capi3101

I would so love to see this fly... I'd join in my own self but the rules specifically say "no mods"...I've got KER default on all my craft these days; it's there whether I want it or not.

I've got a probe carrier en-route in my sandbox game; it will reach Jool's periapsis in three in-game days (it's already in Jool's SOI). It's carrying six "Sandstone" probes - an OKTO2 core w/Z-200 batt, two RTGs, a couple of science packages, an antenna, a 48-7S engine and an FL-T400 tank, 5,697 m/s delta-V in each probe if I've done all the math right - with the goal of putting one probe in orbit of each of Jool's moons plus Jool itself; the carrier will be taking a planetary dive when the job's done. Sounds like I'm trying to do the same thing as the OP; I'm also interested in "how do you do it" and will be comparing notes as time passes. The carrier itself consists of a large RCU w/ Z-4 battery pack, 2 RTGs, an RCS system (due to the mass of the whole thing), 8 BZ-52s w/ 6 stack decouplers (to attach each probe radially; the other two are mountpoints for the engines), 2 LV-Ns w/ a girder setup for attachment to the booster and an orange tank. 5800 m/s of delta-V for the whole shebang. Thrust sucks of course but it did get me to Jool.

Got my six-man Constellation crew off the surface of Duna and drove one of the Hellrider rovers I brought along for the ride a bit. The 'rider's not my favorite rover but it still was more fun to drive it than that damn Hellfury rover I had to schlep along for the sake of the challenge (the 'rider can right itself again if it flips; the 'fury can't). Next up - figure out how to get back to Kerbin from a 85-degree inclined orbit of Duna on a little over 2,000 m/s of delta-V. To boot, there's no Protractor on the craft, so I get to do it the old-fashioned way. No problem.

Well...there's almost always a trade going on between thrust and delta-V; increasing one tends to come at the cost of the other (in general - there may be cases where this is not true). To judge if it's a useful technique in your case, I'd need some data on the lander - its mass, liquid engine, fuel capacity, etc. Without that data, I'm going to have to give you a solid "maybe". It certainly sounds like it might've been a good choice for that Duna lander, though I have to question the need for four boosters...

Let's see...57 tonnes is the payload, right? I'm'a gonna direct the OP to the twin Temstar threads - http://forum.kerbalspaceprogram.com/showthread.php/28248-Is-asparagus-the-best-staging-system-%28might-contain-science%29?p=346702&viewfull=1#post346702 http://forum.kerbalspaceprogram.com/showthread.php/33381-0-20-2-Zenith-rocket-family-%28modernised-for-0-20-x-with-perfect-subassembly%29 Now, in the event that the OP doesn't have all the prerequisite tech for the Zenith series (I didn't catch whether this was a career game or not), I will say that recreating the Zenith VII (which would cover the 57 tonne payload) is reasonably simple affair. You'll need to use the technique Col_Jessep demonstrated in his post and you should not hesitate to use parts clipping to attach the engines (incidentally, an even lower tech way to do that is to use Modular Girder Segments turned on their ends - they allow fuel flow, they're substantially less massive than Tail Connectors and they're Starting Tech; the only downside is that they look like crap). Temstar's asparagus principles (first link) will be of great use here.

At long last, I continued my Constellation Program challenge entry with a 40k rove over the Dunan southern polar ice cap. Apathy, the Hellfury 7 rover sent on the mission, successfully set a new personal total roving distance record - 120 kilometers, breaking the old record held by the twin rovers Malaise and Lack of Prospects (all three, incidentally, were sent to the southern polar cap region on Duna). Lined my crew up for some publicity pics, then started the process of loading them into the return vehicle for the journey back to Kerbin. I ain't done yet, but the important stuff has been finished at this point.

Tier 3...are you referring to General Rocketry/Stability/Survivability, or are you beyond that? I put together a Mün lander just last week that utilized tech from that level... That first image is bogus - I ultimately only had three outboard stacks on the booster. If you go to copy it, you might try four stacks just for good measure/plenty o' thrust.

I've got a bi-elliptic in progress at the moment - its target is not Moho, but rather a heliosynchronous orbit. The big trick of course is to increase the apoapsis of the craft to twelve times the distance that you want for your final periapsis - for Moho, that's about 63,157,659,648 m or higher (roughly out to the orbit of Jool). The plane change burn occurs at the ascending/descending node as normal. Getting the final intersect with Moho - that's the part I have no experience with. I've tried standard Hohmann transfers for Moho - the result of which was pretty much the same as it is for all who attempt that route (vis-a-vis my salad bar). I will say that a bi-elliptic transfer is totally unnecessary for Eeloo - a Hohmann transfer takes long enough as it is. Got a craft en route to there too and I know it'll get there in another year or so. Again, make a plane change at the ascending/descending node.

