natsirt721

Really? That's a pretty poor imitation of reality. In my aircraft performance & design class we typically see Lmax around 15 degrees AoA.

The 0-10 Puff. Poor efficiency and radial attach only. I'd rather use an Ant/Spider or a Spark in nearly every instance that I can think of. The one exception might be an RCS tug for low orbital operations, but I have yet to require one. At least they made it bigger!

I'm not sure about RSS/RO engines in general, but I do know that fuel pumps are not able to remove 100% of the propellant from a tank. I believe LEM's unusable propellant was around 250lbs - granted the tank had a capacity of 18,000lbs so that is a pretty small fraction. It looks like the Titan II used the same fuel blend (hydrazine and nitrogen tetroxide) so I would expect a similar (or at least nonzero) waste proportion. It's interesting that it tells you that you don't have ignitions, because hypergolic fuels like that don't require a designated igniter. But I'm not a propulsion scientist, what do I know? Again, not sure if this (tank waste) is modeled but from what I've read about RO I wouldn't be surprised if it were.

There's a mod for that! I don't know which (FAR?) but I have seen surface landing predictions in some screenshots here.

The mohole starts several kilometers above datum and I don't think it goes negative at all, but I could be wrong. Just out of curiosity, what does the IVA radar altimeter show when you're below datum?

No, not like tidal heating. I remember seeing in a post sometime that the mass ratio between Jool and Laythe is insufficient to supply the necessary energy flux to maintain the surface conditions there.

From what I can tell, the consensus is that Laythe should, by all accounts, not be feasible. Yet it exists, and continues to exist. It somehow has an oxygenated atmosphere that hasn't escaped or reacted away, has enough energy flux at the surface to stay relatively warm, and has exactly the right conditions for liquid water on its surface. This leads to the only explanation I can think of: Laythe is artificial. The nearly global ocean is simply a cover, a layer of insulation for a megastructure that lies beneath the surface. It is heated from within by some arcane power source, and houses...well, something awesome i suppose. I mean, you don't hide anything that isn't awesome in a planetoid. Perhaps Laythe is the original home of the Kerbals, who have been transplanted to Kerbin for reasons unknown? The possibilites are endless!

This sounds like a modding issue rather than a stock gameplay issue. As no planets in stock have unexposed land below datum I'd say this should be a fix packaged with kopernicus or whatnot. HOWEVER I do agree that it should be dealt with.

If I install this over a stock + KAC save, will stuff be messed up?

Thanks! I'm not sure how to do that, but I'll take that to the kopernicus thread.

Plain and simple, this mod would somewhat smooth KSP's current polygonized landscapes into something more realistic, with fewer sharp edges and instantaneous slope changes. Are there any mods out there that do this? I feel like it would be fairly simple to interpolate smooth curves to blend the polygons together at the joints - the landscape can stay fairly planar as long as the joints have smooth transitions. It could either to this on-the-fly (say, in a 100m bubble around each loaded craft) or be run once for each body beforehand and then loaded into the bubble as necessary, acting as another level of detail for the terrain. So, daring modders: is something like this feasible?

Brought a dual sided klaw chassis (no probe core) to allow my fuel truck to attach to the next series of landers. Dropped it on the surface, but forgot to extend the klaws.

Something like XCOM:EU memorial - showing xp, accomplishments, place and vehicle of death.

Ok sure, a cubit is good for knowing approximately how large something is. I can say that my legs are about a meter long - easy to relate to and see, but no better than the cubit for precision. I'm saying you can't base a system of measurement on approximations. Its good enough for technology up to the 18c but no further. And even if society (d)evolves, people are going to keep using the units they know. Why reinvent a whole system of units that you are then going to have to proliferate in order to use?

According to Merriam-Webster, the meter is "...now defined as 1/299,792,458 of the distance light travels in a vacuum in one second." I was incorrect about the definition of the second though, according to NIST it is "the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom". You don't need units to define relative scale. If I say from A to B is 9 and from A to C is 12, you know that latter is further. You don't know whether it's 3 feet further or 3 parsecs further, but it is further. Magnitude, on the other hand, does require a unit; however every unit has to be learned, and with that comes a knowledge of the magnitude it implies. Just because it is easier or more relatable to use a cubit doesn't make it a good unit. The issue with cubits and stones is that they aren't defined rigorously. A cubit can be anywhere between 16 and 20 inches, depending on the person. If you say "this table needs to be 3 cubits wide and 9 long" to 4 different carpenters, you get 4 different tables. If you say "this table needs to be 1 meter wide and 3 long" you get 4 identical tables (within tolerance). The necessary tolerance on a cubit is just too large to be of any use, it doesn't matter if it's relatable or not. Nobody is going to be able to construct anything more complex than simple structures using cubits as base units, unless you use John's cubit or Carl's cubit for every single measurement. But then, if you want to communicate the dimensions to a friend, you still have to send John or Carl to define the cubit to your friend. You can't use a unit that isn't precise, it just doesn't work.

