The Absurdly Energetic Munar Impact Challenge

[Trekkin]

This is a relatively simple challenge, albeit one that requires Mechjeb (and no other mods). Just smack into the moon with as much momentum as possible, as measured in ton-meters per second. Make Jeb proud. 8)

Obviously, since we\'re relying on Mechjeb mass readings and Mechjeb goes offline on impact, some estimation may be necessary if you\'re hemorrhaging mass on your flight in, but a screenshot of a very low-altitude or post-impact screen would be ideal. Highest impact mass x surface velocity wins!

So let\'s see how massive a rocket we can slam into the Mun at multiple kilometers per second, shall we?

Leaderboard:

1. Excalibur, at 5196.9 m/s x 81 tons = 420,949 ton/ms-1.

2. Maltesh, at 10,464.4 m/s x , 5.6 tons = 58,600 ton/ms-1.

[Excalibur]

I LOVE the sound of this challenge.

EDIT: My entry:

http://youtu.be/ttD5E-nw9mA

[zarakon]

I don\'t think I have any actual record of it, but a while ago I hit the moon at over 6000km/s with a fairly large rocket (stock parts). It kind of glitched and went through the surface.

[Trekkin]

Awesome, excalibur! I\'m afraid I can\'t read the speedometer, though; what\'s your impact momentum?

[Excalibur]

Awesome, excalibur! I\'m afraid I can\'t read the speedometer, though; what\'s your impact momentum?

I thought I\'d included that in the video, apologies if it\'s not that clear.

Final velocity = 1504.9m/s.

Impact mass = 160 tons.

Momentum = 240,784 ton/ms-1.

[Trekkin]

Ah, so you had. My apologies, and you\'re on the leaderboard.

[Excalibur]

Cool beans

[maltesh]

I don\'t think I have any actual record of it, but a while ago I hit the moon at over 6000km/s with a fairly large rocket (stock parts). It kind of glitched and went through the surface.

Same thing happened to to me when I hit the Mun at 10.5 km/s on a stock rocket. Clipped through ~300km of Mun.

[Excalibur]

I just tried the challenge again with a different rocket and also clipped through it at ~5km/s.

As a random aside looking at my previous attempt, assuming all kinetic energy was converted to thermal energy on impact the total energy released would have been ~182 PJ of energy.

Thats about 43 kilotons of TNT, or about twice as powerful as the Hiroshima bomb.

Now that\'s Kerbal!

[Cooly568]

Trying this soon.

[Trekkin]

I just tried the challenge again with a different rocket and also clipped through it at ~5km/s.

As a random aside looking at my previous attempt, assuming all kinetic energy was converted to thermal energy on impact the total energy released would have been ~182 PJ of energy.

Thats about 43 kilotons of TNT, or about twice as powerful as the Hiroshima bomb.

Now that\'s Kerbal!

And that is why the challenge is so named. 8)

Now that I think about it, the challenge may be premature without an addition: if you have video of you clipping through the Mun, your mass and velocity when you first start to clip through is also valid, since it\'d be when you\'d crash anyway.

One hopes this gets fixed soon, though. The challenge just isn\'t the same without seeing the Mun-shattering kaboom.

[maltesh]

Fair enough:

This was the spacecraft.

And this is what it looked like at the lowest pre-clip altitude I still have an image at.

(edit: Whoops, thought I had a different engine on it.)

And at that point, the spacecraft consisted of the gimbaled standard liquid engine, four empty liquid fuel tanks, an empty rcs tank, 3 RCS blocks, a stack decoupler, a parachute, and a Mk 1 pod. Hmm. That\'s only 5.6 tons.

So on the 10,464.4 m/s screenshot, the spacecraft has a momentum of 58,600 kiloNewton-seconds. (Or ton-meters per second, if you prefer)

[Excalibur]

I just tried the challenge again with a different rocket and also clipped through it at ~5km/s.

As a random aside looking at my previous attempt, assuming all kinetic energy was converted to thermal energy on impact the total energy released would have been ~182 PJ of energy.

Thats about 43 kilotons of TNT, or about twice as powerful as the Hiroshima bomb.

Now that\'s Kerbal!

I have no idea how I got 182 PJ. My senses must have taken leave of me... I looked at the calculations again and got a more reasonable figure of 181 GJ. :-[

Nowhere near the power of an atomic blast... but still got enough boom to tickle your cockles.

Perhaps you could add an extra prize for the first person to get a kinetic energy on impact of over 1 kiloton of TNT equivalent (4.184 TJ according to Wikipedia).

I do have a new entry though:

Final image taken ~500m above the Munar surface. Engines off, final mass 81tons. Final velocity 5196.9m/s

Impact momentum = 81 x 5196.9 = 420,949 tons/ms-1

Works out about a quarter of a kiloton. It\'ll be hard to get 4 times the energy. I\'ll mean either doubling velocity or doubling mass!

[Trekkin]

Two hundred and fifty tons of TNT is not a bad energy equivalent. ;D

[SunJumper]

If you squeezed out the rest of your RCS, you would have gotten a higher impact momentum.

