arkie87

Is this question not answered to your satisfaction? If it is, please mark it as answered (edit origional post and the click advanced edit)

A summary of topics discussed: For the return from Mun case (which is what this thread is about) we are not discussing optimizing cost. The cost discussion was related to Mr. Horseman's challenge. In this thread, we are talking about for a given craft with high TWR, which transfer method will require least deltaV to get out of Mun's SoI: vertical or horizontal. We are all aware of the potential gains to be made to the total deltaV of the craft by reducing TWR i.e. using a lighter engine, but the focus of this discussion is not on getting the most deltaV of the craft by picking the right engine, but of the deltaV required by the different transfer methods for a given craft design (one with high but finite TWR). That said, LethalDose and I provided theoretical calculations for the deltaV required for the different transfer methods (LethalDose even conducted super-accurate simulations of vertical launch with finite TWR and mass loss), and showed that while horizontal impulse burn is most efficient (it's the lower limit on deltaV), in practice, it is not possible, since there might be terrain to clear and it requires an impulse burn. To that end, I showed that for infinite TWR, if one climbed only 300 m (by accident or to avoid terrain), it would pay to have just departed vertically. For finite TWR, I used LethalDose's excellent results to calculate the height at which point it pays to depart vertical is reduced-- for a TWR of 6, it was only 1 km, and for TWR 3, it was 7.5 km. Thus, even though horizontal burn might be more efficient for infinite TWR (i.e. impulse burns), in practice, there might be terrain in the way or it might be hard to perform accurately enough to keep apoapsis down low enough in order to keep horizontal burn the most efficient transfer method (let alone the fact that it requires an infinite TWR i.e. impulse burn). To that end, you and LethalDose began to discuss the effect of finite TWR on horizontal burn. LethalDose provided the equations necessary to calculate the required burn angle, as a function of TWR, to balance gravity such that the craft could move horizontally across the surface. For low TWR (which you mention above is both cheaper and has more overall deltaV), the angle becomes more vertical, such that more fuel is wasted preventing vehicle from falling, and in the limit of TWR = 1, the angle is completely vertical such that all fuel is wasted to prevent the craft from falling, and none goes into accelerating the craft horizontally. The angle required is theta = arcsin(1/TWR); the wasted energy/sec is TWR(1-cos(theta)). The graph of that function looks like this: For sqrt(2) TWR, approximately 41% of the fuel spent is fighting gravity vs vertical's 100% (however, fuel spent fighting gravity decreases as horizontal velocity approaches orbital due to centripetal lift). So I dont think the solution is immediately obvious in any way... And i think that is where we left off...

Yes, let's resume to discussing munar vertical vs horizonal. We do seem to mis-communicate frequently.... LethalDose wasn't saying that a launch vehicle under 10k$ (at 2k$) is ridiculous. That was me and 5thHorseman since we were talking about his challenge, and not the one outlined by this thread's OP. You might want to go back and reread (i can imagine if everyone is yelling at you, it's hard to remember who is who ) LethalDose was just asking you to quantify what you were saying with equations, math, or in-flight tests, rather than words (since otherwise, you just repeat yourself).

To be honest, when we have miscommunications that last for days, i start to wonder if you are trolling me. But then when we resolve them, i see your point of view and what you thought, and realize you werent trolling, but rather, had a legitimate point or misunderstanding. But when this happens again and again, i start to wonder. To clarify, i didnt mean to call you a troll (i was saying i was starting to wonder), and i dont think you are a troll. It was all jut a misunderstanding.

In my description of the orbit, i said departing from kerbin, but the original description was vague. However, i clarified it twice thereafter.

I have started a new thread to discuss this challenge, since it has derailed other threads, and confused many-a-person. The challenge is as follows: Build the cheapest craft (in terms of Kerbucks) capable of taking the lander described below from Surface of Kerbin to an Orbit around the Mun (Ap and Pe must both be below 30 km). Lander must arrive at mun without any fuel having been drained from it. You do not (and should not) land it or return to Kerbin. As a side note, once you enter Mun's SoI, you cannot leave it (i.e. no fancy-shmancy gravity assists are allowed). The lander consists of the following parts (in order): Mk16 Parachute Mk1 Command Pod TR-18A Stack Decoupler FL-T400 Fuel tank LV-T30 Liquid fuel engine Disallowed Parts: extra torque-providing parts and control surfaces Mods required: FAR (you may disable aerodynamic disassembly) Modded Parts allowed: none Videos are encouraged.

