Fastest transfer possible

[slouis]

Let's say (hypothetically) that you had a very, very fast spaceship. As in, it can accelerate to 0.5c in a minute. (Presumably, this would require some rather hardy Kerbals, but that's besides the point). It runs on Mysterium fuel, which never depletes (it's a mystery why!)

If this spaceship wants to transfer from Kerbin to Duna in the least amount of time possible*, what should the maneuver be? My intuition is that it could burn towards the target the whole time. But perhaps it is a prograde burn, and then a lot of radial burns? What is the fastest flightplan?
 

* Ignoring timed dilation. The point is not the 0.5c speed per se, but rather that the ship is very fast.

[TimePeriod]

I'm more worried about colliding with other objects along the way. Also, stopping.

[septemberWaves]

If you can achieve a constant acceleration of just 1g, that should be able to get you to Duna in only a couple of weeks on pretty much a straight line trajectory (pointed at where Duna will be when you arrive, that is) so long as you don't have to go too close to the sun, and provided that you accelerate for half of the journey and decelerate for the other half (so that you can easily capture into orbit). The longest duration of transfer in this manner would be when a straight line trajectory passes too close to the sun, in which case you would have to do some radial burning but it would still not take too long to reach Duna.

Now, with your ridiculously fast spacecraft that can travel at half of the speed of light and reach that velocity very quickly (let's just assume that kerbals are invincible) it would take mere minutes to get to Duna, and once again the best trajectory would be a straight line.

[popeter45]

you could do a straight non stop burn in the direction of duna and depending on acceleration it could time as little a ~ 2-3 minutes 

[UmbralRaptor]

Assuming constant acceleration, the perhaps slightly misnamed brachistochrone/flip-skew trajectory comes to mind. Especially if you want to stop at the end. With the OP's assumptions of acceleration >> local gravity, burning directly towards the body will work fine, as it simply won't have time to travel very far. If it has moved enough, you could estimate the travel time and burn for that point instead.

 

Incidentally, a straight-line kerbin-duna trajectory with 1 g acceleration takes somewhere between 10 and 33 hours, depending on where they are and if you want to stop at the end. All of them result in speeds of >350 to >1100 km/s, so gravity and relative motions are just minor perturbations.

[numerobis]

There's also the question of when in the orbit should you launch: you want to minimize time, which means you want to minimize the distance -- which is very much not what a Hohmann transfer does.

[Tyko]

Is there a way using the Navball and Maneuver Nodes to chart a course for a constant acceleration burn - where you accelerate to the halfway (turnover) point and then flip and decelerate the other half.

I don't think that just burning towards the target indicator on the navball is really the best because you're always burning towards where it is, not where it will be.

[slouis]

3 minutes ago, tjt said:

I don't think that just burning towards the target indicator on the navball is really the best because you're always burning towards where it is, not where it will be.

Thanks for all of the responses! I think now that a better question (for my purposes) is: Is there a generalized flight plan for the fastest transfer available? I used really big numbers to allow for the possibility that you are dealing with some monstrous torchship and really don't need to think about efficiency. In general, is pointing at where the planet will be the fastest transfer?

 

[Tyko]

18 minutes ago, slouis said:

Thanks for all of the responses! I think now that a better question (for my purposes) is: Is there a generalized flight plan for the fastest transfer available? I used really big numbers to allow for the possibility that you are dealing with some monstrous torchship and really don't need to think about efficiency. In general, is pointing at where the planet will be the fastest transfer?

 

Well, the shortest distance between two points is a straight line. So, if you can calculate where Duna will be when you complete your Burn>Turnover>Burn and point right at that point, it should be the fastest (waiting for a physics / orbital mechanics guy to tell me I'm dead wrong    )

Edited August 18, 2016 by tjt

[foamyesque]

17 minutes ago, slouis said:

Thanks for all of the responses! I think now that a better question (for my purposes) is: Is there a generalized flight plan for the fastest transfer available? I used really big numbers to allow for the possibility that you are dealing with some monstrous torchship and really don't need to think about efficiency. In general, is pointing at where the planet will be the fastest transfer?

 

 

The "fastest" transfer depends pretty much entirely (not quite, but close) on how much deltaV you've got available. The Hohmann is the slowest; to do it faster, you add more deltaV to overshoot your target orbit (and then slow down to it when you reach it). It's not really a single number.

