OhioBob

50,774.6

I just kept burning retrograde until I killed nearly all my velocity (both horizontal and vertical). I then free fell from about 7-8 km AGL until I started a final breaking burn at about 2.5-3 km AGL. The stage I used for final breaking had a TWR of about 3.5. This higher TWR stage was also used at the very end of the suicide burn, killing the last 250-300 m/s. Using this method my total landing ÃŽâ€v was about 3400 m/s. My ascent/reorbit ÃŽâ€v was right around 2640 m/s, for a total of 6040 m/s. I'm sure that if I used a higher TWR for the majority of the suicide burn, and started from a lower altitude, I could get down to your 5900 m/s or maybe even a little less.

I just successfully completed my first manned launch from Eve. It took what essentially amounted to eight stages, giving me a total ÃŽâ€v of about 12.8 km/s. In simulations I determined it could achieve Eve orbit from an elevation as low as 500 m, though I launched it from about 900 m. From that elevation I got into a 120 km orbit with 200 m/s to spare. All I launched was a Kerbal in a command seat and my total liftoff mass was 83 tonnes (the landed mass was 91 tonnes). The total delivered to orbit was about 320 kg. I also did an unmanned practice launch using a smaller design (also capable of 12.8 km/s). The unmanned version delivered a little over 200 kg to orbit with a launch mass of 51 tonnes. In both cases I'm getting a payload fraction of only about 0.004. There are probably others who have done better than that, but the point is, it takes a huge vehicle to launch a small mass from Eve. Each additional m/s of ÃŽâ€v is very, very hard to attain. (edited to add) The 12.8 km/s ÃŽâ€v is based on vacuum Isp. It's more like 12.1 km/s in actual practice.

Important point. Performing a suicide burn at periapsis of an elliptical orbit is really no different in terms of ÃŽâ€v than first circularizing and then performing a suicide burn. The only difference is that in the first case you burn right through the point where the orbit is circularized, while in the second case there is a pause in the action. The advantage or circularizing first, as Warzouz said, is that you can then select any site along your orbital path to land. In the other case you're pretty much committed to landing at the site of the initial periapsis. I think that's true of everyplace but Tylo. Orbital velocity around Tylo is much higher than any other airless body, which means that the duration of the suicide burn is much longer than other places. Unless you have a very high TWR, this can result in a considerable loss of altitude during the burn. The initial altitude must be high enough to account for this. For my one and only Tylo landing, the stage I used for the suicide burn had an initial TWR of only about 1. I started the burn at an altitude of about 47 km and was down to about 8 km when the burn ended. If I had to do it again, I'd consider increasing the TWR.

I generally classify as "light" any launcher made exclusively of 1.25 m parts. As soon as I start adding 2.5 m parts I classify it as "medium". When I start adding 3.75 m parts it becomes "heavy". I consider "super-heavy" to be launchers that start adding additional 3.75 m parts attached radially. When you say "typical mass" do you mean the total launch mass or the mass of the payload (i.e. lift capacity). Although there are no precise lines of demarcation, my light launch vehicles generally have lift capacities of 5 tonnes or less; my medium launchers lift between 6 and 30 tonnes, my heavy launchers between 30 and 80 tonnes, and my super-heavy launchers >80 tonnes. In regard to total launch mass, light is approximately <40 tonnes, medium 40-200 tonnes, heavy 200-450 tonnes, and super-heavy >450 tonnes.

I use the TT-38K whenever I can because it is cheaper and lighter than the others. In cases where I need more room, or I want to extend my footprint for greater stability, I use the TT70. In the past I used the Hydraulic Detachment Manifold for heavy attachments, but I've found that using the TT-38K or TT70 with Separatrons and extra struts is usually lighter and can even be cheaper in some cases. (I think the mass of the Hydraulic Detachment Manifold needs to be rebalanced to make it more viable.) I've never used the Structural Pylon.

