Yasmy

Here's an old table of the cost of going from any of the moon to any interplanetary destination, either directly, or by dropping down to the parent planet and burning from there. As others have said, from low Mun orbit, it is best to burn directly to Duna or Eve. For other interplanetary destinations, burn down to LKO then complete your transfer burn. For Mun -> Duna, you save about 80 m/s. For Mun -> Eve, you can save up to about 110 m/s. These are pretty small. For further destinations though, the penalty for going directly from Mun is very bad. (300-500 m/s.)

I don't know, Slashy. I checked your equations, and they look approximately correct to me. They are approximate though, correct in the limit that R -> infinity, where R is the radius of the SOI. That could account for some errors. In particular, you say that Vesc = sqrt(2) Vorb. (Cute. I never noticed this before.) That's not exactly true in the SOI approximation. Using the vis-viva equation at the SOI boundary of radius R gives the semi-major axis for escape at zero velocity: 0 = v2 = mu (2/R - 1/a) -> mu (2/R) = 1/a Now at some radius r < R, on the same orbit, vesc2 = mu (2/r - 1/a) = mu (2/r - 2/R) = 2 (Vorb2(r)-Vorb2(R)). But of course this is often a tiny correction, particularly for planets. For moons, it might be a more significant error. In fact, for some light moons, you can often exit on a fairly low eccentricity elliptical orbit instead of hyperbolic orbit because the SOI is too small. Having said all that, I still can't say why we get such different results. Perhaps my orbital data is bad or my code has a bug...

EVA Transfer mod. Much simpler than KIS/KAS. Only needs one part, and can link to craft without any modded parts, i.e., all your already in place vessels. Very handy.

I'm a bit late to the party, but here's a chart I made a while ago just for these kinds of questions. My answers differ slightly from Slashy's. 421 m/s direct vs. 464 via dropping down to low Duna. Interplanetary delta-v cost for direct from moon vs. dropping down to parent P.S.: This table is for starting or dropdown passes 30 km above airless moons, 50 km above airless planets, or a multiple of 10 km above bodies with an atmosphere

It looks to me like your coordinates are latitude 3.0322, longitude -101.1182. Pretty near the equator. If that's not correct, then I don't know what you are asking. I.e., 3 degrees north, -101 degrees east. (-101 degrees east = 101 degrees west.) -101 degrees east is also 360-101 = 259 degrees east. Some people split longitude from -180 to 180. Some from 0 to 360. Means the same thing. Add or subtract 360 degrees from numbers below zero or above 180 respectively to convert between the two forms.

Evolution is not completely dependent on sexual reproduction. Evolution through mutation occurs in both sexual and asexual reproduction. Second, females don't entirely reproduce sexually: mitochondria have their own genes, reproduce asexually, and are passed down from mother to daughter (and son) but not from father to child. Sperm don't contribute directly to the mitochondria history or evolution (other than assisting ova to mature into beings capable of reproduction). Some people jokingly call women immortal, because their mitochondria live on in their progeny. It's not too far fetched to say we're a symbiote of mortal and immortal cell lines.

If the starting or final orbits are not very circular, the math is a bit hairy. Here's how you do the simple case of circular orbits. With a Hohmann transfer, you encounter your target 180 degrees away from starting position. It takes time T = pi sqrt(a3/mu) to go 180 degrees, where the semi-major axis is a = (r1 + r2)/2, where r1 is the initial radius, and r2 is the final radius. The target's orbital period is T2 = 2 pi sqrt(r23/mu), so in time T, the target planet travels T/T2 of an orbit, 360 * T/T2 degrees. Thus the phase angle is 180 - 360 * T/T2. The math is the same for a bi-elliptic transfer. It simply takes another step. Let's say the intermediate step has radius r. Then the total transfer time is Tb = pi sqrt(((r1+r)/2)3/mu) + pi sqrt(((r+r2)/2)3/mu). Now instead of meeting your target 180 degrees away from your starting position, you meet it 360 degrees away. Thus the phase angle is 360 (1 - Tb/T2) degrees. Like 5H said, you can often pick r to make the phase angle more or less whatever you want, including whereever the target is at the moment relative to your starting location.

This method of firing a bit on each of the axes (radial, normal, pro/retrograde) is less efficient than burning directly along the maneuver node (ignoring for the moment, corrections at the end of a burn when the maneuver node marker wanders because you did not exactly point at the maneuver node for the entire burn duration). Sometimes you have fuel to spare and efficiency doesn't matter, and it's just easiest to manually 'feather' in a burn. So it's a judgement call. If you need every last bit of delta-v, or prefer not to be grossly inefficient, drop a maneuver node and follow it. If you want to fine-tune an encounter, and have fuel to spare, and feel like eyeballing it, burn on the axes as needed. Let's say you need to burn N m/s normal, R m/s radial, and P m/s prograde. If you do it one axis at a time, the delta-v cost is N + R + P. If you burn along the maneuver node the delta-v cost is sqrt(N2 + R2 + P2), which is always less than or equal to N + R + P. For example, if N = R = P: sqrt(N2 + R2 + P2) = sqrt(3)*N < N + R + P = 3N. Burning along the axes burns up to sqrt(3) times as much fuel as burning along the maneuver node.