[Tyko]

1 minute ago, foamyesque said:

 

The "fastest" transfer depends pretty much entirely (not quite, but close) on how much deltaV you've got available. The Hohmann is the slowest; to do it faster, you add more deltaV to overshoot your target orbit (and then slow down to it when you reach it). It's not really a single number.

The OP was regarding the "fastest" without DeltaV limits. I think it's more about how to plot the course that utilizes constant acceleration.

[foamyesque]

Just now, tjt said:

The OP was regarding the "fastest" without DeltaV limits. I think it's more about how to plot the course that utilizes constant acceleration.

 

That depends on how much deltaV you've got. You could run an ion engine for years at 0.05% throttle via thrust limiting if you wanted, and your trajectory would be a lot different than a sepratrons-with-infinite-fuel craft.

[Tyko]

understood...is there a way to use the Navball and Maneuver Nodes to plot a constant acceleration course? The OP had asked about it, but I'd been wondering the same thing too...say for instance you wanted to make a speed run where a ship had enough DeltaV and wanted to run at 1G acceleration for the entire trip. Accelerate at 1G, turnover at the halfway point and decelerate at 1G to your destination. @UmbralRaptor mentioned 10-33 hours from Kerbin to Duna. I'm asking if you could use built in tools to plot that course.

Edited August 18, 2016 by tjt

[Archgeek]

33 minutes ago, tjt said:

Is there a way using the Navball and Maneuver Nodes to chart a course for a constant acceleration burn - where you accelerate to the halfway (turnover) point and then flip and decelerate the other half.

I don't think that just burning towards the target indicator on the navball is really the best because you're always burning towards where it is, not where it will be.

I'd pick a likely spot on the target orbit, set a node to eject that-a-way, and increase it until an intercept shows up.  Then keep increasing it, such that you dreadfully overshoot.  Then put a second node around halfway along your trajectory, and pull it retrograde until the intercept comes back.  Then adjust the timing and burn amounts until you've "minimized" the needed delta-v and are still within your ship's thrust capability.

[UmbralRaptor]

18 minutes ago, tjt said:

understood...is there a way to use the Navball and Maneuver Nodes to plot a constant acceleration course? The OP had asked about it, but I'd been wondering the same thing too...say for instance you wanted to make a speed run where a ship had enough DeltaV and wanted to run at 1G acceleration for the entire trip. Accelerate at 1G, turnover at the halfway point and decelerate at 1G to your destination. @UmbralRaptor mentioned 10-33 hours from Kerbin to Duna. I'm asking if you could use built in tools to plot that course.

Stock tools seem very poorly geared towards it, given the way they assume instant burns. I expect one could make a maneuver node with a burn that gets you equivalent flight time mainly to get a heading estimate.

[Deddly]

It's not really about KSP, but it is a nice thread about spaceflight, so I moved it to Spaceflight

[GoSlash27]

I had a highly- cheaty craft that could maintain a constant 1G back in KSP .24. I flew so that my prograde was aligned directly with Duna and intercept happened in a matter of days.

You achieve a trajectory that's a virtual straight line in very short order, so the target planet has no chance to get out of the way.

HTHs,
-Slashy

 

[magnemoe]

9 hours ago, UmbralRaptor said:

Stock tools seem very poorly geared towards it, given the way they assume instant burns. I expect one could make a maneuver node with a burn that gets you equivalent flight time mainly to get a heading estimate.

The porkshorp chart in mechjeb works pretty well for 20-50 km/s burns, at this speeds you still have to aim a bit ahead, optimal launch window is then outer planet is a bit ahead of you as you can do an fast turnover. at target and return. If you go much faster you just aim a bit ahead. 
 

[Archgeek]

4 hours ago, GoSlash27 said:

You achieve a trajectory that's a virtual straight line in very short order, so the target planet has no chance to get out of the way.

Yes, but if you're going full brachistochrone, you'll need to start braking midway, giving the planet time to get away from you.

...and that got me too curious, so I started doing some Math.

Spoiler

Assuming a crazed torchship that isn't overly affected by the gravity of the sun, we wind up with a basic kinematics problem: d = .5a*t² + v_i*t + d_i
for any straight-line distance 'd' and ship acceleration 'a'
distance accelerating d_a = .5a*t²

For max speed, we want d_a = d/2.  Sub in a*t²/2 for d_a and we find that a*t² = d, meaning the two burns will be sqrt(d/a) seconds long, and the whole trip will be twice that.
a actually changes as we burn fuel, but we'll ignore that as a fudge factor for ignoring solar gravity.