My launch vehicles I save as subassemblies and I'm not too original at naming them. I just give them a number designation to signify their lift capacity. For example, my 30-tonne launch vehicle is simply named LV-30. I'm not much more creative when it comes to naming my robotic payloads. I typically just name them to indicate the destination and mission type, i.e. Mun Rover, Minmus Lander, Duna Orbiter, etc. I also include a Roman numeral (or a number/letter combination) to indicate the number in a series, or the generation of spacecraft. My manned missions is where I use more colorful names. I generally name them after constellations or stars. I haven't done much with stations and bases other than to deploy them just to fulfill contracts. I generally name stations using Greek letters, i.e. Alpha Station, while bases I name after the planet or moon, i.e. Eve Base I. I've been thinking of trying to come up with a more formalized naming system for all my payloads, but I haven't settled on anything specific yet.

I've made some good progress since I last answered this poll. I've now landed on and returned from every planet and moon in the Kerbol system. 2 + 4 + 4 + 24 + 10 + 16 + 0 + 4 + 120 = 184 / 20 = 9.20

I believe it's called a "Materials Bay" in the contract descriptions.

Regrettably I've found an error in one of my formulae used to derive the periapsis altitudes. I've corrected it and have revised all the numbers in the opening post. These new numbers should produce a more accurate landing site prediction. I apologize to anyone who may have used the old numbers.

Or move your rescue craft to the Kerbal. I still haven't gotten the knack for controlling Kerbals on EVA. Instead, I just slowly maneuver my ship to the Kerbal and, as soon as he's within reach of the ladder, I switch the the Kerbal and immediately grab hold.

The numbers that I generally budget for a Mun mission are, [TABLE=class: outer_border, width: 400] [TR] [TD]Launch to Kerbin orbit[/TD] [TD=align: right]4,550 m/s[/TD] [/TR] [TR] [TD]Injection into trans-Mun trajectory[/TD] [TD=align: right]860 m/s[/TD] [/TR] [TR] [TD]Mun orbit insertion[/TD] [TD=align: right]260 m/s[/TD] [/TR] [TR] [TD]Deorbit, descent and landing[/TD] [TD=align: right]875 m/s[/TD] [/TR] [TR] [TD]Ascent to Mun orbit[/TD] [TD=align: right]700 m/s[/TD] [/TR] [TR] [TD]Injection into trans-Kerbin trajectory[/TD] [TD=align: right]270 m/s[/TD] [/TR] [TR] [TD]TOTAL[/TD] [TD=align: right]7,515 m/s[/TD] [/TR] [/TABLE] As others have said, these numbers can vary depending on how efficiently you perform each maneuver. For example, I'm pretty good at getting to Kerbin orbit, rarely exceeding the 4,550 m/s minimum; however, the technique I use to land on Mun is not the most efficient, though neither is it the least efficient. I use a method that I'm comfortable with and just accept the fact that I must budget a little extra ÃŽâ€v.

No, go is always 9.81 m/s2 (though it is my understanding that in KSP it actually works out to 9.82 m/s2 when back calculated). It has to do with the defined relationship between Newtons and kilograms. It has nothing to do with local gravity. - - - Updated - - - For the record, I have a few simple guidelines that I like to follow when designing a rocket that work very well. I describe those guidelines and provide an example in the following post: http://forum.kerbalspaceprogram.com/threads/107763-Designing-Launch-Vehicles?p=1678756&viewfull=1#post1678756

I use those charts for circular orbits, they work nicely. They were part of my inspiration to develop a similar technique from hyperbolic orbits. The map with the scale on it comes in really handy. I keep a printout of it nearby and refer to it every time I'm doing a reentry from Mun, Minmus, or interplanetary space. The fact that it is graduated in 15 minute increments makes it really easy to use.

That probably wouldn't be too hard to do. I already have the computer simulations to figure it out, it's just a matter of putting in the time. Personally, I've never done a direct entry from deep space on any planet other than Kerbin. Every time I've landed elsewhere, I've always entered orbit first. For precise landings from orbit, somebody else has already figured that out. Before I put any further effort into it, I think it would be wise to wait for the revised aerodynamics in 1.0. Anything I do now will soon become obsolete.

