Maximum cost of a constant thrust trajectory

[Streetwind]

The dV cost to escape Kerbin with a reasonable approximation of an instant transfer burn is well known.

Let's assume I have a very heavy spacecraft in low Kerbin orbit, and I tape a single Dawn to it and set it to thrust prograde. The acceleration would be so low that very little progress is made over the duration of a single orbit, resulting in a scenario that's reasonably close to real-life electric propulsion: the spacecraft cannot just raise its apoapsis and coast there, but rather needs to slowly spiral outwards. I needs to thrust constantly and is forced to raise all sections of its orbit simultaneously, expending extra dV in the process.

Just like the cost to escape Kerbin approaches a lower limit the closer the burn is to an instant velocity change, I suspect that it approaches an upper bound the closer it comes to a 'worst case' constant thrust trajectory.

I'm curious to know how I would calculate what this upper bound is. And also. whether there is an easily memorized relationship along the lines of "x times the ideal minimum cost" that applies to any given celestial body, or whether you have to do the math individually each time.

 

Edited August 28, 2017 by Streetwind

[Cpt Kerbalkrunch]

I know you're looking for a very specific answer (which I couldn't possibly provide), but your question has me wondering about the timing, as well. When you eventually accelerate enough to escape Kerbin (or whichever body), in which direction would you be thrusting? Prograde, to raise your orbit? Or retrograde, to lower your orbit? I suppose that could be calculated, but to also time an encounter (even with Jool) sounds like a job for NASA. These type of thoughts remind me why I never use the Dawn or even the Nerv. I like my rockets dumb and powerful. 

[Human Person]

Interesting thought ... I should give this calculation a try.

My guess is that it wouldn't make much difference if you spiral out let's say 5 times before escaping or 100 times.

 

Wait! We can simulate it, we don't need to calculate!

[Streetwind]

28 minutes ago, Cpt Kerbalkrunch said:

I know you're looking for a very specific answer (which I couldn't possibly provide), but your question has me wondering about the timing, as well. When you eventually accelerate enough to escape Kerbin (or whichever body), in which direction would you be thrusting? Prograde, to raise your orbit? Or retrograde, to lower your orbit?

This is actually irrelevant to the thought experiment. The resultant solar orbit obviously changes, but Kerbin escape velocity is independent of direction. I could do this from a polar orbit and it would not make a difference  

[bewing]

The way I'd do it is to start with a perfectly circular orbit at x meters altitude. Now, to get into a circular orbit at x + deltaX altitude, you have to burn a little at point A, and then burn a little at your new Pe (on the opposite side of A).

So write that equation out O(x + deltax) = O(x) + dv1 + dv2. Where dv1 and dv2 are calculated in terms of deltaX.

Then convert that equation to differentials, and integrate from 70km to 84Mm.

 

[steuben]

1 hour ago, Physics Student said:

Wait! We can simulate it, we don't need to calculate!

Always calculate. You can try sims first to narrow your calcualtions; calculate some, sim some, and calc some more. or calc to determine possibility and then sim to determine practicality. I always keep in mind one of the little paragraphs from the Uplift Storm Trilogy in which Calculus Is Arcane Knowledge... or aleast several times far more arcane than it is for most people.

My reflex answer is infinity for a trivial upper bound. My zero approximation answer is that it will be dependant on the TWR and have an equation that is similar to hyperbolic curve. But, the maths is beyond my ken.

a quick experiment for craft with a twr of 0.19 yields roughly 1kps dv for a kerbin escape velocity for a 100km X 100km orbit.

[Cpt Kerbalkrunch]

2 hours ago, Streetwind said:

This is actually irrelevant to the thought experiment. The resultant solar orbit obviously changes, but Kerbin escape velocity is independent of direction. I could do this from a polar orbit and it would not make a difference  

Yup, I got the question. I know it's off topic but, since an escape is kind of pointless without a destination, I was wondering about a practical application of the answer. Actually using it to plot a transfer. In which case, the ejection angle would matter a bit. 

Edited August 28, 2017 by Cpt Kerbalkrunch

[Norcalplanner]

A few months ago, I tried to calculate how long it would take for the Hermes (from The Martian) to leave Earth orbit at its stated acceleration of 2 mm/s squared. The straight calc is over 20 days, but that doesn't take into account a whole host if factors which would likely increase that by at least 50%. I'm afraid my math skills aren't good enough to be of more detailed help.

[EpicSpaceTroll139]

I was thinking of using an integral based on an infinite number of tiny hohman transfers to find the deltav, but I found this excerpt from Wikipedia:

"Going from one circular orbit to another by gradually changing the radius simply requires the same delta-v as the difference between the two speeds.^[6] Such maneuver requires more delta-v than a 2-burn Hohmann transfer maneuver, but does so with continuous low thrust rather than the short applications of high thrust."

