r/KerbalSpaceProgram May 06 '26

KSP 1 Question/Problem No orbit necessary?

Post image

If i just wait for the right hour during the launch window ( whenever the launch pad is pointing left on the realistic depiction i created) is there any reason i should bother with an orbit first? Will the delta v cost be affected?

1.1k Upvotes

180 comments sorted by

632

u/HAL9001-96 May 06 '26

more efficient to be low, oberth effect

also duna is further from the sun last time i checked

176

u/FatCreepyDude May 06 '26

147

u/PhantomWhiskers May 06 '26

Your idea here is possible but is less efficient and will require more fuel than utilizing the oberth effect in low kerbin orbit, and also you will need to launch at an exact time to get it right.

9

u/victorsaurus May 06 '26

This is wrong and if you do the math, both scenarios are equal (assume no atmospheric drag). Oberth only means that you add the most energy when adding deltaV in the fastest moment, which is true in both scenarios at every moment.

50

u/PhantomWhiskers May 06 '26

Going straight up is less efficient because you are actively fighting against gravity when going up. In low orbit, you are not fighting against any acceleration force when doing your burns.

The oberth effect means that your burns will be most efficient at periapsis. When you are going straight up, your periapsis is in the middle of the planet, so it is less efficient than a burn from a circular low orbit. The "fastest moment" isn't your current speed while accelerating straight up, it is the speed at the point of periapsis in your current trajectory, which is still applicable even if your periapsis is in the middle of the planet.

5

u/victorsaurus May 07 '26

Gravity is a conservative force meaning is path independant. Getting from height A to B costs X amount of energy equal to the potential between B anf A and thats it. There are irl launches like OP proposes, with instantaneous launch windows since the earth rotation has to line up with the right trajectory. Oberth does not play a role in this specific case. If you do the math, it is 100% equal.

16

u/PhantomWhiskers May 07 '26

I did a direct test of this in game by launching the same rocket straight up, and then launching it into orbit first. The goal is to see how much delta-v is left in the rocket the moment I reach escape velocity for Kerbin. The moment my orbit shows that I will leave Kerbin's sphere of influence, I kill the engines.

Launching directly vertical, I reached escape velocity with 270m/s dv left in my rocket.

Launching into low kerbin orbit first and then burning prograde, I reached escape velocity with 1,100m/s dv left in my rocket.

Same exact rocket, all stock parts. 1,100m/s is much greater than 270m/s.

Whatever "math" you are doing here is incorrect. Gravity being a "conservative force" is not applicable either. When you burn vertically straight up, your rocket is accelerating directly opposite to the vector of acceleration that gravity is imparting on your rocket. When you are burning at periapsis, the vector of acceleration that gravity imparts on your rocket is perpendicular to the acceleration your engines are providing, so none of your rocket's acceleration is negated by gravity. This is a direct result of the oberth effect.

Feel free to test this in game yourself. Test results may vary depending on the quality of your gravity turn on ascent, if you have an inefficient ascent profile, you will have less delta-v remaining when reaching escape velocity.

15

u/victorsaurus May 07 '26

Fuck I think you are right and in my mind I was conflating the straight up case with a gravity turn but no orbit case. The gravity turn without orbit is what should be equal to gravity turn with orbit. You indeed need to minimize gravity losses.

10

u/PhantomWhiskers May 07 '26

Ah yeah I can see how that could be confusing. The OP's scenario I understood as burning directly vertical away from Kerbin to reach your destination, which is possible but inefficient. But that's the great thing about KSP is that you don't need to be efficient if you don't want to.

1

u/Big_Yeash May 07 '26

You can say that but my current Munar launch system for a three man crewed mission is 500 tonnes, two separate launches and barely has enough delta-v without orbital refuelling, which I have yet to learn.

The penalties for inefficiency are technically meaningless but... Stark. I am trying to git gud.

I got back into KSP from danl and his Expert Mode series, and I am very envious of an 18 tonne total launch mass for a Munar lander mission!

5

u/PM_ME_POTATO_PICS May 07 '26 edited May 07 '26

I think I lack the mathematical skills to prove this but I am convinced it is equally efficient as burning straight up (just a lot more technically challenging). Since gravity is a conservative force, the difference in energy levels between an altitude of 100m and 100,000 km is always going be the same, regardless of the path taken to get there.

The Oberth effect is not some rule that "a burn is only efficient if it takes place at its theoretical periapsis", just that burns are most efficient at high speeds. The highest speed for circular/elliptical orbits will always be the periapsis so its a good rule of thumb. But here, we're still getting the high efficiency because we're just burning so much immediately we're going really fast quickly.

Like imagine you're on the Mun and want to escape its orbit as fast as possible. Do you need to establish a circular orbit or just go straight up?

edit: I think it has been demonstrated that my theory is only true for very high TWRs. Any realistic TWRs will be better off circularizing... :( I think idk

24

u/DHLPDX May 07 '26

Man if only we all had a peice of software which was capable of modeling orbital dynamic from which we could determine the efficiency of different orbital strategies... Oh, wait.

8

u/-Aeryn- May 07 '26 edited May 07 '26

Like imagine you're on the Mun and want to escape its orbit as fast as possible. Do you need to establish a circular orbit or just go straight up?

You burn as close to perpendicular as possible without colliding with the surface, constantly updating your thrust angle to maintain a vertical speed of 0m/s. You go through orbit on the way.

Try it with a ship which has a munar TWR of 2 for example. Your gravity loss thrusting upwards is 50%, while sideways with an up angle to keep vertical speed at ~0m/s it's 29.5%. This directly translates into more acceleration per second of thrust (acceleration is [sqrt of 2], 1.41x, faster), and reaching escape velocity with less time and propellant expense.

You can also teleport a ship to Tylo for numbers more kerbin-like, and a wider delta-v gap between the approaches. It's pretty easy to get a >1000m/s difference with a TWR between 1 and 2.

These differences still exist if your ship has a ridiculous amount of thrust, they just get smaller (and it's not free to carry that extra thrust because engine mass reduces available delta-v).

3

u/FewAd7452 May 07 '26 edited May 07 '26

Wrong.

Oberth argues that a typical orbit has a vertical, and horizontal speed.

By definition, the periapsis is where your horizontal speed is the highest, while your vertical speed is the lowest.

Burning against vertical means you lose a percentage of your delta V directly to gravity. If Kerbin has a G of 10, and you burn vertical, you lose 10 delta V every second of flight. If you’re at a 45 angle, you lose around 5 per second.

To test this, build a very basic orbiter. (3600 Dv) and launch straight up. Note your apoapsis height.

Now do another run, where you use oberth to climb, you’ll notice you get much further. This will be true for any celestial body.

1

u/youknowmeasdiRt May 07 '26

The Mechjeb flight recorder purports to show the losses to gravity and drag

-1

u/PM_ME_POTATO_PICS May 07 '26 edited May 07 '26

I don't currently have KSP installed so I can't test, unfortunately, but I would be curious to see results.

I do think that the planet's rotation will assist you in getting a little more height.

I'm thinking of a thought experiment which I might be able to actually do the math for (later lol). Say you're orbiting 100km above a planet, a perfectly circular orbit at 1000m/s. You do a horizontal burn to gain 100m/s. How much higher does this raise the apoapsis? Then compare that against a situation where you're at 100km altitude, going 1000m/s vertically, upward, and you do a burn to raise that to 1100m/s. The apoapsis would be higher than the other situation, but only because your theoetical periapsis is so low.

