Arkalius

You're right. I was frustrated at the time I wrote it because I wasn't noticing the truncation of the number I was entering, so I spent an inordinate amount of time trying to figure out what I was doing wrong before I noticed what was happening. My suggestion is maybe you store the value as a double too, in kilometers? Anyway, I'll let you figure it out.

So I downloaded and installed the mod but found it cannot do the simple thing I needed it for. I have a small probe that's diving down toward the sun because I have a contract that requires me to test a part at a very narrow altitude range of the sun, closer than Moho's orbit. It's periapsis is below this point, and I don't want to miss the time it happens. So I wanted to set an alarm based on distance. But, the distance box won't let me enter an altitude in meters larger than about 2.1 billion (I'm guessing it's limited to a 32-bit integer), so it defaults to the periapsis altitude which is not helpful for me. This is kind of a dumb limitation to have in a game dealing with spaceflight and astronomical distances, don't you think? I mean, 2.1 billion meters is well inside Moho's orbit. At the very least I should be able to enter a distance in kilometers.

Well, the biggest tip I can give you is try to design rockets that somewhat resemble real-life rockets. I don't think I've ever seen a monstrosity like that in real life

The asteroid might fulfill the asteroid requirement, but what might be happening is that when you dock to it, the new "vessel" that is created is not considered "newly built for the contract" because part of it, the asteroid, existed before the contract was taken.

I'm apparently averse to time-accelerating more than a day or so without doing something in my career game. I feel like if a craft is going to take a few days to get somewhere, that's time I can do other missions in the interim. I'm slowly breaking this habit as I take on more missions of a long-term nature, but I doubt I'll ever fully overcome it.

So I have a question for which I can't really find a good answer. Mathematically speaking, we say a and b are inversely proportional if the relationship a * b = k for some constant k holds. Here, k is the constant of proportionality. Colloquially, when we say "a is inversely proportional to b", we expect that as a increases, b will decrease by some relative amount. Now, consider what happens when k is a negative number. In this situation, exactly one of a or b must be negative as well. We also note that as the absolute value of one increases, the absolute value of the other must decrease. However, the actual value of both will vary in the same direction. If a goes up (ie if you add a positive value to it), so too does b (it gains positive value). Conversely, if a goes down (subtract from it) the same is true for b. However, to say in this situation that a is inversely proportional to b, while technically correct, suggests the inaccurate scenario that when a goes up, b goes down and vice versa. Colloquially speaking we might want to say they are directly proportional but that mathematical relationship doesn't hold. So how do we describe this relationship in a way that is both mathematically accurate, and which conveys the reality of how it works? For context, I was looking for a way to explain the relationship between specific orbital energy and semi-major axis of the orbit. That relationship is expressed by: E = - mu / (2 * a) Where mu is the standard gravitational parameter of the central body. In this case, we can change the equation to E * a = - mu / 2, and since mu is always positive, we have a negative constant of proportionality. Thus either E or a is negative. E is negative when it is a closed orbit, and a is negative in an unbounded one (hyperbolic). In this situation, if you increase the value of a (ie, add a positive value to it), the value of E also increases (as if you added a positive value to it). Saying they are inversely proportional seems to convey the opposite. So, anyone have any insights in how to best communicate the relationship between E and a? Side Note: for those wondering how you can have negative energy, specific orbital energy is considered the sum of specific kinetic energy and specific gravitational potential energy, which is always negative. This value represents the negative of the kinetic energy per kilogram required to just escape from the body in question (ie be slowed to a velocity of 0 by gravity at infinite distance). Thus, specific orbital energy represents the amount of kinetic energy per kilogram the object has in excess of what is needed to escape the body. So, in a closed orbit, this excess is actually a deficit and thus the value is negative.

If you use LV-Ns with LF/LOX tanks and pull the LOX out, it's not going to be very useful. Your fuel mass fraction has gone down, plus the LV-Ns are quite heavy (3t each) which also reduces mass fraction. So, unless you have a lot of fuel tanks on there, your mass fraction is likely to go down enough to make the gain in Isp insufficient to actually add dV. You're going to want to use LF-only tanks with LV-N's, and best to do only with larger craft, due to the mass of the engines.

I like that design, it's pretty neat. Though, I wonder at it's utility at that size. I'm not sure how disorienting a sustained 4.2 rpm is on the average human's equilibrium. Plus, by my calculations, a person at the 1g section would experience about 1.12g's walking with the rotation, and about 0.88g's walking against it with average walking speed, more than a 25% variation. Not sure how difficult that would be to deal with. Still, it's a neat concept, and could potentially be scaled up to compensate for those issues.

If you "close" a future node (ie double-click away from it in map view), then hover the cursor over the closed node, it will tell you what its dV will be.

