Yasmy

I watched SpaceX's recent ISS resupply mission 3. I seem to recall it throttled back just before MECO and SECO, presumably for the same reason. I might have to rewatch to confirm.

I think the conclusion is that minimizing instantaneous drag or cumulative drag losses before the gravity turn, is not the same as minimizing total delta-v to orbit. Not that your math is wrong.

Thanks tavert and Alchemist, and everyone. (Edit: and rkman who slipped in the thread when I was typing) I was focusing on the effect of deviations from terminal velocity on delta-v to orbit. It doesn't make a huge difference in delta-v to orbit: +/- 20% of terminal velocity amounts to less than +2% delta-v to orbit. Deviations from ideal just don't hurt that much in delta-v. But I completely agree with both of you that having that kind of thrust (always > 2): 1) is not delta-v optimal 2) is no where near payload-fraction optimal 3) is not cost effective (more fuel is cheaper and simpler than more and bigger engines) Of course, a super-aspargussied-up rocket may be delta-v and payload-fraction optimal, but not cost or complexity optimal. Me personally: I flip-flop between the crazy-asparagus mode, or just a simple no-nonsense two stage rocket with reasonable TWR (and maybe some boosters for launch) I'm not advocating overthrusting here.

No need to get fancy. Kerbin is 1.2 Mm in diameter and 12 Mm away from the moon. Thus the angular size of Kerbin from the Munar surface is atan(0.1) ~= 0.1 radians ~= 5.7 degrees. If your solar panels track Kerbol, then you lose 0.1/pi ~= 5.7/180 ~= 3.2% of your Munar daylight to eclipse. If your solar panels do not track Kerbol, then you lose sin(0.1)/2 = 5% of your solar power.

I say just don't worry about it. I flew an unplanned circumnavigation of Laythe and had no problem landing at every island cluster without scouting ahead of time. The beaches present many opportunities for smooth landings.

Here's a neat demonstration of reaction wheels:

This is a nice analysis, and it makes a testable prediction. I think I'll build that rocket tonight and find out. I'm not convinced that the effects are multiplicative though. I would expect something more like (1.1^2 + 1/1.1)/(1 + 1) = 2.12/2 for 6% extra drag. Additionally, I don't think the reasoning is correct around the global minimum: If you go 10% below terminal velocity, air resistance is reduced by 0.9^2 and gravity drag increased by 10/9ths for either a multiplicative savings of 10% or additive savings of 1 - (0.9^2 + 10/9)/2 = 4%. Then either way, your argument suggests it is better to go below terminal velocity, because it reduces instantaneous total drag.

First, I mostly agree with you. Though I was hoping people would try to quantify statements like "highly dependant". Second. My question was explicitly about the lower atmosphere. Let's just say I completely agree with the upper atmosphere stuff. Third, your argument about the functional forms of the drags are relevant but incomplete for the optimization problem. Variations around the global minimum must be quadratic. That is, for some small velocity deviation away from the ideal velocity profile, the delta-v difference from ideal is symmetric. Going a little slower should be just as bad as going a little faster. When this matters then is when your initial TWR is low enough that it takes an excessively long time to get your velocity up near terminal velocity. I don't doubt that 2x terminal velocity is highly sub-optimal. I really intended to discuss say, 10% below terminal velocity to 10% above terminal velocity, just to make up some numbers. I've flown rockets with minimum TWR of 2.2, max 3+ at full throttle into orbit for less than 4400 m/s. (4396 m/s) I'm betting that that is a lot more efficient than most people would expect. The point is not that this is what people should do, but that it's not a bad thing to do. The difference from optimal is small. What I'm hoping to see is that people will demonstrate that being off by 10% of ideal velocity in either direction can result in less than a percent or two difference in delta-v to orbit. I see people are calling doom and gloom on losses to air resistance, without knowing that those excess losses plus the reduced gravity drag, can be quite small for non-trivial deviations from ideal. Additionally, starting with a low TWR (< 1.6) can completely swamp excess air drag of a too high TWR (2.2+). Obviously, a lot depends on the individual rocket. But for most rockets I fly, I find deviations from ideal have a real, but very small effect.

