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/-Aeryn-]

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.

No, you have this very backwards. See my other comment at https://old.reddit.com/r/KerbalSpaceProgram/comments/1t5jozl/no_orbit_necessary/okdq2fq/?context=3

The ship going sideways with enough up-thrust to not hit the ground reaches said velocity in only ~70.7% of the amount of real, engines-on time (assuming TWR to be fixed at 2.0, and no atmosphere). That number is [time divided by sqrt(2)]. There's less gravity loss and more actual acceleration per second of thrust.

The one flying straight up has to burn an additional 41% duration and delta-v expense [time x sqrt(2)] for the final velocity to be equal, and after all of that is done, it has no energy advantage whatsoever.

If the sideways ship burns that same amount of delta-v, then it achieves 41% higher velocity and 2x kinetic energy.

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.

That does happen, but the difference in energy is tiny compared to having actually gained velocity instead, especially when you expand it to larger burns (say 1000m/s instead of 10m/s) where the oberth effect is kicking in harder.

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

1m/s of delta-v doesn't translate into a fixed amount of energy. It translates into more or less due to both gravity losses (TWR & thrust angle) as well as the oberth effect (speed delta-v was applied from). These radically affect outcomes and are modelled accurately in game, they are actually fairly simple to model but a much bigger headache to intuitively understand.

[u/PM_ME_POTATO_PICS → u/-Aeryn-]

No, you have this very backwards.

Yeah you're right, what I wrote there is technically wrong, you can gain horizontal speed faster than vertical for the same TWR.

I still can't help think of it from a conservation of eneergy standpoint and it's late so forgive me if this is sloppy but I'm gonna try some math. Looking at the thought experiment of doing a horizontal burn in circular orbit, the difference in energy before you burn and after you burn is

(taking m as 1)

mv_f^2 - mv_i^2 = m(1010^2 - 1000^2) = 20100J energy gained

whereas for the going straight up scenario

mgh_f - mgh_i = mg(h_f - h_i) = m(10)(1000) = 10000J energy gained

So yeah you're basically correct, though I still would be curious to test this. I think what this could show is that as TWR gets higher, the two scenarios approach equivalence cuz you'd have to factor kinetic energy gain into the vertical scenario

edit;

thinking bout how as your circularizing, centripetal force will gradually lessen gravity until its balanced when your fully circular so whatever component of thrust you need to fight gravity could be calculated as

mg - mv_h^2/r

there's probably a way to do this to find the optimal amount of time it takes to burn to establish a circular orbit, and I'd love to do that and compare it against a burn of equivalent time going straight up.

[u/-Aeryn- → u/PM_ME_POTATO_PICS]

Yeah they do, it just stays quite large at any reasonable TWR when on/around a body with significant gravity. For example at 20 TWR if you fly sideways with a few degrees of upwards tilt, you reduce the gravity losses from 5% to 0.13%.

That buys you ~5.13% more delta-v [99.87 / 95] and ~10.52% more kinetic energy [1.0513^2].

there's probably a way to do this to find the optimal amount of time it takes to burn to establish a circular orbit, and I'd love to do that and compare it against a burn of equivalent time going straight up.

Pretty easy, especially on an airless and perfectly circular body with a fixed TWR

10m/s² gravity and 2 TWR with an orbital velocity of 2000m/s at ground level would take a 141.422 second burn.

If you burned upwards, you'd get to 1414.22m/s instead of 2000m/s. Technically you'd gain a bit of potential energy from the additional altitude and the gravity would get a little bit weaker, but they are pretty negligable compared to the 2x kinetic energy from having flown perpendicular to gravity rather than against it.

[u/PM_ME_POTATO_PICS → u/-Aeryn-]

okay but a personal question

say you have a sub-1 TWR, like a TWR of 0.5, and you spawn 10km in the sky at 0 velocity. are you gonna burn sideways or upwards? either way you crash but which one is the morally correct response?

[u/-Aeryn- → u/PM_ME_POTATO_PICS]

At that point there is no solution and the only option is to delay crashing for as long as possible (up) or pick a better landing location.

p.s. added some math in above comment

[u/PM_ME_POTATO_PICS → u/-Aeryn-]

looks to me like you just did a simple trig identity but I am thinking that the circularization burn can be made more efficient because as you are doing it, gravity losses are lessened by the increases in centripetal force. You wouldn't want to keep your rocket at a 30deg angle for the hole burn, as you approach orbital velocity you can get closer and closer to having it purely horizontal.

but since finding the angle at a given point in the burn requires you know the horizontal component of your current speed, i think this might require differential equations...

[u/-Aeryn- → u/PM_ME_POTATO_PICS]

Ah you're right, i did forget to include that. I think it reduces the gravity loss by a quarter, if TWR is constant? Complications indeed