[Excalibur]

If you squeezed out the rest of your RCS, you would have gotten a higher impact momentum.

Good point. Though I wonder if the extra velocity would cancel out the mass lost of the RCS propellant... I\'m tempted to say no and guess that using RCS propellant as impacting mass would give more of a benefit than as reaction mass. Just a guess though

[CriminallyVulgar]

I would expect that using the RCS is better, just from how powerful it is in general, but it is a lot of mass you\'re trying to accelerate. The only real ways of testing it are functionally the same as just doing the crash again with and without the extra kick.

Also, I\'m pretty sure you\'d have to quadruple the mass for a kiloton, 0.5mv^2 and all that.

Anyway, sounds like a fun challenge, I throw my hat in with a joke lander I built today, turned out to be unnecessarily heavy at 32.2 tons w/o fuel, but that\'s exactly what this challenge needs!

I got it up to 5423.7m/s with the original slightly overpowered (slightly underkerbal?) rocket that got it to minmus with fuel to spare. I\'ll probably try this again with moar power on the final stage. I bet some SRBs would do the job...

Anyway, 32.2 * 5423.7 = 174,643 ton m/s

In Excalibur terms, that\'s about 0.103 kilotons. Increasing the energy seems a more interesting challenge to me, but I\'m not sure how well the physics in KSP work with that goal (I\'m guessing the answer would always be to drain all your fuel for a heavy impact). It means that maltesh\'s entry, while having a momentum less than 1/3 of mine, has an energy of 0.066 kilotons, damn near 2/3 of mine, due to the high velocity. The momentum challenge does certainly seem to favour the heavier crafts.

~~~Edit~~~

I have a new PB, and it\'s a doozy. A high-velocity landing craft consisting of a pointy cockpit (fer piercin\' them Munrocks), 6 LFTs (1 too many, it turned out, I had half a tank left on impact), a liquid engine, and 104 landing wheels (for safety), with total impact mass of 58.2 tons rendezvous\'d with the Mun at a velocity of 8299 m/s.

That\'s 483,001 ton m/s.

Also, the energy on impact was 0.434 kilotons.

It\'s unfortunate I couldn\'t use the last half tank, I reckon I\'d have cracked 500 ktm/s if I\'d started my burn before entering mun SOI, or if I hadn\'t (foolishly) used the vectoring liquid engine. I use the previous stage to line everything up, I really don\'t need the vectoring. The real key with this was the landing craft. What started as a joke ended up being a really densely weighted dart, surprisingly good for this challenge. Unfortunately, it lags the hell out of my game, and it would be utterly uncontrollable at ~1fps without mechjeb. Also, real Jeb was on board, so it couldn\'t possibly go wrong. I really need something more... single-bodied... to smash into this rock.

Anyway, have fun trying to beat this mark! I know I will.

[Excalibur]

Nice work mate! Someone finally beat my score! Looks like ITS ON!

Also, I\'m pretty sure you\'d have to quadruple the mass for a kiloton, 0.5mv^2 and all that.

Nope, to attain four times the energy you only need to double either the mass or velocity since the whole lot is squared.

Example:

0.5*10*10^2=2500

0.5*20*10^2=10000 (or four times the energy)

Be careful with your units, I noticed you mentioned ktm/s - I\'m sure you mean tonsms-1.

In Excalibur terms, that\'s about 0.103 kilotons. Increasing the energy seems a more interesting challenge to me, but I\'m not sure how well the physics in KSP work with that goal (I\'m guessing the answer would always be to drain all your fuel for a heavy impact). It means that maltesh\'s entry, while having a momentum less than 1/3 of mine, has an energy of 0.066 kilotons, damn near 2/3 of mine, due to the high velocity. The momentum challenge does certainly seem to favour the heavier crafts.

Lifted from wikipedia are two relevant equations;

Where Ek is the kinetic energy in joules, m is mass in kilograms, v is velocity in metres per second and p is your momentum in kgms-1.

In our case when inputting our momentum into the second equation remember to convert from tons/ms-1 to kgms-1. When giving my kiloton figures for impact energy I make the assumption that all kinetic energy is converted to thermal energy in the impact.

I had a look over your calculations and the energy of your impact was actually a bit higher than you found at 0.47902 kT (taking 1kT as 4.184x10^12J). Good job, getting closer to the magical 1kT mark there! Even though you beat my momentum by a hair, you smashed my energy on impact (0.26kT); as you pointed out the momentum and energy challenges are in fact two different, but inter-related goals.

In light of this I\'m going to start a new challenge thread I think... Hope to see you there!

EDIT: Here\'s the new challenge thread: http://kerbalspaceprogram.com/forum/index.php?topic=14037.0

[maltesh]

Minor point: kg*m/s and ton*m/s (mass multiplied by velocity) are units of momentum, not kg/ms-1.

[Excalibur]

Minor point: kg*m/s and ton*m/s (mass multiplied by velocity) are units of momentum, not kg/ms-1.

You\'re right, typo there. Got it right in earlier posts! Thanks for pointing it out!