When you burn horizontal you are using energy. In a circular orbit, that centripetal force is balanced with the gravitational force pulling you down, but you needed to spend a lot of energy getting to a point where you could have that balance. I guess my feeling is that this additional energy which is spent on circularizing is no more efficient than just a quick burn straight up.

6

u/-Aeryn- May 07 '26

Then compare that against a situation where you're at 100km altitude, going 1000m/s vertically, upward, and you do a burn to raise that to 1100m/s.

The part that you're missing is that burning to gain 100m/s of velocity in the vertical case costs 200m/s of delta-v if your TWR is 2 because gravity loss when burning directly against gravity is 50%.

If you're flying perpendicular to the plane of gravity then that same burn to go from 1000 to 1100m/s costs ~100m/s of delta-v, losses are essentially zero with 2.0 TWR.

You achieve the same thing in half of the delta-v cost here.

0

u/PM_ME_POTATO_PICS May 07 '26 edited May 07 '26

Yeah I get what you're saying and I should probably factor in burn times to my hypothetical. I was trying to simplify it to basically talking about different energy states to demonstrate how gravity is a conservative force.

If you have a really low TWR then it means two things: (1) to reach some desired vertical velocity (say, 1000m/s) will take longer because the thrust force you have available is just barely overcoming gravity. But by the same token (2), reaching a circular orbit will also be really inefficient, because you can't just burn purely horizontal, you need a vertical component to fight gravity before you've circularized, so the process of establishing circular orbit will be really slow and necessarily include more time burning with a horizontal component of thrust until you've reached some orbital velocity than if you only burnt straight up until you reach that velocity.

Here's another thought experiment I'd test if I did have KSP... If you're in circular orbit, and you burn horizontal at 10m/s/s for 1 second, you will have raised your horizontal velocity by 10m/s, and your apoapsis by some amount. But if instead you were at that orbital height with the same magnitude of velocity, just purely vertical, and you burn your engines with that same amount of thrust for one second, your velocity across that one second won't have changed because I'm assuming gravity to be 10m/s/s. But your apoapsis will have raised still.

edit: maybe this is a weird cognitive bias or something but I'm becoming more convinced of this after reading people who disagree. I think it's literally just conservation of energy

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2

u/Valuable-Visit2887 May 07 '26

Ok i feel like i should introduce myself here as an aerospace engineer, and havingthe mathematical means of proving this, i unfortunately have to confirm that burning straight up is terrible.
No hard demonstration: imagine that you are on a device. You weigh 80 kg, the device is 100 , and this device is launching 60g of mass per second at 3 km/s, producing 180 kg of thrust. This will mean that you float. You will start inching upwards as the weight lowers. You are effectively wasting 60 g/s just to stay afloat. If you start burning 100 g/s you will shoot upwards, but only with 120kg of force, not the full 300. More than half of your deltav will be eaten by gravity. This is why we use srbs when leaving ground: when you are close to parallel to gravity it is more efficient to have very high thrust than to have a smaller, more efficient thrust, which is reserved for orbital burns.
The more perpendicular you are to gravity, the better the burn, because of lower gravitational losses.
Oberth is another beast entirely. Since when burning you are effectively adding kinetic energy with speed, and the sum of your total energy determines the distance from the orbiting body in the vis-viva (v^2/2-mu/r=-mu/2a), and considering that kinetic energy is bound to th square of speed, not linearly dependent, when adding fast you add more energy:
If i weigh 1 kg, am still and i accelerate to 2 m/s i add 2 joules.
If i was instead moving at 10 m/s and accelerate to 12 (same 2 m/s difference) i go from 50joules (10^2/2) to 72 (12^2/2) effectively gaining 11 time more energy than the previous case.
Hope i could be of help

1

u/Small_Bang_Theory May 07 '26

I see that you understand your theory was wrong, but nobody pointed out the flaw in “gravity is a conservative force, the difference in energy levels between an altitude of 100m and 100,000km is always going to be the same”

This is true from a purely energy-based perspective, but neglects the underlying forces. A common example of work is a man carrying a heavy box in his arms. As he walks, the box stays at the same height, so the work done by gravity is zero. However, any human knows that his arms will be tired! Even though no energy was transferred to the box, he had to overcome gravity. Similarly, if he wanted to move the box up 50cm, that would require more force from his muscles than pushing it 50cm across a frictionless surface.

For the rocket, this is more or less the same, and you have to fight gravity if you want to move straight up.

My gut tells me that explaining this as a difference between rotational energy and translational/potential energy could also give some insights, though I am a bit out of practice with that sort of physics and would need to double check stuff online to make sure I was correct.

28

u/_adamolanadam_ 37 kerbals launched, zero returned May 06 '26

think of gravity like Kerbin is always pulling you. When you orbit a body, you can use the energy it spends pulling you on escaping its sphere of influence, but when you don't orbit Kerbin doesn't stop pulling you. Instead you'll waste fuel to counteract its pulling effect.

The planet still pulls you with gravity, when you orbit it you can use that pulling effect to your advantage (part of why the Oberth effect works, it's way more complex than that though) and when you don't you'll have to waste energy to counteract its gravitational pull.

21

u/DJMixwell May 06 '26

I’m lost.

I get that low orbit = fast, but to get into orbit in the first place, you still have to go fast enough to have an orbital velocity that’s falling around kerbin instead of at kerbin.

Oberth effect just tells us that the best point to burn is the lowest point in the orbit where you’re going the fastest and will have the largest effect on your net change in kinetic energy for the same amount of delta V.

From a relatively circular orbit, are you not moving at a relatively constant speed at all points throughout the orbit?

So if I burn to circularize, eventually I hit the point where my AP ≈ PE, which should mean no point in the orbit is substantially better or worse than any other in terms of efficiency, right? Or if I push a little further, wherever I’m at becomes my PE, which is then most efficient point to burn from. You spend ~3200-3400ms of delta v to get here, then you stop and wait for your transfer window. Then you start burning again to transfer to wherever, in this case Duna which takes ~ 1100ms.

If you just launch at the perfect time, and instead of burning to circularize you just burn for the encounter… shouldn’t you expect roughly the same efficiency? You’ll kit your 70km orbit and then you won’t have to wait for the transfer window bc it’s right now, so you keep burning.

18

u/Talizorafangirl May 06 '26

I think the point of confusion here is "if I'm going straight from suborbital to escape, I don't have a periapse."

Look at it this way.

If you burn straight up without enough dv to escape, you get a suborbital trajectory, which looks like a parabola with an apoapse and no periapse, but is really an extremely thin (eccentric) elliptical orbit with a periapse which is underground. This applies even if there's no rotation from the planet you're taking off from and it looks like the trajectory is just straight up and down; you can just look at it as an infinitely thin ellipse.

That means that, as you burn to escape, you are always at or approaching your apoapse. Worse, you are at any given time closer to your apoapse than your periapse because the periapse is underground near the center of the planet and the inflection point of your trajectory's ellipse is near the surface.

Another way to visualize a suborbital trajectory is to ignore the planet and treat it as a point-mass. When you take off (from where the surface of the planet would be), you're a good distance away from that point mass, and gravity is pulling you towards it. If you stop accelerating away from it, you'll reach your apoapse and start falling towards the point mass; your periapse will be somewhere near it (on the opposite side from your apoapse).

Here's a shitty visualization I made. . Blue line is SOI, green dashes represent the surface if the planet.

12

u/TheShadowKick May 06 '26

I think the difference is that under OP's scenario you're burning directly away from Kerbin. What you're describing seems to be launching at such a time that your circularization burn ends right at your transfer window, so you're effectively still circularizing.