Over the course of the last couple months I started to develop and use a method to predict the location of my landing site on Kerbin when returning from deep space. I just recently formalized the technique by producing some useful charts and equations. I figured I might as well share it with the rest of the KSP community. Perhaps somebody else here might find it useful. When a spacecraft is returning from deep space (i.e. interplanetary space) and on course for a direct entry into the Kerbin atmosphere, we have only limited control over where the spacecraft will land. It is nice to have some idea of where we're going to land while still a considerable distance from Kerbin. Being able to predict the landing site lets us know in advance whether we'll be descending into a safe area or a hazardous area. If we know the spacecraft will be descending onto the ocean or a flat plain, then we can relax and not have much to worry about. However, if the prediction shows that we're likely to descend onto the steep slopes of a mountain, then we likely need to alter our trajectory the best we can to avoid potential disaster. The method presented here allows us to deduce and, to some degree, control the location of the landing site while near the edge of Kerbin space and still days or hours away from entry. The method also works for predicting the landing site for returns from Mun and Minmus. The numbers that follow were generated by performing computer simulations outside of the game environment. In the instances where I've been able to put the technique to use in actual game conditions, the results have matched the theory pretty closely. The simulations are based on stock aerodynamics. Although the basic concept and procedural steps should be valid with any aerodynamic model, the specific numbers presented are valid only with stock aero. Step 1 - Compute hyperbolic excess velocity. To determine the landing site we must know at what speed we're approaching Kerbin. We do this by computing the hyperbolic excess velocity, v∞, of our trajectory within Kerbin's sphere of influence. Shortly after entering Kerbin’s sphere of influence, briefly phase the game and note the spacecraft’s altitude, h, and velocity, v. This can be done at any point along the trajectory, and any pair of h-v values will do, as long as they are taken at the same time. The hyperbolic excess velocity is computed using the following equation: v∞ = [ v2 – (2μ/r) ]1/2 where μ is Kerbin’s gravitational parameter, equal to 3.5316×1012 m3/s2, and r is the radius distance, equal to h + 600,000 meters. Skip this step for returns from Mun and Minmus (or any elliptical orbit within Kerbin’s SOI). Step 2 – Determine target periapsis and alter trajectory. The next step is to establish a trajectory that allows us to predict the flight path of the spacecraft through Kerbin’s atmosphere. The trajectory we choose is that in which the spacecraft’s landing site is directly beneath the periapsis of the orbit. This condition can be achieved by carefully selecting and controlling the periapsis altitude of the trajectory. Periapsis altitude, Pe, as a function of v∞ is obtained using the following graph: The periapsis altitude can also be calculated using one of the following equations: For v∞ ≤ 2000 m/s, Pe = 0.0000001416 v∞3 – 0.0007332 v∞2 + 0.02001 v∞ + 21970 For v∞ > 2000 m/s, Pe = 0.00011174 v∞2 – 1.7053 v∞ + 23174 For a return from Mun, use Pe = 22,380 m, and for a return from Minmus, use Pe = 22,090 m. Now perform a burn to set the periapsis of the trajectory to the altitude determined above (as closely as can be reasonably done). In general, to raise the periapsis burn radially outward, and to lower the periapsis burn radially inward. You can also use this opportunity to adjust the inclination if necessary. I typically like to make sure my trajectory’s swing around Kerbin is as close to the equatorial plane as possible. Completing this step means that the spacecraft is now on a trajectory to land on the surface of Kerbin at a point that is directly beneath the periapsis. You may note that the periapsis moves once the spacecraft enters the atmosphere and starts to experience drag forces. Ignore this; the landing site will be beneath the original periapsis before it was altered by reentry. Step 3 – Determine landing site location. Go to the map view and note the Kerbin terrain that is located directly beneath the periapsis. Since Kerbin rotates, this is not the terrain that will be beneath the periapsis at the time the spacecraft arrives. To determine this we must also note the time to periapsis. In the time to periapsis we can ignore the days, because for every integer day that passes, Kerbin will return to the exact same position. Simply note the hours and minutes. Since Kerbin’s rotational period is six hours, the planet’s surface rotates eastward 60o every hour and 1o every minute. This means that the landing site will be west of the periapsis by the amount that Kerbin will rotate in the time between the present and the arrival of the spacecraft. To visually identify the landing site, use the following map. Each white vertical line on the scale is separated by a distance of 15o, or 15 minutes of time. The distance between the slightly longer lines is 60o, or 1 hour. Identify on the map the current location of the periapsis. From that position, count off to the west (left) the distance represented by the hours and minutes to periapsis. If you reach the left edge of the map, move all the way to the right and count off the remaining distance starting from the right hand edge. The new position found after counting off the appropriate distance will be where the spacecraft will land. Step 4 (optional) – Hazard avoidance and trajectory alteration. If Step 3 reveals the spacecraft is likely to land in a hazardous location, there is only so much that can be done. The landing location can be altered to some degree by raising or lowering the periapsis altitude. Raising the periapsis (decreasing the entry angle) will cause the spacecraft to land long (farther east, assuming a prograde entry). Lowering the periapsis (increasing the entry angle) will cause the spacecraft to land short (farther west). There are limitations, however. If the periapsis is raised too much, the spacecraft will pass through the atmosphere and back into space. If lowered too much, the craft will experience suicidal g-forces (if that even matters to a Kerbal). By simulation I have determined how much the periapsis altitude must be adjusted to