arkie87

I'm not saying it's the best method. All i'm saying is: (1) it's not necessarily as bad a deltaV killer as people say it is (2) given how nominally worse it is but how drastically easier it is to perform, it should be considered as an option (3) It's definitely something to consider in career mode early on in the tech tree --> see below Yes. I agree. It's not always about dV. In campaign it's about what parts you have available and cost. In early campaign, it's hard to make a gravity turn since you have no torque or control surfaces as per original post. Furthermore, boosters are great at increasing TWR during launch and are CHEAP in campaign. Thus, boosters and vertical launch go well together.

I did the test and recorded it (many discarded trials). Results were very close but direct "gravity turn" and immediate injection to mun apoapsis wins. Please check my video and tell me if im doing something less efficiently that i could be... But i think the results are close enough such that its the decision between carefully executing the gravity-turn and then transfer vs. just burning straight up and getting a snack is not a trivial one (assuming you have a high TWR)-- after all, kerbals like (read: NEED) snacks. The difference was not anywhere near a km/s (though i didnt include circularization burn). Will post video soon.

10 Units of fuel out of how many? If it's hundreds, then they are a lot closer than i would have thought... What was your TWR?

%_Loss is calculated as (dV_vertical/dV_horizontal - 1)*100%, so 50% loss is about 1650 m/s if 3300 is required total. i might code it better accounting for the fact that before doing a gravity turn, one has to jump up to a certain height first before starting the turn. If that height is 20 km, then 632 m/s is wasted to jump up that high, which is about 20% of total deltaV...

Wait, you did it already, and which method won? Except that this method doesnt work, so ill have haters not fans

So i made a simple program in Matlab that simulates non-impulse burn. You specify TWR, and it will calculate acceleration, velocity, and displacement. Both cases ignore atmospheric drag forces and assume TWR is constant (no mass is lost--sorry, i was lazy). For vertical ascent, it calculates vertical burn starting from surface until it reaches velocity necessary to reach Mun altitude (3348.2 m/s). For horizontal ascent, it calculates thrust angle necessary for thrust to counteract gravity minus centripetal acceleration (once velocity gets above orbital, it switches to zero degrees rather than go negative ) at sea level, then integrates horizontal momentum until the velocity necessary to obtain apoapsis at Mun's altitude is reached. Both burns are timed. The expended deltaV is proportional to time, so small times mean less deltaV. Below are the results of excess/wasted DeltaV of vertical burn relative to horizontal burn as a function of TWR Horizontal is significantly more efficient, unless TWR is high. So, it seems the Kerbals were right when they said to add MOAR BOOSTERS!

Yes, that is true. But you also also not accounting for the amount of time it takes to climb vertically before you can begin gravity turn (which you then waste by turning sideways). And on the other end, gravity gets weaker faster if you burn vertically upward. I wouldn't be surprised if the two approaches are close... and if they are within 250 m/s or so, i'd say vertical has the advantage due to simplicity, but obviously, to each his own. The main point of this thread was to show that a direct mun transfer via vertical burn is easy to do and more convenient when gravity turning is hard due to lack of control, and might even be more fuel efficient (especially in career mode before you have reaction wheels and control surfaces). I'll post a link to the youtube video once i make it Thanks for the respectful discussion guys

You mean assuming non-impulse burns? Yeah, i'm too lazy to do that... i'll just do it in KSP and record it.

I plan on doing just that tonight when i get home from work, and recording the results and posting to youtube Conservation of energy: Kinetic + Potential = Kinetic + Potential: 1/2 V^2 - G*M/R = -G*M/(R+H) Plugging in G, M, R and H, and solving for V results in 3348 m/s

Here are my calculations: Admittedly, this method requires high TWR, which is not addressed in calculations given impulse burn assumption..

But when you burn sideways, you are losing vertical speed at the same rate until your velocity becomes large enough that centripetal lift becomes significant...similarly, when you burn vertically, you lose vertical speed until you get high enough that gravity becomes weaker, plus all of your deltaV is going prograde whereas a gravity turn part of it is wasted vertical height and part of it is horizontal... Using the same calculations (assuming no atmospheric drag), it would take 75 seconds to get up to orbital speed for a horizontal burn during which time you will be losing about 10 m/s per second as well...

