OhioBob

1.03/1.04 fixed most problems, but there are still parts that are producing way more drag than they should. For example, the large reaction wheel produces many times more drag than it should given its small exposed surface area. Heat shields also produce excessive drag during launch despite the fact they are occluded and enclosed in a fairing. There are likely other examples that I haven't noticed. As of 1.04, these issues still haven't been fixed. From what I've seen, this is possibly the biggest bug that still needs to be addressed. The bug makes launch vehicles using these parts excessively draggy, far more than is certainly intended.

There's also this: Landing Site Prediction on Return from Deep Space Unfortunately that page hasn't been revised since version 1.0. Early testing suggests that, with the new aero, the targeted periapsis altitude should be about 9-10% lower than before; however, I don't have enough results yet to verify that with confidence. Furthermore, there are now additional variables in the game (drag coefficient & air density) that make it impossible to come up with an accurate 'one-size-fits-all' solution to the problem.

Jool's temperature has been revamped since version 1.0; it is now quite cold. And, suprisingly, Jool's atmosphere really isn't all that dense. Although it's atmospheric pressure is very high, the molecular weight of Jool air is only 2.8 g/mol. At zero elevation, Jool's atmospheric pressure is 15 times Kerbin's sea level pressure, but the air is only about twice as dense.

I have some information on the other planets as well, though I don't yet have it in a presentable form. I'll probably post something eventually.

The formula for be found here: http://www.braeunig.us/space/orbmech.htm#maneuver Equation 4.73 for a simple plane change and 4.74 for combined plane change and altitude change.

^This^ is an important point. If you depart Minmus along a trajectory that is parallel to the equatorial plane, your AN and DN will be very close to Kerbin. Instead, when you depart Minmus, you must incline your trajectory so that you cross Kerbin's equatorial plane when you are still far from Kerbin. If Minmus is below Kerbin's equatorial plane when you depart, you should leave Minmus orbit along an inclined trajectory that will take you northward. If Minmus is above Kerbin's equatorial plane, you should leave along an inclined trajectory that will take you southward. In this way you'll reach the ascending node (when traveling northward) or descending node (when traveling southward) relatively soon after leaving Minmus and while you are still a long distance from Kerbin. However, you don't want to incline your trajectory too much. Although you gain by performing the plane change farther from Kerbin, you lose by having to perform a larger plane change. Somewhere in the middle is an optimum balance, which is something you'll have to figure out.

The Stayputnik also lacks SAS capability, so adding reaction wheels doesn't help because there is no way to control them. The only way to steer a rocket with a Stayputnik is engine gimballing or control surfaces. On the other hand, neither the QBE nor the OCTO2 have reaction wheels, but they do have SAS capability. In this case adding separate reaction wheels will provide attitude control. The HECS and OKTO have both reactions wheels and SAS, so separate reaction wheels aren't needed unless the extra torque is required.

Orbital Vagabond, I'll grant you that the units in the SI system are more intuitive and easier to learn, but that's not what I was talking about. The point I was making was about problem solving. I'm assuming that a person is already familiar with the particular units that are customary in his/her particular field or industry. In that case it is no more difficult to solve a problem in imperial units than it is in SI units. Each system has base units, which are typically what you use in problem solving. In SI the base units are meter (length), kilogram (mass), and second (time). Force is a derived unit, the Newton. In the imperial system the base units are foot (length), pound (force), and second (time). Mass is a derived unit, the slug. That is really all you need to know. If you are given a quantity in other units, you just have to convert it to the base units. Although there are many different units in the imperial system, the number that a person has to become familiar with in normal practice with is rather few. And if imperial units are what you work with every day, the units become second nature.

How so? Either system requires proper training and good habits. Most of the confusion with imperial units is the result of poor training and habits. That has nothing to do with the system of units itself. Using meters, newtons and kilograms is no different than using feet, pounds and slugs. Just different names for the same thing.

