Red Iron Crown

The number you are looking for is Thrust-to-Weight Ratio (TWR). A TWR greater than 1 will have enough thrust to lift off, with greater acceleration the larger TWR gets. A craft with a TWR of less than 1 will sit on the pad burning fuel until its mass is reduced enough to bring the TWR up to 1, at which point it will start to slowly lift off. Unfortunately, there is currently no TWR calculator built into KSP. You can calculate it manually by adding the thrust of all your engines, then dividing by the sum of the masses of all your parts multiplied by 9.81. Alternately, you can install a mod that calculates it for you, MechJeb and Kerbal Engineer Redux are popular.

It's confusing because you're taking gravity out of consideration, which removes the potential energy of altitude. The increased kinetic energy will be converted into greater potential energy (i.e. greater altitude) at the other end of the orbit.

No calculus required to calculate delta-V of a craft, just a bit of algebra. It is tedious, though, which is why I use the mods you mention.

While that would be more elegant, it's never going to happen. We'd have to rework the metric system to retain its advantages, most currencies are decimal and would need to be reworked, and finger counting becomes more difficult.

I don't think the science system is meant to be realistic. If you think about it, it's a bit silly that learning more about various celestial bodies unlocks better rocket parts. In reality you spend money on engineers, testing, and materials science to get better rocket parts. The science system is meant to give a reason to put craft on those other bodies. The transmit penalty is meant to give a reason to return those craft. Crew reports and EVA reports are valuable to give a reason to send manned craft.

The important thing to remember is that when completing a burn and expending dV, it's not the velocity change we're interested in per se, it's the change in the orbit's total energy. At any point in an orbit, the orbital energy is constant. It is the sum of the potential energy (energy stored as altitude above the gravity well) and kinetic energy (the energy stored as velocity). This sum doesn't change as long as no dV is expended. Much of the potential energy is converted to kinetic energy as the craft falls from apoapsis to periapsis, then that kinetic energy converts back to potential energy as the craft climbs to apoapsis. When we spend dV in the prograde or retrograde direction, we are adding or subtracting kinetic energy to the orbit. Burning prograde increases total orbital energy, burning retrograde decreases it. The Oberth effect allows us to get more kinetic energy change from the same amount dV spent, and it is the kinetic energy change that is important. Think of dV as the currency spent to buy a change in kinetic energy. The Oberth effect allows you to buy at a discount.

No, it's better to go East for both scenarios. The dV saved by building on Kerbin's rotational speed rather than working against it is beneficial whether you're heading inbound or outbound. The only difference is where in your orbit you do your interplanetary burn (angle to prograde).

Treaties have the force of law in the US once they are ratified. Corporations like SpaceX are definitely bound by them.

Ooops, you're right about the mass used, how embarrassing. Let me rerun the numbers: A 1-ton craft is in an eccentric orbit with an orbital velocity of 1000m/s at periapsis and 100m/s at apoapsis. There is enough fuel to complete a 100m/s dV burn. The fuel to complete the burn masses 0.1 tons. Prograde example: If the burn is done at periapsis, the kinetic energy change is as follows: delta-Ek = Ek(final)-Ek(initial) = 1/2mvf2-1/2mvi2 = 1/2*900*11002 - 1/2*1000*10002 = 544,500,000 - 500,000,000 = 44,500,000 If the burn is done at apoapsis, the kinetic energy change is as follows: delta-Ek = Ek(final)-Ek(initial) = 1/2mvf2-1/2mvi2 = 1/2*900*2002 - 1/2*1000*1002 = 18,000,000 - 5,000,000 = 13,000,000 J Retrograde example: If the burn is done at periapsis, the kinetic energy change is as follows: delta-Ek = Ek(final)-Ek(initial) = 1/2mvf2-1/2mvi2 = 1/2*900*9002 - 1/2*1000*10002 = 364,500,000 - 500,000,000 = -135,500,000 J If the burn is done at apoapsis, the kinetic energy change is as follows: delta-Ek = Ek(final)-Ek(initial) = 1/2mvf2-1/2mvi2 = 1/2*900*02 - 1/2*1000*1002 = 0 - 5,000,000 = -5,000,000 J Sorry for the wall of math, I should have done it right the first time.

That wording really shows how Oberth is win-win. The only loser is your exhaust!