http://en.wikipedia.org/wiki/Delta-v Mathematically, delta-V is calculated with the Rocket Equation Specialist290 mentioned - it takes the form: delta-V = ln(M/Mo)* Isp *go where delta-V is the change in velocity (a vector that's a dot-product of the scalar component of speed with the vector component of direction), ln is the natural logarithm function (look for it on a scientific calculator, or use =LN() in MS Excel), M is the full mass of the rocket stage, Mo is the dry mass of the rocket stage (i.e. what it weighs when all its fuel tanks are empty), go is standard gravity (9.81 m/s2regardless of what body you're orbiting/launching from) and Isp is the specific impulse of the engine (a way of measuring the engine's efficiency). It's importance, as has been mentioned, is in determining the total magnitude of the changes the rocket may make to its velocity before it runs out of fuel; in the process it determines where a rocket may go given a certain mission profile. There are three main ways of increasing a rocket's delta-V: 1) improving propellant mass fraction (i.e. moar fuel) 2) increasing specific impulse (by selecting an engine combination that increases this value - the main reason nuclear engines are recommended for interplanetary flight) 3) staging (shedding mass that's no longer needed, which has the effect of improving the propellant mass fraction)

Not necessarily...

Hell Kraken sucks - I've lost half a dozen craft to that dirty S.O.B. Only one of them was manned, but that was one too many...

So brute force it, eh? Very well. Set up a spreadsheet to solve the equation for x given values of y, where y is incremented in values of 0.0625 tonnes (the dry mass of an FL-T100 fuel tank, the smallest tank for which the 9:1 assumption holds - I tend to think of fuel requirements in FL-T100 equivalents; e.g. an FL-T400 is 4 FL-T100 equivalents, an FL-T800 is 8 FL-T100s, etc.). And I'll just go ahead and report the results rather than post screenies of the spreadsheet. Okay...for the case where the non-engine dead mass is 0.6 tonnes, the tipping point is somewhere between 14-15 FL-T100 equivalents (so with 14 FL-T100s, you're better with the 48-7S; with 15, you're better with the LV-909). 14 FL-T100s equivalents = a dry mass 0.875 tonnes. 15 FL-T100 equivalents = 0.9375 tonnes. 16 FL-T100 equivalents = 1 tonne, our initial test case. For the case where there is no non-engine dead mass, the tipping point is somewhere between 20-21 FL-T100 equivalents. For the case where the non-engine dead mass is 1.25 tonnes (a Mk1 Cockpit), the tipping point is somewhere between 9-10 FL-T100 equivalents. For the case where the non-engine dead mass is 2.5 tonnes (a Mk2 Lander Can), the tipping point is somewhere between 3-4 FL-T100 equivalents. For the case where the non-engine dead mass is 4 tonnes (a Mk1-2 Command Pod), it's always better to go with the LV-909. So it looks like the higher the non-engine dead mass, the more likely it is that the LV-909 is the better choice. Not a surprise really. The highest non-engine dead mass for which I'm detecting any delta-V advantage whatsoever with the 48-7S is 3.098639 tonnes (3.098640 tonnes favors the LV-909). 21 FL-T100 equivalents is 1.3125 tonnes, 11.8125 tonnes fuel, and 7,330.07 m/s for the LV-909. I should so totally go through this whole exercise with nukes. Anybody want to see that, and what engine should I compare it against?

So in my original derivation, this was the last step I had correct: ln(M48-7S/Mo, 48-7S) = 1.11429*ln(MLV-909/Mo, LV-909) So if I take e of both sides and do it correctly, I should now have: M48-7S/Mo, 48-7S = (M, LV-909/Mo, LV-909)1.11429 If you then sub in the M = 9Mo relationship, you have: 9Mo, 48-7S/Mo, 48-7S = (9Mo, LV-909/Mo, LV-909)1.11429 And now take out the deadmass in each part of the equation: (9Mo, 48-7S + Mdead, 48-7S)/(Mo, 48-7S + Mdead, 48-7S) = ((9Mo, LV-909 + Mdead, LV-909)/(Mo, LV-909 + Mdead, LV-909))1.11429 Set Mo, 48-7S = x and Mo, LV-909 = y to simplify and you have (9x + Mdead, 48-7S)/(x + Mdead, 48-7S) = ((9y + Mdead, LV-909)/(y + Mdead, LV-909))1.11429 Now that should be as far as that can go without putting in specific terms. Okay. So now for the case where Mdead, 48-7S = 0.7, Mdead, LV-909 is 1.1 and y = 1: (9x + 0.7)/(x + 0.7) = ((9 + 1.1)/(1 + 1.1))1.11429 (9x + 0.7)/(x + 0.7) = ((10.1)/(2.1))1.11429 (9x + 0.7)/(x + 0.7) = 5.75519 (9x + 0.7) = 5.75519(x + 0.7) (9x + 0.7) = 5.75519x + 4.02863 3.24481x = 3.32863 x = 1.02583 Let's verify: ln(10.1/2.1) * 9.81 * 390 = 6008.95 m/s ln (((9*1.02583) + 0.7)/ (1.02583 + 0.7)) * 9.81 * 350 = dV ln (9.93249/1.72583) * 9.81 * 350 = dV dV = 6008.98 m/s The difference can be attributed to rounding errors, so this derivation is correct. Hallelujah. For the case where y = 0.5 tonnes, x = .42875 tonnes, so the "tipping point" is somewhere between 0.5 tonnes and 1 tonne. Now I suppose the trick is going to be to figure out exactly where the "tipping point" is - the point at which it becomes better to use one engine over another given cases of x amount of dead mass. Should be a matter of setting x=y, right? Let's try it for the same case: (9x + 0.7)/(x + 0.7) = ((9x + 1.1)/(x + 1.1))1.11429 Oof. Who wants to tackle that equation?

