LN400

Ok, one more question. I am doing a bit of concept study and need a small time interval dt. Below is the code so far. The dt part is in the main loop at the bottom. There you will see 2 lines commented out containing dt. Those are my to-do steps but I need some help in finding the optimal way for what I hope can be a blistering fast code. Also please shout out if you spot any mistakes, typos, errors etc etc. Gracias. EDIT: Rearranging hoping that will get the values for h and m to match better while making the code slightly shorter by baking a into the last expression:

Thanks. Just had to ask in case kOS did things a bit differently.

Given two numbers h and R when h is updated regularly and R is constant, which one of these two expressions is the quickest for kOS to work out? ((h+R)/R)2 or (h/R)2 + 2h/R + 1 I would say the first but I need to make sure.

Lazy Ones: Do what you want, if you can be bothered. Congo Bono: Help U-2 break big in central Africa. Street Fighter II: The Word Warrior: If you can't beat them, talk them out of it. The Blues Bothers: You're in a rut but you're gonna get out of it. Ole Position: Try find a new one. Who Framed Roger Rabbi? And what name is Roger for a rabbi anyway? Can you solve the mystery?

I have dotted down a few more equations but had to fiddle a bit with scale height and got to roughly 5400. For fixed thrust as a function of height ASL I have: Fvac - FASL = FVA F(h) = Fvac - FVA e-h / 5400 For variable thrust, then the total thrust is F(h) = T *(Fvac - FVA e-h / 5400) [1] where T is a function value ranging from 0 to 1 and is represents the throttle. For Isp Isp vac - Isp ASL = Isp VA Isp(h) = Isp vac - Isp VA e-h / 5400 [2] d/dt mfuel(t, T) = T * Fvac / (Isp vac* g0) [3] ... I think.... From [3] it seems to me that over a very short time period dt, the mass loss d mfuel(t, T) = T * Fvac / (Isp vac* g0) dt and the fuel mass mfuel (t, T) = m0 - integral [T * Fvac / (Isp vac* g0)] dt from t = 0 to burnout time tB [4] Since both F and Isp in [4] are constants, then the equation can be rewritten to mfuel (t, T) = m0 - Fvac / (Isp vac* g0) * integral [T] dt from t = 0 to burnout time tB [5] If TWR = R is constant then d/dt R = 0 /end of brainwork for now, back to it when I regain my senses.

Thank you for your input. Would it be reasonable to assume that before one can get down to business with TWR, one should have a mathematical model of the trajectory? A proper gravity turn is of course going to play a huge part in this and here is where the entire rocket design comes in. How quickly will it rotate at a given speed at a given altitude with a given TWR when the initial rotation was a given angle? This links the TWR and fuel flow directly to the trajectory. I don't think rocket scientists are overpaid Edit: The bit about A being the area of the nozzle, that part has me wondering. Isn't the area of a nozzle on a given engine fixed or is this nozzle a mathematical nozzle different from the physical part of the engine? Or is this the actual physical part but with an equation to figure out the area which I will need in the calculations, seeing KSP doesn't mention A anywhere but for a real rocket scientist, every detail of that engine would be on a technical paper? You model F here as a function of h and t but the expression on the right hand side involves mf which itself is depending on F which is what we try to solve for. How does one go about this? EDIT 2: Another thing that has me wondering. On a multi stage rocket, the first stage starts with velocity being zero. The next stages do not. The dv doesn't change (I think) but the increase in kinetic energy would depend on that initial velocity and how would that affect the launch profile? (or am I running around in the wrong field here?)

I'm sorry but how does the Vis-viva equation relate to a launching rocket? Isn't that equation more appropriate for a spaceship already in orbit? As for the rocket equation as we see it here often, that is related to dv more than rate of change of mass. Isp however does relate to rate of change of mass since it relates to the exhaust gas velocity.

Yeah but where even to begin formulating those equations? EDIT: One of my former physics teachers once mentioned that each engine of a particular kind was controlled by computers running and solving more than 30 differential equations, several times per second. I don't want to punish my laptop trying to do the same but then again, theory is a lot less costly than actually sending a rocket anywhere.

