Red Iron Crown

I'm not defending strategic bombing against civilians, but how do you figure that the nuclear bombing of Japan didn't hasten their surrender?

Your math looks good to me (and while I prefer proper superscripts I can read your notation just fine ). It really does take a tremendous amount of energy to change the velocity of something so massive moving so quickly. The good news is, if you're just trying to deflect it from striking Earth and you can get the bomb to it when it's far enough away, only a very small change of velocity is required. If, however, you're trying to capture it and put it into an orbit of your choosing, you're gonna have a bad time.

Actually, I think you're both right. Oberth effect is only related to speed as you say. As MarvinKitFox implies, it is easier to get to a higher speed when orbiting a more massive body than a smaller one. Remember that orbital energy is the sum of potential and kinetic energy, and as the craft moves through its orbit it exchanges kinetic energy for potential energy as it rises and slows to apoapsis, then exchanges potential energy for kinetic energy as it falls and gains speed. The orbital energy remains the same, but the distribution between kinetic and potential changes. So if you add a given amount of energy to an orbit, it will change the speed only at the point the burn was made but will change speed and altitude at every other point. Because there is no energy gain in the total system. Oberth doesn't add any energy to the system, it just allows more energy to be exchanged between ship and exhaust for a given burn. Let's say that a 10m/s dV burn will increase a craft's kinetic energy by 1000KJ if done at periapsis and 100KJ if done at apoapsis. If done at periapsis, the craft gains 1000KJ of kinetic energy and the exhaust loses 1000KJ of kinetic energy, no net change in the kinetic energy of craft + exhaust. If done at periapsis, the craft gains 100KJ of kinetic energy and the exhaust loses 100KJ of kinetic energy, no net change in the kinetic energy of craft + exhaust.

A couple of things I'd like to clarify as I've been thinking about the Oberth effect and its consequences. The Oberth effect isn't some isolated phenomenon that needs to be separately programmed into KSP. It is a consequence of kinetic energy being proportional to the square of speed in Newtonian physics. You cannot have Newtonian physics without also having the Oberth effect. So the Oberth effect must be in KSP, since its physics model is Newtonian. From the rocket equation it can be seen that the only factors that affect dV are starting mass, ending mass, and Isp. If none of those factors change, then dV cannot change. So it should be obvious that speed has no effect on the amount of delta-V for a given craft, i.e. the Oberth effect cannot add or subtract any delta-V. Similarly, delta-V is a measurement of potential acceleration. If a craft increases its speed by 10m/s at a given point in its orbit, it necessarily has spent exactly 10m/s of dV to gain that speed. It doesn't matter at what altitude or speed this occurs, a 10m/s dV burn will always add 10m/s of speed. So the Oberth effect cannot increase the amount of speed gained by spending a given amount of dV. So, if the Oberth effect doesn't increase the amount of dV or get more acceleration from dV what is its advantage? Energy. The orbital energy of a craft is the sum of its potential energy (from altitude) and its kinetic energy (speed). If no dV is spent, this sum is constant at all points in the orbit. Kinetic energy is converted into potential energy and back again as the craft moves through its orbit. To change orbits, energy must be added (or subtracted) to the orbital energy. This is done by spending delta-V. But remember, delta-V is not a measure of energy, but of potential acceleration. It is this distinction that makes the Oberth effect possible. 1 m/s of dV always results in 1 m/s change in velocity, but not all m/s are equal in terms of energy. Because kinetic energy is proportional to the square of speed, the faster you are going the more energy is added by each m/s of speed gained. The speed gained is the same but the energy gained is greater. So, if we look at the kinetic energy change for a 1-ton craft spending 1m/s of dV to accelerate by 1 m/s at different initial speeds using the formula Ek = 1/2mv2 we get the following (I have assumed no mass change from the burn to keep the math simple): 0m/s -> 1 m/s = 0.5KJ energy added 10m/s -> 11m/s = 10.5KJ energy added 100m/s -> 101m/s = 100.5KJ energy added Each of these burns consumes exactly the same amount of dV, but the amount of kinetic energy added to the orbital energy is vastly different. This is the heart of the Oberth effect: dV spent at high speed changes orbital energy by more than spending dV at low speed.

Thinking about the Wiki quote, this occurs to me: When doing a burn, the thrust, mass and therefore acceleration are the same whether the burn is done at high speed. The mechanical power (in Watts, or J/s) is higher at higher velocity due to Oberth. So for any duration burn, the change in delta-V is the same but the amount of energy gained is different the depending on speed at which the burn is made.

