arkie87

You would be absolutely correct if this work integrated up to escape velocity. But this work is concerned with the most efficient way to get up to orbital velocity, and so, integration stops after orbital velocity is reached i.e. the instant theta = 0 (since once you reach orbital velocity, you no longer need to spend any thrust fighting gravity, and can aim prograde, regardless of TWR).

Yeah, basically, the game tells you to put a base on a planet/moon. But unless you bring those kerbals back, they will remain noobs forever....

Ok, so what i meant is that they will need more fuel to get into orbit than what they had rationed after reducing deltaV/fuel due to increasing Isp... simple misunderstanding. Ok, let's analyze this example. For Tylo, V_orbital @ sea level is 2170 m/s. To have DVR = 1 @ Isp = 300, the m_wet/m_dry = 2.09, and FMR = 0.5216. To have DVR = 4 @ Isp = 300, the m_wet/m_dry = 19.09, and FMR = 0.9476. Thus, the second craft carries a higher fraction of fuel with it. However, before the fuel is burnt, it is basically a payload (it doesnt matter that it is fuel). The rocket equation states that: deltaV_expended(t) = Isp * g0 *ln(m(t=0)/m(t)) To burn the same amount of deltaV, since Isp and g0 are constant, all that matters is the ratio between initial mass, m(t=0), and the mass at the end of the burn, m(t). The FMR does not effect deltaV. Thus, the TWR at the end of both burns (and, in fact, throughout the entire burn) will be the same, since the ratio of the initial and final masses will be the same. The reason TVR effects efficiency was given in the OP, but is as follows: Let's assume a craft is designed for DVR = 1 on Tylo. With a low Isp engine (say 300), an FMR of about 0.5 will be needed to achieve this, and so, TWR will nearly double during the flight. Thus, even if one starts out with TWR = 1, by the end, TWR might be around 2, and so, average TWR will be between 1 and 2. Since increasing TWR results in increased efficiency, low Isp engines do not require as high initial TWR, since their TWR will naturally increase significantly over the flight. However, with a high isp engine (say 4200), for DVR = 1 on Tylo, a FMR of nearly 0.05 is needed. During the burn, only 5% of the mass will be lost, and so, TWR will icrease by the same factor. Thus, the average TWR will stay around 1, and so, an initial higher TWR is needed for the same efficiency.

I am not really defending it.. it's just the third independent variable... but it's also good to know it doesnt effect deltaV (it obviously effects fuel though).

I was supporting you, against the guy you quoted

If the player cannot bring them back i.e. they die prematurely (and tragically), then who cares if he/she can level them up?

You could send them to jool, do loads of science, get into orbit around and land on each and every moon, and they dont gain experience until you bring them back to Kerbin???

Basically what title says: Kerbals should level up as soon as they gain enough XP, not only upon being recovered. Otherwise, I have to bring them back home before I get any of the XP benefits of what they did...

I didnt think my resposnes were handwaivy at all. I was referring directly to the equation. When the equation is written like this: m(r'' - V_theta^2/r) = sum(F) The m*V_theta^2/r term is on the inertia side of the equation To treat it like a force, we have to move it to the force side. i.e. we add m*V_theta^2/r to both sides, we arrive at: m*r'' = sum(F) + m*V_theta^2/r Here, m*V_theta^2/r is positive in sign and is on the force side of the equation. It therefore acts as a positive force i.e. lift. That is what i was trying to say. In general, we can say: sum(F) - m*a = 0 Since, m*a is negative and on the force side of the equation we can say m*a, i.e. the inertia term, acts as a negative force to oppose motion. This is D'Alembert's principle.

I'll triple what Numerobis and Slashy said here. if DVR affected efficiency, then the relationship would be more complicated. It could be the case that increasing DVR decreases efficiency faster than DVR increases deltaV, such that, you could never get into orbit. It's good that isnt the case... I said fuel and deltaV. You are splitting hairs, or, i think you need to re-read what i said. Therein lies the issue. DVR does not effect TWR, and TVR doesnt affect TWR either. They are all independent variables. In practice, when designing a rocket, if a player adds fuel, he/she reduces TWR. Thus, the player has two choices (a) add more engines to maintain same TWR or ( look up efficiency with new (reduced) TWR. The model doesnt care how you arrive at the given TWR, TVR, and DVR, it just provides the efficiency if you manage to obtain it. I think the confusion here is that it takes the same amount of deltaV, not the same amount of fuel. DeltaV is more "dimensionless" in that respect, since it scales with the mass of the rocket. Also, before you were arguing that it's obvious that efficiency should be independent of DVR... now you are saying there is an error in my model since it shows this independence?