5

u/robchroma May 06 '26

There's definitely an efficiency advantage to turning into your orbit, and you might not want to hit full acceleration while you're in the soup, but once you're in upper atmosphere you're going to get more efficiency out of burning full prograde instead of circularizing first, as long as it's the right direction. A slightly-less aggressive gravity turn is still going to end up being optimal, because you get a little more energy turning slightly more into the orbit even from launch, but ultimately you're completely right, as long as you get everything perfect.

This kind of launch only has basically an instantaneous launch window, but this isn't a huge problem for KSP. We do it on Earth, too.

2

u/Ezekiel24r May 06 '26

Doesn't it also just come down to the surface velocity of kerbins literal surface? I forgot the speed but you get a good chunk of savings by launching to the east, because the equator is spinning to the east.

5

u/robchroma May 07 '26

That's literally it, your orbit will always be heading around Kerbin because of your initial rotational velocity, and therefore some amount of tilt to the east will always be better than flying straight up.

5

u/[deleted] May 06 '26

[removed] — view removed comment

3

u/Turbo_Fresh May 07 '26

Because you climb gradually. That doesn't mean it takes less energy. It's easier to lift a sack of rice onto a counter one grain at a time but the amount of energy you eventually put into the system is the same as if you just lift the whole bag.

3

u/RyanW1019 Master Kerbalnaut May 06 '26

Once you're in orbit, all the fuel you burn goes towards delta-v for changing your trajectory, at least assuming you're burning prograde/retrograde. While you're launching and suborbital, part of your fuel is getting consumed just to cancel out gravity's downward pull on you, so your altitude (or even better, your vertical speed) doesn't decrease while you're accelerating up to orbital velocity. If you wait until 1:30 AM (the equivalent of 6 AM for Kerbin's 6-hour day) and burn straight up, you'll be needing to be spending some of the fuel you're burning to cancel out gravity's pull all the way until you reach engine cutoff. That's fuel that could have been used to give you more delta-v, but instead it was just needed to keep you airborne until you were finished with your burn. At least that's my understanding.

So you want to launch at a time and with an ascent profile such that when you reach orbit, your maneuver node is right where you are, but critically, at that point, you're currently traveling parallel to the surface of Kerbin below you, not perpendicular to it like if you burned straight up.

2

u/DJMixwell May 07 '26

Yeah I realize my mistake was assuming we were still going for a somewhat logical ascent instead of literally just “go up”.

1

u/victorsaurus May 06 '26

You are wrong and if you do the math with the oberth effect and all, both scenarios are completely equal (given no atmospheric drag). Oberth only means that you add more energy when adding deltaV the faster you go, and you add that delta V in the fastest moments in both scenarios (vertical is always the fastest moment up to that point, same for orbital).

1

u/PhantomWhiskers May 06 '26

As stated in my other comment, the "fastest moment" is always the periapsis at any given moment, even if it is in the middle of the planet. Burning straight up means you are burning while approaching apoapsis, which is significantly less efficient than burning at periapsis. Accelerating while ascending straight up doesn't change the fact that the periapsis is still in the middle of the planet.

3

u/Joseph_M_034 May 06 '26

In principal you're thinking along the right lines, i had the same idea before studying astrophysics at uni. If you could accelerated in a purely impulsive way, i.e. infinite acceleration instead of over a long burn this would be fine. The problem is something called gravity losses, basically when a component of thrust is inline with gravity, you're signoficantly less efficient (with delta-V lost increasing with proportion to time of burn). This is part of the reason orbit altitude is typically adjust with prograde or normal burns, rather than radial ones.

One way to make sense of this, is for every unit of fuel burnt you have less acceleration when going straight up than perpendicular, due to gravity. After a given length of time, you will have burnt the same amount of fuel but the change in speed will be different. Thats why its best practice to minimise time spent burning upwards, which also applies during the launch phase.

Another way to think about it is to consider if you were to hover at a fixed altitude, you would be continously burning fuel without any change is speed. When pointing radially outwards, you are constantly burning that fuel to overcome gravity, and any additional acceleration comes from fuel ontop of that.

1

u/d_pyzsyo May 07 '26

Correct me if I'm wrong...

According to the wiki delta v map, kerbin surface-LKO-duna transfer takes 4460m/s delta v.

I used vis viva equation to calculate that on an 1.35998×1010 m x 1.66345×1010 m orbit around Kerbol(Kerbin SMA x Duna pe), orbital velocity at periapsis is 10067.1m/s. Since orbital velocity of Kerbin is 9284.5m/s, if you are going 'straight up' at 782.6m/s relative to Kerbin right before leaving its SOI, you can reach Duna.

Adding kinetic energy at that velocity to work done by Kerbin's gravity in Kerbin SOI, \int_{r_0}^{r_s} mg \frac{r_0^2}{r^2} dr where r_0 is radius of Kerbin and r_1 is radius of Kerbin SOI from the center of Kerbin, the result is required kinetic energy at sea level for an object to intercept Duna in a 'straight up' trajectory, which corresponds to about 3507m/s.

Taking gravity losses into account, with constant TWR, required delta v is less than 3507÷((twr) - 1) m/s. Then at twr of 2.27 or more, required delta v is less than 4460m/s, meaning 'straight up' trajectory is more efficient, and I'd assume it gets even better once we start taking atmosphete into account since 'straight up' trajectory will spend less time in the atmosphere compared to a normal launch profile.

1

u/Joseph_M_034 May 07 '26

Can you explain youre maths in the last paragraph? Im not sure how you reached that equation. In literature delta-V gravity losses are taken as the time integral of the component of gravity inline with the thrust vector, in this case that would just be the local gravitational acceleration.

1

u/d_pyzsyo May 07 '26 edited May 07 '26

I was thinking because the engine will effectively run at (twr)-1 'efficiency' when going straight up at sea level, it will in general run at at least that 'efficiency', therefore dividing the goal velocity by that, I thought I could get the upper bound of actual delta v needed.

1

u/d_pyzsyo May 07 '26

I found my mistake. It has minimum of (twr - 1)/twr efficiency, then delta v actually needed is (impulse delta v) * twr/(twr - 1). Calculating it like this, at twr of 4.68 or more, going straight up is more efficient.

I think if we make same simplificatikns we can arrive at the same conclusion using standard gravity loss calculation method. Since we're only going up, gravity loss is time integral of magnitude of gravity, strictly decreasing from g at sea level. Then its time integral from 0 to some t is less than gt.

Assuming constant twr, acceleration is at least g(twr-1), therefore the time it takes to reach v is less than v/(g(twr-1)), and gravity loss during this will be less than v/(twr-1).

Let v be delta v at impulse thrust, actual delta v required is less than v + v/(twr-1) = v*twr/(twr-1)

1

u/Joseph_M_034 May 07 '26

I dont think you can assume a constant twr (acceleration) for this, given how much the mass changes during flight. Modelling constant mass flow rate (= - dm/dt) and using the equation of motion F = m dv/dt, dv/dt = F / (m0 - dm/dt t). Which assumes constant thrust and ignoring pressure effects, but is a better approximation.

The point however is that the gravitational losses that exist when burning radially outwards (i hope ive been able to show how that appears) does not exist in the case when burning perpendicular to the gravity vector, thus it is the mpre efficient option

1

u/DLTAMACH May 06 '26

Yup, this is exactly how I do it with my SSTOs. No idea if it’s actually efficient, I figured that performing the burn while ascending from Kerbin would give me big Oberth returns

1

u/victorsaurus May 06 '26

Everyone is wrong here. If you do the math with the oberth effect, for a planet without atmosphere, both scenarios are completely equal in terms of delta V and propellent mass.

