700NitroXpress

No actually, Jeb is on his way to Vall with Bob. Bill is currently on Duna in charge of my Duna Command Center and surface base of operations. This lucky kerbal is Macdon Kerman.

Yeah, with rendezvous, there's no question about it otherwise you're going to hit each other at the same speed. Really, Eeloo? In the video Scott Manley even says that an object like that can't exist in reality as a natural object.

So where in the solar system would it actually be noticed then? Tylo?

Therefor, it's the same no matter what direction you come in from.

Let's go to Mun. Counter Clockwise Mun orbit. Counter Clockwise Mun Landing. Clockwise Mun orbit. Clockwise Mun landing. Notice that there is an altitude difference in the landing zones by about 2000 meters. This accounts for the small difference in fuel consumption. Both craft killed all orbital velocity and came straight down on top of the ground. No counter burns were needed. Fuel difference 1.5% which is attributed to altitude difference in both the orbit and the landing zone.

Here is orbital speed canceled from a 100,000 meter orbit. Altitude accurate to within 10 meters, orbital speed is 4.6m/s on engine shutdown. Clockwise Orbit. Here is the maximum speed achieved while falling to the surface under the force of gravity alone before the air resistance kicks in. Here is right before it hits the ground. You can see by the prograde marker on the nav ball that the horizontal velocity is 0

You're horizontal surface speed changes once your gauge switches from orbit to surface. In orbit the retrograde marker was right on top of the nav ball. The orbital speed was canceled out. Now if I were going in for a landing, the atmosphere deceleration would take care of the horizontal surface speed for me and I would hit the ground at terminal velocity.

Ok, everyone who likes physics, here is your proof, this is a clockwise orbit and a counter clockwise orbit with the same craft. I slowed the craft accurately down to 20 m/s from the orbital speed of 2288.5 and 2285.8 So, if you give or take for the error of speed and the orbital altitude, and piloting for the deorbit burn and what time the vessel entered the rim on the atmosphere during the burn. The fuel consumption for both orbits is exactly the same. And the craft accelerates towards the surface at the same rate in both instances because we have a gravitational constant. Craft on the launch pad: Clockwise orbit: Clockwise orbit fuel consumption to kill orbital velocity to about 20 m/s vertical. Counter Clockwise orbit: Counter Clockwise Orbit Fuel consumption to get about 20 m/s vertical:

Yes, that change is to equalize my relative speed to the conveyor. If the conveyor is going faster than I am, 2.5 and I'm going 2, I need to increase my speed by 0.5. Now I'm going at 2.5 and it is going at 2.5, 2.5 - 2.5 = 0 If it's going slower than I am, then I need to slow down 0.5 so now I'm going 1.5 and it's going 1.5. 1.5 - 1.5 = 0 The change in speed for both is the same in order to make the relative speeds equal.

I changed my speed by 0.5 because I was moving at 2 to begin with. If I stop moving I'm now going 0 Both conveyors are going 0.5. I now go 0.5 and stand on the conveyor.

No, what speed are you going to begin with? 1000 m/s 1000 - 1000 = 0 Your target is moving at 50 m/s in one direction so you need 50 m/s to catch up to it

In this equation they don't equal 0. 2 = 2. So you are taking the relative speed which is the 2.5 and 1.5 Then you add you're compensation speed which is -0.5 and +0.5 So now your relative speed in relation to it is 2 and now you kill your relative speed so you're moving at the same speed which is -2 So now you're speed in relation to the target is 0.0

My bad I was writing too quickly. So yes in terms of relative speed, they are different, but they are different in the same way. You have to compensate the same amount for both. If you subtract 1000 m/s from this to equal +50 and -50, you still have to compensate for 50 on both.

[quote name= 0 = (2 + 0.5) - 2.5 0 = (2 - 0.5) - 1.5 else your equation is wrong, because you would be saying 0 = 2 ... Third law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction to that of the first body. What does this have to do with anything (here)?

OMG, no, the surface speed is the same. The planet is not accelerating which is the only thing that would cause the speed to be different. Are the speeds of the conveyors different? No, they are both going at 0.5 m/s are they not? You are going at 2m/s are you not? If you stop going 2m/s, how fast are the conveyors going? I think they're still going at 0.5 m/s. In order to go the same speed that they are going, I have to increase or decrease my speed 0.5 m/s for both, do I not? Therefore lets look at this: Conveyor A is going 0.5 m/s and Conveyor B is going at 0.5 m/s in the opposite direction. I'm going at 2 m/s in the same direction as conveyor A. Therefor, my speed in relation to Conveyor A is 1.5 m/s and my speed in relation to conveyor B is 2.5 m/s 2 + 0.5 = 2.5 2 - 0.5 = 1.5 In order to compensate for the speed of both conveyors, I need to make my speed in relation to the conveyor 0. I have 0.5 units of fuel to do this. 0 = (2 + 0.5) - 0.5 0 = (2 - 0.5) + 0.5 In order to go at the same speed and land on go on both conveyors I need to compensate for the equal opposite speed due to Newton's third law. I used the same compensation speed of 0.5 to equalize both. Any questions?

Thank god someone here understands Newton's Third Law. You would have to slow down 2m/s in order to land on both conveyors. Both conveyors are moving at the same rate. So it doesn't matter if you land on one or the other. If you compensate for the rate of the conveyors, then you have to either speed up or slow down 0.5m/s in order to land on both without being swept away. This means that you have to either speed up or slow down at the same speed no matter which direction it's going in order to land on a specific point in relation to you.