This makes sense, because as the number of loops increases without bound, the ∆v is expended closer and closer to the horizontal (resulting in less and less being spent fighting gravity), so its limit would be the difference between the initial velocity and the final velocity.

Update: I did the math

∆v required should be found by the following equation:

∆v = sqrt(μ/r0) - sqrt(μ/(rSOI))

μ = Kerbin's gravitational parameter = 3.5315984×10^12m³/s²

r0 = initial orbit radius (I'll assume 75km orbit so that's 600000 + 75000) = 675000m

rSOI = Kerbin's SOI radius = 84159271m

 

∆v = sqrt(3.5315984×10^12/675000) - sqrt(3.5315984×10^12/84159271) ≈ 2082.5m/s²

 

Edited August 29, 2017 by EpicSpaceTroll139

[Streetwind]

17 hours ago, Cpt Kerbalkrunch said:

Yup, I got the question. I know it's off topic but, since an escape is kind of pointless without a destination, I was wondering about a practical application of the answer. Actually using it to plot a transfer. In which case, the ejection angle would matter a bit. 

Counterintuitively enough, ejection angle matters less than you think. It definitely matters less than with high thrust propulsion. 

All you're going to be trying to do is exit Kerbin's SOI either directly prograde or directly retrograde, and you can do so by raising your orbit to skirt the inner edge of the SOI, then waiting for the right time to push it over the edge. You end up in a solar orbit that's nearly identical to Kerbin's, and then you're back to what you were doing before: very, very slowly modifying that orbit with constant thrust. You don't need to hit your target squarely in the face in the last split second of a powerful burn. You can patiently creep up to it over the course of your travel, and comfortably settle in next to it in a slightly lower (if going outward) or slightly higher (if going inward) solar orbit. Then you just wait and let the target SOI pluck you out of solar orbit automatically on the next close encounter, at which point you perform your capture burn. This is exactly what Dawn did with Ceres IRL, for example. How precise you are with your timing merely determines how long you have to wait for the required close encounter. If you are very precise, the close encounter coincides with your carefully sidling-up to the target's orbit, and it looks the same as a traditional encounter would. If you are less precise, then you may have to wait a bit until you catch up with the target (or the target catches up with you).

 

10 hours ago, EpicSpaceTroll139 said:

I was thinking of using an integral based on an infinite number of tiny hohman transfers to find the deltav, but I found this excerpt from Wikipedia:

"Going from one circular orbit to another by gradually changing the radius simply requires the same delta-v as the difference between the two speeds.^[6] Such maneuver requires more delta-v than a 2-burn Hohmann transfer maneuver, but does so with continuous low thrust rather than the short applications of high thrust."

This makes sense, because as the number of loops increases without bound, the ∆v is expended closer and closer to the horizontal (resulting in less and less being spent fighting gravity), so its limit would be the difference between the initial velocity and the final velocity.

Update: I did the math

∆v required should be found by the following equation:

∆v = sqrt(μ/r0) - sqrt(μ/(rSOI))

μ = Kerbin's gravitational parameter = 3.5315984×10^12m³/s²

r0 = initial orbit radius (I'll assume 75km orbit so that's 600000 + 75000) = 675000m

rSOI = Kerbin's SOI radius = 84159271m

 

∆v = sqrt(3.5315984×10^12/675000) - sqrt(3.5315984×10^12/84159271) ≈ 2082.5m/s²

A most useful answer!  I have not empirically verified this result, but intuitively it feels right. I already went and made a table with all of the KSP celestial bodies listed.

Now all that I need to answer my second question is the cost of a normal Hohmann style escape burn starting from the initial orbit, so i can compare it. I will have to figure those out ingame. But as a first order approximation, I checked the community dV map, which includes "low orbit to SOI edge" figures for the four different planets that have satellites.

It gave me the following results:
- Eve: 2.18x normal cost
- Kerbin: 2.19x normal cost
- Duna: 2.27x normal cost
- Jool: 2.28x normal cost

Given the fact that the low orbit altitudes in that map are not the same as mine, and the values are rounded, this is looking pretty good. It may very well be that we're looking at a predictable relationship in the form of a roughly 2.2x-2.3x cost multiplier for a worst case constant thrust trajectory over an ideal instant velocity change.  

I will know for sure once I determine the input the proper values for all the celestial bodies. I reckon there will still be slight derivations, especially on the very small bodies where initial orbit altitude matters a lot more than on for example Tylo. But it should be precise enough to derive a useful relationship.

Edited August 29, 2017 by Streetwind

[Rodhern]

2 hours ago, Streetwind said:

... It may very well be that we're looking at a predictable relationship in the form of a roughly 2.2x-2.3x cost multiplier for a worst case constant thrust trajectory over an ideal instant velocity change. ...