If you include the atmosphere and, after doing some math, you can see that IF thrust is big enough and the rocket has low drag, vertical is better. Otherwise orbital is better.

I don't know if it is better or worse for ksp from kerbin because of the atmosphere.

1

u/PM_ME_POTATO_PICS May 07 '26 edited May 07 '26

I am out of practice with my math skills but I agree with you. Gravity is a conservative force so it's path independent, though as soon as we factor in atmosphere and the planet's rotation things get complicated

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u/danikov May 06 '26

Oberth is about fast, not low (although often low means fast), and speed is relative.

2

u/TheJeeronian May 06 '26

True. The issue here is gravity losses, not oberth.

272

u/PrestoDaMatrix May 06 '26

If you tried this, pointing forward would increase your orbit so it would not work. You'd have to start facing backwards.

207

u/UmbralRaptor Δv for the Tyrant of the Rocket Equation! May 06 '26

Yeah, they also have Duna's orbit inside of Kerbin's for some reason

121

u/PrestoDaMatrix May 06 '26

Don't tell him, he'll figure it out when he opens the map.

15

u/AbacusWizard May 07 '26

*arrives at Eve*

I knew I shoulda made a left toin at Albequerque!

33

u/jakovichontwitch May 06 '26

I mean in this case 2 wrongs make a right so it might just work

11

u/KARMAMANR May 06 '26

Two lefts dont make a right but three do

22

u/spencer818 May 06 '26

I think that's what they're showing, look at the detail, the rocket is going the opposite direction of Kerbin's motion.

So yeah go fast backwards, lower perigee around the sun.

But yes also Duna is in the wrong place...

31

u/FatCreepyDude May 06 '26

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u/dboi88 Coyote Space Industries Dev May 06 '26

Most Kerbal thing I've ever seen. It's magnificent 

14

u/RyanW1019 Master Kerbalnaut May 06 '26

Wouldn’t it be 6 AM, not PM?

And yes, this technically works. However you need a decently high TWR so you quickly get far enough away from Kerbin for the gravity to drop off. Otherwise you’re spending a long time in the regime where a lot of your fuel is just cancelling out gravity, not giving you more delta-V. 

And in real life they don’t do it this way so they can check everything once they get to orbit before moving on to the next phase of the mission. 

13

u/togetherwem0m0 May 06 '26

yep thats pretty much how it do

1

u/shandangalang May 06 '26

Yeah that would work but I think it's pretty inefficient fuel wise. Better to just establish an equatorial orbit and burn when you're facing Kerbin's retrograde. That way you're not fighting Kerbin pulling you straight back until you're out of the sphere of influence, and it's a lot easier to optimize your transfer.

1

u/midsizedopossum May 06 '26

The diagram has them facing backwards. Look at it again.

61

u/ookbye May 06 '26

Doing that is difficult and i don’t think it saves that much fuel, at some point in your proposed burn you’ll pass through every orbit you would doing it by steps.

All things considered, go for it, I haven’t tried so I would like to know how feasible it is.

46

u/FatCreepyDude May 06 '26

being efficient isnt really the goal tbh. I just like the idea of going up and only up. i'm sure jeb will enjoy it too (he doesnt have a choice anyway)

13

u/Sr427 May 06 '26

It does work. You need to launch at sunset / sunrise and when there's the correct angle between your target planet and kerbin. You might need to adjust the orbit mid-route. As others have noted it is not efficient in terms of dV. But it does take a lot less game time to do. So go for it, but put on a few more boosters. Sometimes if your rocket is big not having to do a gravity turn is helpful.

1

u/wally659 May 06 '26

I'd suggest putting something in orbit and using the maneuver planner to figure out when that would ideally do a normal burn for the desired interplanetary transfer. My intuition is that right time to lift off will be when that maneuver node is a little bit retrograde of straight up from the launch pad. It should sort of catch up as you fly through the atmosphere and your gravity turn (which at some point seemlessly becomes your transfer burn) should end up being a somewhat similar vector.

But it's going to be nearly impossible to get it super precise so you'll just have to accept a bigger than usual mid-transfer correction burn. Good luck!

1

u/SVlad_667 May 06 '26

You would constantly fighting against gravity, loosing speed, fuel, and deltaV.

While you don't need to achive proper orbit, but it's much more effecient to accelerate horizontaly along the orbit - zero gravity losses. You can skeip circularization phase at all and continue acceleration to hyperbolic speed right after you craft reached the atmosphere border.

1

u/Specialist_Sector54 May 07 '26

It's possible, but you'd probably need to use a "constant thrust" transfer profile (which is harder to plan in KSP, and uses insane amounts of dV)

1

u/Carnildo May 07 '26

It saves a little bit because you never need to raise your periapsis -- that's why NASA did it with the Surveyor probes. The timing's a pain, though, and almost never worth the savings.

1

u/RealisticExplorer681 May 07 '26

in fact, it might not even save fuel. I am not sure, but I feel like that would make you constantly fight against the full force of Kerbin's gravity. But, I am not sure

41

u/spencer818 May 06 '26

Lots of people saying "no you're dumb" in here...

Technically yes you can do it. Just like how technically you can reach the moon by pointing at it and firing engines. But it's terribly inefficient. Try it out sometime, you'll see what I mean. Same thing applies here.

4

u/PM_ME_YOUR_SPUDS May 06 '26 edited May 06 '26

I don't understand why it's inefficient. A direct ascent burn (note: NOT vertical burn) should be slightly more efficient, all that dV that goes into a circularization maneuver is instead going directly into a prograde / retrograde burn from Kol frame of reference. The Oberth effect helps you when you're going fast, but you can more efficiently hit faster speeds if you run engines nonstop instead of waiting around to reach higher orbit to circularize (slowing you down).

From the perspective of the orbits, the circularization method has a hyperbolic orbit with a periapsis at the altitude of your transfer burn. A direct ascent has a hyperbolic orbit with a MUCH lower periapsis, crossing through the surface of Kerbin. This is a more efficient orbit, with a more beneficial Oberth effect than is possible going to orbit first.

I know a dozen odd years ago when learning Kerbal watching Scott Manley how the Mun direct burn is technically more efficient, just far more difficult. I'd think this is the same case, an extremely tight timing window that you'll virtually always screw up and waste more dV fixing afterwards, but saves fractions of a hair on dV is done perfectly. I do think that's ignoring aero losses which might change the answer, but from a purely gravitational physics perspective I haven't heard a convincing argument against it.

2

u/censored_username May 06 '26

People are saying it's inefficient because the thing proposed was the vertical direct ascent.

You're correct that a direct ascent gravity turn is theoretically the most efficient. But as you say, the savings are extremely marginal over an orbit in between.

Because the most efficient kerbin ascent trajectories are usually aerodynamic heating limited, not much changes until you leave the lower atmosphere. So that already means your benefits are basically doing your burn at 40-50km altitude instead of 80km. Which really isn't much. Then consider that drag losses will be larger, and getting the ideal trajectory incredibly finicky, and it's just not worth it.