Ha! Changing the probe name from "Oberth 1" to "Oberth's Glory".

Delta V is not a measure of specific energy. Simple dimensional analysis shows it: Delta V is in m/s, specific energy is in J/kg, which is the same as m2/s2. How you can write that and not see that Oberth reduces dV requirements is beyond me. Changing the amount of fuel spent is changing the amount of dV spent.

OK, I did a much simpler test that demonstrates the effect a bit better. I put a probe into a 100x1000km orbit. I then set maneuver nodes to raise the apoapsis to various altitudes (which is equivalent to a transfer burn to a target at that altitude). Here are the results: [table=width: 300] [tr] [td][/td] [td][/td] [td]Target Ap[/td] [td][/td] [/tr] [tr] [td]Burn at[/td] [td]10,000km[/td] [td]100,000km[/td] [td]1,000,000km[/td] [/tr] [tr] [td]Periapsis[/td] [td]427[/td] [td]513[/td] [td]526[/td] [/tr] [tr] [td]Apoapsis[/td] [td]799[/td] [td]925[/td] [td]939[/td] [/tr] [tr] [td]Ratio[/td] [td]0.53[/td] [td]0.55[/td] [td]0.56[/td] [/tr] [/table] I think I'm seeing Oberth at work here. The savings are decreasing as the target grows more distant, too. It's amusing that this probe has no science parts but is actually doing real science.

I must report that this test doesn't work as I expected. I put three identical probes in circular orbits at 100, 1,000 and 10,000km. Set up the transfers and here are the results: [table=width: 250] [tr] [td]Altitude[/td] [td]Speed[/td] [td]Transfer dV[/td] [/tr] [tr] [td]100km[/td] [td]2,248m/s[/td] [td]1,035m/s[/td] [/tr] [tr] [td]1,000km[/td] [td]1,485m/s[/td] [td]767m/s[/td] [/tr] [tr] [td]10,000km[/td] [td]571m/s[/td] [td]578m/s[/td] [/tr] [/table] Obviously my understanding is flawed. The higher total energy of the higher orbits is pretty clearly dominating the Oberth effect advantage of the lower orbits. I guess to see the effect properly, they shouldn't be in circular orbits but eccentric ones with the same orbital energy. I think this may be beyond my piloting skills as the apoapsis would have to be at the correct angle to prograde to make the transfer burn. Back to calculating...