I understand. But I made it clear i was talking about the case with 5thHorseman. And i repeated the important parts of the problem statement, and for the reasons i gave above, it was relevant to this discussion since it motivates/gives a good practical case where you might have high TWR (since SRB's are so cheap).

I understand. I take partial responsibility, but I DID repeat for you the problem statement (nearly in full): And in my defense, this case is a perfect motivation behind the TWR debate (which is present in this thread as well), since optimizing for cost leads you to use SRB's (in FAR), and optimal SRB usage leads to staging for horizontal ascent, and high TWR for vertical ascent without having to change the craft design. It is therefore not an unrelated topic/case by any means.... Should i make a new thread to discuss the 5thHorseman challenge?

When i referred to 5thHorseman's challenge (the 10k$ vehicle that you said you could do for 2k$), I am talking about this, as i have said multiple times: I am not confused by my different cases/questions/scenerios/threads. If you would read what i wrote, thought about it, made sure you understood it, instead of replying so fast assuming i was saying something wrong/saying something other than i was saying... Admittedly, it's partially my fault for entertaining the Kerbin-to-Mun debate in the wrong thread. I would create a new thread, but then I would get yelled at for creating too many new threads... Perhaps 5thHorseman wants to create a new thread with his/our challenge? I would happily do it if people dont wouldnt get annoyed at me for starting too many threads about "seemingly" the same topic... clearly, they arent the same topic...

If you launch with twr of one, the angle above horizontal from eastward is 90 degrees

You really need to read what I wrote. Lander from surface of kerbin to orbit around mun. Not surface of mun to kerbin. I honestly can't tell if you are just a troll...

I think you really need to read what I wrote rather than responding so quickly. The challenge is to take a lander from kerbins surface and get it into orbit around mun. If you can do that for 2000 kerbucks, then I and 5thhorseman will be shocked.

If you burn up with TWR = 1, nothing happens. If you burn horizontally with TWR = 1, you accelerate sideways while falling down. If you are at ground level, you cannot burn horizontally at TWR = 1. You need TWR >1, and obviously, the greater the TWR, the greater the angle you can use. And it's obviously most efficient to have a higher angle away from vertical. Thus, you have to burn at an angle to counteract gravity. He is calculating that angle....

This is my source too Yes, I am aware. Escape velocity was given at sea level. It's a useful sanity check to compare our numbers to. Yes, i was going to point this out, but i think you actually mentioned that in your post Your absolute minimum burn 775.6 m/s is below escape velocity at sea level, which is only possible since you leave the SoI of Mun. If Mun were not orbiting Kerbin, You would need 807 m/s. (I know you know this, but I am explaining it for others).

Many people (including myself) have made this argument already.... I dont know why you keep on bringing it up. This isnt a design question (yet... see below); it is a hypothetical question of what to do if you find yourself in X scenario. I actually think the case 5thHorseman suggested is a perfect test of this: We have a lander craft that must be lifted to orbit around the Mun. This craft is specified it cannot be changed. The challenge is to design the lifter stage(s). He has 5 SRB's in two stages (3 first, then 2) and will launch horizontally from Kerbin into LKO followed by periaps transfer to Mun; I will have 5 SRB's as well, but in one stage burning vertically. We will both be using the same craft, but with different staging. Same potential TWR, same mass, same cost, same everything just the optimal staging for each burn. This design by 5thHorseman is the result of trying to minimize cost (<10k$). And we see, that by minimizing cost, we are led to a design with SRB's that has a potentially high TWR. Furthermore, we can manipulate this design (without adding or removing parts--just staging) to have a high TWR-- optimal for vertical ascent-- or low TWR--optimal for horizontal.

