Frederf

Definitely look to the FAR mod. It makes a lot of adjustments to aerodynamics and allows much tweaking of control surfaces.

I'm not sure if it is mathematically perfect, but if your gravity losses and drag losses are equal then you are probably near optimum solution. To reduce your gravity losses you have to go faster which increases drag. To decrease drag you have to go slower which increases gravity losses.

bitbucket, thanks so much. I thought it might be something like this. For my 300km orbit (had a plane change to do) it was about 45 degrees past side position and then a burn to escape. The delay tip is indeed useful as the ejection arrow doesn't point orbit retro if you burn at the side position. Now I just have to do the reverse to arrive at the Mun and I'll really reduce my dV requirements.

I thought I would share my method for a Kerbin-Minimus injection which relied only on a stopwatch and looking out the window of the IVA spacecraft. I found it both satisfying and effective. Initial conditions: Space craft in Kerbin circular orbit, approx 1000km in this example. Orbital plane matches destination object, in this case Minimus. Orbits are same direction for simplicity but not required. Concept: Using simple orbital math a time can be derived between seeing an object at the periphery of Kerbin's disk and when a rendezvous burn is initiated. Step 1. Phase angle and transit time In this example assume the selected transfer orbit has a transit time of 48 hours. A successful arrival requires that our burn takes place on the opposite side of Kerbin from the position of Minimus 48 hours in the future. Minimus will travel 48/T of its orbital path in this time. Let's say the orbital period is 320 hours. 48/320 is 0.15 or 15% or 54 degrees. If you burn when you are located opposite of this future position of Minimus then you will arrive at 54 degrees in the future of Minimus's orbit and so will Minimus. Step 2. Time from opposition Assume you knew the exact moment you were directly opposed to Minimus's current position. Given that information how long would you wait until you began your burn? That is the goal of this step. The spacecraft in the parking orbit (and the opposition direction) is constantly changing its orbit angle. Minimus is also orbiting and changing its angle but more slowly as it is distant. If a space craft is orbiting 360 degrees in 18 minutes, then it is orbiting at a rate of 20 degrees per minute. If Minimus is orbiting at 360 degrees per 320 hours then that is 0.01875 degrees per minute. The result is that the space craft is generating 19.98125 degrees of relative orbit angle per minute. Going back to Step 1, we want a 54 degree angular difference so we must wait 2.7025 minutes after the spacecraft is directly opposed (opposite of Kerbin) to Minimus before beginning our burn. Step 3. Time from sighting However we don't know the exact moment of opposition, yet. What we can know, by looking out the window, is when Minimus appears out from behind Kerbin. From this we can learn how far in the past (before the sighting) Minimus was in opposition. When Minimus is sighted, it must be some distance along its orbit behind the current opposition point. At an altitude of 1000km I am 1600km from the center of Kerbin along the axis which passes through the spacecraft, Kerbin's center, and the current point of opposition of my orbit (the last one by definition). Minimus at an altitude of 46,400km is 47,000km from Kerbin's center approximately along this same axis. Assuming the angle to be small, Minimus is 48600km away from me. The slice of Kerbin which Minimus emerging from behind is 600km lateral from the axis. Minimus is an unknown lateral distance from this axis and is what we currently wish to find. Using the properties of similar triangles 600/1600 = X/48600 where X is the lateral distance (or in another approximation the orbital path distance) of Minimus. The tangent of (X/47000) is the number of degrees orbit angle between the current position of Minimus and the current position of the opposition point at sighting. In this case the angle is 21.195 degrees. At 19.98125 degrees per minute this means 1.061 minutes ago, Minimus and the spaceship were on opposite sides of Kerbin. Step 4. Putting it all together The last step is straightforward. If we want to burn at 2.7025 minutes after opposition, but by the time we actually see Minimus it will be 1.061 minutes past that, we must burn 2.7025-1.061 minutes after we see Minimus. 1:38 after Minimus is visible will be 2:42 after opposition and will be the required 54 degrees ahead such that when you arrive in the spot two days later, Minimus will be there as well.