Not sure if i agree with that comparison. Either (a) you design in a spreadsheet: realize the answer wont work, use your intuition to guess at a new solution, and then redesign the spreadsheet/modify the equations to tackle this new problem or ( you design in KSP VAB itself: you realize the answer wont work, use your intuition to guess at a new solution, and then build it in the VAB using the existing spacecraft as a template. I think ( should be faster! I can agree with this, and this alone is a huge advantage of using spreadsheets

Did you do any calculations? I think a quantitative argument holds more weight than an unproven qualitative one. My calculations show that they are really close, despite the taboo of "vertical launches being inefficient due to gravity"... and potentially vertical ascent is better depending on the shape of the gravity turn...

Oh, I understand what you mean now. Not sure how relative velocity can be reduced to zero though via gravity assists though... Not sure I agree with the reason they are dealing having anything to do with Oberth effect. The reason they are different is due to the fact that relative velocities are different at periapsis before the circularization burn at 30km. I havent done the math yet, but my hunch is that the 375 m/s case is moving slower at periapsis, and so, needs a smaller burn whereas the 540 m/s is moving faster (since it initially has larger relative velocity), and therefore, needs a larger burn to slow down. This effect is non-linear though, since falling faster will result in LESS deltaV than falling slow due to conservation of energy (i suppose you might consider this the Oberth effect). Yes, this was the simplification i.e. rather than getting into an orbit around the Mun, i merely got co-orbital to eliminate the actual encounter's effect on things (such as gravity assist and the like). My math should be the same as yours. Instead of combining the two equations for conservation of energy and Kepler's law of Area, I left them as is and let EES solve them simultaneously. I have since substituted one into the other and also looked the formula up (it is explicit and pretty simple, actually) so i will use that one from now on. This is very true and something important to consider; however, bear in mind that this is true for the horizontal ascent as well since as you mention below, the horizontal ascent goes vertically for most of the way out of the atmosphere... This might be true. But the main difference is that the horizontal case turns and loses vertical velocity and starts from scratch to build up velocity for Oberth whereas vertical keeps this velocity and fully utilizes oberth since it never burns away from prograde.

I think what you are implying is that circulaization step isnt necessary, since a gravity assist can be used to help. If we dont count circulaization step, then vertical is 4 m/s better with no atmosphere Inside atmosphere, it's probably a few hundred m/s better...

More Heavily Commented EES Code: {!Constants} {Distances} R=600e3 "Kerbin Radius" A=100e3 "LKO Altitude" H=12000e3 "Height of Desired Orbit--in this case: Mun's" {Gravitational Parameters} mu=3.5316e12 [m3/s2] "Kerbin's General Gravitational Parameter: G*M" {!Vertical Ascent} {Initial Climb} -mu/R + 1/2*V[0]^2 = -mu/(R+H) "Conservation of Energy" {Circularization} V[1]^2 = mu/(R+H) "Momentum Balance" {Delta-V's} dV[0]=V[0] dV[1]=V[1] {!Horizontal Ascent to LKO then to Mun} {Ascent to LKO} -mu/R + 1/2*V[2]^2 = -mu/(R+A) + 1/2*V[3]^2 "Conservation of Energy" R*V[2] = (R+A)*V[3] "Kepler's Law" {Ascent to Mun} -mu/(R+A) + 1/2*V[4]^2 = -mu/(R+H) + 1/2*V[5]^2 "Conservation of Energy" (R+A)*V[4] = (R+H)*V[5] "Kepler's Law" {Circularization} V[6]^2=mu/(R+H) "Momentum Balance" {Delta-V's} dV[2]=V[2] dV[3]=V[4]-V[3] dV[4]=V[6]-V[5] {!Horizontal Ascent from Surface} {Ascent to Mun} -mu/R + 1/2*V[7]^2 = -mu/(R+H) + 1/2*V[8]^2 "Conservation of Energy" R*V[4] = (R+H)*V[8] "Kepler's Law" {Circularization} V[9]^2=mu/(R+H) "Momentum Balance" {Delta-V's} dV[5]=V[7] dV[6]=V[9]-V[8] {!Summing Up Different Approaches} dV_vertical = dV[0]+dV[1] dV_horz = dV[2]+dV[3]+dV[4] dV_horz_surf = dV[5]+dV[6]