That's not true if you use imperial units correctly. The base unit of mass in imperial units is the slug. You just have to get into the habit of converting pounds-mass (or tons, or kips, or ounces, or whatever unit you're given) into slugs. Once that is done, solving a problem in imperial units is no different than solving it in SI units. The only thing I find inherently easier about SI it that units are in powers of 10. It's much easier to convert metric tons into kilograms than it is to convert imperial tons into slugs. The one thing that I do have a major gripe about when working with imperial units is that often an equation is modified to include the pound-mass to slug conversion. This is done so that pounds-mass can be entered directly into the equation without first converting to slugs. I think this is a really bad practice that leads to bad habits and a misunderstanding of how to properly use imperial units. People need to be taught that the proper unit of mass when solving problems in imperial units in the slug, no exceptions. Once that becomes engrained into your thought process, imperial units are easy.

Isp is the ratio of the thrust (measured in N or lb) to the flow rate of the weight ejected (N/s or lb/s), standardized to earth gravity (go). The units of force cancel out and we're left with seconds. Multiplying Isp by go gives effective exhaust gas velocity (m/s or ft/s). Multiplying effective exhaust gas velocity by the mass flow rate (kg/s or slug/s) gives thrust (N or lb).

I studied all the engines and came to the conclusion that, when factoring in both ISP and TWR, the best ones are, in order 1) S3 KS-25x4 "Mammoth" Liquid Fuel Engine 2) LFB KR 1x2 "Twin-Boar" Liquid Fuel Engine 3) RE-M3 "Mainsail" Liquid Engine 4) T-1 Toroidal "Aerospike" Liquid Fuel Engine Mammoth and Twin-Boar have about the same TWR, but the Mammoth has better ISP. Twin-Boar and Mainsail have about the same ISP, but Twin-Boar has better TWR. Aerospike is a close second to Mammoth in ISP, but its TWR sucks.

Your premise is wrong right from the start. Under the old aero model, every vehicle, regardless of its size and configuration, had the exact same ballistic coefficient (it was a universal constant in the game)*. Also, the air density at any given altitude on any given planet was a constant. It was therefore possible to compute an exact and consistent value of the terminal velocity at a given altitude. This is no longer true. Ballistic coefficient is now dependent on the shape, size, and speed of the vehicle - it is highly variable. Furthermore, air density is now a function of temperature, which varies by latitude and time of day. All these variable factors now make it impossible to produce a table or graph of terminal velocity with any reliability. As Murph explained, the scale heights listed in the Wiki article are just a rough approximation, and not even a very close approximation in some cases. Scale height now varies with altitude. It is no longer possible to compute pressure using the old method. If an approximate value of pressure is close enough, I recommend the following: Each planet's Wiki article now contains a table of altitude (Z) and pressure (P). Create a graph in which you plot LN(P) versus Z. Fit a trendline to the data points so that you now have LN(P) as a function of Z. The pressure is simply the exponential of LN(P), that is, P = EXP(f(Z)). (edit) * Correction, ballistic coefficient was a constant for a given drag coefficient. However, drag coefficient varied very little from one vehicle to another, with a value of about 0.2. For most computations it was assumed that Cd = 0.2, thus the ballistic coefficient was also assumes to have a constant value of 625 kg/m2.

I had the same problem yesterday, I keep getting the "You cannot warp when the vessel is under acceleration" message. Still haven't figured it out.

Imperial units are easy if that's what you used your entire life.

^ That + Precise Node

That has changed since v1.0. It's temperature at the datum level is now a very cold 200 K.

It is usually only the big pump-fed engines used in launch vehicles that use regenerative cooling. Many spacecraft engines use ablative cooling. This is a very common practice with pressure-fed hypergolic systems. Some very small engines even use radiation cooling.