I am on a quest: To find the maths that describes, for a given rocket with a given engine: The TWR profile during a launch through the atmosphere. To calculate pre-launch the necessary double derivative of decreasing throttle to maintain a pre-set TWR To find a mathematical function of time and/or altitude that will give me the necessary throttle setting to maintain the set TWR. To be able to have a detailed picture of the entire launch in terms of Acceleration at any moment in time during the launch. Thrust at any given time whether it's for a fixed TWR or floating TWR profile. There are several complications: The "constant" g is not constant. I know the formula for a given height above the planet's CoM, it is glocal = g0 / (h / R)2 where (for Kerbin) g0 = 9.80665 m/s2, h is the height above the center of Kerbin and R is the radius of Kerbin = 600km. This change of g affects the local TWR. The thrust of an engine increases with increasing altitude to a maximum. This thrust function of altitude is one of my unknowns. The Isp increases with increasing altitude to a maximum. This function is also an unknown. The mass is changing so Newton's 2nd law can not be applied without modifications (N2 as described in most school text books, F = ma only applies to constant mass problems, not to varying mass problems). This is not entirely unknown but for this topic, regard the correct expression as an unknown to me. The mass of the rocket is constantly changing but the rate of change is not constant because of: Increasing thrust means higher TWR which means thrust must be lowered continuosly to maintain a set TWR. Decreasing thrust means a decreasing fuel mass loss per second. g is decreasing with altitude which will increase the TWR. Luckily, the rate of fuel mass change is only depending on thrust, or one can say the throttle setting for a given engine with a known thrust. Isp translates to velocity but I am only after the TWR. The trajectory will affect how much time is spent in low, dense atmosphere with lower thrust and higher fuel consumption. Precicely how the trajectory affects time I can not explain mathematically. What I hope to see here, unless some bright soul solved this problem with pre-calculated thrust profiles, is a discussion on how to tackle this. Where to start? What did I not think of? and finally and hopefully a conclusion with the equations one would need. This topic is meant as a maths and physics topic, not a discussion on how mods and spreadsheets can make life easier in KSP. It is the maths behind it, and the physics that is the topic.

I had a look at it. Try this: Place a Science Jr somewhere near the CoM. It doesn't have to be spot on but somewhere near is good. Rotate it to face the right way. Now use the "Tool Move" button in the SPH and drag the part to CoM. Hold down shift while dragging for fine tuning. I managed to get it facing right side up, bottom flush with the bottom of the chassis, smack dab in the middle of CoM.

Some "proper" MS Paint work: A spitfire: Incidentally, this is a parody I first drew some 40 years ago when the teacher bored me at school.