To use a manuever node here, we kinda want dv, which is conveniently equal to at, or d/t.  Subbing in sqrt(d/a) for t, we get dv = d/sqrt(d/a).  Roots are annoying though, so let's square everything:
dv² = d²/t² = d²/(d/a) = da

So it seems the delta-v needed for both burns is the square root of the straight-line distance times your ship's acceleration, and the time of the trip is twice the square root of the distance over your ship's acceleration.  Interesting.

With this in hand, you can find the straight-line distance, get the trip time from it, and then know exactly where to aim based on the known orbital parameters of the target body.  Then you just set a node for the given dv, burn it, flip retrograde, and burn for the same time period, and you'll roll up to the target at about the relative orbital velocity of it to your starting point.

So if Duna's at its periapsis of  21,783,189km  and kerbin's at its constant height of 13,599,840.25km, and they're for some reason aligned, then that's a straight-line distance of 8,183,348.75 km.  If our ship accelerates at 10m/s², we get a burn time of sqrt(8,183,348.75km/10m/s) = 28606s = ~7.95hrs, and a trip time of 15.89 hours.  Duna's orbital period is 17,315,400s, which is about 0.075 seconds of arc per second, so it'll move around 71 minutes of arc over the trip, or just over a degree, which is about negligible.  So, we aim straight for it, burning at sqrt(8183348750*10) or about 286.1km/s, then flip around and do it again.  Talk about a torchship!

If we have a weaker drive, say 1m/s², then we get a burn time of 90,461s = ~25.13hrs, a trip time for 50.26hrs, 113 arcminutes of Duna movement, and since a's conveniently 1, clearly 90.461km/s dv expended either side of turnaround.

So, Slashy's entirely correct, you can ignore planetary movement when doing a brachistochrone transfer in a torchship.

[PB666]

On 8/18/2016 at 2:11 PM, slouis said:

Let's say (hypothetically) that you had a very, very fast spaceship. As in, it can accelerate to 0.5c in a minute. (Presumably, this would require some rather hardy Kerbals, but that's besides the point). It runs on Mysterium fuel, which never depletes (it's a mystery why!)

If this spaceship wants to transfer from Kerbin to Duna in the least amount of time possible*, what should the maneuver be? My intuition is that it could burn towards the target the whole time. But perhaps it is a prograde burn, and then a lot of radial burns? What is the fastest flightplan?
 

* Ignoring timed dilation. The point is not the 0.5c speed per se, but rather that the ship is very fast.

The fastest is transfer is one where you wait, wait until the two planets are on the same radial, then transfer. Kerbols gravity would be immaterial. If the two planets are already aligned with the host star, then you don't need to wait.

But I should point out the essential oxymoronic nature of this argument, because if you can travel at 0.5c and you generally don't care about gravity, the only time you would want to wait is when the star is between you and the other planet. You travel roughly at the the other planet anytime except those times with the star is in your way or will cook you. (also occasionally eve might be in your way) because you really don't care whether it would take you 170 seconds or 20 seconds to get there, just go. In addition if you miss duna say by a second at 0.05c (1/100th the amount of energy of the first trip) you can correct course to your target, as long as you stop when your course is tangential (which in almost all cases is a free-fall trajectory into the planet).

[purpleivan]

On 18/08/2016 at 9:43 PM, eloquentJane said:

If you can achieve a constant acceleration of just 1g, that should be able to get you to Duna in only a couple of weeks on pretty much a straight line trajectory (pointed at where Duna will be when you arrive, that is) so long as you don't have to go too close to the sun, and provided that you accelerate for half of the journey and decelerate for the other half (so that you can easily capture into orbit). The longest duration of transfer in this manner would be when a straight line trajectory passes too close to the sun, in which case you would have to do some radial burning but it would still not take too long to reach Duna.

Now, with your ridiculously fast spacecraft that can travel at half of the speed of light and reach that velocity very quickly (let's just assume that kerbals are invincible) it would take mere minutes to get to Duna, and once again the best trajectory would be a straight line.

I've actually done the Duna run (surface to surface) in 16 days in KSP for a challenge.

Here's a shot showing my trajectory once up to 20km/s

This was back in the 0.90 days, so I was able to do some serious aerobraking/capture into the Duna atmosphere before making a landing.

For a similar challenge but this time to Eve, the trip took just 9 days, due to its orbit being that much closer to Kerbin.

 

Edited August 22, 2016 by purpleivan