Yasmy

Great vid EtherDragon. Very helpful.

Maybe they are using Mach 2 as something like a transonic/supersonic boundary, rather than a more conventional 1.2ish. I wonder if there is a similar change in the heating model at a subsonic/transonic boundary and a super/hypersonic boundary (perhaps at Mach 3.5?). From Physics.cfg: (thanks NathanKell) machConvectionStart = 2 machConvectionEnd = 3.5

You can start your turn much earlier. Just don't ever turn your rocket too far from prograde. Make small, slow adjustments. See Mr. Manley

It would be interesting to see these heat profiles on a per-part basis as a function of distance from the front. For example, have the front of a rocket be a stack of several empty liquid fuel tanks with nothing attached to them, and plot the heat fluxes vs. Mach number for each tank. (Might want to pick something with higher max temperature than a fuel tank, but something light and of constant cross section.)

When you perform your transfer burn or mid-course plane change burn, focus on the target planet. Then you will see your trajectory in the planet's frame of reference. Looking at the SOI exit marker on your trajectory, you can tell if the orbit is prograde or retrograde. Then make tiny adjustments to set up your desired encounter. (When you are done, hit Backspace to return the camera to your vessel.) <ninja'd!>

Clockwise assumes you have defined some oriented reference plane, for example, the direction pointing away from a wall clock. If the wall and clock are transparent, then looking from "behind" the wall, the hands are moving counterclockwise. Normal clocks only go clockwise by convention of looking at them from the front. The same goes for planets. 1) Find the axis of rotation. (Nothing to do with magnetic fields. Look at the rotation of the body). 2) You can look at the planet from either end of the spin axis. Looking back along one pole, the planet rotates clockwise. Looking back along the opposite pole, the planet rotates counterclockwise. Clockwise is a matter of perspective. 3) The oriented reference plane mentioned above is the plane through the equator of the planet, and by convention, the up direction (the orientation) from the equatorial plane is the direction along the spin axis from which, if you looked back down at the planet, the planet rotates counterclockwise. For the Earth and the Sun and most planets, we call that direction North. Not magnetic north, but true north. 4) If the angle between a planet's north spin axis is within 90 degrees of its north orbital axis, then the planet orbits and rotates in the same sense. If you looked at Uranus from north of the ecliptic, the surface would appear to (very modestly) rotate counter to the direction of its orbit.

There are plenty. Did you look? Examples: Propellers that work in all atmospheres. Inflatable ships. Jets that run in non-oxygen atmospheres. There are tons of mods that support gameplay on Eve. Many people, self included, have done stock missions just fine on Eve. Eve is one of the hardest places to return from, but it is definitely possible and fun with stock parts. I don't know what you are talking about with stock parts blowing up. Perhaps you should try Laythe next. I think all the planets are fun to land on and return from (excluding Jool - that's flirting with disaster). But Laythe is somewhat more similar to Kerbin than Duna, and exists in an exotic environment. Plenty to see and do around Jool.

<Deleted because Red Iron Crown beat me to it.> Or you can use a mod like Kerbal Attachment System (KAS)

Seconded. Cheers!

I understand what you mean by "take", but I disagree with never. Correct. Steering losses are not delta-v losses. No delta-v is lost, as I said in my first post. Imagine this scenario. My rocket has a stability problem. Instead of pointing prograde when commanded to point prograde, it precesses around prograde like a top. On average it points prograde, but it never actually points prograde. I don't lose any delta-v, but I do uselessly spend some of my delta-v budget by not pointing where I want the rocket to point. That is a steering loss. You need to expend more delta-v to perform a maneuver than a rocket with better attitude control would have to expend. Anyone who's ever built a floppy rocket in KSP has experienced steering losses, and anyone who played KSP before the improvements to ASAS has experienced steering losses. If you want to say that intentional non-prograde burns are not steering losses because those burns were intentional use of delta-v in the non-prograde direction, I would argue that you are redefining the standard definition of steering loss. Is it the word "loss" that you object to? In this case (intentional non-prograde burns) I would agree that you don't "take" a loss. You don't lose anything. But you still have a non-zero steering loss. It's just not relevant.

Steering loss is speed lost due to steering, i.e., not thrusting in the direction of motion. If you thrust prograde, your change in speed is the delta-v spent, ignoring gravity and drag losses. If you thrust at an angle to prograde, the change in your speed is less than delta-v. (Though the magnitude of your change in velocity is actually still equal to delta-v.) If you thrust at angle alpha away from your prograde vector, your steering loss is proportional to (1-cos(alpha)). This means that instead of your speed increasing by dv, your speed will increase by approximately dv * cos(alpha). For alpha < 8 degrees, your steering losses are less than 1%, since (1-cos(8)) = 0.0097. Your total steering loss is the time integral of thrust/mass * (1-cos(alpha)). The time integral of an acceleration is a delta-v. Note that thrust, mass and your steering angle alpha are generally all functions of time.

