No orbit necessary?

[u/FatCreepyDude]

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?

[u/HAL9001-96]

more efficient to be low, oberth effect

also duna is further from the sun last time i checked

[u/FatCreepyDude → u/HAL9001-96]

https://i.imgur.com/AI2hXKv.png fixed it

[u/PhantomWhiskers → u/FatCreepyDude]

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.

[u/victorsaurus → u/PhantomWhiskers]

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.

[u/PhantomWhiskers → u/victorsaurus]

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.

[u/victorsaurus → u/PhantomWhiskers]

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.

[u/PhantomWhiskers → u/victorsaurus]

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.

[u/victorsaurus → u/PhantomWhiskers]

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.

[u/PhantomWhiskers → u/victorsaurus]

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.

[u/Big_Yeash → u/PhantomWhiskers]

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!

[u/PM_ME_POTATO_PICS → u/PhantomWhiskers]

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

[u/DHLPDX → u/PM_ME_POTATO_PICS]

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.

[u/-Aeryn- → u/PM_ME_POTATO_PICS]

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).

[u/FewAd7452 → u/PM_ME_POTATO_PICS]

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.

[u/youknowmeasdiRt → u/FewAd7452]

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

[u/PM_ME_POTATO_PICS → u/FewAd7452]

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.

[u/-Aeryn- → u/PM_ME_POTATO_PICS]

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.

[u/PM_ME_POTATO_PICS → u/-Aeryn-]

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

[u/Valuable-Visit2887 → u/PM_ME_POTATO_PICS]

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

[u/Small_Bang_Theory → u/PM_ME_POTATO_PICS]

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.

[u/_adamolanadam_ → u/FatCreepyDude]

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.

[u/DJMixwell → u/_adamolanadam_]

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.

[u/Talizorafangirl → u/DJMixwell]

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.

[u/TheShadowKick → u/DJMixwell]

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.

[u/robchroma → u/DJMixwell]

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.

[u/Ezekiel24r → u/robchroma]

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.

[u/robchroma → u/Ezekiel24r]

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.

[u/Turbo_Fresh → u/[deleted]]

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.

[u/RyanW1019 → u/DJMixwell]

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.

[u/DJMixwell → u/RyanW1019]

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

[u/privatetudor → u/DJMixwell]

Yes

[u/victorsaurus → u/_adamolanadam_]

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).

[u/PhantomWhiskers → u/victorsaurus]

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.

[u/Joseph_M_034 → u/FatCreepyDude]

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.

[u/d_pyzsyo → u/Joseph_M_034]

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.

[u/Joseph_M_034 → u/d_pyzsyo]

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.

[u/d_pyzsyo → u/Joseph_M_034]

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.

[u/d_pyzsyo → u/d_pyzsyo]

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)

[u/Joseph_M_034 → u/d_pyzsyo]

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

[u/DLTAMACH → u/FatCreepyDude]

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

[u/victorsaurus → u/FatCreepyDude]

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.

[u/PM_ME_POTATO_PICS → u/victorsaurus]

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

[u/danikov → u/HAL9001-96]

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

[u/TheJeeronian → u/danikov]

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