First, let me just say this is not about cost efficiency or dv efficiency as such. The topics are: How to design a rocket based on pre-considered performance demands in terms of TWR and dv. How to calculate payload mass for a given engine or cluster, given the demands for TWR and dv. This topic is all about the maths that go into calculating this before hitting the VAB, for no other reason than the fun of it. The main topic though is: If you have any maths that is useful in building, that goes beyond the rocket equation, and that might be new to someone, then please share. I have just started out looking into this. First I wanted to look at what limitations a given engine has in terms of performance. In particular, I wanted to know what the maximum payload for a given engine is, when considering performance requirements. For my example, I have a 909 driving an upper stage + payload on top. The specifications for the 909 (vaccuum) are: Thrust max: 60 kN (T) Isp: 345s Mass: 0.5t For my upper stage, I decide on: Required dv: 2000 m/s (dv) Required TWR: 1.8 (t) From T, one can calculate maximum total mass (wet mass W) for the stage + payload that will give the required TWR. W = T / (t * 9.80665) = 3.3990540... I will use W = 3.399. The rocket equation says dv = Isp * 9.81 ln(W/D) (equation 1) where D is the dry mass. Furthermore, we have that W = mpayload + mengine + mfuel tank + mfuel (equation 2) D = mpayload + mengine + mfuel tank (equation 3) The rocket equation can then be rewritten dv = Isp * 9.81 ln [1 + mfuel / (mpayload + mengine + mfuel tank)] (equation 4) We set r = dv / (Isp * 9.81) and solve the equation with respect to the ratio W/D (R): R = e^r (equation 5) Substituting R with 1 + mfuel / (mpayload + mengine + mfuel tank) we can rearrange and solve this equation with respect to mfuel mfuel = (mpayload + mengine + mfuel tank)(e^r - 1) (equation 6) In the stock game, the standard tanks have a standard ratio between the mass of a tank and the mass of fuel in a full tank. The mfuel tank of the 4.5t tank is 0.5t, or 1/9 the mass of a full tank. The mfuel is 8/9 the mass of a full tank. The mass of the tanks increase/decrease linearly as you go from one size to another. If we call the 4.5t tank a 1/1 tank, then each smaller step in size is half the size of the bigger one so that in stock game, the 1.25m tanks come in these sizes. 1/1, 1/2, 1/4 and 1/8 with the masses of 0.5t * 1/1, 0.5t * 1/2, 0.5t * 1/4 and 0.5t * 1/8. In my game, I also have access to a 1/16 tank (mod is MRS) and my calculations are based on that. So, this tank has a mass of 0.5 * 1/16 = 0.03125t. One can not have anything other than whole tanks, so the total mass of 1/16 tanks becomes: m\fuel tank = n * 0.03125t (equation 7) where n is a positive integer. Each tank's fuel level can be adjusted in 10% increments. For the 1/1 tank, the total mass of fuel is 4t. For the 1/16 tank the fuel mass is therefor 0.25t. 10% increments means an increment in mass of 0.025t. This is the smallest amount one can adjust the fuel level, using standard series tanks (Oscar-B and ROUND-8 have a different ratio). This means the total mass of fuel must be an integer k times 0.025t. m\fuel = k * 0.025 (equation 8). Back to equation 6, we can now rewrite it as k = (mpayload + mengine + mfuel tank)(e^r - 1) / 0.025 (equation 9) The total wet mass W is the sum of payload mass, fuel mass, engine mass and tank mass (ref equation 2). That gives us that the payload mass mpayload = W - mengine - n * 0.03125 - k * 0.025 (equation 10) Substituting for mpayload in equation 9 and cancelling out adding and subtracting terms, we get k = (W - k * 0.025)(e^r - 1) / 0.025 (equation 11) Solving equation 11 with respect to k gives k = W(e^r - 1) / (0.025 e^r) (equation 12) k is then rounded up to nearest positive integer. Since a tank can hold no more than 100%, for k = 10, this means that for a k from 1 to 10 requires 1 tank, k from 11 to 20 requires 2 tanks, k from 21 to 30 requires 3 tanks and so on. One can therefor use the value of k to determine n in equation 7 . k from 1 to 10 -> n = 1 k from 11 to 20 -> n = 2 etc With k and n known, we can then solve equation 10 to find the payload mass that will give us the required dv and TWR. Conclusion: For the 909, I found that k = 61 from equation 12. n = 7 from equation 7. W = 3.399 Payload mass: 1.15525t rounded to 1.155t from equation 10. Since n is the number of 1/16 tanks, I then substituted 4 of them with a 1/4 tank and 2 with a 1/8 tank. The remaining 1/16 tank had 10% fuel in it. These calculations were done before building. The result of the build was: dv = 2015 m/s TWR = 1.80. These steps can be used when you want to look at the capabilities of a given engine or if you simply want to nerd out and design a rocket taylored for a particular payload. Most of the time though, you know the payload and the challenge is to find the right engine and the right amount of fuel that will give you the required dv and TWR.

The huge difference between base anything systems as we know them, and Roman numerals is we have a positional system and the Romans, at best, had a pseudo-positional system. That's the difference that makes it possible for us to do calculus. Any Roman trying to invent calculus would have be droven mad. Another thing is, there are societies today where the people traditionally had no concept of numbers greater than say 3. To them, the entire number universe consisted of these numbers: One, two and three. Any number greater than 3 was not considered a number but was to them "many" a bit like we use infinity. Negative numbers, don't be ridiculous. Fractions, forget it, has nothing to do with numbers, they would think. They are still modern humans but their culture took some turns that eventually led to this. EDIT: Perhaps another interesting question would be: Are positional systems the way to go, or could there be other systems that inherently includes the properties of a positional system while also possessing other properties that could be extremely useful in describing the world around us?