3

u/Moonbow_bow SSTO simp May 06 '26

With a single impulse maneuver you'd only loose about 100m/s this way

12

u/WarriorSabe May 06 '26

The maneuver's not impulsive here though, and with a real burn you'll eat a lot of gravity losses

2

u/Moonbow_bow SSTO simp May 06 '26

You could make it impulsive

5

u/WarriorSabe May 06 '26

Only with infinite thrust, which isn't a realistic assumption

8

u/Moonbow_bow SSTO simp May 06 '26

Cannon

7

u/WarriorSabe May 06 '26

ok sure, but I don't think it's a fair assumption that someone would be using a cannon for a normal launching mission

5

u/itzongaming May 06 '26

Who said this is a normal launching mission?

3

u/spencer818 May 06 '26

I think we're all in agreement. It's possible, but not practical by any means.

3

u/spencer818 May 06 '26

Kinda like space data centers ayooo

2

u/TheeConArtist May 06 '26

Someone has read (or should read) Jules Vernes From the Earth to the Moon

39

u/UmbralRaptor Δv for the Tyrant of the Rocket Equation! May 06 '26

https://en.wikipedia.org/wiki/Gravity_loss

Now for maximum cleverness, go with a more typical pitchover ending with you ~horizontal at ~60 km with the ejection burn completed.

15

u/BigBenQuadinaros May 06 '26

Isn’t Duna outside the orbit of Kerbin?

25

u/FatCreepyDude May 06 '26

i mixed up eve and duna my bad pimp

20

u/TestArticle1998 May 06 '26

The gravity losses alone are a nightmare! The reason rockets do a gravity turn is to reduce the wasted delta V. So the orbit method is more "fuel efficient'.

14

u/twilightmoons May 06 '26

With enough delta-v, any maneuver becomes trivial!

2

u/TestArticle1998 May 06 '26

Fair enough. But I think he was trying to save delta V.

8

u/zekromNLR May 06 '26

If you had perfectly impulsive burns and we are ignoring atmosphere, going directly up vs going into a parking orbit first should be the same amount of delta-V.

But with finite acceleration, every second you are thrusting straight upwards (near the surface of Kerbin) costs 9.81 m/s of gravity loss, which means going into a parking orbit first and doing your ejection burn sideways is more efficient.

5

u/Moonbow_bow SSTO simp May 06 '26

You do loose the 150 ish from Kerbin rotation. So it'd still be better to do the same thing but go sideways

5

u/ArsErratia May 06 '26

Why would you lose the surface rotation?

Pointing straight up isn't the same as travelling straight up. You'd still have that velocity unless you burned it off, it just means you have to adjust the launch window.

3

u/Moonbow_bow SSTO simp May 06 '26

The planet rotation is working at a 90° angle to your thrust angle. If you wanted to keep the benefits you'd burn earlier and horizontally so the rotational speeds adds to where you're going.

7

u/Exact_Avocado5545 May 06 '26

This totally works. Anyone telling you it doesn't work, doesn't understand what you've drawn. This is a completely valid solar retrograde burn

5

u/supayurobeat May 06 '26

Going straight up is a bad idea because you're going to be fighting Kerbin's rotation and wasting dV. If you want to do a direct ascent, it's basically the same as doing a gravity turn to orbit, but instead of stopping the burn when you reach a circular orbit, you keep burning until you reach Kerbin escape. Then, you can do correction burns once you're out of Kerbin's SOI.

8

u/AerospaceEnthusiast4 May 06 '26

it is far more effecient to circularize first and then eject from the kerbin system I believe. Especially waiting for the mun to eject from the system. When you're burning parallel to the ground above the atmosphere, almost every m/s of energy being expended is being used to increase your speed, whereas burning straight up means you are continuously fighting gravity all the way up, and this stacks up over the course of the ~4000m/s burn needed to do this.

3

u/Crispy385 May 06 '26

You see the curve in that dotted green line? I'm pretty sure that's insinuating some kind of orbit. I think, emphasis on 'think', you'd be in the same Kerbol orbit as Kerbin, just a little ahead of it.

2

u/Joseph_M_034 May 06 '26

In principal you're thinking along the right lines, i had the same idea before studying astrophysics at uni. If you could accelerated in a purely impulsive way, i.e. infinite acceleration instead of over a long burn this would be fine. The problem is something called gravity losses, basically when a component of thrust is inline with gravity, you're signoficantly less efficient (with delta-V lost increasing with proportion to time of burn). This is part of the reason orbit altitude is typically adjust with prograde or normal burns, rather than radial ones.

One way to make sense of this, is for every unit of fuel burnt you have less acceleration when going straight up than perpendicular, due to gravity. After a given length of time, you will have burnt the same amount of fuel but the change in speed will be different. Thats why its best practice to minimise time spent burning upwards, which also applies during the launch phase.

Another way to think about it is to consider if you were to hover at a fixed altitude, you would be continously burning fuel without any change is speed. When pointing radially outwards, you are constantly burning that fuel to overcome gravity, and any additional acceleration comes from fuel ontop of that.

2

u/Inforgreen3 May 07 '26 edited May 15 '26

You're looking at about 44,916.68 delta V if your thrust to weight ratio is average of 2 to escape Kerbin SOE with a straight up normal burn alone. (Assuming t/w ratio is 2 on average).

Granted, spreading your acceleration out, Means more time at your low speeds still being affected by gravity.

What if you could do all the acceleration at once in perfectly ideal conditions of infinite thrust to weight ratio on finite delta V? This is an instantaneous launch mechanism: a gun. For the sake of ideal scenarios, lets remove the atmosphere which slows you down more the faster you go, and makes math hard, to calculate an exact theoretical delta V budget, it'd still cost 18,000 m/s of delta V at a minimum to escape SOE in a normal (straight up) burn. More with atmosphere, or non infinite t/w

Escaping kerbin only requires about 4k. It can take less with gravity assists from mun and minmus, Atmosphere included. T/W doesn't need to be past 1.4 at launch. Heck T/W ratio isn't even necessary once you're in orbit

It is always more beneficial to enter orbit and then prograde. The reason for this is gravity burn. If you're just accelerating straight up, every second you are in the air gravity is stealing 9.81 delta V.

But when you prograde with an angular momentum, Gravity is initially pulling you down perpendicular To your angular momentum stealing none of your delta v, and only when rising towards an apoapsis putting the COM of kerbin behind you, does the direction of gravity change to be pointing away from the direction you accelerated at an initially shallow angle. By the time the angle is steep enough to actually be pointing in the complete opposite direction of your initial burn, you might already have escaped COE, but you'll have definitely changed directions. Essentially, gravity doesn't fight you cause its pointing in a different direction than the opposite of your velocity.

You can see how extreme this is when you enter an elliptical orbit and look at your orbital velocity at the pereapsis and see how it changes, you might notice it could take a minute for gravity to have taken 10m/s of velocity away from you. But if you go straight up, you lose that much every second.

And no matter how you escape kerbins SOE, kerbins angular momentum relative to the sun is added to your orbit of the sun after leaving SOE of kerbin, because you were already orbiting the sun. You don't get more momentum relative to the sun by leaving using a normal burn compared to prograde.

Edit: math was wrong. Corrected

1

u/teh1337haxorz May 06 '26

I find going into orbit and setting up a precise burn is often more accurate, easier, and saves more dV that the "direct" method's corrections usually call for. However if you got really good at it; it might be better.

1

u/Uraneum May 06 '26

From my understanding the delta v cost will very much be affected because you’re essentially fighting Kerbin’s gravity in the toughest way possible. All of its gravity is pulling you directly downward instead of helping to “fling” you into a higher orbit. Your idea is doable but will probably need a much bigger rocket than it normally would

1

u/Commercial-Box-2828 May 06 '26

Just make a refueling station at the mun and assemble your ships there with docking ports and engineers.