Howdy folks. This should be fun... The premise: In KSP, many people have independently verified that while in the thick soupy lower atmosphere, the optimal ascent velocity is terminal velocity. Additionally, this optimal ascent velocity in the lower atmosphere can be followed with a TWR of 2. The optimal ascent profile for any particular vehicle depends on the details of each stage. Let's assume the above three statements are true. Given the premise, here are two questions: 1) Which is better in the lower atmosphere: a) TWR <= 2, start each stage with TWR < 2 and end with TWR >= 2, or c) TWR >= 2? 2) How much does your TWR profile effect your ascent delta-v? The reason for this question is that I think that many people know approximately what is best, but don't know by how much. Similarly I think some people are wrong about what is best, but really are not too far from correct from a practical standpoint. Thus the second question: how much does you TWR profile really matter? If you disagree with the premise, that's a different discussion. Both discussions are welcome, but please indicate clearly if you are arguing against the premise rather than the questions. - - - Updated - - - Hmm. I tried to reply to my own thread, to keep the question and my answer in a different post, but the forum combined them into a single post... The following are my opinions. I'm open to being corrected (via data, theory, or pointers to other threads). I've done some testing to support it. I may post images later. I would be happy to see other people's tests first, so it's not just me filling the entire first page of this thread. 1) From best to worst: a) Terminal velocity ascent is best. A theoretical perfect KSP rocket would follow terminal velocity at least until the gravity turn. TWR > 2, but thrust limited to terminal velocity is second best. It is not cost effective, but it is fairly delta-v effective. c) TWR which keeps you near terminal velocity is third best. Split over and under terminal velocity based on changing TWR or ease of control or staging. d) TWR > 2, and velocity above terminal velocity. (Even up to 20% to 30% above terminal velocity. Maybe higher. More testing required.) e) TWR too low to reach terminal velocity f) TWR much less than 2. (1.2-1.5?) Now here's the kicker: What's the difference between a) and : < 10 m/s delta-v spent to orbit a) and c): < 15 m/s a) and d): < 25 m/s a) and e): < 50 m/s a) and f): could be very large These come from modifying a set of rockets to maintain certain TWRs. I don't have enough data at the moment to fully support this. The reason why going above terminal velocity is not that bad on your delta-v budget is because it reduces your gravity drag. The total drag goes up because atmospheric drag goes up faster than gravity drag goes down, but the total doesn't go up that quickly. The faster you get into your gravity turn, the less total gravity drag you pay. Additionally, for many engines, the Isp is worse at lower altitude. The faster you get out of the soup, the better that engine performs. 2) How much does the TWR profile matter for your final delta-v? For TWRs from 1.8 to 2.5, limiting yourself to terminal velocity or not, the delta-v difference from optimal ascent should be at most a few percent. For many people, I posit that that delta-v difference is insignificant. This is a gameplay style question more than a delta-v question: What is more important to you: optimizing your ascent delta-v or cost or part count or beauty or awesomeness or (lack of) time spent on design or science points gained? The delta-v difference is so small that only people who care about efficiency need to think about it at all. (Once you can perform a half-way decent gravity turn.)

Sure, sure. Perfect efficiency is not what I was going for. A little wastage can be just fine. We're not talking about optimal ascent here. Just making sure we can return from some other surface. Edit: If you want to talk about how bad a non-ideal TWR is, come join me.

A TWR of about 1.8-2.2 is about what you want for takeoff. That TWR is relative to the local gravity. Options: 1) Set MJ or KER or whatever mod you use for info to display TWR for your vehicle relative to some other planet or moon. 2) Find the TWR relative to Kerbin and multiply it by the surface gravity from the wiki and divide by 10 (Kerbin's approx. surface gravity.) Once in a stable orbit, TWR is less important. Low TWR in space just means longer burn times.

We're not forgetting the insertion burn. That is ÃŽâ€v2 above. We're just debugging the transfer burn calculation at the moment. Just for kicks, the above formula for the insertion burn at Moho from Kerbin gives about 2.4k m/s for injection to a 25km circular Moho orbit. Assumptions/idealizations/approximations: 1) circular orbits 2) zero inclination orbits 3) spacecraft exits/enters planet at the planet's orbital radius around Kerbol, rather than +/- the impact parameter (the distance between your SOI exit line and the planet prograde line) So, YMMV, by a lot of dV. This was just a quick lesson on how to compute combined escape + Hohmann transfer + insertion.