I fail to see how the removal of 'precision measurement systems' versus 'traditional human systems' provides any value. The reason we don't use cubits anymore is because their definition was arbitrary and changed with the measurer. Metric units are grounded in real-world constants. A kilo is a liter (cubic decimeter) of water at 4C, and meter and second are based on the speed of light in a vacuum. Those values don't change, won't change, ever. I know that the length my thumb joint is about an inch, but I also know that the width of my pinky is about a centimeter. Neither are correct, but where is the benefit from using inches? Without precision measurement tools in any units, you're still unable to apply them.

Take a look at this page. You can ignore the perturbations equations as KSP assumes a spherical body and no outside forces.

Yeah, having the L5 engineer is a huge boon. I made one similar to Foxster's with 2 Junos that had 15 minutes of powered flight on 300 ore.

Does anyone have any good data on the atmospheric density curves? I found a thread from 2012 that suggested that Kerbin's atmosphere followed a curve with the same scale factor as pressure, that is: p=p0 * exp(-h/5000) Where p0 is density at 0m. I assume it is 1.22 kg/m3 as IRL, but I could be wrong. Is this equation still accurate? While the wiki has scale curves for the atmospheric bodies, it doesn't give surface density. Does anyone have surface densities for those bodies?

I had a save in 0.24-25 with a dozen or so mods, including KAC, KER, KIS, RPM, Karbonite+ and a few others. The game was incredibly buggy, and after having my Minmus surface base implode on load-in a few times I finally scrapped the idea of mods, or at least running a bunch together. In my current game the only mod I have is KAC, because it is the most useful thing ever. I have considered reinstalling KIS/KAS for some base building, and maybe OPM once I flush out the base system. I also have a custom LV-N motor with slightly better thrust (75kN) and Isp (900s) which also functions as an electric generator. The tradeoff is a mass of 7t. Stats are loosely based on NASA's bimodal NTR for a Mars mission. EDIT: Didn't really answer the question: I don't use mods because I'm lazy and waiting for them all to update after each rev change is too much of a pain for me to bother with.

Those are some good names! Some more: Slung: Flyby Mun and reach Duna orbit with the same craft. Nice Shooting: Take off from Kerbin, orbit the Mun, and land on the Runway at KSC A Less Civilized Age: Find the alternate space center The Grand Tour: Orbit every body with a single craft Crusin': Cover 100km with a single rover Down the Rabbit Hole: Explore the Mohole Bonneville: Reach ground speed equal to liftoff velocity on the Minmus flats Calculated: Complete Nice Shooting, but return with <1% fuel remaining Veteran: Have a kerbal reach level 5 Requiescat in Pace: Kill a kerbal Ultralight: Reach orbit with a very low mass vehicle G-LOC: Have a kerbal pass out from over-gee Buzz the Tower: Pass within 10 meters of the Mun surface going faster than 400m/s One could create a plethora of achievements simply for reaching certain geographical features. Sadly I think Squad has said that they will not implement Steam achievements. Might be an outdated announcement but idk.