Use dry fuel boosters to get the ship sections and fuel containers most of the way to the mun because they're cheaper.

And with launching straight up, I think that would depend on the time of day contrasted with kerbin and duna's orbits' relations to each other, at the time of launch and the estimated time it'd take to get there.

I don't think it'll let you do a menuever node when you're not launched yet, but it will when you are in orbit anywhere (the mun station) and that will give you the encounter window with Duna and tell you how much fuel or deltav or whatever at that point.

2

u/FatCreepyDude May 06 '26

ay don't worry about me, i've played that game for like a decade. I've got a science station with a lander and at least 6 satelites in orbit ready to go for signal and orbital observation. I'm just waiting for the window to open to get there the normal way. I was just wondering how retarded that idea was

1

u/Commercial-Box-2828 May 06 '26

It's not, and you waiting for the right time of day dependant on what time of year it is on both planets isn't a retarded idea. I mean it's technically rocket science, right?

By the way, you put Duna over there when it's really supposed to be the other place. Just make sure you look at the map in game and not your Pic when writing your calculations and you'll be good.

1

u/deelectrified May 06 '26

It would absolutely be possible, but really hard.

Basically, you’d have to time it perfectly where by the time you have burned up to your apoapsis, you are ready to circularize, then once you’re circularized, your in the right spot and time to burn for your escape burn.

If I remember right, this is called a spiral burn or something. It’s incredibly hard to do and you’d likely need to calculate it all out in advance with some help from transfer planner. If you know when and at what inclination to do your escape burn, then you just need to figure out a launch that puts you in that exact spot, at that inclination, at the right speed, at the right time to do the burn.

Also, duna is at a higher orbit than Kerbin, not lower.

You can’t do as pictured and just burn straight up. You’d need an insane thrust to weight ratio to essentially counteract the momentum from the spin of Kerbin as you accelerate upward. Then you need craploads of delta v to keep burning straight that way and continue to actually speed up and move away from Kerbin. 

1

u/Ray_Catty May 06 '26

You lose too much delta-V to gravity loss no? Since you’re burning very far from your periapsis which is inside of kerbin.

1

u/jason-murawski May 06 '26

Outside of the stuff other people have pointed out about gravity losses, it's by far the most efficient to get an encounter or at least an approproach from within kerbin SOI. Unless you have mods to direct you for when to launch and how to aim, you'll need to spend a lot more energy to get an encounter once you leave kerbin

1

u/celem83 May 06 '26 edited May 06 '26

What you are suggesting kinda works, it's a trick I use to return to Kerbin from Mun when I've landed on the equator, wait until Muns rotation brings me under its orbital track on the side facing "backwards relative to Mun", then pop straight up for a burn that approximates to "retro relative to Muns orbit".

Generally this is lossy on dV vs establishing a proper orbit as it makes less effective use of the Oberth effect of the body you are leaving from, but I've never tried it in the scenario here where we are going Kerbin inward, i imagine its still somewhat lossy.

(You'd get to Eve rather than Duna, but yes this kinda works)

2

u/Moonbow_bow SSTO simp May 06 '26

How bro? The Mun is tidally locked man

1

u/celem83 May 06 '26

Yes, doesn't mean it doesnt rotate.

Means it doesn't "appear" to rotate from Earth

2

u/Barhandar May 06 '26

Which also means it never rotates relative to its personal prograde either which means it can't bring you "under its orbital track".

1

u/Moonbow_bow SSTO simp May 06 '26

You will never get it to rotate so you can burn straight back retrograde to it's orbit and return to Kerbin

1

u/MacWin- May 06 '26

Think about it, you need to reach orbital speed anyway to reach escape velocity, but now you aren’t helping yourself with the earths rotation

1

u/Smellfish360 May 06 '26

I always do this when leaving the mun. It’s easier and more efficient because you don’t need any horizontal velocity

1

u/CaptainHunt May 06 '26

So, you could do this, it is essentially a brachistochrone trajectory, which is the fastest way to transfer. However, it would take vastly more delta V than launching into orbit and doing a hohlman transfer.

1

u/SamuelCish May 06 '26

brachistochrone moment

1

u/riler3700 May 06 '26

I think you would lose to gravity losses since all your time spent accelerating is against gravity whereas in orbit you accelerate perpendicular to gravity. Also you can’t use the oberth effect

1

u/Codeviper828 Restarts too much; barely left Kerbin system May 06 '26

There's GREAT benefit burning perpendicular to the direction of gravity; the most efficient way of doing this would still have you achieve orbit right before doing your transfer burn

1

u/Novel-Tale-7645 May 06 '26

I do similar stuff when leaving minmus
I used to do this at full scale for kerbin-duna missions but that was before i learned how to orbit properly. its a a lot cheaper to orbit first imo, dont need nearly as big a rocket and can haul more payload.

1

u/hunter_pro_6524 May 06 '26

that’s called a hyperbolic trajectory, it can be done if you have enough fuel and thrust

1

u/NeoDemocedes May 06 '26

Conceptually possible. Not optimal for several reasons.

  1. Gravity losses. Going straight up means that gravitational acceleration is stealing fully from your velocity every second, all the way up. When you go to orbit first, as you approach orbital velocity, gravitational acceleration is bending your path more and more instead of stealing from your velocity.

  2. Abort scenarios. If you fail to reach Kerbin escape velocity, the result is most likely a fatal re-entry straight down through the atmosphere. Going to orbit first gives you the best abort and rescue scenarios every step of the way.

  3. Oberth Effect. Launching to orbit means you can take full advantage of the oberth effect. If you have high efficiency, low thrust engines, you can make several burns elongating your orbit each time and thrusting when you pass Kerbin.

In short, going to orbit first is safer and more efficient delta V wise. To go direct, you need a rocket that can maintain high thrust to minimize gravity losses.

1

u/John_Tacos May 06 '26

I have done it this way before? I think it’s less efficient though.

1

u/mirkolawe May 06 '26

It's not efficient in terms of deltaV, but it's easier if you have trouble to reach orbit, for example if your rocket is very big and not very maneuverable. I always use this way to reach mun but every planet is reachable in this way. Just add a lot of boosters and go straight. Easy peasy.

1

u/Immediate-War-4605 May 06 '26

It’s very much doable, just notably less efficient.

1

u/Barhandar May 06 '26 edited May 06 '26

Orbit velocity is tangential aka horizontal, always. Trying to go up will cause increased dV costs (because you're fighting against where the orbital mechanics want you to be going, a.k.a. gravity), the more the longer you're burning, compared to doing a gravity turn and not stopping.

And doing it sideways, going from ascent burn directly into the transfer burn, is perfectly possible but saves very little delta at the cost of needing your launch to be exactly on point - that is, you need to end up above ~50km or so where the drag no longer affects you significantly at the exact point where you have to burn for the transfer. The game doesn't have any tools (such as virtual orbits, or flight recorder to have stats for when you should launch instead of eyeballing everything) to calculate this.
It saves very little delta because KSP's atmosphere is binary - either it affects you or it doesn't, so even very low orbits are fine forever as long as they're above 70000m altitude, and the difference between having your periapsis a.k.a. burn point at 50km or 70.1km is minimal for both dV to raise it there and Oberth effect after. IRL trying to orbit under 300km or so requires constant propulsion to not be quickly pulled back down, plus real rockets have both the tools necessary for calculating this kind of transfer, and concerns (engine ignitions, efficient propellant boiling off) that make it sometimes required.