1) See above where I've fixed a mistake. That second sqrt() shouldn't have been there. Sorry about that, and thank you for checking my math. My initial post on how to derive the delta-v was correct (if somewhat simple) though. 2) Careful how you enter into wolframalpha. You made a couple mistakes. Here ya go: Idealized Kerbin -> Moho transfer burn. And let me know if you have any more problems.

Thanks. I was proud of that one. And I highly recommend Iain Banks. Start anywhere. The Culture series is not a linear series. Pick any one.

Stuff a <- root b -> stuff c -> not-root d -> stuff e To make d the new root (if it can be a root part): Take it apart into a, b, c, d and e. Delete part d. Delete part b. This forces you to choose a new root part. Choose a new part d. Put it all back together.

I either just make up ridiculous names, or make them reminiscent of Iain Bank's Culture series ship names. Bob's Fuel Depot (not the best name, but it had a long and glorious run) Pearl of Heaven, Explosive Class (lived up to it's name) LCU Fullness of Time LSV Getting Warmer (you can guess where this one went) LCU Frozen Entree (unplanned moho return vehicle with a Kerbal strapped in an exterior command seat. very sad year in KSP history) O. Cautionary Tale O. Unintended Consequences Quarrelsome III Zod needed a refuel before leaving Kerbin, so I sent up Fuel Before Zod. Dammit, Jeb (He pulled off some amazing stunts, but had to be rescued a few times.) Double Dammit, Jeb A pair of mini-rovers: Tiny Dancer and Tony Danza Let's see, acronymonious ships: Dres Remote Equatorial Survey Eve Verticality Examiner

Start with a circular orbit of radius r around the equator. At t = 0, T, 2T, ..., kT, you are back at (x,y,z) coordinate (r,0,0): x = r cos(Ét) y = r sin(Ét) z = 0 É = 2À/T = sqrt(μ/r3) Now rotate this orbit by the inclination angle α about the x axis: x -> x y -> y cos(α) - z sin(α) z -> y sin(α) + z cos(α) x = r cos(Ét) y = r sin(Ét) cos(α) z = r sin(Ét) sin(α) Now the orbit is correctly inclined, but the longitude of the ascending node is zero. So rotate the entire orbit about the z axis by Æ0, the new longitude of the ascending node: x -> x cos(Æ0) - y sin(Æ0) y -> x sin(Æ0) + y cos(Æ0) z -> z Thus x = r cos(Ét) cos(Æ0) - r sin(Ét) cos(α) sin(Æ0) y = r cos(Ét) sin(Æ0) + r sin(Ét) cos(α) cos(Æ0) z = r sin(Ét) sin(α) θ(t) = asin(z/r) = asin(sin(Ét) sin(α)) Æ(t) = atan(y/x) = atan((cos(Ét) sin(Æ0) + sin(Ét) cos(α) cos(Æ0)) / (cos(Ét) cos(Æ0) + sin(Ét) cos(α) sin(Æ0))) But note that you should use a 2-argument inverse tangent function atan2(y,x), because atan2(-y,x) is different from atan2(y,-x), which atan(y/x) can't tell. And as mentioned above, if you want the coordinates relative to the surface of the body you are orbiting, subtract the rotation of the body: Æ(t) -> Æ(t) - Ébt.

The above applies to planets in circular or nearly circular orbits. It also applies to transfers between two moons orbiting the same planet. And, of course, if body B has an atmosphere, you can avoid paying most of ÃŽâ€v2.

Ninja'd again, Dkmdlb! I'll simplify the above so people can plug in numbers: ÃŽâ€v1 = sqrt( (μS/RA) (sqrt(2RB/(RA+RB)) - 1)2 + μA (2/rA - 2/sA) ) - sqrt(μA/rA) ÃŽâ€v2 = sqrt( (μS/RB) (sqrt(2RA/(RA+RB)) - 1)2 + μB (2/rB - 2/sB) ) - sqrt(μB/rB) Pleasingly symmetric, as it should be. What it means is that you have to spend the Hohmann transfer ÃŽâ€v plus the planetary escape ÃŽâ€v burn minus your orbital velocity around the planet at each planet. N.B.: I've edited the above to change sqrt(2/rA - 2/sA) to (2/rA - 2/sA). Likewise for the second equation.