First, I'd like to thank you for your work! Question: for the two horizontal cases, I assume you are using the standard orbital equations (not performing a simulation) since I have matched the values you have given using these equations. It is worth noting that these values will be slightly worse in the real world due to non-infinite TWR i.e. assuming maneuvers are performed with impulse burns. However, doing these simulations would be complicated and rather annoying... I used the equation and got 776 m/s (i assume that's just a typo or maybe rounding error). Escape velocity is 807 m/s, as a reference. This is actually a very important point. Parking orbit altitude (regardless of whether it is obtained by vertically climbing or climbing by slingshotting around the planet) can make vertical burn more efficient than horizontal burn. I have repeated your calculations done and presented them in graphical form: From the graph, direct horizontal burn is always most efficient, but not always practical for two reasons: (1) You might be below nearby terrain, and therefore, have to climb. Climbing at a shallow angle before velocity is comparable to orbital will not result in centripetal lift and will waste fuel. (2) You cannot perform an impulse burn, and therefore, must aim slightly upwards, even if you could avoid all terrain. (3) You are not Mechjeb and will not perform perfectly. The graph shows that if you climb up 300 m (that's VERY low), turning 90 degrees and burning horizontally and keeping on going vertically will require the same deltaV. However, in the real world, you do not have infinite TWR, and the break-even point will change. For a TWR of 6, from LethalDose's result, we need 830 m/s; that relates to roughly a 1 km climb being the break even point. For a TWR of 3, from LethalDose's result, we need 910 m/s; that relates to roughly a 7.5 km climb being the break even point.

Really? are you really arguing about this? I said this in my own post, that 10km is a bit exaggerated, and its better to aim slightly upward if you arent near the crater wall... I am going to make a graph of at what altitude it pays to keep going vertical rather than turn as a function of TWR.

Sorry for putting words in your mouth then Though you do admit that you have said that in the past-- but now you realize that it is less efficient. Your solution to stage the SRB's sounds better. So my guess is, my craft will likely look like yours, except while you stage the SRB's and go horiztonal, I will put all 5 on at once, and go vertical... I am still waiting for FAR to come out for 0.90... And can you verify cost of your craft has not changed in 0.90 (Squad did some balancing, no? so part costs might have changed).

Let's talk about airless body, since that is what this thread is supposed to be about. As a side point, "prograde" is a bit vague since launching vertically is "prograde" as well. I prefer to say "horizontal" vs. "vertical" rather than "vertical" vs. "prograde". For such a low TWR, of course, burning vertically will not pay. The discussion must involve high TWR (even if you would have had more deltaV had you chosen a lighter engine and lower TWR). Even for an infinite TWR launch (i.e. impulse burns), as you pointed out, horizontal is probably better or, at worst, equal to vertical launch. I have admitted this. However, on the Mun, you cannot perform a horizontal burn unless you are a daredevil... That is the basis for this thread! For the Mun case, unless you land on a hill, you probably want to burn vertical first to avoid crashing. I've seen that 10 km is a safe altitude, so if you first burn vertically to 10 km, and then turn over and burn horizontally, you will end up using more fuel, as i have calculated in OP. Now, granted, shooting straight up to 10km is inefficient; it is more efficient with the horizontal approach to burn horizontally the whole time and raise apoapsis via centripetal acceleration. In addition, 10km is a bit high, since you can burn horizontally with a slight inclination and visually inspect to avoid hills. And the closer you are to the ground the entire time, the less fuel you will use, but also the more dangerous. A better result might be to know the balance point i.e. if you first climb vertical, at what altitude will it use less fuel to continue going vertically than to turn horizontal and burn to LMO. I am working on this now....

This is perfect. That is exactly what i want and expect By all means, explain all concepts to me. Sometimes I am already aware of them, sometimes i'm not... The reason the thread isnt closed is because we are still debating it... For the Mun case, my math suggests that with high TWR, direct vertical ascent is more efficient since you have to climb first. If you want to refute it, either try both cases yourself, or perform your own mathematical calculations. Logical arguments do not trump math or experiment. Those were GoSlashly's opinions. I am skeptical of deltaV calculations in KER (though it might just be the result of ISP varying with altitude or something), but you cannot argue with the similar fuel consumptions, which is independent of anything KER outputted. My fuel consumption results indicate that the results are similar for Kerbin case (and by the way, what we are discussing now is the Kerbin case, which is another reason i was confused which case--mun or kerbin-- you wanted to discuss). I dont see where I am telling you I you are wrong? I am debating; I am presenting counter arguments. If you want me to accept your opinion as fact without being allowed to present counter arguments until I either understand the flaw in my reasoning, or convince you of mine, then please stop debating with me. That is not scientific. Calling my reasoning "circular" and saying "remember now? Your numbers are wonky" is antagonizing, belligerent, condescending, and rude. If you dont mean to be rude, the i recommend adding a kerbal-smiley face...