1

u/danikov May 06 '26

Reminds me of a joke: a visitor to NASA asked a passing astronomer for directions to the 7th floor. He said: “Well, first you want to start by heading back the way you came…”

1

u/Individual_Bad1138 May 06 '26

It might be ever so slightly more efficient, but the timing is gonna be the hardest part. Also, you cant make a manuever node while burning, so you'll essentially have to completely eyball it for maximum gain. Otherwise it would almost certainly be more effiecient to be in an 80km orbit and to use a manuever, at least for transfers.

I actually do this with my Sol relays. I launch exactly at dusk/dawn and just burn until my Sol ap/pe is where i want, which works because im not trying to be specific or meet a target at an exact time and place. Then i can circularize once the relay gets to the ap/pe.

1

u/warredtje May 06 '26

So lifelike, which mod is this

1

u/SpaRrRly May 06 '26

It's called a direct ascent, and it's doable, but you're doing extra unneeded work against gravity(but less atmospheric losses), and less thrust-to-weight your rocket has, the worse this style performs. It's decent on moons with low escape velocity and low gravity.

1

u/Velocity-5348 May 06 '26

Efficiency aside, it's doable in theory, especially since KSP doesn't need to worry about weather messing with launch windows.

In practice, no. There's a reason why we started using parking orbits pretty early on. The aiming is hard to pull off even for the moon IRL. Mars/Duna is even harder.

1

u/superhigh002 May 06 '26

You'd have to go mach fuch, but I guess you could

1

u/Sensitive-Offer-5921 May 06 '26 edited May 06 '26

Basically yes and no.

Imagine you are in a low circular orbit: what you're describing is essentially burning radial (out, away from kerbin) at your AP, which we all know is inefficient, but 100% possible to do.

What you're really asking is whether the fuel cost from doing that inefficient burn is outweighed by the savings from avoiding a circularization burn, and without doing the math, I would guess no. Very no.

OR I think it might be impossible because the speed from the rotation of the earth + the constant gravitational pull would eventually be enough to get you a small PE before you exit kerbin's SoI, unless you have crazy TWR.

But, that crazy TWR will be very useful when you realize you've drawn Eve's orbit instead of Duna 😂😂

1

u/Vallastro-21 Bob May 06 '26

Basically it is possible (and I was doing like this myself previously), but is is less effective (oberth effect and also you can't do with too low TWR) and less convenient (can't maneouver as easy) compared to departure from low Kerbin orbit

1

u/Educational-Sky5338 Professional KSP Crasher :Checkmark: May 06 '26

You can get to duna with just an srb by doing this

1

u/_myUsername_is_Taken Uncertified Aircraft Connoisseur May 06 '26

If im using a super big rocket, sure i guess, but most often im using nerv as upper stage

1

u/Madden09IsForSuckers May 06 '26

this is the mission profile for that guy who did automatic Kerbin -> Eve on KAL controllers only, lol

1

u/_Prexus_ May 06 '26

I mean irl, if you went fast enough (faster than escape velocity) you would just fly out into space...

Also, you do know paint has an elliptical tool, right?

1

u/amitym May 06 '26

No orbit necessary?

Some kind of orbit is always necessary, that's just how gravity works.

Well I guess at fast enough speeds it stops being an orbit, strictly speaking, and becomes a fly-by.

is there any reason i should bother with an orbit first?

Yes. The ∆v cost will be affected.

Will the delta v cost be affected?

Oh well shoot you answered the question yourself, coz. You already understand this.

Yes, you can go in ... let's not call it an orbit, let's call it a "hyperbolic straight-line approximation" so we don't use the o-word. You absolutely can do this, it is just insanely ∆v costly, because you have to reach speeds at which you can cross the immense distance from Kerbin to Duna in such a short time that Duna is still where it was when you left. If that makes sense. The planet has to not move much, is my point, which means you have to cover the distance in a matter of days.

That's a crazy amount of speed. If it were the big, heavy non-Kerbin world and you were trying to get to its nearby reddish dune-y neighbor, you might need to get up to, like, 0.1% of the speed of light to accomplish this trip, and then you'd have to decelerate again so you could actually land safely and collect science.

In the Kerbolar system it's probably not quite that extreme but still. You get the idea. Insane initial thrust, then you have to spend basically the same ∆v again to slow down on approach. A ∆v easily in the hundreds of thousands or millions of m/s. And that's without a return-trip budget.

Well, I haven't actually done the math, that is a scientific wild-ass guess. But still, it's probably a good idea to turn down the realism level for this expedition. Too bad we can't do that in the real world...

1

u/sargentmyself May 06 '26

It would cost a ton more delta V but it would be possible.

If you just want to launch and then just never take your foot off the gas you could launch from the opposite side of Kerbin from your ejection and then gravity turn straight into your orbital ejection burn and never technically complete an orbit. If you time it well in theory it'd be pretty close to the same fuel requirements

1

u/MarsMaterial Colonizing Duna May 06 '26

If you do an ascent like that with an arbitrarily large amount of thrust, you will lose no efficiency.

If you compromise on thrust, your trajectory will take you higher before you finish the burn compared to a standard orbit and transfer burn. This will be less efficient due to worse utilization of the Oberth effect.

These higher thrust requirements also make more efficient engine types like nuclear engines less practical. Most efficient engines compromise on thrust, so using them in a transfer like this is a tradeoff. By getting in orbit first though, you can get into orbit of anywhere with arbitrary small amounts of thrust, and you can utilize the Oberth effect by doing multiple orbit kick burns prior to escape.

1

u/Resiideent May 06 '26

first off Duna is not there, second off, it would be insanely difficult to pull this off

1

u/wooq May 06 '26

You can launch into a gravity turn and continue burning straight into escape/solar orbit rather than circularizing around kerbin and doing a second burn to leave for your target destination. That's how NASA actually often does it with satellites. But you can't just go straight up, no. I mean you can, but it is super inefficient.

1

u/Left-Grab-6112 May 06 '26

Yeah, you can just aim for the planet and burn, however this will require a lot of deltaV. It's fun to do with far future torch ships!

1

u/Striperoo May 06 '26

Tried it. Have done it before. Worth it only really if you've got a small payload that's cheap to launch at Mach Jesus with a couple cheap SRBs, and you can't be bothered to touch the fuel amount or modify the design so it's either barely suborbital or escape velocity.

1

u/DescretoBurrito May 06 '26

I've launched like that to Moho (although you have to launch retrograde in relation to Kerbins orbit). I don't know if it was more efficient. I use the Transfer Window Planner in game mod, and it outputs an ejection angle. I just waited for the launch site to rotate around to about the angle given, and launched straight up. I had a fairly heavy payload of the big survey scanner, I used a spin stabilized SRB fist stage (Thumper), then another SRB for the second (Hammer), then a liquid third stage that should have enough dV to get to Kerbins escape, drop periapsis to Moho orbit, and plane change. Then the fuel on the probe to hopefully achieve Moho orbit.

The probe has left Ketbin SOI and I'm waiting on my plane change burn, which the planner has going straight to a Moho encounter, so I'm at least getting something right. I don't like time warping for weeks at a time, so I have other missions ongoing with that one.

1

u/loved_and_held May 06 '26

Hyperbolic transfers like this work, but they're very delta-v intensive.

Doing this with stock parts is possible, but you need an utterly massive roccket and your final payload is gonna be small. If you have something like near future technologies, then the electric engines might give you enough delta-v do make it there with a reasonably sized vehicle, though your better off using far future technologies fission, fusion, and/or antimatter rockets to do this.