And what happens if you flip the orientation of the delta wing? For now, the wing parts are supposed to be used in their default orientation. Hopefully this will change in 1.0.

ftp, gopher and email on a library system in 89 or 90, but that was pretty uninteresting. My first interesting encounter was the college computer network in 91. We had an instant messaging system called zephyr which you could use to message anyone logged in on campus or any particular machine. People didn't have to be running a client; it would just pop a message window up on their screen. Used finger a lot to locate people back then too.

Backup the "saves" directory in your KSP install directory. It has everything you created and all you game state.

Ok. After reading MartinKitFox's (absolutely correct) post, I went back and reread what you were trying to do: Get to low Kerbol polar orbit. I missed that. Though correct, I didn't think MartinKitFox's post was relevant because you said you didn't want to burn way out of the way (ie, to Jool) to reduce delta-v costs. Turns out his post is actually on the money because for a Kerbin to low Kerbol orbit, you are already on an extremely elliptical orbit at apoapsis. With no fancy slingshots, just a single ejection burn: 1) LKO to LKerbolO. Use the vis-viva equation. AP = radius of Kerbin's orbit PE = radius of Kerbol + 1000 km v = sqrt(mu (2/AP + 2/(AP + PE))) v = 1807 m/s So when you exit Kerbin's SOI, you want to be going 1.8 km/s on a 90 degree inclined orbit. (Might want to check my math.) 2) To get to a Kerbol polar orbit going 1.8 km/s at Kerbin, you need to shed Kerbin's 9.3 km/s (9284.5 m/s) orbital velocity. So before crossing Kerbin's SOI, you need to be going 9.3 km/s west, 1.8 km/s north = 9.5 km/s inclined by 169 degrees. 9.5 = sqrt(9.3^2 + 1.8^2), 169 = atan2(1.8,-9.3) Just trig. So the plane change is cheap (9.5 - 9.3 = 0.2 km/s) since we combined it with a huge change in periapsis. 3) To get to the target velocity at SOI, you need to be going the target SOI velocity plus Kerbin escape velocity from LKO of approximately 1km/s. So launch into a Kerbin orbit inclined at 169 degrees. Burn approximately 9.5 km/s + 1.0 km/s = 10.5 km/s. 4) If you want to circularize, ouch. At Kerbol periapsis you'll be going: v = sqrt(mu (2/PE - 2/(AP+PE))) = 93.6 km/s and to circularize you need to be going v = 66.8 km/s dv = 26.8 km/s Yup. 26.8 km/s circularization + 10.5 km/s LKO ejection = 37.3 km/s. Ouch. ------------- Instead of 1000 km PE, let's change it to 1,000,000 km PE, the limit of low Kerbol science: PE = radius of Kerbol + 1e9 m. v = sqrt(mu (2/AP + 2/(AP + PE))) = 3.8 km/s outside Kerbin's SOI. Inside Kerbin's SOI: v = sqrt(9.3^2 + 3.8^2) km/s = 10 km/s, inclined at 157 degrees. Adding LKO escape velocity, 11 km/s delta-v along a 157 degree inclined orbit. At PE, delta-v = sqrt(mu (2/PE - 2/(AP+PE))) - sqrt(mu (1/PE)) = 41.2 km/s - 30.5 km/s = 10.7 km/s Total delta-v from LKO: 21.7 km/s. Save you about 16 km/s over 1000 km PE. ------------- That's the cost of going direct. Not cheap. I'd be inclined (heh) to use another method. - - - Updated - - - Third option, still with the inefficient direct method. Don't completely circularize. Shrink your orbit to 1000km x 1000000km. This would only cost 10.5 km/s ejection from LKO + 7.6 km/s apoapsis shrinking.

More or less both. Kerbin surface velocity changes the angle slightly, adding a bit of prograde velocity to your orbital velocity relative to surface velocity. So to compensate, launch a bit more westerly than your intended orbit. Probably a couple degrees. Just use Pythagoras's theorem if you care to do the math. Launch into an inclined orbit which is already aligned with your ejection burn. More or less northwest instead of the usual easterly launch. Upon crossing Kerbin's SOI, your trajectory should switch from inclined at 135 degrees to inclined at 90 degrees. Edit: This is for a Kerbol orbit of the same size as Kerbin's Kerbol orbit. Not what you were looking for. More below in a bit...