Why? If you thrust horizontally and up enough to cancel gravity at the current velocity, your velocity increases, and now, your apparent gravity is even less, so you should rise...

Yeah, i really like scientific computing and numerical methods. 'Stuffs fascinating to me Interesting. While retracting the wings might decrease drag coefficient and definitely decreases area, I would seriously doubt if it changed the overall drag coefficient/drag force of the craft that significantly, since most of the drag is experienced by the fuselage, no--wings are designed to be low drag, after all. But this is beside the point... Let's pretend the entire craft is a wing: a sensor that retracts the wings will essentially be changing the drag force vs. velocity curve. Thus, before it retracts its wing, it has one terminal velocity; after it retracts, it has another. It does not simultaneously have two roots i.e. two terminal velocities. Retracting the wings in this manner is like attaching wings to decouplers. A better example is the wings-on-springs, since the deformation would be continuous (i assume by wings-on-springs, you mean the wings deform like a spring proportional to the drag force) and inherent to the system, rather than a sudden change. Even so, I dont think there would be two roots based on the reasoning in this example: Consider a craft going at terminal velocity, Vt, with drag force = weight i.e. D = W and "wings on springs" (which sounds awesome by the way). If we try to push the craft negligibly faster, the drag force will increase slightly and the wings will recede slightly. If the wings receding decreases the total drag force to below the terminal velocity drag force i.e. our initial value (which is what is required for their to be two roots), the wings will bounce back to their initial condition/displacement at terminal velocity. The point being, its not possible for the drag to decrease with increasing velocity due to the wrings receding, since if that were to occur, the wings would "un-recede" since receding only occurs with increasing drag force... If by wings on springs you just mean some arbitrary controller that recedes the wings at will, then the spacecraft will have two (or more) roots, but the control system will be basically changing the spacecraft. A command pod has two terminal velocities-- one with its parachute stowed and one with it deployed... but that clearly never was the case we were discussing I will check it out

You have made these logical arguments before, but I'd prefer mathematical ones or experimental trials. Since you propose an experiment later on, let's focus on those so as not to derail this conversation, ok? While you might not have mentioned anything regarding an atmosphere, 5thHorsemen and I were discussing the Kerbin problem, so i wasnt sure which problem you were discussing. So maybe i should have insisted he discuss it with me in the proper thread. I think mathematical theory and experiment are both valid. I would agree experiment is preferred though, since theory might make certain simplifications or assumptions, but in no way is a mathematical theoretical discussion "foundationless". I really dont have the patience to debate with someone who will tell me that i should look back up a few pages to see what you were talking about. I spent hours debating with GoSlash, who did that, and it took us a few days to finally come to an agreement. If you want to have a productive and constructive debate, please cite what you are referring to (like i am doing by citing you now). Otherwise, misunderstandings will compound. Please cite where i said this... because i agree with you, saying that would be stupid. I think you are misunderstanding me... I also wouldnt recommend saying something like that, as it could get the negative attention of the forum moderators. Try to keep it civil. Hypothesis is more like: if TWR is high, launching vertical might be comparable or better (i never said always) than an LXO-to-X burn, depending on the intended target. In this case, its SoI change and Kerbin-capture. I would be happy to do this test. I can use hyper edit to get craft to proper location, i hope...

You seem to be more of a mathematician than an engineer i.e. you seem to know more theory than me I have lots of experience with numerical methods (i've dont a lot of numerical work for my PhD), but not so much of the theory, especially when it comes to stability. If drag is some arbitrary function, then of course, a priori, we do not know its behavior. But since it is a physical function, it has to make physical sense, and it wouldnt make physical sense for an increased velocity to result in less total drag force (less drag coefficient is fine though). Thus, we should know that it is monotonically increasing with velocity, and so, there is only one root.

Thanks!!! I DIDNT NOTICE THAT!