Mechjeb's rendezvous algorithm works by using a known LPA. The LPA is calculated based on the previous launch. It also only works equatorial plane and I think only eastward at that. You haven't said how you're using the rendezvous (what ship on what body) but I'm assuming this is a post-munar landing rendezvous with the CSM? Try quicksaving on the surface, doing a munar liftoff, noting what the LPA turns out to be in Mechjeb. Then manually enter this number for your second try quickload. Alternately look up the transit time for a Hohmann transfer from surface to destination and figure what that is in fractions of an orbit that is for your destination craft. The should be very close to your LPA. To find my burn time for my last Kerbin-Minimus injection I did the following: 1. Find the transit time from parking orbit to Minimus orbit. 2. Find the angular orbit speed of my parking orbit and of Minimus 3. Find the orbit angle Minimus traveled in the time from #1. This is my LPA. 4. Find my effective orbit speed by subtracting Minimus's. 5. Find the time after opposition to achieve LPA from #4 #3. 6. Calculate the orbit angle of Minimus when it appears to come over Kerbin's disk. 7. Use #6 and #4 to find time since opposition to sighting. 8. Subtract time since opposition to sighting from time after opposition to burn to get time from sighting to burn. 9. Observe Minimus emerge from behind Kerbin, start timer. 10. At #8 time later, begin injection burn. I intersected with Minimus's rather small SOI quite well using this method and all it took was looking out the window which was surprisingly precise time-wise. The higher your parking orbit the lower your orbit angle rate which will enhance precision.

I've started thinking in terms of "leaving out the back door." Imagine the Mun is a car and your exit burn is jumping from the car. For return to Kerbin you want to reduce your forward speed as much as possible by jumping against the direction the Mun/car. If the Mun is orbiting counter-clockwise (eastward) with respect to Kerbin then you want to exit its SOI going westward. I'm still not sure what's the best way to exit a SOI. Should I make a long elliptic lobe facing Kerbin and start my escape burn near the edge or should I try to point the lobe opposite to the Mun's travel and burn to escape with excess speed equal to the Mun's orbital speed.

What I said is true, just not handy. Since a L of fuel has the same mass any other (in KSP) they are exchangeable with each other and also with fuel weight. Specific impulse expressed in any of those three ways is the same measure, just not necessarily the same units of measure. A physicist such as myself can barely be bothered with such trivia as units. Indeed the convention in KSP is to express it in Newton-seconds-impule/Newtons-fuel (aka seconds). What I wanted to show is the origin of the phrase "specific impulse" so it had more real meaning than "a crazy science term that I just have to memorize."

Impulse is just the total momentum change. 10 newtons over 10 seconds, 100 newtons over 1 second, 1 newton over 100 seconds. All are the same total 100 Newton-seconds impulse. Specific impulse means "amount of impulse you get, per unit." In this case it is per liter fuel. An engine that delivers more thrust x time or mass x speed on the same amount of fuel is sad to have "a higher specific impulse." It's a complex sounding term but broken down it is simple. Another "specific" example is "specific heat capacity." 20 lbs of iron will have twice the heat capacity as 10 lbs, but their specific (per pound) heat capacity will be the same.

Assuming you start on the edge of the Mun's SOI with no speed you have escape energy. You want zero energy at the surface. To get rid of this energy you have to retro burn. You get rid of more energy per unit fuel at a high speed. Thus you want to do "just in time" landing burns (highest speed). Adjust technique for fault tolerance and safety.

No, slingshot is a conservation of momentum thing with three bodies that doesn't involve engine thrust at all. It's a lot like pushing off your friend in the pool. Your speed is at the expense of his. Oberth effect is simply that rockets are more efficiently increasing craft energy at high speed. A given unit of fuel will increase the craft's speed by some amount. The amount of kinetic energy added to the craft by each unit speed is proportional to the speed. It's the "V-squared" term in the equation KE = 1/2 m v^2. The orbital technique is to apply thrust at high speed (deeper in the gravity well) to gain the most energy per fuel spent. Notice that this logic holds true even for a constant mass rocket. There's a secondary benefit to burning deeper in the gravity well and that's not losing energy lifting fuel up high in the first place. It's as surprising that the Oberth effect is noticeable in KSP as orbits are possible. Despite its name, it's just a noticeable feature of basic mechanics and not wizardry.

Digital computers work in finite precision values. There is a window where two values can be exactly the same in KSP.

Spasm here as well with unfolding hinges with strut arms for satellite. It is ok at "home" position but has spasms in orbit (only orbit) when moved away from home rotation position. Also when reloading flight it forgets hinge position (but not position of thing attached to thing attached to hinge.)

I don't mean anything that fancy, just a "buffer" part so that the engine isn't the senior-most part in the debris.

Is it possible to simply "double decoupler" so that the debris engine has a decoupler parent that came with the rest of the debris? Really I'm stuck trying to make a semi-sophisticated remote probe. All I can think of is to have the probe be the main craft and the launch vehicle the debris. In which case I don't really need to remote the launch vehicle since it's disposable.

