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Well, just to share some of my experiences of driving on the moon. I've had my Armadillo APC on the moon for a while now, it's way heavier than this ATV you have and it uses the same wheels. Getting down a steep crater is tricky, but if you use the rear and center tire to reverse all the way down and disable the front wheels, it can be done if you go slow enough. I didn't go faster than 7 m/s. I also used the RCS system to provide downforce to the tires using the translation controls. This pushed the vehicle down into the ground, which gives the tires more traction in the low gravity. When climbing up a steep incline, I switched the drive train to front wheel drive with the center tires also and disabled the rear tires that way it doesn't flip over. Also, you have vertical takeoff engines. So, if you also have rear thrusters, you can effectively use the rocket propulsion to your advantage to climb out of steep craters. So my advice is: Use the drive train to your advantage depending on rather you're going uphill or downhill, don't go in excess of 10 m/s downhill, use your propulsion systems to your advantage when moving over difficult terrain. Here is the Armadillo APC and it's driven half way around the moon across all of the terrain, up and down steep craters and all over the place. This was the inspiration vehicle for this challenge. But compact designs are really good too. So otherwise, this video also confirms that you can preform a wheelie and a stoppie in your ATV so that bonus challenge is satisfied.

I have 2 questions. 1. Can I refuel this craft in orbit at my orbital refueling station? 2. After the SSTO achieves orbit, can I fly it to another planet with an orbital tug, land on that other planet, take off from that other planet, use the same tug to get back to Kerban and land at the KSC?

Here was my first SSTO flight on Duna back in the day.

That's the smallest SSTO Plane I've ever seen... pretty cool.

Here's my first design for my HS MK - III, top speed in atmosphere is 1716 m/s without the afterburners on the sides. Once you shut down the main engines and fire the aerospikes it has SSTO capability. This is my new design, HS MK - V It has docking ports on the top of the afterburners now so I'm making a tug to take it out to Laythe.

You don't have to wait for the planets to align in just the right way in order to get to them, it helps a little if you do, but otherwise you just want to eject from Kerban's orbit around the sun going either prograde for the outer planets and retrograde for the inner planets. As far as transfer stages go, anything with more than 3 nuclear engines is overkill and a waste of fuel. You can easily get craft of over 150 tons out to Jool, with just 3 nuclear engines. This is my personal favorite ship design to use at the moment. It has an apollo style design. So the lander on top can undock and transfer one kerbal to the surface and then return to orbit. It has enough fuel to go to Moho and return, it also comes with my Leviathan MK - I super max payload lifter. Easily gets anything up to and over 200 tons into orbit. Or if you would like to try out my Duna Command Center, you can fly it out to Duna and land it on the surface as a permanent base. All you have to do is put the refit engines on it. Download it here: https://drive.google.com/file/d/0B-vTRL2n8wvzTmVCdWM3RElYZHM/edit?usp=sharing Download the subassembly for the Leviathan lifter here: https://docs.google.com/file/d/0B-vTRL2n8wvzQkEtUTBCNHpIMW8/edit Download the Duna Command Center Here: https://drive.google.com/file/d/0B-vTRL2n8wvzalY4b0VZc0Fsdk0/edit?usp=sharing Download the Refit Engine Here: https://drive.google.com/file/d/0B-vTRL2n8wvzMldXempWQ1BiMUU/edit?usp=sharing Duna Command Center Here's a video to show you how to use the Leviathan lifter: Here's a video of how to setup and land the Duna Command Center on Duna: If you try these out or if you found the designs helpful, let me know. Hope these help in some way. Have fun!

Oh man! Does it have docking capability with the transfer stage then? Does it have SSTO capability on Kerban?

Yeah, that should be good enough for Duna.

Yeah, even minus the 360 liquid fuel tank, I had enough to put the ship on a direct collision course with Kerban, and still have two engines to help brake for the landing. All other fuel tanks were completely depleted. I almost thought that the parachutes would rip the ship apart, but she held together and I let out out a sigh of relief. All 19 kerbals made it to the ground safely. Even though if wasn't at the KSC, they're all happy to be alive.

What's all in that thing? It looks cool, but what can it do?

So is 216 tons the max payload then?

Kill horizontal and vertical equally as you come in for a landing. Since you're traveling in an arc, if you point the nav marker behind the retrograde marker, then it will kill the horizontal velocity with the vertical velocity and you'll come in straight down. If you gain any horizontal velocity while angling your craft during landing burns, using the RCS translation controls of JKLI you can counter the horizontal movement.