Typical that I make an amateur mistake with my units whilst telling someone to watch their units!

[CriminallyVulgar]

Sorry dude, but I think you\'re using the equations wrong. 0.5mv^2 means (0.5)*(m)*(v*v), not, as you seem to be saying (0.5*m*v)*(0.5*m*v).

Even in the case of (p^2)/2m, because p = mv, it becomes, (m*m*v*v)/2*m, the 2nd m on the top cancels with the one on the bottom and you\'re left with the same eqn as above - linear in mass, quadratic in velocity. When I use either eqn, I still get 0.434 kT, so I\'m not sure where 0.479 comes from.

And in the case of ktm/s, I was just intruducing a scaling factor, using kilotons instead of tons, it didn\'t occur to me to actually look up a shorthand notation, I just used what first came to mind ( t for tons, T makes a lot more sense now that I\'ve seen you use it)

I will say, though, I competely forgot the 2nd version of the kinetic energy eqn existed, a mistake I frequently make, it\'s a much quicker way of getting the energy when you already have the momentum...

I\'m sorry if there\'s a tone of assholery in any of the above, it\'s just that I use these relations on a fairly regular basis, so if I\'m interpretting them wrong, it\'s really, really bad for my degree prospects.

The energy challenge looks fun though, I have a plan for it that might make for a good entry in both challenges, and maybe also another one I\'ve been looking at...

[Excalibur]

Sorry dude, but I think you\'re using the equations wrong. 0.5mv^2 means (0.5)*(m)*(v*v), not, as you seem to be saying (0.5*m*v)*(0.5*m*v).

Even in the case of (p^2)/2m, because p = mv, it becomes, (m*m*v*v)/2*m, the 2nd m on the top cancels with the one on the bottom and you\'re left with the same eqn as above - linear in mass, quadratic in velocity. When I use either eqn, I still get 0.434 kT, so I\'m not sure where 0.479 comes from.

And in the case of ktm/s, I was just intruducing a scaling factor, using kilotons instead of tons, it didn\'t occur to me to actually look up a shorthand notation, I just used what first came to mind ( t for tons, T makes a lot more sense now that I\'ve seen you use it)

I will say, though, I competely forgot the 2nd version of the kinetic energy eqn existed, a mistake I frequently make, it\'s a much quicker way of getting the energy when you already have the momentum...

I\'m sorry if there\'s a tone of assholery in any of the above, it\'s just that I use these relations on a fairly regular basis, so if I\'m interpretting them wrong, it\'s really, really bad for my degree prospects.

The energy challenge looks fun though, I have a plan for it that might make for a good entry in both challenges, and maybe also another one I\'ve been looking at...

No assholery detected at all! I do actually enjoy being proven wrong, because when I am, it means I\'ve learnt something.

I must apologise for my earlier post - you\'re obviously an intelligent fellow and I must have come across as more than a little condescending. Thanks for being patient and polite.

I think I was half right in saying doubling either mass or velocity would quadruple the energy as velocity is indeed quadratic. But as you say mass is linear so doesn\'t apply there and that was my mistake. Looking back at my earlier recalculations it looks like I was indeed squaring the product of mass and velocity. :-[ (I hope I understand it now..?)

It\'s been a while since I did any physics classes and my maths has always been a bit wobbly (probably why I didn\'t pursue a degree in physics - I knew I wouldn\'t be able to hack the mathematics).

Thanks for pointing out where I went wrong, though there is one caveat... would you terribly mind me messaging you in the future to check my maths before I go shouting my mouth off in the forums?

Once bitten twice shy as they say.

[CriminallyVulgar]

Don\'t worry about it. I\'m new to the forum, and might not be around 100% reliably, but I certainly wouldn\'t mind checking stuff from time to time if I\'m about.

Anyway, new record! I\'ll enter it into the energy challenge too. I think I\'m kind of at the limit of my patience with this 'landing gears on a rocket' strategy. My latest iteration has 156 landing gears on it, and makes KSP run at 1fps even when I\'m down to the impact stage. I\'d really rather not use mods, but otherwise this challenge is a bit... irritating...

Anyway, The final section had a mass of 83.9 tons, and impacted at 8016.3 m/s.

Final momentum: well, zero, impact and all that. But fractionally before that, 83.9 * 8016.3 = 672,567 ton m/s

I\'m very pleased to have broken 500 kTm/s, even if it was frame-by-frame in the end!

The energy was 0.584 kT, but I\'ll keep that to the other thread for brevity\'s sake. The two challenges feel very seperate now that I\'m reaching my limits with my current strategy.

[JellyCubes]

Unfortunately, landing gears don\'t actually have any mass. You could change the landing gear mass in the cfg file to 100,000 and it would still be massless.

[CriminallyVulgar]

I did wonder why it wasn\'t harder to accelerate... assumed I was just overthinking it, because I redesigned the rocket each time. Consider my entries moot until I have a less cheaty (and hopefully less laggy) entry.