I'm happy to discuss this further, my understanding of the Oberth effect certainly could be flawed. I'm not a rocket scientist by training or profession, I only play one in KSP. The math I posted is certainly simplified, as it assumes an instantaneous burn among other things. But it is a valid approximation of the Oberth effect. It really is simple vector addition. Certainly, when at the other apsis of the orbit, the speed will be very different than when the burn occured, but, and this is the important part, the total orbital energy of the craft will be the same. Orbital energy is the sum of the potential energy (a function of mass, gravitational force, and altitude) and kinetic energy (a function of mass and the square of speed). So whatever energy is added at one apsis will still be present at the other apsis, it will just have a different distribution between potential and kinetic. Careful here, you need to remember that delta-V is not a measure of energy, it is a measure of potential velocity change. Delta-V is in units of m/s, energy is in units of Joules, which is shorthand for kg*m2/s2. Obviously, these are not interchangeable units. It is possible to consume delta-V without changing orbital energy, by burning in a direction perpendicular to prograde (normal, antinormal, radial, antiradial and all points between). The eccentricity or inclination will change, but the sum of potential and kinetic energy will not. I'm sure you are familiar with the Tsiolkovsky rocket equation for calculating delta-V, and that the only factors used in calculating delta-V are initial mass, final mass and specific impulse. You'll note that speed does not appear in that equation, so delta-V obviously does not vary with speed, ever. So it cannot be that the Oberth effect, which is speed related, changes the amount of delta-V in a given craft. It can be easily demonstrated that the amount of energy extractable from a given amount of delta-V varies with speed, and this is the heart of the Oberth effect. Consider the following: Two identical 1-ton craft perform a 100m/s prograde burn consuming 0.1t of propellant, with one (F) having an initial speed of 200m/s and the other (S) having an initial speed of 100m/s. Let's calculate their change in kinetic energy: Delta-KEF = EF-Final - EF-Initial = 1/2mFinalvFinal2 - 1/2mInitialvInitial2 = 1/2*0.9*3002 - 1/2*1.0*2002 = 40,500 - 20,000 = 20,500 KJ Delta-KES = ES-Final - ES-Initial = 1/2mFinalvFinal2 - 1/2mInitialvInitial2 = 1/2*0.9*2002 - 1/2*1.0*1002 = 18,000 - 5,000 = 13,000 KJ From this, it should be clear that the faster a craft is going, the more kinetic energy is extracted from a given amount of delta-V. This is because of the non-linear, geometric relationship between speed and kinetic energy. Let's refine that example. Instead of two craft, make it one craft in an eccentric orbit with a speed 100m/s at apoapsis and 200m/s at periapsis. The same calculations above apply. But, you may ask, how can the energy delivered by the burn vary for the same craft in one orbit? Where does the energy come from? Doesn't that violate conservation of energy? The answer is that our fuel contains more than just chemical energy. It contains potential energy, kinetic energy, and chemical energy. Let's look at these more closely: Chemical energy is the most familiar, and is a fixed quantity. This is extracted to complete the burn. Potential energy is based on altitude, and cannot be changed in our instantaneous burn, so we cannot use this energy. Kinetic energy is based on the square of speed, which can be utilized via the Oberth effect. At apoapsis, more of the fuel's total energy is "locked away" in potential energy, where it cannot be used to change our orbital energy. Less of it is kinetic energy, which can be used. At periapsis, less of the energy is in potential energy where it cannot be used and more of it is kinetic energy, which can be. So, while total energy remains the same at both apses, the energy that can be used to change orbital energy is greatest at periapsis, when the largest fraction of the fuel's total energy is kinetic. I agree that my math is not complete, as I was trying to make an example in which all the numbers were simple and no calculus was involved. However, it's generally good form when criticizing someone's math to provide better math to replace it. All you've provided is some vague handwaving about vector math (and claiming that vector addition is not vector math!). So I will ask you respectfully to either provide better math for my example or accept it as a reasonable approximation. Oberth most certainly does change the amount of delta-V required, that's the whole point! Here's the thing: delta-V doesn't change your orbit directly, it changes your speed, which affects your total orbital energy. This is an important distinction, because differing amounts of energy are contained in a m/s depending on how fast you're going (see my math example above again). This is trivial to test in KSP. Put identical ships into circular orbits at differing altitudes (and thus differing speeds). You will agree, I'm sure that the higher orbit is a higher energy orbit (it cost more dV to get there) even though its speed is lower. Now plot an interplanetary transfer to the same target from both orbits and you'll see that the higher orbit requires more delta-V, even though it's a higher energy orbit and is attempting to transfer to the same target. The reason for this is that more of its fuel's energy is trapped as potential energy which doesn't help us, while the lower orbiting ship's fuel has more kinetic energy which does help us. (Incidentally, this also demonstrates that Oberth is alive and well in KSP) You might be wondering how delta-V maps work if transfers cost different amounts from different orbits. The answer is to look closely at the map. It always specifies the altitude of the low orbit that is the first node away from a body. Which means the velocity is known, and thus the Oberth effect can be taken into account. You'll also notice that there are no circularization nodes between two bodies, only transfers and escape/captures. The map is assuming you will make your entire transfer burn from the low orbit to fully harness Oberth's advantage. You are assuming that a larger change in kinetic energy equals a larger change in velocity, which is not so. Two equal changes in velocity can have differing changes in kinetic energy, depending on the initial velocity, because kinetic energy is proportional to the square of velocity. The fuel consumed in both changes is the same, though. Sorry for this very long and wordy post, but you obviously put a lot of time and thought into yours and I felt I should do the same. If you see where I've made an error or obvious misstep, please point it out. It is entirely possible that I don't understand this as well as I think I do.

One LV-N is always more efficient than two LV-Ns and will always deliver more delta-V. Multiple LV-Ns are only good for increasing TWR to permit landings or less painfully long burns.

*Sigh* Oberth works for both prograde and retrograde, and Oberth does not change delta-V but instead increases kinetic energy change from expending delta-V. This happens because the relationship between speed and kinetic energy is not linear, being "in motion in that direction" has nothing to do with it.

I've never seen a numpad with a shift key.

Electrons residing in microchips.

You really should give links to the sites you're copying/pasting these articles from, and maybe just post an excerpt rather than the full article text. It's generally considered bad form to post someone else's content in its entirety, and is considered copyright infringement in some jurisdictions.