Looks good, very nice design! So please post any additional screenshots in your current post as edits. Or make a screenshot library on Imgur and link that for photos of the rest of the bonus challenges if you intend to do them. From these screenshots, your ATV meets the core requirements and is confirmed to both run and drive on the Mun and on Duna. So that's 2 bonus challenges that are confirmed completed. Good job!

Don't use serial or non-feed staging. If you're using boosters, then at the end of the boosters, they flameout and cause sudden deceleration. With slack tanks, there is no deceleration ever, because you just drop dead weight and not something that was providing thrust. You could use this in combination with serial staging, but basically this is just a method to get things into orbit, so that way your serial or non-feed staging rocket can start out in orbit and not on the surface of Kerban. That way, since it's in space already, you get maximum efficiency out of your craft because you don't have to fight your way through the atmosphere first. Also this method works very well for transfer stages for interplanetary travel, maintain thrust with the nuclear engines and then drop the empty slack tanks as you go without loosing thrust.

After Jeb's initial survey of the target LZ, he discovered the anomaly in the photo's and radioed back to the mothership. "LZ is clear bring'er in!" Bob - "Rodger that, mothership is on the way!" The mothership has landed!!! Smile for the camera boys! Leaving Vall and Jool behind. That looks good enough... Good, we will land on land and not water. We don't need those. Reverse thrusters depleted, parachutes deployed! Mission Complete.

Total mass is 15.21 tons. Liquid fuel: 1072 Oxidizer: 1310 Electric charge: 210 I don't use Mech Jeb or any mods at all. Is it Tylo capable? Well, because we know that Tylo's gravity is 0.8, I would say that it is capable of landing and takeoff. However, I don't know if it could preform both in one mission. But if you made a fuel tank and landed it on Tylo as well, it would be able to refuel on the surface and then launch back into orbit. I know that it can operate on Vall and Moho because I just landed Jeb on Vall as part of the Vall exploration challenge. Vall and Moho have about equal gravity and I have more than enough fuel to achieve orbit and dock. If you wanted to take it to Tylo, my idea would be to make a bottom attachment for it that has enough fuel and power to handle the descent to the surface, then detach it about 1000 meters or less above the surface and use the lander to go the rest of the way. Then you would be able to get back into orbit with just the lander. It's primary purpose was to land a kerbal safely on a moon with gravity equal to Moho and then achieve orbit again and dock. So for the purpose of which it was designed, it preforms perfectly.

First landing on Vall!!! I think I got pretty close for my first attempt. 206.3 meters from the radius of the anomaly. Jeb is the first to set foot on vall and the first to plant a flag right in the center of the anomaly.

Yeah, the thing wasn't using physical time warp either so after a set amount of time it just decides to fall off. Mission is still a go tho. Public service announcement for all those that use command seats, remember to switch to the unmanned pod and select control from here after your kerbal gets in it, if your craft has one.

This is only in relevance in a universe without mass and gravity. We do not live in such a universe. If something is traveling at a speed in one direction over something else, is this object in orbit? Also, does an object that rotates produce a different gravity than one that does not? Does gravity act differently on two objects in orbit? What forces are at work on an object in orbit as opposed to an object on the ground? Do objects of the same mass that are released fall at the same speed? You aim two guns on the equator opposite each other pointed East and West and fire them, also you drop the same bullet from a stationary elevated height at the exact micro second that the two guns are fired and at the same height that the two guns fire from. Which bullet hits the ground first? Which bullet travels the greatest distance? Which bullet of the two fired has the greatest range? Answer all of these questions correctly if you can.

Right so if you slow down a set speed in either direction that speed is the same. This means that you will hit the surface at the same time. The angle at which you are going in the parabolic arc is equal opposite on both sides and it can't be any other way. This means that you would angle your craft in the equal opposite direction to your velocity to the surface. Which is... guess what, shown on the nav ball. This means that by coming into an object rotating, you would angle your craft and burn to zero the velocity which is the same in both cases because V = gt The rotation speed of the planet is it's angular velocity, which is the same on all points along the equator of a planet. Angular velocities are either equal at the poles or equal opposites. That means that it takes the equal opposite amount of fuel to decelerate and land given the rotational direction and the angle of approach. This is because angular velocity is a pseudovector. As described here: http://en.wikipedia.org/wiki/Pseudovector. Equal magnitude, but opposite direction.

Horizontal velocity In direction D is the same as the opposite horizontal velocity to B. If you have two craft in orbit and are 100 meters apart and you are in between them. You decide to you want to go to D at a speed of 5 m/s. You are also getting further away from the craft C at 5 m/s because that was your change in velocity. This means that you are traveling to D at +5 m/s and leaving C at -5 m/s. Thus you are using the same amount of fuel for the burn to change your velocity 5 m/s total. The velocity of the ground has nothing to do with this. By the time you reach D, time will have passed for 10 seconds. You will now be at D and you will be 150 meters away from C. The opposite is true if you decide to go to C instead of D.