I have computed for this case: TVR: 9.81*290/sqrt(1.69*210000) = 4.775 Running model for TWR = 1.01, TVR = 4.775 yields: efficiency = 0.6455 So you are slightly below, as expected, since let's be honest, you are no Jeb

Yes, but if you have low TWR (which in general, should be cheaper in terms of funds, since it requires fewer engines), you cannot wait until you reach apoapsis to burn horizontal, since you cannot accelerate fast enough. Thus, you probably have to start burning horizontally before, and once you reach apoapsis, then pitch up to the angle necessary to keep "time to apoapsis" as close to zero seconds as possible Also, yeah it is scary. Sometimes you pass over a hill by only a few meters Gets the blood pumping...

mu is less vague, since i've only seen mu in this context mean standard gravitational parameter capital G usually means gravitational constant, not acceleration due to gravity, which is lowercase g.

Capital G is vague -- could be referring to gravitational constant or gravitational acceleration... perhaps we should use: V_orbital = sqrt(mu/R) where mu is standard gravitational parameter G*M?

Why on earth would i have to provide this? That would defeat the whole point of non-dimensionalization! The player can calculate what their TVR is for the engines they are using and the body they are departing from. It might be more useful to use ISP and then have separate plots for each body in KSP, but it is definitely more convenient to have one chart, which can be used to generate all those graphs for all those bodies in KSP, and in RSS, and for any new planets that are made etc... Besides, the definition of TVR teaches us something about the physics: that a craft with ISP=400 on a planet with V_orb = 1000 m/s will have the same efficiency as craft with ISP=800 on a planet with V_orb = 500 m/s. That's not the definition of DVR. DVR is the ratio of deltaV your craft carries if it were floating in space in the absence of any gravitational field to the amount of deltaV required to get into orbit using an impulse burn i.e. minimum possible. It could easily have been the case that carrying more deltaV would effect efficiency and require higher TWR. Regardless, it is good to know and show it doesnt. We are talking past each other. I agree, they will need less mass of fuel-- that is what increasing ISP does. However, knowing this, players are likely to decrease the mass of their fuel accordingly (!) such that they will have the same total deltaV, not realizing that they need more deltaV to get into orbit due to reduced efficiency (assuming they also adjusted their engines to maintain the same TWR).... That is true but not relevant at all. My model requires inputs of TWR, TVR, and DVR. The players can get these numbers from their craft and use it to assess whether their design has enough deltaV. It might not be possible to get any arbitrary combination of TWR, TVR, and DVR from the parts available in KSP, as you pointed out. The player either (a) supplies these numbers using current design to assess current design, ( uses the results to generally guide TWR selection (i.e. keep it 1.4 < TWR < 1.6) or © uses the results to try to improve design, but has to cleverly tradeoff TWR, TVR, and DVR to improve efficiency since adding removing one part might improve TWR, but reduce TVR and/or DVR... My formula assumes nothing about the craft, except its TWR, TVR, and DVR. If you can get those values, then the model will predict results accurately. For the case you described above, where the only mass of the ship is fuel, then DVR = infinity since m_wet/m_dry = infinity since m_dry = 0. Thus, in order for my model to "make" those assumptions, I will have to input DVR = infinity. Otherwise, it assumes realistic values for DVR.

I recommend designing for TWR >>1, using hyperedit to get into orbit around desired body, using infinite fuel to land with high TWR, and then turning infinite fuel off once landed and limitting throttle to desired TWR (just make sure your craft has enough deltaV to get into orbit given it's predicted efficiency -- ive made that mistake more than once!) Thanks for testing! Thanks again!

I dont think coasting to apoapsis is fuel efficient, particularly for low TWR. You have already spent fuel to get apoapsis up, and bought yourself time to burn horizontally, I would think you should do that...

Can you revise to say Initial Thrust to Weight Ratio (to be clear it's not changing with time or mass) Also, can you change v_0 to v_orb and maybe define v_orb as sqrt(g*R)

Yeah, i was just working on revising graph, myself... but yours is better... hehe Thanks!!

First, can you tell me what altitude your Mun tests were conducted at? I can adjust my model to see what prediction is for higher altitude (since effectively TWR increases)

First, i think by periapsis, you mean apoapsis? EDIT This paper mentions it doesnt include the effects of terrain. However, unless you are right next to a vertical ridge, there is almost always likely a relatively horizontal ascent approach. I'm not sure what you mean by a "short burn to go sub-orbital". If you mean burn vertically to go a few km up, this is a really bad approach (unless you are literally next to a vertical wall) since all deltaV used to jump up will be wasted when you start to burn horizontally. You can do that math assuming infinite TWR (impulse burns) and jumping up even a short distance is much worse than aiming as horizontal as possible. I had another thread http://forum.kerbalspaceprogram.com/threads/102947-Vertical-Ascent-vs-To-LXO-First that tried to see if there was a single instance in the Kerbol solar system of this being practical, and the only place i found was Minmus (since there are regions which are surrounded by tall hills).

Oh, given our previous exchanges, i wasnt sure if you were being sarcastic...

What i am trying to say is this:

Maybe my post was vague... If you were to interpolate for TWR = 2 TVR = anything, what would efficiency be? Less than 0.9 or greater? The way matlab plots, it is greater. The line of efficiency equals 0.9 is exactly on the border between colors, not in the center of its color...

What is your initial altitude? I can revise model to adjust g based on altitude. On the Mun, landing 5 km above sea level results in 5% change in gravity...