That depends on what other assumptions we make. Do you mean ten times larger diameter while having the same mass? Ten times larger diameter while having the same density? Ten times larger diameter while having the same surface gravity? The latter of these scenarios would be the most lifelike, so let's go with that one. In that case many of the planets/moons would be large enough to retain atmospheres, even Duna. The minimum molecular weights for each body become, [TABLE=width: 250] [TR] [TD]Jool[/TD] [TD]0.33 g/mol[/TD] [/TR] [TR] [TD]Tylo[/TD] [TD]3.3[/TD] [/TR] [TR] [TD]Eve[/TD] [TD]3.5[/TD] [/TR] [TR] [TD]Laythe[/TD] [TD]3.9[/TD] [/TR] [TR] [TD]Kerbin[/TD] [TD]5.9[/TD] [/TR] [TR] [TD]Vall[/TD] [TD]22[/TD] [/TR] [TR] [TD]Duna[/TD] [TD]30[/TD] [/TR] [TR] [TD]Eeloo[/TD] [TD]38[/TD] [/TR] [/TABLE] Any body not listed is too small to have any chance of retaining an atmosphere. (ETA) This assumes that each planet is subjected to the same solar constant. If the sun is also scaled up in size, then I'm assuming the planets' distances from the sun are scaled up to compensate.

Although Mars is 5 times more massive than Eve, because of the unrealistic planet densities in KSP, Eve's escape velocity is just slightly less than Mars (4.8 km/s vs. 5.0 km/s). Performing the same calculation for Mars I come up with a minimum MW of 22 g/mol, which makes sense because Mars' atmosphere contains nitrogen (28 g/mol). I'm also not suggesting that Eve would have an atmosphere. I'm just saying that mathematically it has enough gravity to hold onto some heavier gases. I believe that what atmosphere Eve could have initially retained would have likely been stripped away by the solar wind, in the same way that Mars lost most of its atmosphere. If the bodies in KSP were real life worlds, the only one that I think would have an appreciable atmosphere is Jool.

I didn't put Duna on the list because after Kerbin it's not even close. The minimum MW for Duna works out to be 298 g/mol. Vall is actually next best with MW = 223 g/mol. Being close to the sun, Eve obviously has a high black body temperature (327 K), but it also has the second highest escape velocity in the game (after Jool). The high escape velocity is what allows it to hold onto some heavy gases. If Eve where the size of Kerbin, then its minimum MW would be 69 g/mol.

KerikBalm is correct; it's escape velocity that is the key. The rule of thumb that I've heard is that the average speed of a gas molecule has to be no more than about 10% of the escape velocity to be retained over an extended period of time. If the average speed is much more than that, there will be enough molecules exceeding escape velocity that, over billions of years, the gas will slowly be lost. The average speed of a gas molecule is given by the root-mean-squared velocity, which is function of temperature and molecular weight. If we know the escape velocity and temperature, we can compute the minimum molecular weight molecule that a planet can hold onto. For temperature, we can use the sun's luminosity of 3.16×1024 watts and calculate the black body temperature. I estimate the following minimum molecular weights: [TABLE=width: 250] [TR] [TD]Jool[/TD] [TD]3.3 g/mol[/TD] [/TR] [TR] [TD]Tylo[/TD] [TD]33[/TD] [/TR] [TR] [TD]Eve[/TD] [TD]35[/TD] [/TR] [TR] [TD]Laythe[/TD] [TD]39[/TD] [/TR] [TR] [TD]Kerbin[/TD] [TD]59[/TD] [/TR] [/TABLE] Clearly Jool can retain a thick atmosphere. Tylo and Laythe might be able to retain some heavy gases, such as carbon dioxide (44 g/mol). Although Eve can theoretically hold onto heavier gases, it's closeness to the sun also means it must deal with a strong solar wind. I doubt Eve would retain an atmosphere. Kerbin really has no chance.