Yes, i stated that in the assumptions. It helps both, but i think it helps gravity-turn more. The code is commented out, but basically works as follows: Each paragraph/group contains one approach. Each approach goes through each burn, calculates deltaV's and then sums them up. For each burn, there are two equations that relate distance 1 and velocity 1 to distance 2 and velocity 2, and those are conservation of energy (the ones involving velocity squared terms) and kepler's (2nd?) law (distance_1*velocity_1 = distance_2*velocity_2). The only other equation is V^2/d = mu/d^2 which simplifies to V^2 = mu/d and is used to calculate the velocity needed for a circular orbit at distance d. EDIT: it might also help to know that EES is a numerical solver of systems of non-linear equations. Thus, rather than use a and b to calculate c, i can specify 3 equations with a, b, and c in them, and it will calculate all three at the same time. So, each line doesnt have to calculate something, but rather, give a relationship between variables. It's a really powerful software. Highly recommended

I'm not sure if i follow. I think more words or perhaps mathematical relations would help me understand what you are trying to say. Vertical ascent gets you to Mun's SoI with 0 m/s horizontal velocity relative to Kerbin, and takes 530 m/s to burn horizontally to get into a circular orbit around Kerbin at Mun's altitude. Not sure if thats what you understand me to mean. If so, then disregard this...

I'm using FAR (the only mod relevant to physics) and stock parts. I can send you the craft files if you really want, but they are kinda difficult to fly with a gravity turn (no torque). I would recommend making your own simple craft with about 7 km/s deltaV -- orange fuel tank + mainsail + reaction wheel + control surfaces and small fuel tank + lv909 and command pod should be close? If you make a video, mention me for giving you the idea! I want to be famous!! Edit: Now i kinda want to try this myself.... and make a video too, but that will have to happen later tonight...

I hear what you are saying, but those effects should be potentially equal depending on the encounter. Regardless, the best comparison we can make is to get both ships into a circular orbit with same altitude as the Mun and compare.

EES Code: {!Constants} {Distances} R=600e3 A=100e3 H=12000e3 {Gravitational Parameters} mu=3.5316e12 [m3/s2] {!Vertical Ascent} {Initial Climb} -mu/R + 1/2*V[0]^2 = -mu/(R+H) {Circularization} V[1]^2 = mu/(R+H) {Delta-V's} dV[0]=V[0] dV[1]=V[1] {!Horizontal Ascent} {Ascent to LKO} -mu/R + 1/2*V[2]^2 = -mu/(R+A) + 1/2*V[3]^2 R*V[2] = (R+A)*V[3] {Ascent to Mun} -mu/(R+A) + 1/2*V[4]^2 = -mu/(R+H) + 1/2*V[5]^2 (R+A)*V[4] = (R+H)*V[5] {Circularization} V[6]^2=mu/(R+H) {Delta-V's} dV[2]=V[2] dV[3]=V[4]-V[3] dV[4]=V[6]-V[5] {!Horizontal Ascent from Surface} {Ascent to Mun} -mu/R + 1/2*V[7]^2 = -mu/(R+H) + 1/2*V[8]^2 R*V[4] = (R+H)*V[8] {Circularization} V[9]^2=mu/(R+H) {Delta-V's} dV[5]=V[7] dV[6]=V[9]-V[8] {!Summing Up Different Approaches} dV_vertical = dV[0]+dV[1] dV_horz = dV[2]+dV[3]+dV[4] dV_horz_surf = dV[5]+dV[6]

Gravity turn is *most efficient* for getting to LKO If your ultimate goal isn't LKO, then of course, something can be more efficient than it...

Assuming Kerbin has no atmosphere and is perfectly spherical so LKO is at 0 m altitude and zero rotation: Burning vertically upward to Mun's altitude takes a total of 3348 m/s. If you include circularization burn once you get to to altitude, it takes an additional 529.4 m/s, totaling 3878 m/s. Burning horizontally upward to Mun's altitude from surface of Kerbin takes 3352 m/s (4 m/s more!). If you include the circularization burn one you get to altitude, it takes an additional 369.8 m/s, totaling 3722 m/s. Thus, horizontal beets vertical by 155.8 m/s, but only if we assume a transfer from 0 m altitude. Since even horizontal approach must burn vertically upward for a bit (I'd venture to say way more than 155.8 m/s), I'd say vertical ascent method is probably more fuel efficient...

I said "burning vertically upward" so actually that is what i meant I dont see how Oberth effect has anything to do with this -- you utilize oberth effect by doing gravity turn and by burning vertically upward the whole time since both instances you are burning prograde. The only difference is how you raise your apoapsis, and i've heard, it's more effiicent to burn sideways than vertically. I am running some numbers now and i'll get back to you.