land –15o, +15o and +30o from the landing position found in Step 3. The following table gives the vertical distances that the periapsis should be lowered to land 15o short, and raised to land 15o or 30o long. These distances are a function of v∞. Please be advised that this is only theoretical; I just recently produced the table and haven’t had the opportunity to test the results in game situations. [TABLE=class: outer_border, width: 500] [TR] [TD=align: center]v∞ (m/s)[/TD] [TD=align: center]Short -15o (m)[/TD] [TD=align: center]Long +15o (m)[/TD] [TD=align: center]Long +30o (m)[/TD] [/TR] [TR] [TD=align: center]0[/TD] [TD=align: center]-20,130[/TD] [TD=align: center]+7,535[/TD] [TD=align: center]+9,735[/TD] [/TR] [TR] [TD=align: center]500[/TD] [TD=align: center]-20,380[/TD] [TD=align: center]+7,535[/TD] [TD=align: center]+9,705[/TD] [/TR] [TR] [TD=align: center]1,000[/TD] [TD=align: center]-21,040[/TD] [TD=align: center]+7,535[/TD] [TD=align: center]+9,605[/TD] [/TR] [TR] [TD=align: center]1,500[/TD] [TD=align: center]-21,950[/TD] [TD=align: center]+7,520[/TD] [TD=align: center]+9,465[/TD] [/TR] [TR] [TD=align: center]2,000[/TD] [TD=align: center]-22,940[/TD] [TD=align: center]+7,480[/TD] [TD=align: center]+9,280[/TD] [/TR] [TR] [TD=align: center]2,500[/TD] [TD=align: center]-23,900[/TD] [TD=align: center]+7,415[/TD] [TD=align: center]+9,085[/TD] [/TR] [TR] [TD=align: center]3,000[/TD] [TD=align: center]-24,740[/TD] [TD=align: center]+7,330[/TD] [TD=align: center]+8,885[/TD] [/TR] [TR] [TD=align: center]3,500[/TD] [TD=align: center]-25,460[/TD] [TD=align: center]+7,240[/TD] [TD=align: center]+8,695[/TD] [/TR] [TR] [TD=align: center]4,000[/TD] [TD=align: center]-26,060[/TD] [TD=align: center]+7,150[/TD] [TD=align: center]+8,515[/TD] [/TR] [TR] [TD=align: center]4,500[/TD] [TD=align: center]-26,530[/TD] [TD=align: center]+7,055[/TD] [TD=align: center]+8,345[/TD] [/TR] [TR] [TD=align: center]5,000[/TD] [TD=align: center]-26,920[/TD] [TD=align: center]+6,965[/TD] [TD=align: center]+8,195[/TD] [/TR] [/TABLE] Note that the farther we land long of the position found in Step 3, the more sensitive the final landing position is to small changes in periapsis altitude. Furthermore, unpredictability due to variations in the drag coefficient is magnified when the spacecraft follows a longer path through the atmosphere. We should, therefore, expect greater error in the predicted landing position when attempting to land long. Also note that for the -15o scenario, Pe becomes negative for v∞ greater than about 1,135 m/s. In this case the periapsis is below ground level and will not be visible in the map view. Given the above, we have what amounts to a 45o long landing footprint (about 500 km). It should be possible to find a suitable landing site within the available footprint. For returns from Mun and Minmus we have much greater control over the landing site. This is because we can always delay our departure until Kerbin has rotated into a suitable position, thereby allowing us to place our periapsis over whatever terrain we desire. In this way it is possible to obtain landings very close to Kerbal Space Center. Acceleration Loads In the stock game, g-forces are not a concern because there are no detrimental effects from high acceleration loads (or heating). Nonetheless, below I provide the peak acceleration loads that will occur in each scenario. Note that for very high values of v∞ the g-forces are extreme. If you are playing a modded form of the game in which acceleration loads matter, then please take notice and use the necessary caution. In the +30o scenario we have a simple form of a "skip" reentry. That is, the spacecraft descends into the atmosphere and then, for a brief period, begins to rise in altitude before making its final descent. This results in two peaks in the acceleration curve (the larger is given). [TABLE=class: outer_border, width: 500] [TR] [TD=align: center]v∞ (m/s)[/TD] [TD=align: center]Short -15o (g)[/TD] [TD=align: center]Norm 0o (g)[/TD] [TD=align: center]Long +15o (g)[/TD] [TD=align: center]Long +30o (g)[/TD] [/TR] [TR] [TD=align: center]0[/TD] [TD=align: center]8.1[/TD] [TD=align: center]4.2[/TD] [TD=align: center]2.2[/TD] [TD=align: center]2.4[/TD] [/TR] [TR] [TD=align: center]500[/TD] [TD=align: center]8.4[/TD] [TD=align: center]4.4[/TD] [TD=align: center]2.2[/TD] [TD=align: center]2.4[/TD] [/TR] [TR] [TD=align: center]1,000[/TD] [TD=align: center]9.3[/TD] [TD=align: center]4.8[/TD] [TD=align: center]2.4[/TD] [TD=align: center]2.4[/TD] [/TR] [TR] [TD=align: center]1,500[/TD] [TD=align: center]10.9[/TD] [TD=align: center]5.5[/TD] [TD=align: center]2.8[/TD] [TD=align: center]2.4[/TD] [/TR] [TR] [TD=align: center]2,000[/TD] [TD=align: center]13.0[/TD] [TD=align: center]6.6[/TD] [TD=align: center]3.5[/TD] [TD=align: center]2.8[/TD] [/TR] [TR] [TD=align: center]2,500[/TD] [TD=align: center]15.8[/TD] [TD=align: center]7.9[/TD] [TD=align: center]4.3[/TD] [TD=align: center]3.5[/TD] [/TR] [TR] [TD=align: center]3,000[/TD] [TD=align: center]19.1[/TD] [TD=align: center]9.6[/TD] [TD=align: center]5.4[/TD] [TD=align: center]4.5[/TD] [/TR] [TR] [TD=align: center]3,500[/TD] [TD=align: center]23.1[/TD] [TD=align: center]11.6[/TD] [TD=align: center]6.7[/TD] [TD=align: center]5.7[/TD] [/TR] [TR] [TD=align: center]4,000[/TD] [TD=align: center]27.7[/TD] [TD=align: center]13.9[/TD] [TD=align: center]8.3[/TD] [TD=align: center]7.2[/TD] [/TR] [TR] [TD=align: center]4,500[/TD] [TD=align: center]33.0[/TD] [TD=align: center]16.5[/TD] [TD=align: center]10.1[/TD] [TD=align: center]8.8[/TD] [/TR] [TR] [TD=align: center]5,000[/TD] [TD=align: center]38.7[/TD] [TD=align: center]19.5[/TD] [TD=align: center]12.2[/TD] [TD=align: center]10.8[/TD] [/TR] [/TABLE] EXAMPLE Let’s say we are on a return from Duna. After passing into Kerbin’s SOI, we note that the altitude is 83,400,000 m and the velocity is 920 m/s. We compute v∞ as follows: v∞ = [ 9202 – (2 × 3.5316×1012 / (83,400,000 + 600,000)) ]1/2 = 873 m/s We next compute the periapsis altitude of the desired trajectory (or read Pe from the graph), Pe = 0.0000001416 × 8733 – 0.0007332 × 8732 + 0.02001 × 873 + 21970 = 21,520 m We now perform a course correction burn to change our periapsis altitude to 21,520 m. We go to the game map view and identify the terrain features beneath the current periapsis, and note that the time to periapsis is, say, 3 days, 4 hours, 45 minutes. We identify on the following map the current periapsis location and count off 4:45 (285o) to the west to find the projected landing site.