Those are some interesting observations. It's really not that hard to figure out how much pressure should vary with height for each atmosphere. All that's needed are the following information: 1) Planet radius 2) Surface gravity 3) Surface pressure 4) Molar mass of atmosphere 5) Temperature-height profile All this is given, though the temperature-height profile varies based on latitude and time of day. I derived an approximate global mean temperature-height profile for each body. From the surface conditions I extrapolated the pressure-height profile as if we were dealing with real life planets. As an arbitrary upper boundary, I stopped when the pressure dropped to 1 Pa. In cases were the atmosphere extended beyond the limit defined in the game (e.g. 70 km for Kerbin), I assumed the temperature remained constant at the last known boundary value (assuredly not true in real life). Below are the altitudes at which the atmospheric pressure becomes 1 Pa. [TABLE=width: 250] [TR] [TD]Eve[/TD] [TD]41.3 km[/TD] [/TR] [TR] [TD]Kerbin[/TD] [TD]85.5 km[/TD] [/TR] [TR] [TD]Duna[/TD] [TD]beyond SOI[/TD] [/TR] [TR] [TD]Jool[/TD] [TD]2,570 km[/TD] [/TR] [TR] [TD]Laythe[/TD] [TD]107 km[/TD] [/TR] [/TABLE] As you can see, although Eve has the second deepest atmosphere in the game, in real life it would actually have the shallowest atmosphere. Eve's scale height is very small due in part to its high surface gravity (1.7 g) and to its high molar mass (43 g/mol). Also, despite Eve's high surface temperature, its upper atmosphere (above ≈20 km) it quite cold. The real oddball is Duna. For some reason Squad gave it a very unrealistic molar mass of 14 g/mol. In real life a small terrestrial planet like Duna could never retain such lightweight gases. This low molar mass, coupled with its low gravity, means that Duna has a very large scale height – about 50 km at its surface. This gives Duna a very deep atmosphere. I extended the atmosphere all the way to Duna's sphere of influence and the pressure had dropped to only 2.8 Pa at this extreme distance. Obviously in real life Duna could never have an atmosphere this thick – the solar wind would surely strip most of it away. If we gave Duna a more Mars-like molar mass of 43 g/mol, its atmosphere would extend to 180 km (1 Pa level), which would still be the second deepest in the game. If we also gave Duna a Mars-like surface pressure of 6.36 mbar, then its atmosphere would extend to about 97.5 km.

The equation is a cubic Hermite spline, which has the basic form p(t) = (2t3 - 3t2 + 1) po + (t3 - 2t2 + t) mo + (-2t3 + 3t2) p1 + (t3 - t2) m1 Between the first and second keys we have po = 300 p1= 270 mo = 270 - 300 = -30 m1= ((270-300)+(0.001-270)/(7-1))/2*(1-0) = -37.4999 and, t = (pressure-0)/(1-0) For example, if the pressure is 0.35 atm, we have t = (0.35-0)/(1-0) = 0.35 p(0.35) = (2*0.353-3*0.352+1)*300 + (0.353-2*0.352+0.35)*-30 + (-2*0.353+3*0.352)*270+(0.353-0.352)*-37.4999 = 290.0972 Between the second and third keys we have po = 270 p1= 0.001 mo = ((270-300)+(0.001-270)/(7-1))/2*(7-1) = -224.9995 m1= 0.001 - 270 = -269.999 and, t = (pressure-1)/(7-1) For example, if the pressure is 2.6 atm, we have t = (2.6-1)/(7-1) = 0.26666667 p(0.26666667) = (2*0.266666673-3*0.266666672+1)*270 + (0.266666673-2*0.266666672+0.26666667)*-224.9995 + (-2*0.266666673+3*0.266666672)*0.001+(0.266666673-0.266666672)*-269.999 = 204.4535

According to Physics.cfg, the solar constant at Kerbin is 1360 W/m2. From this we can determine that Kerbol's luminosity is 3.16×1024 watts. Using the size of our Sun's habitable zone as a guideline, and adjusting for the difference in luminosity, we can estimate that Kerbol's habitable zone extends from roughly 10 million km to 20 million km. This is roughly from the orbit of Eve to the orbit of Duna. The only planet with enough mass to have a thick atmosphere is Jool. Based on their mass and temperature, Eve, Tylo, and Laythe might be able to retain a very thin atmosphere of heavy gases (e.g. carbon dioxide). Although Kerbin is larger than either Tylo or Laythe, its warmer temperature makes it less likely to be able to hold onto an atmosphere.