You are forgetting one very important detail that you drew that contradicts what you said about it. The velocity on the surface for B, is the same as the velocity in orbit for A. These two arrows cancel each other out because they are both moving at the same speed at the same time. This means that now you have three arrows for both halfs of the sphere. When A is stationary like that the projected point D and the projected point C are not moving faster or slower than each other. This is because the rotational speed of the surface is the same in one point as it is in another. When you are stationary above point B, you move at the same speed it moves. That point moves across the surface at the same speed that all the other points move across the surface. This is because the surface speed is constant. The surface does not accelerate ever. This is what's known as the Coriolis effect. The person on the ground gets the illusion that A is moving faster towards it than D, than the same person on the ground would see A moving toward C. All objects of the same mass fall at the same speed of gravity rather or not they are launched or dropped. Object A gets launched in both directions from it's position above B. V = GT both objects arrive in the same position at the same time. Or if you still don't believe that, Mythbusters says the same thing:

Due to the conservation of energy, they will have the exact same amount of energy and thus, you need the same amount of counter energy to stop them. For every action their is an equal opposite reaction.

Right, so objects released from the same point will arrive at the same time. The ship stopping at 75000 meters will hit the ground at the exact same time that a ship traveling in an arc with the planets rotation will hit the ground when it's descent begins at 75000 meters. Both ships will land on the planet at the same time. That's because the force of gravity on the two objects is the same and constant. Even if you take it the other way around, a ship descending at an angle of 20 degrees both to and against the planets rotation will arrive on the ground at the same time. This is because the planet's rotational speed has 0% impact on the object that's falling to the ground. Physics for my rocket chair Total Mass = 15.21 tons 13798.3 kg Starting Height = 75000 meters The force of gravity, g = 9.8 m/s2 Gravity accelerates you at 9.8 meters per second per second. After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on. Time to splat: sqrt ( 2 * height / 9.8 ) It's the square root because you fall faster the longer you fall. The more interesting question is why it's times two: If you accelerate for 1 second, your average speed over that time is increased by only 9.8 / 2 m/s. Velocity at splat time: sqrt( 2 * g * height ) This is why falling from a higher height hurts more. Energy at splat time: 1/2 * mass * velocity2 = mass * g * height If object A is descending at a 20 degree prograde arc to the planets rotation it's final speed when it hit's the surface will be 1212.44 m/s and it's descent time will be 123.72 seconds If object B is descending at a 20 degree retrograde arc to the planets rotation, it's final speed when it hit's the surface will be 1212.44 m/s and it's descent time will 123.72 seconds This means that both chairs will hit the ground at the exact same point in time, traveling at the same speed. This means that you would have to slow down 1212.44 m/s before you hit the ground. Therefore, the deceleration amount for both objects is the same, which means you use the same amount of fuel. If object A is rotating around the planet at 2000 m/s counter clockwise, and Object B is rotating around the planet clockwise, they are both orbiting at the same speed. The planet is rotating at 174 m/s Those speeds are constant because the planet isn't accelerating and the force of gravity is equal on both objects. If Both object A and B decelerate 1500 m/s they are going to head towards the surface at the same speed and arrive at the same time. Thus it is possible for a person standing on the surface to see both objects come down towards him at exactly the same speed and land at the exact same time and they will both appear to have fallen from the exact same height. Thus, both pods can land in the exact same spot right next to each other at the exact same time. This is because both pods are moving at the same speed. This is because orbital speed and surface speed are independent of each other. That means that one has 0% effect on the other. In conclusion, objects that deorbit from the same height and orbiting at the same speed will hit the ground at the same time. This means that their Velocities to the ground in both cases are equal. That means that they both have to decelerate the same amount in order to land. That gives us the wonderful conclusion that a craft will consume the same amount of fuel to decelerate the same velocity to ground in both cases because again V = GT

I just didn't think of Eeloo, that's all. In the video, that planet can't exist because the rotational speed is greater than what the planet's gravity can hold together. The thing would fly apart as Scott Manley points out, he says it would be an artificial construct and not a natural body. Actually, if you could tell me the exact amount of time it would take to descend to the surface of Kerban from an orbital altitude of 75,000 meters. Factor this value as if Kerban has no atmosphere just for the sake of negating the effects of aerobraking and wind resistance. Then I can show the point. For your last question, that answer is no. And I never said it was the same for the ascent. When you are on something, you're moving at the same speed it is moving. When taking off from Kerban, you're already moving at 174.6 m/s, you're not starting from 0. If Kerban was stationary than you would take off at the same speed.