Prograde, in almost any context, means the direction of your relative velocity around the central body of your orbit. When used to describe an orbit itself, it means an orbit of less than 90 degrees equatorial inclination. The context is almost always enough to understand what is meant when the term is used.

I've had situations where it seemed like my wheels (or at least the nosewheel) was "stuck" inside the runway and thus I couldn't lift off until going off the end of it. Does the back of your plane bounce up in the air a bit as it spawns on the runway? That's usually a sign this is happening. I'm not sure exactly how to fix it, but I found moving the nose gear around a bit seemed to solve the problem for me when I was having it.

It's more like 3.5km/s in 1.0. You can get it down to maybe 3.3 or 3.2 if you're really efficient with your ascent. More awkward loads may require more as a result of extra drag or difficulties in flying them efficiently.

One thing to note is that you have several fuel tanks stacked up on that bottom stage. They will drain top to bottom which causes your CoM to shift to the bottom of the rocket, which makes it more unstable (the drag at the top applies more torque). This CoM shifting would happen even with a single large fuel tank, but it's more pronounced when you use several smaller ones. You can try and disable the fuel feed from the top ones and enable them as needed in order to resolve this (or even have the CoM shift upward, increasing stability). You can also just do manual crossfeeds.

For all these values, atmosphere is unchecked in KER. I like to allow for about 3500 m/s for getting into LKO. Then you need about 1200 m/s to get to the Mun and insert into a low orbit. Landing and getting back into Mun orbit is about 600m/s each (more if you haven't perfected your Mun landing skills or have to hunt for a safe spot). Then you need like 320 or so to escape the Mun into a direct re-entry back to Kerbin. That's a total of just over 6200 m/s. That's all assuming you can fly the mission with decent efficiency with your maneuvers. You'd probably want a little extra wiggle room there. You could probably do it with less if you're really good. With the 6295, is entirely conceivable that you were simply inefficient at some point and ended up short some delta-V in the end. Efficient ascent to LKO is probably where most people struggle the most. Mun landing can also come with a good bit of waste, especially if you're trying to land in a specific spot. As for the atmospheric selection on KER, I usually only use that to check my TWR with my launch stage. The delta-V values with that mode selected aren't terribly useful unless you plan to do a lot of flying around a specific altitude range in the atmosphere.

Well the 400km orbit is part of the issue. This will create a higher angle of reentry, and a higher speed at periapsis. What kind of periapsis are you using for reentry? You may want to go for 30-40 km. Also, be sure to try and hold a fairly decent angle of attack with your space plane. This will create more drag early on, bleeding more speed before you get into the deeper, thick parts of the atmosphere.

From a mathematical perspective, the diminishing returns comes as a result of the natural logarithm in the rocket equation. Logarithms increase exponentially more slowly as the term you provide to them increases. In fact, if you consider that as you add more fuel mass, you also must add some dry mass for the additional tankage, the ln term in the rocket equation ends up having an asymptote. We have dV = Vex * ln (m0/m1) Look at the ratio m0/m1. That's full mass over dry mass. Let's assume we have a fixed payload of 10 kg, and that for every 10 kg of fuel we add, we need 1 kg of tankage dry mass. So, dry mass m1 = 10 + x/10, and full mass becomes m0 = m1 + x = 10 + x/10 + x = 10 + 11x/10. If you then take the ratio of m0/m1 and simplify, you end up with something like (100 + 11x)/(100 + x). As x grows significantly large, you can begin to ignore the 100's and just have 11x/x, which simplifies to 11. That means as x grows large, the ratio approaches a value of 11 but never exceeds it. Thus, the natural log of that ratio can never exceed ln(11). The whole equation develops an asymptote - a value it approaches, and to which you can get arbitrarily close, but never reach or exceed. Basically as you continue to add more fuel, more and more of what you add is being expended to lift what you added instead of the payload. Of course, this ignores the fact that adding mass to the craft lowers your TWR which has its own problems. Yes, you can add more engines, but that is more dry mass on its own, thus reducing delta-V. . This is what is known as the tyranny of the rocket equation.

First, figure out how much delta-v your mission requires. This can be tricky, but once you have it, you're in a good place. The Tsiolkovsky rocket equation tells you the delta-v of your rocket, which will be Isp * 9.81 * ln( m0/m1 ) Where Isp is specific impulse in seconds, m0 is full mass of the craft, and m1 is the dry mass (ie all fuel used up). To figure out how much mass of fuel you'd need, you would have: m1 * ( e ^ (dV / (Isp * 9.81) ) - 1 ) However, this isn't super useful when designing a spacecraft. For one, you can't just simply add more fuel mass on its own. If you need more fuel, you need more tankage, which adds dry mass. Second, you're likely to build a rocket in stages, dropping dry mass over time as you go. This is why a mod like Kerbal Engineer Redux is useful. It will tell you how much delta-v you have in your rocket in each stage, helping you build what you need and plan it out. You can see how the numbers change as you move things around and add more or less fuel etc.