Subscribed. Love these projects, wish I had the money, time and talent to try it myself.

Nuclear thermal rockets really are a different breed than chemical rockets, in that what you use for propellant is orthogonal to your energy source, whereas in a chemical rocket the propellant is the energy source. This lets you choose a propellant based on things other than the energy it stores. If you want the highest Isp, LH2 is a good choice (and no heavy oxidizer is required). If you want higher fuel density to reduce tank volume, water is a good choice as it's cheap, dense, and has other uses (life support, can be electrolysized to produce H2 and O2 for life support and powering auxiliary vehicles). If realism was the goal, the LV-N should be able to use liquid fuel by itself for propellant, or oxidizer or monopropellant for that matter. Would certainly make jet/rocket hybrids easier to design as no oxidizer would be required in any mode. architeuthis, thanks for the correction, I had forgotten that NERVA wasn't bimodal.

As I understand it, under thrust the propellant in a nuclear rocket cools it as it passes through. When not thrusting, it is put in a lower power state (control rod position) to generate electrical power in a closed coolant loop, with excess heat being removed with radiators.

That would just be splitting the transfer burn into two separate burns, compared to burning from an 80kmx80km orbit. The burn to make your orbit 150kmx80km would be identical to the start of a transfer burn, the second burn at periapsis would be the same as the finish of a single burn, so I don't think there'd be any net gain. Unless, I suppose, your TWR is low enough that a significant part of the transfer burn would be away from periapsis, then it might be worthwhile.

32-bit apps on 64-bit Windows are still limited to 4GB of virtual address space, so the /3GB switch should be helpful no matter which OS is used. So KSP would have more memory to work with if /3GB is used, even on a 64-bit system. Edit: I did a bit of reading, turns out 64-bit Windows automatically gives a full 4GB of address space to 32-bit apps which are large address aware. My mistake. Delighted to learn that KSP is large address aware, learn something new every day.

I take it you mean burn until your apoapsis is at the altitude at which the satellite orbits? In effect a single direct burn from launchpad to transfer? A few reasons I can think of: - It's far more difficult to time correctly. - Much of a direct burn would be in the atmosphere with drag losses. - Your capture burn will be much more delta-V, spent where your velocity is much lower and hence less efficient. - Inclination burns do not benefit from Oberth, in fact they are best done at apoapsis if possible.

The virtual address space Windows allocates to each 32-bit application is 4GB, of which 2GB is reserved for the OS. So the most space KSP itself can use is 2GB. The /3GB switch can be used to reallocate this to 3GB for the app and 1GB for the OS, but I haven't tested to see if KSP can take advantage of this, but I doubt it as that's usually used for server apps. For the OP, lag is caused by two things, both of which can be improved with a better PC. Low framerates are caused by too high a resolution/too many visual effects, this can be reduced by using a better GPU (or reducing resolution/effects). Lag is also caused by the physics calculations, which increases with part count. This can be mitigated with a more powerful CPU (or reducing part count).

To be fair, I'm not asking for greater timewarp levels, just access to the existing ones without altitude restrictions. It doesn't break the game, as you can switch to another craft at a greater altitude and warp to your heart's content.

Don't forget that Minmus is rotating, so your orbital velocity when landed is not zero. It will take a bit less dV to achieve the desired orbit if you launch in the direction of rotation, a bit more dV if you launch opposite the direction of rotation.

I'm starting this thread for suggestions that are probably not significant enough to warrant their own thread. If you have a small change that you feel should be incorporated into the game, post it here! If it is a major change to game mechanics or is controversial, it probably deserves its own thread. My own modest suggestions: Key Bindings: Add alternate key bindings for those keys that may not be present on some keyboards, such as laptop keyboards or tenkeyless keyboards. While these can be added by the user in the settings having default alternates is a simple change that makes the program more device agnostic. The controls I see that could benefit from this are: - Zoom Controls (Numeric keypad + & -) - Staging Controls (Insert, not present on many keyboards) - Toggle Navball (Numeric keypad .) Time Warp Allow an override for current maximum on-rails timewarp. If I have an interplanetary craft in LKO it would be nice to be able to warp to the transfer window at maximum warp without the workaround of switching to another craft. Maybe Ctrl + . with a warning dialog similar to physics warp? Being able to warp in the tracking center would be great, too. What small details would you like to see changed?