation works so long as atmosphere is a bigger factor than TWR.

Anyhow, onto derivation. Let us consider a purely vertical assent in which we attempt to escape the atmosphere. First approximation is that by the time gravity really starts to change, atmosphere is no longer a factor. So we will assume that g is constant. This is one of the very few approximations here that works even better in the real world than in KSP. This problem has an exact solution for quadratic drag. One must ascend at terminal velocity. Proof for constant density is trivial. The reason it's a bit more complicated with decreasing density is that you have to increase TWR past 2 to account for acceleration. Nonetheless, it's possible to prove that prescription of ascent at terminal velocity is still an optimal one. (Using Euler Lagrange Equation.)

We consider special case of exponential atmosphere. This approximation is almost exact for KSP except for cutoff. It's considerably worse for real world. We can then write v(h) = v_t exp(h/(2H)), where h is altitude. Clearly, we can look at it as a differential equation h'(t) = v(h(t)), and it has a relatively simple solution.

h(t) = -2H ln((H - v_t t / 2) / H)

This is where things start getting exciting. You'll note that this equation diverges at t = 2H/v_t. In other words, it takes that long to escape completely to infinity if you continue maintaining terminal velocity. Of course, you'll be moving infinitely fast, and that will require infinite acceleration and infinite fuel, but we aren't really trying to escape out to infinity. We just need to clear significant atmosphere. And naturally, acceleration diverges to infinity just as you clear atmosphere, meaning that nearly all of this t = 2H/v_t is spent escaping the atmosphere.

Final observation is that amount of throttle you need to apply to follow this trajectory is F(t) = 2mg + ma(t), where 'a' is acceleration and 2mg are your losses to drag and gravity. (Equally split at v = v_t.) In vacuum and zero gravity, Integrating F(t)/m would give you your final velocity. So that's your real delta-V. However, the velocity you are actually getting is integral of a(t). That means integral of 2g over time it takes you to escape is your loss. And we know the amount of time it takes to escape. It's right above. So the total loss is 4gH/v_t.

There are several approximations up there, but they are all reasonable, and one can even do some estimates on the errors, and they do work out to be very, very small for KSP, and reasonable for real world. Hand-waving starts here. We don't have a full solution for gravity turn. There are numerical models, but no analytic solution we can take. At this point, I wouldn't call it an assumption. The conjecture is that gravity + drag losses during optimal ascent through gravity turn are approximately the same as during vertical escape. I do not have a satisfactory explanation for why it should be so. But it agrees with all of the numerical simulations on ascent, with all of the empirical numbers from KSP, including ascent from tallest peaks of Eve, and is even in the ballpark for real rockets.

Edited January 13, 2015 by K^2

[Kerano]

Thanks for the detailed derivation K^2, that's very interesting indeed. I think I'll go with the 4gH/v_t estimate at least as a first approximation for atmospheric delta-v costs in the model I'm planning in Matlab. Better to start with something relatively simple and see if it even vaguely resembles a reasonable answer.

So, thinking it through a bit more... for a given body of mass M and radius R, g will be a constant. To a first approximation H should also be a constant, assuming the mean atmospheric temperature and mean molecular mass stay roughly the same while we're pumping particles in/out of the atmosphere. So for a very simplified model, atmospheric delta-v should be proportional only to v_t at the surface.

The first question, then, is how to model v_t in a simple way as we vary the surface pressure (and radius) through a wide range of values. Any suggestions here would be welcome.

Thinking through the factors in the equation for terminal velocity...

Mass of the spacecraft on the surface will depend on the delta-v required to get back to orbit, which we're trying to solve for. Could be a slight problem due to the circular requirement, but running through a couple of iterations to get a ballpark figure might work.

Surface gravity is a constant.

Surface air density is proportional to surface air pressure, we just need to assume a temperature and a mean molecular mass of the atmosphere.

Projected area of the spacecraft I guess can probably be fudged based on the total mass of the spacecraft - maybe proportional to the cubed root of spacecraft mass?

Drag coefficient is complex and obviously needs some sort of fudge factor simplification to proceed, but I'm not sure what sort of values might be reasonable across a wide range of surface pressures (and gravities). I'm guessing C_d ~ 0.1-0.2 for rockets on Earth? But I presume that would change with a thicker or thinner atmosphere. Any advice?

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If you're optimising for delta-V, a thin atmosphere and massive amounts of parachute in most cases. All that parachute will be heavy, increasing the amount of fuel required. Even if you dump them before takeoff, you still had to transfer it to the planet in the first place.

True, but there's a limit to how big a parachute (and/or how many parachutes) you could realistically fit on a descent stage without running into design problems. An arbitrary cutoff point might be a good idea to simplify the situation - maybe limiting parachutes to making up no more than 10% of the descent stage mass.

Which is why delta-V is generally a silly thing to optimise for. Cost is ultimately what counts but is complicated to determine. Payload fraction is easier to determine and more closely relates to cost, though you would need to assume an engine performance.

Sure it's not necessarily that useful to optimize for delta-v. Nonetheless, I think it's interesting as a thought experiment - and worth at least a simple modelling.

Edited January 14, 2015 by Kerano