I don't know if the following is what you're looking for, but I just performed an experiment in which I simulated an interplanetary injection from low Kerbin orbit using different thrust-to-weight ratios. I assumed starting from a 70 km circular orbit and maintained the thrust vector pointed in the prograde direction throughout the duration of the burn. I stopped the burn when the hyperbolic excess velocity reached 1,000 m/s, which is about the median for a trip to Duna or Eve (a little less for Duna and a little more for Eve). I assumed a specific impulse of 390 seconds, though Isp really doesn't make that much difference. Below are the results: [TABLE=width: 300] [TR] [TD=align: center]TWR[/TD] [TD=align: center]ÃŽâ€v (m/s)[/TD] [/TR] [TR] [TD=align: center]∞[/TD] [TD=align: center]1101.5[/TD] [/TR] [TR] [TD=align: center]2.00[/TD] [TD=align: center]1102.6[/TD] [/TR] [TR] [TD=align: center]1.50[/TD] [TD=align: center]1103.2[/TD] [/TR] [TR] [TD=align: center]1.00[/TD] [TD=align: center]1104.9[/TD] [/TR] [TR] [TD=align: center]0.75[/TD] [TD=align: center]1107.3[/TD] [/TR] [TR] [TD=align: center]0.50[/TD] [TD=align: center]1113.8[/TD] [/TR] [TR] [TD=align: center]0.30[/TD] [TD=align: center]1133.6[/TD] [/TR] [TR] [TD=align: center]0.20[/TD] [TD=align: center]1167.9[/TD] [/TR] [TR] [TD=align: center]0.15[/TD] [TD=align: center]1209.2[/TD] [/TR] [TR] [TD=align: center]0.10[/TD] [TD=align: center]1301.7[/TD] [/TR] [/TABLE] A TWR of ∞ is, of course, a theoretical instantaneous burn. As you can see, for TWRs greater than about 0.50, the losses are very small. However, for extremely small TWRs, the losses start to become significant. Another problem with small TWRs is the issue of obtaining a good intercept with the destination planet. Low-thrust burns are so long in duration that there can be considerable error between the planned trajectory, based on the maneuver node, and the actual trajectory, based on the injection burn. After performing a high-thrust burn it is typical that only a small correction is required to fine tune the trajectory. Low-thrust burns can require rather significant course corrections, which must be added on top of the losses we see in the table above.