As a thought experiment, imagine the following: Two identical ships are docked together, with one facing prograde and the other facing retrograde. An explosive charge is placed between them and detonated at periapsis thrusting the ships apart with equal force in the prograde and retrograde directions, with each ship's kinetic energy changing by the same amount. Each ship perceives the other as fuel it has expended to change its velocity. The ship headed in the prograde direction, you will admit, gains extra kinetic energy due to the Oberth effect. Due to conservation of energy, the ship headed in the retrograde direction must have lost exactly the same amount of energy as the ship going prograde gained, so it must have experienced the Oberth effect as well. The Oberth effect is symmetrical. The exhaust of the ship gains or loses exactly the same amount of "extra" energy as the ship itself.

Sure. A 1-ton craft is in an eccentric orbit with an orbital velocity of 1000m/s at periapsis and 100m/s at apoapsis. There is enough fuel to complete a 100m/s dV burn. If the burn is done at periapsis, the kinetic energy change is as follows: delta-Ek = Ek(final)-Ek(initial) = 1/2mvf2-1/2mvi2 = 1/2*1000*9002 - 1/2*1000*10002 = 405,000,000 - 500,000,000 = -95,000,000 J If the burn is done at apoapsis, the kinetic energy change is as follows: delta-Ek = Ek(final)-Ek(initial) = 1/2mvf2-1/2mvi2 = 1/2*1000*02 - 1/2*1000*1002 = 0 - 5,000,000 = -5,000,000 J So you can see, that even though the same delta V was expended, the change in kinetic energy is an order of magnitude larger when burning at periapsis due to the Oberth effect.

Oberth doesn't gain you any dV, it just lets you use your dV more efficiently. An example: A 1-ton craft is in an eccentric orbit with an orbital velocity of 1000m/s at periapsis and 100m/s at apoapsis. There is enough fuel to complete a 100m/s dV burn. If the burn is done at periapsis, the kinetic energy change is as follows: delta-Ek = Ek(final)-Ek(initial) = 1/2mvf2-1/2mvi2 = 1/2*1000*11002 - 1/2*1000*10002 = 605,000,000 - 500,000,000 = 105,000,000 If the burn is done at apoapsis, the kinetic energy change is as follows: delta-Ek = Ek(final)-Ek(initial) = 1/2mvf2-1/2mvi2 = 1/2*1000*2002 - 1/2*1000*1002 = 20,000,000 - 5,000,000 = 15,000,000 J So you can see, that even though the same delta V was expended, the change in kinetic energy is an order of magnitude larger when burning at periapsis due to the Oberth effect.

Hydrazine.

Except that it doesn't actually explain the Oberth effect, and introduces the idea that you gain energy on the way down and lose it on the way up, which is incorrect.

Ugh, a car analogy and a bad one at that. Cars become harder to accelerate the faster they are going, due to greater aerodynamic and mechanical friction losses. And they aren't propelled by shedding mass, so Oberth wouldn't apply anyway.

It should be made clear that Oberth effect is related to speed, not altitude. Once in orbit, the two are closely related, but it matters for a launch when there is low altitude combined with low speed so there is little Oberth effect. If you are in a circular orbit your periapsis and apoapsis are so close that the difference between them speed wise is negligible or actually zero, so it doesn't really matter where in the orbit you make the burn; Oberth will be about the same at all points in an orbit. WRT lower parking orbits versus higher ones, lower is better. To get from a lower orbit to a higher one requires raising the apoapsis at some point on the lower orbit, and you're right in thinking this part would be identical to starting a transfer burn from the lower orbit. However, the circularization burn and remaining part of the transfer burn would be done in the higher orbit where the Oberth effect would be smaller. It is generally best to do a transfer burn all at once from the lowest possible parking orbit.

Clever! I wonder how much payload can be orbited that way.

Make me think of the steam-powered spider in Wild Wild West.

Interesting example here of Oberth effect reducing the delta-V needed for an orbital capture: http://www.clowder.net/hop/railroad/Oberth.html It compares a rendezvous with Phobos using Oberth versus a rendezvous with a Martian Trojan asteroid.

He posts on here: http://forum.kerbalspaceprogram.com/threads/28428-Orion-aka-Ol-Boom-boom

Useful for fine tuning interplanetary transfers, too, when you want to make very small adjustments to your velocity.

As Rhomphaia notes, when burning retrograde you are trying to subtract energy from your craft, so you want more of the energy to end up in the exhaust. It is the same "something for nothing" you get when burning prograde.

I agree that intelligence + tool making is the ultimate in evolutionary success (so far, at least), but the point I was trying to make is that they're not inevitable. Your statement "On the long run, another species would have evolved that also has intelligence as a key skill." is not necessarily true. The dinosaur example was meant to illustrate that, as they had a much, much longer run in which to develop them but never did.

Are we looking at the same thing? Distance is a component of the first equation: delta Ek = F * s where F is thrust and s is distance Whether F is positive or negative, the magnitude of delta Ek increases as s increases.