I think the predictable constant you might be looking for is 2.41, as in 1/(sqrt(2)-1) = 1+sqrt(2) .

Spoiler

Long-winded comment of the same thing:

@EpicSpaceTroll139 noted that the big budget burn is ∆v = sqrt(μ/r0) - sqrt(μ/(rSOI)) .

I am going to call this v₀ = sqrt(μ/r0) (approx).

Wikipedia has a nice formula for an elliptical orbit (Wiki-formula of sqrt 2). v = sqrt(μ*(2/r0 - 1/a)), where a is the SMA.

I am going to call it v₁ = sqrt(2)*v₀ (approx).

So by approximation the small budget burn is (v₁-v₀) = v₀*(sqrt(2)-1) ,

and the big budget is (v₀-0) = v₀ .

So by division 1/(sqrt(2)-1).

 

Edited August 29, 2017 by Rodhern
added "1+sqrt(2)" for easier reference; spelling; text format

[Streetwind]

1 hour ago, Rodhern said:

I think the predictable constant you might be looking for is 2.41, as in 1/(sqrt(2)-1) = 1+sqrt(2) .

Hmm. I can't follow your derivation entirely in my head, but I'm not the best at math, so you're probably right.

Still, you made approximations, so the result has to be an approximation itself, right? Like, dropping the second term in v0 introduces an error greater than 5%, from the math that I've done. Based on that alone, you'd be looking at (probably at least) 2.41 +/- 0.15, so a result of 2.26 is still within the possible span of correct results. And then there's the second approximation you made, which I'd have to repeat on paper to understand the margin of error.

[Rodhern]

1 minute ago, Streetwind said:

Hmm. I can't follow your derivation entirely in my head, ...

I didn't want to write a long sequence of formulas - most of the time they are just a nuisance to people. Most readers skip them; some do the math themselves (because it is fun), and almost nobody runs through the entire thing in their head . I will try to explain the main point a little better below.

 

1 minute ago, Streetwind said:

Still, you made approximations, so the result has to be an approximation itself, right?

Yups, the result is an approximation, but at the same time kind of a worst case upper bound. Let us pretend we are in a low circular orbit with speed v₀. Also let us assume that the SOI is really really large (like e.g. 100 Gm).

By EpicSpaceTroll139's post, the worst case budget is almost the same as v₀, because of the assumption that rSOI is a really big number, and the speed in an orbit that wide is really slow (almost zero).

The best case budget, used in the delta-v maps, is to increase the speed at the periapsis so that we get the apoapsis to reach rSOI. Again, because rSOI is a really big number, so is the semi-major axis length, and we are left with just the v₁ = sqrt(μ*(2/r0)) part of the speed equation. That is, v₁ = sqrt(2) * v₀.

So if the SOI is really large, then we might end up with close to 2.41 as the factor.

As you have already shown with examples, if the SOI has a more modest size, we seem to be better off. And indeed your examples show that it is by a non-trivial margin. An interesting exercise none the less.

[Streetwind]

Ah, I see now what you mean. If it is an upper bound depending on SOI size, that makes a lot of sense.  

I was already wondering if I'm going see a trend where the number increases slightly with every step going out from the parent body, like the four sample numbers I've tried seem to indicate. This would seem logical, as SOI size is dependant on the distance to the parent body; moving Kerbin further inwards without changing anything else would shrink the SOI. And if an infinite SOI results in the multiplier approaching 2.41, such a progression should be more or less expected.

 

Edited August 29, 2017 by Streetwind

[EpicSpaceTroll139]

Now that I think about it, the equation I gave might not be the actual maximum ∆v for a constant thrust trajectory. It might be that the worst case is some certain finite number of loops in the spiral, since with fewer loops, you expend less of your ∆v raising all parts of the orbit, but also spend more of your ∆v directly countering gravity.

I however have no idea how to figure out whether this is true

[Pand5461]

4 hours ago, EpicSpaceTroll139 said:

Now that I think about it, the equation I gave might not be the actual maximum ∆v for a constant thrust trajectory.

Well that was a bit above it. Thing is, the higher you go, the larger orbital period becomes. So, what was a "low thrust" maneuver at low orbit becomes a "high thrust" maneuver at higher orbit. For example, imagine a day-long burn. In a low-earth orbit, it's a long maneuver because orbital period is about 1.5 hours. At the height of Moon orbit, however, this may be considered as almost point maneuver due to orbital period of 28 days.

Therefore, what starts as a tight spiral unwinds quicker and quicker each orbit until it turns into hyberbola.

Theoretically.

When I applied this reasoning for Dawn probe (which, to my knowledge, is the only one spacecraft to leave LEO using electric propulsion) with 0.1 N thrust and 1000 kg mass, the distance where "low thrust" becomes "high thrust" turned out to be larger than radius of Earth's SOI.