I'm having a problem where my components launched with this are all immediately activated, including engines. What's even more weird is when I rejoin the "RC" modules the engine sound is going but the engine/fuel components are grayed out. However the whole thing is activated. What's causing this "activate everything else" behavior on decoupling components containing a RemoteControl from the main craft? Testing shows that if I decouple a RemoteControl set of components they lose ALL of their staging information. What is the way to deploy a craft such that once controlled it has intact staging of parts and unactivated parts?

Very important mod function, best of luck!

None because KSP/unity can\'t handle my joystick once calibrated as calibrated resulting in the entire setup unusable.

I can see what SOPA is doing conceptually. It\'s moving the task of policing piracy from the victims (IP holders) to the infrastructure providers. Previously if you wanted to prosecute the crime you\'d have to go to the criminal which was hard because they were many and distant. So the IP holders were finding the task impossible. If the responsibility can be transferred to others then it is now someone else\'s fault when they don\'t succeed at the impossible task. They can march in lawsuits and demands and such of this new party made responsible. It\'s rather cunning, choosing an easier target to nail without regard of the real perpetrator. That and the method of implementation is bad.

I have a fetish for closed-form analytical solutions so color me disappointed. But... I\'m rooting for you! I\'m very interested to see what the optimized throttle profile for this simple system is. Remember that gravity drag is time spent in the gravity well times the strength of the well at that particular height. Mass is decreasing at a throttle-fraction-specific rate. Don\'t feel bad just throwing in a Cd of 0.3, drag increasing as a function of velocity to the 2.5 power, and just getting a taste for what the answer might be for very assumed aero drag conditions.

I thought we already had evidence of faster than light information travel in certain experiments? 60 ns was something like 0.0026% of the travel time? That\'s not huge but it\'s not super small as billion dollar experiments go.

It all makes sense when you think in terms of energy. In an eccentric orbit you have a fixed total energy consisting of kinetic (T) and potential (U). As you reach peak altitude you have the maximum U (well, least negative) and the minimum T. As you fall back down again into the gravity well you lose U and increase T. To have escape energy your T+U must be greater than the gravitational well's value at infinite distance (usually assigned as 0). If you're in a high orbit you have a lot of U so you only need a little more T to have T+U over the threshold. If you're at the bottom of the well (large negative U) you need a lot of T such that T+U is enough. The Oberth effect is simple, if an increase in speed costs the same amount of rocket fuel no matter if you're going from 1m/s to 2m/s or from 1000m/s to 1001m/s, then you want to maximize how much T that burn gives you. Since kinetic energy is the classic Newtonian T= 1/2 mv^2, your increase in T is better the bigger v is. Your speed is highest at the bottom of the gravity well because you speed up as you fall. If you compare T for 1m/s with 2m/s you see that you've increased T by 1.5m. If you compare 1000m/s with 1001m/s you've increased T by 1000.5m. In fact the calculus of it works out that doubling your speed at burn doubles how much kinetic energy you receive. Going from 500m/s to 501m/s in an increase of 500.5m in T.

I was thinking about this as challenge. Pod/LFT/LFE is an excellent testbed for liftoff methods. Your peak one-dimensional altitude must involve one technique that is absolutely the best for that purpose. Your only real control is the throttle. The faster the throttle the better Oberth and less gravity drag but more aero drag. Since I'm a little lazy with the math I'll go for T+U increase from point 0 at launch to point 1 at burnout as the yardstick for success. U0=0 U1=m1*(GM/(x1+600,000)^2)*x1 T0=0 T1=m1*v1^2 / 2 62.5*200,000 Ns of thrust-time 5,500-3,300 kg weight dV=A(t)=Thr/m(t)=Thr / 5500-2200t/62.5 =int[0,62.5] 200000 / 5500-2200t/62.5 = -5681.82 log(5500-35.2x) [x=0,62.5] = 5681.82 log (5500/3300) = 2902.42 x1 = int(int(A(t)))[0,62.5] = 5681.82 x - 5681.82 x Log[5500. - 35.2 x] + 887784. Log[-5500. + 35.2 x] (0,62.5) OK I'm getting tired of the math now. And my A(t) is missing gravity drag not to mention aero drag. I did manage to get 45,343m or something with the keep it at 100m/s until 20km though. Much better than my constant throttle 23/24km runs. Getting the general equation for optimized launches (at least in one dimension to start) would be a great accomplishment.