Newton says that you're wrong. http://en.wikipedia.org/wiki/Gravitational_constant

Your picture is physically impossible. Your orbital speed and altitude is fine. But you can't have a different velocity going one way as opposed to the other due to Newton's Third Law "To every action there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions." Your orbital speed of 2290 m/s at 75,000 meters is the same clockwise as it is counterclockwise, and lets not forget that gravity is constant. When you are in orbit, the force of friction isn't acting on your vessel. The rotation of the planet isn't acting on your vessel. Therefore there is no extra force to cancel. Thus, there is no additional fuel consumed to land from orbit. When you are taking off from a planet, you inherit the planets rotational speed because you are on the surface. As soon as you leave that surface, that surface no longer is causing any force that is acting on you, but you are still moving at the surface speed in the horizontal direction, but you don't notice it because both you and the surface are moving at the same speed. When you launch from a planet you are already going at the speed of the rotation. This means that your horizontal velocity while stationary on the surface of Kerban is equal to it's rotational speed of 174 m/s. However, your vertical speed is 0 and your speed in relation to a specific point on the ground is also 0.

A man moving on a motorcycle falls at the same velocity of 1g on Earth rather he's moving at 100 m/s or not at all because GRAVITY IS CONSTANT! He hit's the ground at the same time because the same force is pulling him to the ground in both cases. Friction has absolutely no effect on an object that isn't in contact with another object. They both fall at the same rate. When you fall to a planet, you fall at the same rate. If this man on a motorcycle was in space and orbiting Minmus at 100 m/s he would not touch the ground because he is in orbit. Objects in orbit are falling towards the surface still because they are being acted on by gravity. When you stop going 100 m/s, you are no longer countering the force of gravity. You then fall at the rate of gravity of the object. Again Newton's laws.

If I'm in orbit around Kerban at 75,000 meters traveling at 2300 m/s and I decide to come to a stop and land on the surface. I kill the 2300 m/s horizontal velocity. Right now the only force acting on the craft is GRAVITY. Gravity pulls me to the surface of Kerban at a rate of 9.8m/s/s. During the descent I deploy my parachutes and softly land perfectly straight up right on the ground. This is because the rotation of the planet doesn't affect the craft's velocity at all. Newton's Laws.

NO, your descent velocity is from you to the surface, that velocity is caused by one force and one force only GRAVITY and GRAVITY IS CONSTANT therefore your decent velocity to the ground is the same. There are two forces at work here: Force A = orbital velocity, Force B = Gravitational velocity Force B is constant and cannot be changed Force A gets changed to 0 from a deceleration burn. Therefore only Gravity is acting on the craft. Your speed is only going in two directions horizontal and vertical. The rotational speed of an object doesn't affect it's gravity or the distance between you and the ground. There is absolutely no force of an objects rotation that impacts distance or gravity. If it take 100 seconds to descend 10km then it takes 100 seconds to descend 10km. If it takes 100 units of fuel to stop and land because you're countering your vertical velocity, then it takes 100 units of fuel to stop. Rotation speed means nothing, it's Newtons laws!

Now when we're talking about SOI, it is highly important which way you exit an SOI, exiting prograde to a planets rotation around the sun will eject you outward from the planet's orbital path and ejecting retrograde will put you inside the planets orbital path. This is because you have to think about the planet's position in it's orbit around the sun as a periapsis. To get to Moho and Eve from Kerban, you go retrograde and to go everywhere else, you go prograde.

True, but as Kasuha stated below, this bonus is only really relevant for takeoff of Kerban and Eve. Otherwise it doesn't matter. But since my statement was directed towards takeoffs from Kerban, you're right, the bonus is there and we all want to save fuel.

No, because you are going at a specific velocity to the ground in relation to your altitude. A craft orbiting a moon at 1000 m/s would need to slow down 1000 m/s to stop rather or not it's going clockwise or counter clockwise. Also, that same craft would need to descend to the ground. So if you have an orbital altitude of 100km and you kill the horizontal velocity, that velocity is the direction you are traveling and has nothing to do with the ground because you are not in contact with the ground. So the amount of fuel it takes to slow down going 1000 m/s and land for a powered landing from 100 km is the same in both cases because your are going to descend at a set velocity and that velocity is dependent on the objects gravity and the gravity alone because there is no other force acting on the object in orbit. If wouldn't matter if the moon was rotating 1000 times per second or 1 time per day, the gravity of the moon is dependent on mass and the moon's mass is the same. Therefore the gravity is the same, and thus you're downward velocity is the same no matter which way it rotates or you rotate to begin with. And that is Newtons laws of physics which are fact and the physics that this game uses, so you sir are wrong. If you still disagree, then I suggest you read this: http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation

I don't need fancy shmancy math. I know the the 1500 power engines have the most power, so If you have 9 of them on one payload then you have a huge amount of power. Then you figure that any asparagus staging that has engines that flameout cause sudden deceleration which hurts your structural integrity and your TWR. So if you eliminate this, you get a TWR that increases at a linear rate, and 0% loss in velocity. This means that you can throttle down at a linear rate and therefore use less fuel to keep the same TWR. By the time the craft is in space, you can use half to 1/4 throttle for light craft and full throttle to 3/4 throttle for ultra heavy craft. This lifter design is a one size fits all solution for just about everything so I don't have to redesign a lifter every time I build a new craft. If you're using light craft, then just redesign it for the smaller engines and smaller fuel tanks. The power of the current version can be increased if needed and more fuel can be added without major changes.