Same here. I would also like to see that option for propellant transfers.

I can't get 50% either, I have to settle for 49.5 or 50.5. The lowest I can get is 5.5%.

None of the above. My first post/thread was to report an error and make a suggestion in the "Suggestions and Development Discussion" forum. Since the post included some "Sciency Stuff", I've selected the second option in your poll.

For a single stage? The highest possible mass ratio for just a propellant tank, without engine or payload, is 9. This gives a theoretical ÃŽâ€v of a little more than 17 km/s. Of course this is impossible to achieve because we must add the mass of the engine and at least a probe core. From my experience, I believe a mass ratio of 5 is about the highest that can be practically achieved. With an LV-N engine, a mass ratio of 5 yields a ÃŽâ€v of 12,630 m/s; however, this also gives a paltry TWR of only 0.27. If we consider just the weight of the LV-N engine and fuel tank, and assume a fuel tank propellant fraction of 8/9, we can compute the ÃŽâ€v and TWR ratio for any given mass ratio. [TABLE=width: 500] [TR] [TD=align: center]Mass Ratio[/TD] [TD=align: center]ÃŽâ€v (m/s)[/TD] [TD=align: center]TWR[/TD] [/TR] [TR] [TD=align: center]2[/TD] [TD=align: center]5,440[/TD] [TD=align: center]1.19[/TD] [/TR] [TR] [TD=align: center]3[/TD] [TD=align: center]8,622[/TD] [TD=align: center]0.68[/TD] [/TR] [TR] [TD=align: center]4[/TD] [TD=align: center]10,880[/TD] [TD=align: center]0.42[/TD] [/TR] [TR] [TD=align: center]5[/TD] [TD=align: center]12,631[/TD] [TD=align: center]0.27[/TD] [/TR] [TR] [TD=align: center]6[/TD] [TD=align: center]14,062[/TD] [TD=align: center]0.17[/TD] [/TR] [TR] [TD=align: center]7[/TD] [TD=align: center]15,272[/TD] [TD=align: center]0.10[/TD] [/TR] [TR] [TD=align: center]8[/TD] [TD=align: center]16,319[/TD] [TD=align: center]0.04[/TD] [/TR] [/TABLE]