Linear Momentum or 'momentum' to layfolk is the product of mass (kg) and velocity (m/s). Units are kg*m/s. By nature, linear momentum of a system is conserved. Angular Momentum is the product of a rotating body's moment of inertia (kg*m2) and its angular velocity (rad/s). Units are kg*m2/s. Angular momentum of a system is also conserved. It is important to note that angular momentum is measured with respect to an axis, and for simplicity we take that axis to be that of the flywheel. From a strictly dimensional analysis standpoint, we can see that these two terms do not describe comparable quantities, but that glaring error aside, let us consider the simplest system possible, that is a 100,000 kg flywheel-tether apparatus and and 10,000 kg spaceship braking on the tether. This scenario also assumes that at no point propellant is expended to change the velocity of either the flywheel or the spacecraft. The flywheel is moving east at 8,000 m/s and the ship east at 7,000 m/s. The linear momentum of the ship is 10,000 kg * 7,000 m/s = 70,000,000 kg*m/s. Likewise, the flywheel has linear momentum of 100,000 kg * 8,000 m/s = 800,000,000 kg*m/s. The total momentum of the system is 870,000,000 kg*m/s. The angular momentum of the flywheel requires the moment of inertia, which in this case is m*r2/2 for a cylinder rotating about its axis. For a 5 m cylinder at 100 rad/s the angular momentum is 125,000,000 kg*m2/s. The angular momentum of the ship is a little different, but it boils down to the linear momentum of the ship times the tangent distance to the axis of rotation. This distance is going to be the radius of the tether spool, which for simplicity we will assume to be the same as the flywheel. For a ship moving (with respect to the flywheel, our reference axis) at 1,000 m/s west, the angular momentum is -1,000 m/s * 10,000 kg * 5 m = -50,000,000 kg*m2/s. The total angular momentum of the system is 125,000,000 - 50,000,000 = 75,000,000 kg*m2/s. Both of these quantities must remain constant, because the laws of physics say so. No matter what you do with changes in geometry, friction, or any other internal forces, both linear and angular momenta will stay the same for the system. Now, lets examine another state involving the same system. We want to accelerate the ship by 1,000 m/s to match the velocity of the flywheel , so we trade angular momentum of the flywheel for angular momentum of the ship. The final angular momentum of the ship = m * v * r = 0, because then the ship and the flywheel will have the same velocity and therefore velocity relative to the axis is 0. So the ship gains 50,000,000 kg*m2/s in angular momentum (which is transferred to the flywheels via the tether), and the flywheel loses 50,000,000 kg*m2/s, bringing the new totals to 0 kg*m2/s for the ship and 75,000,000 kg*m2/s for the flywheel. Total is still 75,000,000 kg*m2/s - check; however this doesn't work out. If the ship and flywheel are now moving at 8,000 m/s, then the linear momentum of the flywheel is still 800,000,000 kg*m/s, but the linear momentum of the ship is now 10,000 kg * 8,000 m/s = 80,000,000 kg*m/s. The total linear momentum of the system is 880,000,000 kg*m*s, or 10,000,000 higher than it was. You've created linear momentum seeming out of nowhere, even though for a closed system linear momentum is constant. This is impossible. What actually occurs is, during the angular momentum exchange the flywheel will decelerate by 100 m/s in order to keep linear momentum constant. If we try this again but include the linear momentum exchange, we see that the final relative velocity of the flywheel and ship is now +100 m/s, because the ship is going 8,000 m/s and the flywheel 7,900 m/s. The final angular momentum of the ship is 5,000,000 kg*m2/s, a net gain of 55,000,000 kg*m2/s. As before, the flywheel takes this up, bringing the new totals to 5,000,000 kg*m2/s for the ship and 70,000,000 for the flywheel, a total of 75,000,000 kg*m2/s - check. The linear momentum of the ship is 80,000,000 kg*m/s as calculated before, and the linear momentum of the flywheel is 100,000 kg * 7,900 m/s = 790,000,000 kg*m/s. Total linear momentum is 870,000,000 kg*m/s - double-check. Both linear and angular momenta are conserved. As you can see, even using rotational energy from a flywheel to increase the kinetic energy (and therefore linear momentum) of the spacecraft still results in a deceleration of the flywheel itself, because the total linear momentum of the system is independent of of the angular momentum of the flywheel, but dependent on the linear momentum of the spacecraft. If the spacecraft accelerates, the station has to decelerate, and there is no way around that.

Man, I thought the death of the Dean Drive in the 50's really nailed this point down but boy was I wrong.

Ding ding ding, this is correct. For those less well versed on orbital mechanics, here's what those are: True anomaly (theta) is the actual angular position along an orbit, measured from the primary focus. This is the thing that we want. Mean anomaly is a 'rough guess' term that amounts to 2*pi*time / period - not really useful for anything other than circular orbits and for a baseline comparison between orbits. Eccentric anomaly (E) is the angular position, measured from the center of a circle with radius equal to your semimajor axis (a). While it is possible to geometrically solve for theta as a function of time directly the equation is, as we say in the biz "pretty gross"; however, solving for E and then converting to theta is much nicer. Equations are as follows: Kepler's Equation: (2*pi*t) / T = E - E*sin(e) where T is the orbital period and e the eccentricity. While Kepler's equation isn't super useful on its own, it does allow us to do some geometry and get another equation: cos(E) = (e + cos(theta)) / (1 + e*cos(theta)) or tan(E/2) = sqrt((1 - e) / (1 + e)) * tan(theta/2) If the intersection altitude is known, using geometry the true anomaly can be determined. From that, you can use the above equation to find eccentric anomaly, and from there, multiply Kepler's equation by T / 2*pi to solve for time t. This does require the eccentricity of the transfer orbit to be known, so this is not a one-shot deal.

Kepler says that orbiting bodies move on ellipses, but that's really just an approximation. In an n-body system, every body pulls on every other body, not just the sun and its satellites. This creates disturbances, irregularities in the otherwise elliptical orbits of the satellites. Sometimes these irregularities have positive feedback - they tend towards configurations that are 'unstable' (eg; bodies ejected from a system or dramatic changes of orbit). Sometimes they have negative feedback - they tend towards configurations that are 'stable' (eg; orbital resonance between bodies). Whether a system is 'stable' or not depends on, well, everything. Gravity, radiation pressure, tidal forces, non-spherical bodies, the list goes on. Sometimes resonant systems are also unstable, and sometimes otherwise unstable configurations are actually periodically stable. There's no good way to tell just by looking at a system, and therein lies the problem. I think p1t1o had it spot on when they said: