hydropos

I just re-installed RSS, but with the default config. Firstly, it appears no earth/moon models come with RSS anymore? Secondly, Kerbin's SoI is still only 78,000 km, IE, smaller than the orbit of the Mun. No one else is getting this issue?

According to this formula and the masses/distances in the config file, kerbin's SoI should be about 840 thousand km, but in practice it is only 78 thousand km. Not sure where the discrepancy is happening.

I've been playing with JSimmons' 10x kerbin config file, but for some reason kerbin's SoI is smaller than the orbit of the Mun. It makes maneuver nodes useless, which makes navigation too complicated for my liking. Could this be a compatibility issue of some kind? Or is this a bug with the config file? Also, if there's any simple config edit to fix this, I'd be interested.

Can anyone recommend a simple mod to extend the craft load distance? IIRC, there's a weapons mod (Lazor Systems?) that does this, but I don't know how much impact that will have on the rest of the game.

I just set up a spreadsheet to calculate ∆V for different engines, and the poodle loses almost across the board. It only beats the LV-T30 when you get below a TWR of 0.5, and it's marginal even then. The same can be said for the LV-909, and both are outclassed by the aerospike for every TWR. The spreadsheet also showed that the NASA engines were clear choices for anything with a TWR > 1. I've been playing the Real Solar System mod using stock parts, so this relates to me. I don't use mod parts for two reasons: First, modded parts put many people off (I like posting my craft). Secondly, balance comes down to which mods I use, which feels like cheating.

Ironically, based on how useful they are, I thought that price was not unfair. I was a bit mad about the cost of canards. $3000 a pair? This is madness!

In advance of the update, I've started taking cost into consideration when building my rockets. I've noticed that the costs of many parts seem inconsistent for their relative size or utility. If you look at fuel tanks, some of the smallest liquid tanks cost the same amount as the largest SRB's. One big orange tank costs more than double one the white half-tanks. I can't remember all of the examples I found, but there were a lot. I found myself making counter-intuitive and bizarre design choices to save on cost. I guess I'm wondering, have the Dev's have addressed this, and are there plans to adjust these costs prior to 0.24?

I was so aligned that just rotating the capsule with the SAS would move the ascending node around. It turns out part of the problem was setting the Moon as target. If I just used the targeting system to align planes, then unset target, things got better. Although I still had to repeatedly click the node position to get the encounter to show up, even for very low (7km) periapses.

Does anyone else have problems finding an encounter using the maneuver nodes? I've been trying some moon missions, and the game will frequently deny any encounter until I actually fly into the moon's SoI.

I'm not. I'll try that. Thanks!

Anyone else having issues with the new parts? Sometimes Engineer detects them and calculates ∆V, and others it doesn't seem to know they exist (save for mass). I had a ship that worked fine with Engineer the first time I launched, but when I "reverted to the VAB", none of the new parts were adding to ∆V. Other times, it will detect the new liquid engines, but not the boosters. I just re-dl'd and re-installed, and that didn't help.

Well, the flaw in my logic was facepalmingly obvious. If you need to gain both radial and angular velocity, you would not burn up then sideways (even if you had infinite TWR). If you burn at a 45° angle, you have to burn only 1.4 times the fuel as you would need in either direction, but you get both components of velocity. Not sure how to take this into account mathematically, as it essentially means ∆V efficiency isn't just dependent on velocity, but also rocket orientation.

Well, I figured out why my logic failed. You need to use circular coordinates (r, θ) instead of x and y. The radius of Kerbin is sufficiently small that the "x-velocity" you pick up at 20km is pointing in a distinctly different direction than the "x-velocity" you get at 40km. I've had a few too many beers tonight to be working the actual math out, but I am confident in this. EDIT: I just started doing the math. It's hideous. Parametric equations everywhere... EDIT2: For clarification, what I meant was that to accurately represent the effect of gravity, you need to consider the angle between the velocity vector and the radial position vector. You'll also need the angular velocity to determine the centrifugal offset to gravity. Both of these will require you set up relations for r and θ as a function of x and y.

Follow-up question: are there any regions/trajectories that would result in a craft switching back and forth between kerbin and the Mun to act like a Lagrange point?