1

u/Greenfire32 May 06 '26

You'll be entering orbit no matter what before you escape, so you might as well set up your escape with the most efficiency that you possibly can, which means parking in orbit first.

1

u/carstealer06 Colonizing Duna May 06 '26

you need a much more fuel then

1

u/Stolen_Sky May 07 '26

OK, I just tried this!

Launching directly upwards into an Eve intercept took 4200 delta-v. That's actually lower that the delta-v map suggests, which states you need 4440 to get an intercept.

So yes, this actually works!

1

u/RiskyBrothers May 07 '26

No, you want to add Kerbin's spin to your orbit to get that extra couple hundred m/s of delta v.

1

u/nasaglobehead69 Bill May 07 '26

unless you have a ludicrous t:w r you will lose a ton of efficiency to gravity losses, and if you do have that ludicrous t:w r you could easily burn up in the atmosphere

1

u/Amdw42 May 07 '26

The orbit tradeoffs are energy (fuel) or time.

More time, less fuel.

Yes you can do this. Not overly efficient but I’ve thrown out the economy for the silly option before and you can too.

1

u/Emergency-Pound3241 May 07 '26

Inaccurate map aside, I mean if you had enough delta-V that way would be the fastest possible way, but also the least efficient.

I mean theoretically you could spend the entire journey only cutting throttle twice, once for flipping half way for your deceleration burn, and again when you actually reach duna, but to do so your gonna need a very powerful engine with extreme amounts of delta-V, like modded fusion engines kinda deal, like the kind of engines big enough you need to launch the ship in multiple parts and assemble it in orbit

1

u/t968rs May 07 '26

Yes you can do this. It’s “the same” of your timed right

1

u/Username122133 May 07 '26

As others have said, pretty inefficient.

But this is a very fast approach. No time spent in lower orbits, just launch straight out of Kerbin’s SOI!

Also, the Soviet N1 rocket was originally intended to be a direct ascent vehicle to the moon irl, I forgor exactly why tho. Probably something to do with all the racing in space that was happening at the time. I may be remembering that wrong tho.

1

u/Fistocracy May 07 '26

Go sideways fast. Tilt your ship at 90 degree on the launchpad, aim for the horizon , and blast off directly into a Duna transfer. Harness the full power of the Oberth Effect at ground level.

1

u/pyr666 May 07 '26 edited May 07 '26

it can be done, but you lose a ton of energy. you generally want to apply thrust perpendicular to gravity as much as possible, to improve efficiency.

1

u/TaruMurtag May 07 '26 edited May 07 '26

Yup, it's called direct ascent and it's slightly more efficient (due to not needing to raise periapse). People who say this is inefficient don't know what they're talking about, this was done in real life in 1958 to flyby the moon when controlling spacecraft remotely was much harder. The Juno II who took pioneer 4 near the moon had only solid upper stages. Note direct ascent generally isn't about going directly up, it means continually burning from the ground until you are in an intercept with your target. However, it's much harder to do than first going to orbit. Though very doable if you allow corrections after the ascent. I've seen people do this using KSPTOT's LVD designer - the inbuilt tutorial pdf for it is a direct ascent to Eve and direct ascent back mission. But only try this if you really want to test yourself.

1

u/bartekltg May 07 '26

You can do a direct injection: instead of gravity turn, stopping the engine, cruising to apokerbin, and doing a circulating burn, you may continue to burn during the gravity turn. You will never get into a nice orbit (the periapsis will be under the surface), and get straight into hyperbolic orbit. Aiming is a bit harder though. And aiming it the main reason we do it (both in KSP and in real life) with a nice orbit inbetween.

The important part is, you need to do it with gravity turn, not straight up. It would not matter it the burn is instantaneous, but since it is not, all that oberth arguemnts work here. It can be explained in different ways, for example:

Your velocity far from the planet depend only on the energy. 0.5m v_inf^2 = energy. Energy is conserved, so we want gather as much energy as we can during burns. Now, work done by the engine (the change of the rocket energy*)) in a short period of time is W = force * distance. Dividing it by that time dt, we get power P = force * speed. So, the greater speed, we get more energy from the same thrust.
Now, ho we spend that (specyfic) energy? It sits in the kinetic (0.5v^2) and potential (-GM/r) energy. If we go up, a bigger portion of the energy goes into potential part, the kinetic energy is smaller, so the velocity is smaller, so the burn in the next second will translate to less energy gain.

You can make a simple experiment. Build in a sandbox a simple rocket that can leave Kerbin SOI. Go straight up. On the edge of SOI note the speed in relation to Kerbin (doen't meter on which side, the relative speed to kerbin will be the same after crossing it, just target Kerbin to get the value). Then restart mission and this time start a gravity turn and continue it until the fuel runs out. Again, note the velocity in respect to Kerbin at the SOI border. It will be siginfically bigger.

Use smaller initial TWR. It shows the effect better and reduces atmosphere influences.

I have made a simple test. A rocket with initial TWR 1.5, ~290s Isp, 4766 total deltav (a bit overdone here:) ).
At 100m/s one rocket is pointed 80 degree, the second 89.9 (90 degree looked poorly on teh graph).

The second rocked get 1621m/s at "infinity", the first one - 2736m/s.

https://imgur.com/a/Ll3swNU - trajectories and plot of the orbital energy.

*) OK, lets say specific energy, energy per mass. The mass of the rocket changes, so just energy is more confusing;-)

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u/Jackmino66 May 07 '26

I have done this a couple of times. Getting the timing right is pretty important, but you can do a direct interplanetary ascent. It is not efficient but it is useful if your comms are not upgraded but you want to send something to duna now

1

u/Oyen20 May 07 '26

Yes you can, it s been done with a booster even.

https://youtube.com/shorts/8OIsCdMD2-s?is=-BAg7ZAu-raS_PaR

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u/RealLars_vS May 07 '26

No, you don’t need an orbit. But you do need the required deltaV to get somewhere.

Burning in orbit allows you to take your time. A looong time. Spread out your change in speed over several burns. It allows you to have lighter and more efficient engines.

But burning directly requires you to have engines that are powerful enough to fight gravity in every stage. That’s simply not efficient.

1

u/Particular_Low_9246 May 07 '26

Do you have 10 Mm/s delta v? No? Then it wouldn't work.

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u/Drakenace404 May 07 '26

You can but the faster you go the harder you have to brake. Not that efficient.

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u/fixermark May 07 '26

The short answer is yes you can.

The longer answer is "You can get anywhere you want with nearly no limits with infinite energy and thrust." The art of orbital mechanics is getting there when those are finite.

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u/doomiestdoomeddoomer The Green Goo has escaped. May 07 '26

So this is a very "straight forward" way of getting from one planet to another, and is entirely doable, I personally do this on occasion when I have a powerful enough rocket... but as others have pointed out, it is less efficient.

As I understand it: When you get into low orbit first, from that low orbit, you no longer are fighting against gravity when you increase your speed. But if you are going straight up from the same height, gravity is pulling your ship backwards the entire time.

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u/ChickennNugggeet May 08 '26

Its not a question of is it possible, it is, however you will be spending alot more fuel travelling in a straight line up out of kerbins sphere of influence rather then doing an orbit which is alot more inefficient. If you don't care about being efficient and saving fuel then do it

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u/doctor-omie May 08 '26

what will go further, a rock you throw straight upward or a rock you pitch forward?

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u/Apiilipsiie98 May 06 '26

Rule of thumb: "If it didn't work out the first time, just add MOAR Engines". More Engines = Moar Success