Finally had a chance to have a look at this. Rodhern is correct, this is for rovers. (someone tell me how to "mention" names here, with that blue background and all) I did try 2 of the modes: Hdt and Fine. The directional PID for the Hdt needs tweaking. That could be a challenge as tweaking is on a case to case basis but perhaps one could include sliders for a DIY tweaking. The Fine mode seems to work fine. One needs to manualy steer but if the rover has enough motor power and traction, then it holds the speed fairly well. A bit of tweaking would work but it's a far better choice than Hdt, the way I figure. Only thing is, to maintain speed, it appears that the computer will not use brakes but run the wheels in reverse. This drains power unneccessarily but all things given, for game balance it actually works pretty well. You gain some, you lose some and a higher power consumption is a fair price. Perhaps if it would use the brakes and drain the batteries at a higher rate than not at all could work. The turn function in Fine is a bit strange and I'm not entirely sure what use it would have. Basically, from what I figured, you set the wheel slider to whatever turn radius you want, in percentage, then in the turn box, type in a turn angle between 0 and 90 degrees. Hit enter and it performs the turn. This function cancels any other program you have running so if for instance you have a program running to drive straight ahead for 10km at 15 m/s, then hitting the turn function will cancel the drive program and initiate the turn. After the turn, the computer is idle with the original drive program erased. All modes it seems apply parking brakes after the program ends.

Funny thing is, it has always been like that for me, ever since the demo. Mods, no mods, on 2 different machines, none of that ever mattered. I took it for intended behaviour after a while. Annoying but hey, not the end of the universe For a bug, this one is strange indeed.

Thanks but this really goes against what I see happening in this game. Not only is the ship's orientation not fixed but it can be waaay off when I return to a ship. 90*, 180*, no problem. When I look at the orbit, that orbit is fixed in space. The sun's apparent movement, Kerbin's movement, they are all irrelevant, I understand that but here is the latest example from real life gaming: A space station I have in polar orbit was pointing anti-normal. In an orbit, that should be a constant heading relative to the stars. When I returned to it later, it had rotated to point prograde. That is not possible unless the ship rotates when on rails, or am I totally wrong here?

There is one thing that has always ticked me off ever since I got the demo some years back: The tendency of a space craft (SAS on) to rotate while in orbit and out of the physics bubble so that say, solar panels no longer get the amount of sunlight they could get. Even after aligning the panel with the sun, going to the space center for a few hours then coming back to the ship, it has rotated so the ship no longer gets the energy it needs, and give it enough time, it will have depleted its batteries. Of course, this is only an issue with static panels and can be sorted by using self aligning panels but they are heavier and costlier than the static ones, and they are only available later on. Is there a way to make sure the on-rails-ships keep their orientation? If not then that would be a much welcome change in any upcoming version of KSP.

My 2.14632... cents: A1: I don't think any number is more natural than others, What it comes down to is notation and only that. The value of a number is independent from whichever symbols we choose. A base number is there for no other reason than to allow a consistent system of notation. A2: If the physicists are correct then there is a system. By that I mean there are laws of nature that are the same wherever you go. That's system right there. The maths in science is not about describing some kind of absolute reality but to allow us to predict future events when the initial conditions are known. They create models of the real world, models with a limited field of applications. So far, physics have shown some staggering promise and many mathematicians and physicists are themselves amazed at how well maths can be used to describe and predict. This speaks for a systematic universe. A3: 1 and 2 are not special in base 2. 0 and 1 are. There are 2 symbols for a very good reason. If you think of them not as numbers but 2 different logical statements, then you have here the minimum number of logical statements that still allows conveying of information. Everything can be expressed using 2 and only 2 symbols, the complete work of Shakespeare can be written in 0s and 1s. Less than 2 statementss and it is impossible to convey information. 2 is therefor the lowest number of digits one can construct a consistent number system from. It doesn't mean it's more natural or easier to use though, A4: To look at integers, one need to look at 2 other values first: The Unit and no units. An integer is nothing more than a value constructed entirely of units. If pi is your base, then pi/pi becomes the unit and since that is constructed of units, then it is an integer and 5 isn't. Pi/pi can then be written using the symbol 1 but 5 would be a symbol for an irrational number just like pi is in base 10. Integers aren't more special than that but once you have a set of integers, then they will share some properties not shared by any onther, non-integer number. Another thing is, we have created a universe of mathematics based on how we percieve the world. We see 1 apple, 34 sheep. We count units because that's how our brains and sensory organs are wired. It allows us to function in a society. We understand that 1 sheep is not 2 sheep, we understand that if we have some sheep and lose a few then we will have fewer sheep than before. We count units and we created systems that allow us to note down on clay or paper how much of this and that we have. One such system is the base 10 system we use, with symbols slowly getting acceptance before being agreed on they work for our purposes. Anyway, just my take on it.