I will start by admitting that I skimmed most of this thread after about page 4, so please let me know if I missed something. This and other threads on the topic seem to focus on simulating different paths and rocket constructions. What I sought out to do was find the minimum ∆V ascent trajectory to Low Kerbin Orbit (LKO) purely mathematically. This ignores anything about mass, thrust, or ISP. I caught some inspiration yesterday and appear to have something. The two most notable findings were this: 1) You should only ascend at terminal velocity until you hit 440 m/s, then stay at that speed until your gravity turn 2) Any gravity turn under 30km will cost you ∆V Note, this also produces the result that the "ideal" gravity turn should be instantly 90°, and assumes your TWR is absurdly large (~10), which may explain why most launches do not work this way. Here is my reasoning. Let me know if I'm missing something: 1) It's already been established that the most efficient ascent at low altitudes is totally vertical at terminal velocity. However, since all that vertical velocity will eventually be gone (assuming a roughly circular orbit) continuing to accelerate to the exponentially-increasing terminal velocity must become wasteful at some point. Here is my math: lost ∂V/∂t = gravity + drag + net acceleration where gravity and drag are their corresponding accelerations, rather than raw forces. At terminal velocity (VT), gravity and drag will be equal. If you're trying to maintain terminal velocity, then net acceleration will be the derivative of terminal velocity (∂VT/∂t). I also rearranged things in terms of lost ∂V per altitude gained (∂y) so that the solutions at V = 0 didn't muck up my calculations. This means dividing the right side by the velocity (assumed to be VT): lost ∂V/∂y = (2*g + ∂VT/∂t)/VT where VT = 6.01*10â·*exp(y/10000)/(600000+y) {assumes ship drag = 0.2} This function has a minimum value at a height that depends on drag coefficient, which is about 15.1km for liquid fuel rockets (drag 0.2) and 17.1km for solid fuel rockets (drag 0.3). However, they both happen at a velocity around 440 m/s. After that minimum, any additional acceleration is going to be wasted ∆V, since all that vertical velocity must eventually go to zero. Thus you should stay at 440 m/s until the gravity turn. 2) The cheapest ∆V way to gain altitude in the absence of atmosphere is to burn horizontally. With atmosphere, the centrifugal acceleration benefit of turning sideways must outweigh the added drag. Their accelerations are as follows: centripetal acceleration = -V²/(Kerbin's Radius + Current Altitude) drag ≈ V²*0.0011/exp(altitude/5000) I use a negative sign in front of the centripetal force, since that quantity is ultimately being subtracted from gravitational acceleration. Also note that V is only the horizontal component of velocity. You can factor the V² out to get the net effect: acceleration = V²[0.0011/exp(y/5000) - 1/(6*10âµ+y)] Only when this term becomes negative is it beneficial to have any non-vertical velocity (32km). Note this drag equation is based on a mass-average drag value of 0.2, which is the case for liquid fuel rockets. For higher drag vessels (say solid rockets with drag 0.3) this altitude is just over 34km. Final Notes I did some numerical integration for both the low and high drag cases to toss out some values for the minimum ∆V. From launch to the critical terminal velocity of 440m/s should require 1935 or 2361 m/s for liquid and solid rockets respectively. From launch up to the "gravity turn" should require 2392 or 2816 m/s for liquid and solid rockets respectively. After this point you no longer have to spend ∆V to fight gravity and drag is next to nothing, so I would add to these the ∆V of LKO, 2293 m/s. If you've launched eastward, then you can subtract around 180 m/s from Kerbin's rotation. This gives: minimum LKO ∆V: 4505 m/s for liquid fuel rockets and 4929 m/s for solid fuel rockets. You can get pretty close you this in practice because most of these functions are aren't steep near the minima. In other words, you don't need a TWR of 10 to do well. If you've ever wondered if it's possible to build an all-solid fuel SSTO, you now know the answer is no, but only by about 150 m/s ∆V.

Can you clarify this? I just combed through that. There was a lot of good stuff, but nothing quite like I was hoping for. This, OTOH, is almost exactly what I was looking for. Imma be honest, I TLDR'd at page 5, but it looks like they got stuck where I did. I think I'm going to continue my efforts in that thread. This is very true. I am prepared to work under the assumption that I have infinite TWR at my disposal. I suspect you really don't need more than about 3g's of thrust for a "perfect" kerbin launch. Sort of. It's much more complicated with realistic aerodynamics.

Can you direct me to any plots or tables of the numerical solutions? I've tried searching, but the only thing that comes up are a lot of "well I usually do this" posts. The closest thing to a rigorous approach were some folks who used mechjeb and different turn parameters on a given ship. However, my math has already shown that any circumferential velocity prior to 34km up is actually bad for your ∆V, so those are unlikely to be good approximations.

For the last 3 days I've been trying to develop an equation for the lowest ∆V launch trajectory to LKO. Specifically, I'm looking for the optimal radial and circumferential velocity as a function of time or altitude. I've filled pages with differential equations and assorted calculus, but I can't seem to put all the pieces together. Has anyone calculated this already, and if so, where can I find that?

I've been trying to calculate a few things myself. It seems like you're always better off using a rocket design than a spaceplane. Maybe on takeoff from a really high gravity/atmosphere planet like Eve, things would be different, but it's not looking very likely.

Is this force acting in the direction orthogonal to the wing surface, or is this only the vertical component? If the former, then high AoA's would be bad, since a substantial part of the wing force would be fighting the engines.