Well, since you are not overly concerned with efficiency, just be direct about it. Kerbin orbits Kerbol at 9284.5 m/s prograde, and you want to go, say, 9284.5 m/s north from Kerbin instead. Add in about 1000 m/s of Kerbin escape velocity relative to LKO. Using good old Mr. Pythagoras, you want to burn approximately sqrt( 9284.5^2 + (9284.5 + 1000)^2 ) = 13855 m/s. Call it 14 km/s. The angle you want to burn is atan2( 9284.5 + 1000, -9284.5 ) = 132 degrees counter-clockwise from prograde, ie, pretty close to north-west. So launch from the surface Kerbin into a NW orbit and burn 14 km/s delta-v.

Well, this isn't infinite TWR of a modded zero-mass, infinite fuel ship, but it is pretty fun. Stock, unmodded, but with infinite fuel hack, Kerbin to Jool in under 2 hours:

I think no one mentioned the 2nd best reason to use the OX-STAT (liberally): They can really make your vessel look snazzy. Particularly if you lack fancy parts from fancy mod packs.

0) Prediction is already done and in game via the trajectory lines plus Pe markers. 1) Control: When you are far away from the target SOI, drop a maneuver node. 2) Hit Tab a bunch of times, changing your focus to the target. Now you can read your periapsis at the target. 3) Gently tug on the maneuver node handles to adjust your arrival. 4) If 3 doesn't work satisfactorily, move the maneuver node somewhen else. 5) Hit Backspace to target your ship again. A maneuver node mod like PreciseNode or the one in MechJeb makes it very easy to design the perfect burn for your desired intercept.

First, that's no epsilon. ξ = xi ε = epsilon Let's call the missing angle beta, β. Then β = 90 - Æ- ξ or β = À/2 - Æ- ξ if you prefer angles in radians. Just add up all the angles in a 90 degree or 180 degree section around point V. As far as I can tell, that is what you asked for. If you need the value of ξ in terms of r, (Sat2-S) and Æ, check out PakledHostage's answer. You know two sides of a triangle and the angle between them.

In addition to high delta-v, for direct intercept, you probably also need a high TWR. Given enough thrust, all (non-singular) gravity wells are shallow. Perhaps then you won't need the intercept markers. What happens if you target Minmus and just burn until the prograde marker and target indicator are aligned? With high thrust, high delta-v, I say wing it.

The notion of relativistic mass is, thankfully, losing favor. Unfortunately most textbooks still use the concept. I'm in the camp that says there is no need for the concept. Just let the energy and momentum scale like gamma, and don't try to assign meaning to the quantity m*gamma, where gamma = 1/sqrt(1-v^2/c^2). Suppose you are moving at velocity v in the x direction. Then the force laws are fx = m gamma^3 ax fy = m gamma ay fz = m gamma az This suggests, if you believe in relativistic mass, that the mass is somehow different in the direction of motion than perpendicular to the direction of motion. The notion of relativistic mass comes from trying to make the mechanics look analogous to Newtonian mechanics (F = ma, p = mv) by letting m suck up the relativistic corrections. The analogy breaks down though, as above, where in the direction of motion there are a couple extra factors of gamma in the force to acceleration equation. Note I generally frown on appeals to authority. It is not a good logical argument. But I'll just put this quote here from Einstein in 1948: "It is not good to introduce the concept of the mass M = m / sqrt(1-v^2/c^2) of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the 'rest mass' m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion." (And to comply with the quote, the formulae for the energy and momentum are: E = m c^2 / sqrt(1-v^2/c^2) and p = m v / sqrt(1-v^2/c^2).)

"Why" is a philosophical or religious question. It requires faith to believe the universe exists the way it exists for some reason. We can never prove the existence of that reason. As much as physicists like to think that certain theories are "natural", physics really can't explain why anything is true. Additionally, we can't even prove that theories are true, but only that they seem to work very well over the regime we can test them in. We can demonstrate that some theories are false though, by showing that some experiments contradict the theory. "How" on the other hand, is the question that physics answers. "How is the speed of light a limit?" is a very simple question to answer, if you believe (faith) that the theory of Relativity holds. The answer has already been given: it would take an infinite amount of energy to accelerate a non-zero mass particle up to the speed of light. According to Relativity, the energy of a particle of rest mass m, moving at velocity v, is E = m c2 / sqrt(1-v2/c2). If you plot that function, you see that the energy goes to infinity as the velocity goes towards the speed of light. Note that we don't actually know that c is a fundamental limit. We have just never seen it violated. We know it takes incredible energy to accelerate tiny mass particles to near the speed of light, and we know we have never seen anything massive move at or above the speed of light.

The game devs provided the mod system. Playing with mods is intended and supported behavior, and Squad often incorporates modders work directly into the game. I don't see any problem with people choosing to play stock. But saying that using a modded game is somehow different from playing the game the devs' intended or provided is clearly false.