KER seems to be a bit better about displaying the info it displays. But, KER is purely informational. MechJeb offers fancy autopilots. If all you need is data and info, I would go with KER but some people are used to MechJeb and would stick to it even if not using the autopilots.

Service bay doors are notorious for their ability to screw up ships. You need to give them lots of room to move.

1.1 to 1.3 TWR seems really low. You're trading drag losses for gravity loses. Until you've gained some decent horizontal speed, you're bleeding 9.8 m/s of delta-v to gravity losses every second. Of course, if you push too hard you'll bleed off a lot of delta-v from atmospheric drag. There's a balance, and 1.1-1.3 TWR seems way too low. I try to stick around 1.7 or so. If you ignore transonic drag, a TWR of 2 is ideal, because you'll approach terminal velocity, where gravity drag and atmospheric drag are equivalent (at least initially). The main reason to not go quite that high, though, is because of the aforementioned transonic drag. You'll end up in the transonic region fairly low in the atmosphere at that level of TWR which generates too much extra drag.

If you just want to compare acceleration potential of various ships, sticking with a Kerbin-based TWR (or whatever body you desire really) gives you an apples-to-apples comparison. I only say this because KER gives you TWR and not thrust-to-mass ratio.

It's funny, after posting this, I went to bed, and then figured it out while lying there thinking about it. I was originally envisioning it as what fraction of the chemical potential energy in the propellant only. But now I realize it is a combination of that and the kinetic energy. Clearly when moving faster than the exhaust velocity, some of the kinetic energy will be left behind in the exhaust, with no way to recover it, thus the reduced efficiency. I also realized why the value is 0.8 at both 1/2 and 2x exhaust velocity. At 1/2 speed, the propellant has 4x more chemical potential energy than kinetic energy, and you can obtain all of the chemical energy on the craft, but none of the kinetic. At 2x speed, you can get 3/4 of the kinetic (which is 4x the chemical) energy and all of the chemical, so in both cases you get 80% of the total energy involved. I didn't read through your equations in detail, but I'm assuming they more precisely explained the concepts I just summarized. I almost feel silly for posting a question the answer to which dawned on me merely minutes later, but at least others can learn from it

So I've been reading a lot of things about the various aspects of the physics of spaceflight. I've recently stumbled on something regarding the propulsive efficiency in rocket engines in regards to energy utilization and the Oberth effect. I understand how the effect works. The thing that confuses me, though, is how propulsive efficiency is capped. The equation I've seen is: With c as exhaust speed, and v is speed of the spacecraft. With this equation, efficiency caps out (at 1.0) when v = c, and then starts decreasing, but this doesn't make sense to me. In a static firing of the engine, where the craft doesn't move, all of the (non-wasted) energy from combustion is imparted on the exhaust shooting out the back. When the craft is moving at half the exhaust velocity, all of the energy is given to the craft. As an example, say exhaust velocity is 100m/s. The craft is moving 50 m/s. The reaction mass changes from moving 50m/s with the craft to 50m/s away in the opposite direction, thus its kinetic energy hasn't changed, so all of the produced energy is with the craft. Why this turns out to be 0.8 all the time, I'm not sure... I'd think the value would depend on the specific chemical energy of the propellant being used, but anyway that's not as important. When the craft is moving at the same speed as the exhaust velocity, not only is it getting all the energy of combustion, but it's also stealing all of the kinetic energy of the exhaust, which will have a velocity of 0. With the 100m/s example, that's an extra gain of 5kJ/kg. However, it seems to me the efficiency should continue to climb as the craft gets faster. Say it is moving 200 m/s. The exhaust will be moving 100m/s in the same direction after leaving the craft. It's specific kinetic energy would go from 20 kJ/kg down to 5 kJ/kg, a difference of 15 kJ/kg, 3 times more than in the above situation. So why doesn't this translate into even greater propulsive efficiency for the spacecraft? According to that equation, the 2:1 velocity to exhaust velocity ratio has the same 0.8 efficiency as the 1:2 ratio (with 1:1 being 1.0 efficiency). Anyone able to explain this?

It's probably got lower Isp than other engines. Having it off lets you consume more fuel with more efficient engines first, thus improving total delta-v. However if this gives you a poor TWR, then it may end up costing you efficiency in the end.

A Kolniya orbit, as it's called (It's official, I've gotten a contract to put a satellite into one) can be sun-synchronous but not continuously. Since KSP uses 2-body physics, and spherical gravity, there is no way to use gravity to precess the longitude of the ascending node so that the orbit stays continuously sun-synchronous.