Let's go from planet A to planet B, where planet B orbits farther from the sun S. Let small first letters denote the ship, and capitol first letters denote the planets: rA = the ship's radius around planet A RA = the planet's radius around the sun S vA = the ship's velocity relative to planet A VA = planet A's velocity relative to sun S = sqrt(muS / RA) Additionally, let sA be the radius of the SOI of planet A. 1) Start at circular orbit around A: vA^2 = muA / rA 2) Burn dv1 up to velocity v1. We will determine dv1 and thus v1 later. v1^2 = (vA + dv1)^2 = muA (2/rA - 1/aA) or 1/aA = 2/rA - v1^2 / muA 3) Just before you leave planet A's SOI, you have slowed down to: v2^2 = muA (2/sA - 1/aA) v2^2 = v1^2 - muA (2/rA - 2/sA) 4) Once you exit A's SOI, (if you exit prograde) you gain planet A's orbital velocity VA: v3 = v2 + VA 5) Now pick your velocity v3 outside the SOI so that you intercept your target half way around the sun: So you are on an elliptical orbit around the sun with semi-major axis (RA + RB)/2, with velocity v3 at RA: v3^2 = muS (2/RA - 2/(RA+RB)) 6) You can solve for dv1 now, your transfer burn from planet A. Find v3. Solve for v2: v2 = v3 - VA Solve for v1. v1 = sqrt(v2^2 + muA (2/rA - 2/sA) dv1 = v1 - vA 7) Now do the reverse at planet B. Solve for the circularization burn at planet B. Before entering the SOI at planet B you are going v4: v4^2 = muS (2/RB - 2/(RA+RB)) 8) Entering the SOI, you lose B's orbital velocity: v5 = v4 - VB 9) Inside the SOI, find aB, the semi-major axis of your hyperbolic orbit of planet B: v5^2 = muB (2/sB - 1/aB) 10) At planet B periapsis, use 1/aB to find your velocity: v6^2 = muB (2/rB - 1/aB) = v5^2 + muB (2/rB - 2/sB) 11) Circularize at B periapsis: vB^2 = muB / rB dvB = v6 - vB

Check the wiki: http://wiki.kerbalspaceprogram.com/wiki/Parts

Practice docking a small ship. It makes everything so much easier.

Hehe. I get caught trying to rotate images or videos too... Of course you can throttle down to keep your TWR where every you like, but it is less efficient. If you don't care about efficiency, don't sweat it. Some people care about such things. Some don't. As always, play the way you want to play.

I'll write up the derivation in a tutorial in a day or two. Like I said in the other post, all you need is the vis-viva equation. The numbers should make sense though, if you think about it. At Minmus, you are nearly at Kerbin escape velocity. Thus you don't need much of a boost to get to Eve or Duna, compared to the extra dV it takes to get back to low Kerbin periapsis. But here is the clearest example of the Oberth effect I can offer: For an interplanetary transfer from Minmus, your starting velocity is about 150 m/s, iirc. Direct interplanetary transfers need like 10-20 m/s to escape Kerbin. After that you are using the escape Kerbin and then burn method, more or less. Much more expensive than direct from Kerbin. On the other hand, it costs about 100 m/s to get back to a 100 km Kerbin periapsis, and when you get there, you'll be going, (I don't remember off-hand) 3 km/s? So compare the Oberth effect at 150 m/s vs 3 km/s. Now these are large maneuvers, so you can't use the small dV approximation: de = v dV. (For a small dV, the change in specific orbital energy is approximately equal to your velocity times the small dV spent: de = 1/2 (v+dV)^2 - 1/2 v^2 ~= v dV.) For a small burn, you get 20 times (3000/150) the specific orbital energy on a slingshot past Kerbin from Minmus that you do from a burn at Minmus.

You have that backwards. Look at the table I posted earlier. If you are at Minmus and want to go to distant planets, you can double the cost by going direct. It's not quite that bad for Eve or Duna.