I don't know of any mods that allow you to pre-designate a set of coordinates on a target body; really a scouting mission is your best bet. I know KER includes surface coordinates as part of the data it provides. So you can install it, add a KER flight computer to a small probe lander and use it to get a feel for where you're at while you're landing. Stick one on a rover and you can monitor your position as you travel along; the rover becomes the ground marker when it arrives at its destination.

Ah....thank you. I've been blowing my mind for the last hour or so trying to figure out why nothing was working out right. Gonna take a break and try again later.

Alright - time for a specific case. Let's just go with a basic setup - a Mk1 Lander Can (0.6 tonnes), the fuel tank and the engine. As a reminder, the 48-7S has a mass of 0.1 tonnes and the LV-909 has a mass of 0.5 tonnes. That means Mdead, 48-7S = 0.1 + 0.6 = 0.7 tonnes Mdead, LV-909 = 0.5 + 0.6 = 1.1 tonnes Let's set Mdry tank, 48-7S = X and Mdry tank, LV-909 = Y, and substitute in our masses. We have: (9X + 0.7)/(X + 0.7) = (27.42651Y + 3.04739*1.1)/(Y + 1.1) Alright. Let's multiply through by the divisors: (9X + 0.7)(Y + 1.1) = (27.42651Y + 3.35213)(X + 0.7) Multiply through: 9XY + 9.9X + .7Y + .77 = 27.42651XY + 19.19856Y + 3.5213X + 2.34649 Now, let's get all the terms involving X on one side of the equation and everything else on the other side: 6.3787X -18.42651XY = 18.49856Y + 1.57649 For the case where Y = 1 tonne 6.3787X -18.42651X = 18.49856 + 1.57649 -12.04781X = 20.07509 X = -1.66629 I've got a math error somewhere...let's do some verification. For the LV-909 case with one tonne of dry mass: dV = ln ((9+1.1)/(1+1.1)) * 9.81 * 390 = 5160.83 5160.83 = ln ((9M + 0.7) / (M + 0.7) * 9.81 * 350 4.49552 = (9M + 0.7) / (M + 0.7) 4.49552(M + 0.7) = 9M + 0.7 4.49552M + 3.14686 = 9M + 0.7 4.50448M = 2.44686 M = .543206 So full mass is 9M = 4.88885 Okay - definite error there. I think I see where...over did my first equation.

Both true. The five engine solution is viable - you'd just have less fine control over the throttle. Probably the best that can be done before you have ducts (unless you want to a setup like what I had earlier). It's a good suicide burn setup, I guess.

That's easy - it's the Save directory under wherever you keep your KSP instance. Just copy the files from one machine to the other; everything should be covered. If you're working with mods, you'll have to reinstall the mods of course. Of course, you could just copy over the whole shebang; it'll still work if you do (provided the new machine can run KSP, of course).