Alright, thanks for the info, guys. Looks like one better mine for science then before 1.2 comes out (provided the save games will still work). Hugging the walls did the trick. They really enjoy hiding these biomes, don't they?

Well, although it's pretty hard to imagine anything we are utterly unaware of, one can of course try (and fail abysmally): I can't think of any evolutionary steps that would produce wheels for "legs" but then again, I can't rule out the possibility that an alien life form has just that. Or caterpillar tracks. Perhaps in a low gravity environment, they have "jet propulsion", like squids here on earth, with nozzles at the end of some appendix with "air" being thrown out. Or what if these life forms were aquatic, or what would pass as aquatic there? Silly as it may sound, who am I to rule out the possibility? Just as we would collapse on a high gravity surface, so would they perhaps collapse if stuck here on earth but for their home environment that brought on that solution, it would work. And why this need for exactly 4 appendixes for moving around/manipulating stuff? The reason we and all those other animals have 4 appendixes for that is because of evolution as it happened on this one planet. Other planets would have their own evolution, blissfully unaware of any evolution on a random remote planet nobody there knows exists. These life forms could very well have a swizz army knife layout with all sorts of organs for manipulating, sensing, moving, allowing things we humans need machinery for.

Best as in how we determine best. It does have some drawbacks though. 2 legs require a more sophisticated brain that requires more energy, it requires a sturdy skeleton where other solutions could get away with a less resource hungry skeleton, more muscles just to stay upright. We stumble and fall if we're not careful, one leg out and you have a disadvantage. From an energy and resources point of view, 2 legs may not be all parade and confetti. Since the existence of aliens is purely speculative, one can speculate that there are other possibilities that we just can't envision since our ability to envision is limited by our experiences here on earth.

Some time ago, I remember coming across, totally by accident, a corner lab on R&D counting as a separate mini biome. Now I have looked up all the mini biomes but in the current game (science mode), there is only 1 (R&D) biome there. Moving around trying to write EVA reports, taking measurements, samples, you name it, they all are from R&D, none from the mini biomes. Can someone tell me why this is? Have the mini biomes been removed? Are the mini biomes only available in career mode? Do I need to hug one particular loaction on the walls of the buildings? Anything else?

I'm no rocket scientist but a few bolts doesn't sound unreasonable. The strap-on boosters are not the only ones fireing so the bolts are not carrying the full weight of the center core/payload, more like they are strapped on to keep them on a leash, stopping them from running around like lunatics. That would, I imagine, demand less in terms of strength.

One reason could be the same reason we squash flies whereever we see them. Flies don't mean much to most of us, we might be as significant to any alien as flies are to us. It's not like we are hellbent on squashing flies, it's just that we don't care and if a fly gets in our way or is of any niusance, then we squash it. We use insect repellants and poison to get rid of mosquitos, they might use whatever they find practical to get rid of us for the same reasons we have. Humans are not necessarily significant in the grand scheme of things. Food source? Doubt that. Evolution on other planets have no obligation to follow the carbon path and the substances that make up living things here might be deadly poison to other life forms. Heck, even the oxygen rich atmosphere here might be a poisonous hurdle impossible for them to overcome. Resources: See the first point. Other than that Because it would look as awesome as in the movies. Because they are interstellar psychos, a travelling Bates' Motel. Because our existence goes against what their beliefs dictates. Because they're clowns who don't know when to quit.

Looking at the KSP Wiki to see what it has to say about Kerbin: At the start of the game (UT = 0), Kerbin is at its apoapsis [mean anomaly (at UT = 0) = pi] (eccentricity = 0 but the apo is still defined apparently), in its orbit around Kerbol. For Kerbin's orbit, the Angle to Ascending Node is 0 degrees. That means, if one draws a straight arrow from the center of Kerbol through the center of Kerbin, pointing away from Kerbol, that arrow is the reference direction, since the Angle to Ascending Node is the angle between the reference direction and the ascending node. This does not add up to what KER says about the AtEAN nor to what my Mk I Eyeball sees. If KER is correct, then the reference direction would be the opposite direction. Either KER is wrong, or the Wiki is wrong or I need to go back to elementary school. If I am right so far, then it is possible that either KER or the Wiki wrongly uses pi as the mean anomaly at periapsis. EDIT: Thinking about it, it could be a workaround for the situation where e = 0 leaving AtAN and AoP undefined. Still confusing trying to work with this.