I agree with everything Tex_NL said, but additionally want to point out that if you are starting from low munar orbit, and want to go to Duna or Eve, it is cheaper to go directly from the Mun, than it is to dip down to LKO and burn there. That's only for Duna and Eve though. For other destinations from the Mun, it is significantly cheaper to drop down to Kerbin and perform the interplanetary burn there. (You can see this in the table in the linked thread one post up.)

Here's an old table of approximate delta-v costs from various moons to interplanetary destinations, showing the direct transfer cost vs. the cost if you dip down to low orbit around the parent body. Also shown for comparison is the rough cost starting in low orbit around the moon's parent.

I agree that this statement was imprecise. Local gravity always accelerates you vertically by the same amount no matter your current velocity. But like 5H said, at nearly orbital velocity, due to the curvature of the body, that gravitational acceleration does not add significantly to your total velocity, therefore it doesn't add significantly to your total energy. In the extreme case of a stable circular orbit, gravity constantly accelerates objects towards the center of the body you are orbiting, but doesn't increase your vertical velocity at all, because the local direction of vertical has changed. Additionally, your total energy has not changed. Issues of accuracy are a different consideration, and may be the deciding factor for many people, but are not the issue at hand.

Don't forget you can speed-waddle: 4x physical time warp, and holding down Shift to sprint.

What people mean by efficient is that it is about the cheapest delta-v landing you can make with a low TWR vehicle. Using a pure retrograde suicide burn trajectory with a low TWR vehicle will cost more in gravity losses and more in total delta-v (but less in cosine losses) than a constant altitude trajectory, because you spend more delta-v to counteract a large vertical velocity. Yes, burning off-retrograde is less efficient at changing your total orbital energy than burning retrograde, but not by as much as you may think. Burning 10 degrees off retrograde is only about 1-cos(10) = 1.5% less efficient than burning retrograde at reducing your retrograde velocity. That's tiny. Even at 20 degrees off retrograde, you are still getting 94% of your thrust directed retrograde. Gravity losses are kept low by keeping your vertical velocity low. Gravity losses are proportional to the time integral of your vertical velocity times the difference in your vertical thrust and local g. Going to longer times, but with lower vertical velocity and vertical thrust means lower total gravity losses. And a low TWR vehicle is usually a lighter vehicle, which means much cheaper launch cost. You didn't define efficiency. For some people on these forums, total cost or part count or total mass are considered the primary factors in efficiency. Of course, once you are orbiting something with some particular vessel, then efficiency is usually in terms of delta-v. But don't ignore the considerations that caused you to be in orbit with a vessel with a certain TWR vs fuel budget.

I assume by efficiency, you mean in terms of delta-v. You haven't specified. You also have not specified what you mean by "finite burn compared to an equivalent impulse burn". Equivalent in what manner? Without specifying the details of your finite burn, there is no way to know. You stipulated that constant acceleration is an OK assumption. (While not always realistic, it is an OK starting place.) Now fill in the rest of the details of the circularization process. Once you do define the problem, you still won't find an equation for your efficiency measure. The equations of motion for a rocket undergoing arbitrary thrust in a gravitational field are not analytically solvable. You have a couple options: 1) simulation. You could do a bunch of tests in KSP or you could numerically integrate the equations of motion a bunch of times for different thrust profiles. The equations of motion are pretty simple to specify in the restricted two body system of KSP: dv/dt = (1/m(t)) (FThrust + FGravity). You can solve this in two dimensions if the thrust is always in the plane of the initial trajectory. Bold quantities are vectors. m(t) is mass as a function of time. You need to specify the direction and magnitude of FThrust as a function of time. 2) approximation: Instead of a single impulse burn, approximate the finite burn as a small number of impulse burns. This is a bit hairy, but doable. Instead of the equations of motions, you use the vis-viva equation and equations for the geometry of conic sections. In practice, simulation is more accurate, and in fact, easier, than doing the approximate calculation. (I have been working on a related calculation in my spare time, and it's seriously ugly. Many pages of equations, even without showing the intermediate algebra. It took me way less time (< 1/10th) to write a trivial numerical simulator and run some quick tests.)

A suicide burn is the most efficient landing, and the higher the TWR, the more efficient, in terms of delta-v. Note that a proper constant altitude landing is a suicide burn. You touch the ground just as you bring your surface speed to zero. But like you said, if you want to do a quick, high TWR suicide burn, you have to pay the upfront cost of heavier engines, more fuel and larger launch platforms. The reason why high TWR suicide burns are more efficient is because they allow you to spend less time fighting gravity. The constant altitude burn is designed to likewise minimize gravity losses by preventing gravity from accelerating your rocket vertically. If you can't beat a constant altitude burn, then I would argue that your TWR is really not all that high.