Okay, so you're trying to find situations where the delta-V for an LV-909 equals that available from a 48-7S, right? First thing's first then: a 48-7S has a mass of .1 tonnes and a vacuum Isp of 350. For the LV-909, it's .5 tonnes and a 390 Isp. And of course we have the rocket equation: dV = ln(M/Mo) * Isp * Go If we're looking for situations where the delta-V from the two engines is equal, we can just set two rocket equations equal to one another for the same case. Thus: dV48-7S = ln(M48-7S/Mo, 48-7S) * Isp, 48-7S * Go dVLV-909 = ln(MLV-909/Mo, LV-909) * Isp, LV-909 * Go Where dV48-7S = dVLV-909, ln(M48-7S/Mo, 48-7S) * Isp, 48-7S * Go = ln(MLV-909/Mo, LV-909) * Isp, LV-909 * Go Go cancels out, so we're left with ln(M48-7S/Mo, 48-7S) * Isp, 48-7S = ln(MLV-909/Mo, LV-909) * Isp, LV-909 We have the Isp of our engines, so we can plug that data straight in - we're left with: 350*ln(M48-7S/Mo, 48-7S) = 390*ln(MLV-909/Mo, LV-909) Since we have a constant on both sides of the equation, we can simplify - we get: ln(M48-7S/Mo, 48-7S) = 1.11429*ln(MLV-909/Mo, LV-909) Take the inverse natural logarithm (e) of both sides! We get: M48-7S/Mo, 48-7S = 3.04739 *(MLV-909/Mo, LV-909) Now, the key thing about the mass factors in the rocket equation is that there are actually two components to it - a part that handles the mass that changes (accounting for what's changing it mass - fuel) and a part that handles the mass that doesn't (everything else); that's called dead mass, of which the engine's mass is a part. We need to split off those bits from one another; we get: (Mfuel, 48-7S + Mdead, 48-7S)/(Mdry tank, 48-7S + Mdead, 48-7S) = 3.04739 *((Mfuel, LV-909 + Mdead, LV-909)/(Mdry tank, LV-909 + Mdead, LV-909)) You can then use the assumption that Mfuel = 9Mdry tank (which is true for all liquid fuel tanks in KSP except the Round-8 and Oscar- and carry through the constant on the right hand side. You get: (9Mdry tank, 48-7S + Mdead, 48-7S)/(Mdry tank, 48-7S + Mdead, 48-7S) = (27.42651Mdry tank, LV-909 + 3.04739Mdead, LV-909)/(Mdry tank, LV-909 + Mdead, LV-909)) That's about as simple as you can get it without putting in specific terms. I'm out of time right now, but I'll see if we can't apply this equation to a specific case in the near future.

Just pointing something out - if OP doesn't have Fuel Ducts, he has access to neither the 24-77 nor the 48-7S engine. You get the 24-77 with Precision Engineering, a Tier 5 tech, and the 48-7S comes with Fuel Systems, a Tier 4 tech (and the same one that has Fuel Ducts). The -909 really is his best bet at his current tech level; it's got the highest available Isp, ergo the most bang for his buck given x amount of fuel. Only problem with it is that it doesn't generate electricity when it's running.

Drafted into a family for the duration of the week; y'all try not to have too much fun without me. See you guys probably next Monday or so. Last thing I did was test out a Tier 2 Mun lander design. Wound up 30 m/s short of a Kerbin return, traced the problem to a screw-up of the booster design.

Hit Electrics, Fuel Systems, Space Exploration, Advanced Flight Control in that order - with solar panels, you can transmit more and actually afford to screw up your steering a bit more. Small batts and lights are a nice bonus as well (particularly for making landings). Fuel Systems is a consideration not for RCS - you want that one next for Fuel Ducts (allowing asparagus staging and shorter, wider lander stacks); the 48-7S is a nice bonus. Space Ex and Advanced Flight Control have the Pegasus I Mobility Enhancer (ladders) and Probodobodyne OKTO (the first probe core available) respectively; Space Ex also comes with the Thermometer, a Science Part, which is why I recommend it first. All the others you have available are fluff - even Advanced Rocketry (nothing you can do with an FL-T800 that you can't do with two FL-T400s and I just don't care much for the Mk55).

You'll need to do a plane change at the ascending or descending node. I recommend burning just enough to escape Kerbin's SOI, targeting Eeloo, making the plane change and then making a prograde burn with radial adjustments as needed. When you get the encounter, kick back and relax, 'cause it's going to take a while to get out there. I have probes en route; they will get there - just not any time soon.

Alright - some pics. This is the design I proposed above. Had to turn on parts clipping to put it together and then it turned out too flimsy. Also turned out to be wholly unnecessary - three outboard stacks proved sufficient for the task (though I'd go with four just to be safe - as I'll explain momentarily). You can see the remains of the booster in this next pic. My problem there was that I only made the outboard stacks five tanks long instead of six - a mistake that ultimately forced me to scrub the flight entirely. Had there been six tanks, all of the booster engines would've run out of fuel simultaneously. More importantly, there would've been sufficent delta-V that I could've avoided having to finish the orbit burn on the transfer stage engine (which meant I didn't have it to deorbit at the Mün, which ultimately put me 30 m/s shy of being able to finish the mission). The lander - I got the impression that was the confusing bit. Putting the legs out on girders widens the base, which makes it less prone to tipping over on landing. That little trick works best with angled I-beams - you can use that kind of setup to sling small rovers underneath the lander. No batts or panels with this design, of course, so you've got to watch your steering - especially once the mission is solely on LV-909s. That's another thing to consider.

Shouldn't be. Might want to aim the other way, of course...not essential but helpful.