I think that's the point when you should start to think about staging, though I've found it practical to run that percentage to about 20% before staging really becomes necessary. Most of my small launchers are in the 25-30% range, but the larger ones (say >100 t total launch mass) are about 20%. Another reason for staging is to limit the acceleration loads, though this is more of a real-life issue than it is a KSP problem. Let's say you have a single stage launcher that is 90% propellant and 10% dry mass. If your initial thrust-to-weight ratio is 1.2, you're TWR will max out at 12 by the time the engine burns off the last of the propellant (assuming fixed thrust). That high a g-load will likely cause your crew to blackout. Staging allows you to not only drop unneeded dry mass, but to also switch to lower thrust engines to manage the acceleration loads. Since all engines in KSP are throttleable, you can avoid this problem by simply dialing back the thrust. Most real-world engines don't have a throttle.

The given properties of Kerbin's atmosphere are inconsistent and couldn't exist together in real life. This makes it impossible to derive a definitive value for the speed of sound. For example, below are two different methods for computing speed of sound. Both methods should yield the same answer but do not. The first method uses the given sea level values of density, pressure and temperature. à= 1.2230948554874 kg/m2 P = 101325 Pa T = 293.15 K From ideal gas law the average molecular weight is M = ÃÂRT/P = 1.2230948554874 * 8314.4621 * 293.15 / 101325 = 29.422 kg/kmol where R is the universal gas constant. Given this molecular weight the atmosphere is almost certainly nitrogen-oxygen, giving us a specific heat ratio of 1.40. Therefore the speed of sound is C = (γRT/M)1/2 = (1.40 * 8314.4621 * 293.15 / 29.422)1/2 = 340.56 m/s The second method uses the scale height of the atmosphere, which for Kerbin is 5000 m. Scale height is given by the equation H = RT/(Mg) Therefore, by substitution we get C = (γHg)1/2 Again assuming γ = 1.40, we obtain a speed of sound at sea level of C = (1.40 * 5000 * 9.81)1/2 = 262.05 m/s The problem appears to be that the game developers selected the scale height as needed to produce the desired pressure-height profile without regard to internal consistency. If we are to believe the sea level values of P, T and ÃÂ, then Kerbin's sea level scale height would have to be 8,445 meters, not 5,000 m as given. We run into the same inconsistency at other altitudes as well. KSP atmospheres are simply not real-world possible.

Depends on your definition of failure. I consider it a failure anytime a primary mission goal it not achieved. This can be as simple as trying to fulfill an orbit contract but failing to get the funds because I did something stupid like leaving off the antenna. I don't consider it a failure if something goes wrong in a test flight, because identifying potential problems and finding the solutions to them is the purpose of a test. I usually revert my test flights back to the VAB after I've gained the data I need. I count it as a failure only if something goes wrong when I'm flying a mission "for real". I also don't need to crash or blow up to have a failure. If I complete 4 out of 5 mission goals, then not completing the 5th is a failure. I don't count game glitches or bugs as failures because those are out of my control. To me failures can be in either design or execution. Failures in design rarely happen for me, perhaps 1 in 100 (after testing). Since I got past the initial learning curve, I can think of only a couple instances were I made a design error or miscalculation. For me, most errors occur in execution. For example, I have a hard landing and sustain damage, or I accidental stage when I wasn't supposed to. If I can work around the problem and still complete my goals, then I don't count it as a failure. However, if the problem causes the termination of the mission, or the scrubbing of some part of the mission, then clearly it's a failure. Although I usually use reverts to correct these mistakes, for the purpose of this poll I'm counting those as failures. In that case I'd say that I make some sort of mission critical execution mistake on perhaps 1 in 10 missions.

Although I can probably handle more if necessary, so far I've maxed out at close to 200 parts. Most of my missions are in the 50-150 range.