Delta V calculations accuracy

[Sans Solo]

Hello guys I've been playing KSP for about 1 year and 5 months and lately I've been taking it seriously, I am currently building a rocket to the Mun and have been calculating my delta V so that I can efficiently travel there. But, unfortunately my calculations seem to be off (even if I use the Vac and ASL values for each engine/stage) and the speed indicator value is always less than what I have calculated. Is there something that I do not know yet or have forgotten to factor like aerodynamic drag?

The two formulas I use are:

The Tsiolkovksy Rocket Equation

delta V = V_Exhaust x ln( Mass_INITIAL / Mass_FINAL )

And

V_Exhaust = g (I_sp)

[Rocket In My Pocket]

Cough...cough...

 

[dafidge9898]

2 hours ago, Sans Solo said:

Is there something that I do not know yet or have forgotten to factor like aerodynamic drag?

delta V = V_Exhaust x ln( Mass_INITIAL / Mass_FINAL )

V_Exhaust = g (I_sp)

If you are calculating total Dv, including during the ascent, then yes, you need to take the atmosphere into consideration. The amount of Dv on your ship decreases if you're in the atmosphere, because when in the atmosphere, the exhaust has to push out against the air. I don't know how to compensate for drag/atmospheric pressure, but it probably involves a lot of calculus because of the all the different and changing factors like speed, gravity losses, atmospheric pressure, etc. Oh, there's also the issue of gravity losses. The lower thrust you have during ascent, the more gravity losses there are. Your best bet is to not bother with calculating the first two stages of your rocket (the ones that get you into orbit), and whatever fuel and tanks that are left in them you jettison. Just calculate the Dv of your transfer stages and whatever stages you are only going to use in space. That way you won't have to worry about the atmosphere. 

 

Edit: Or, you could just follow above's advice and save yourself the trouble and math

Edited July 25, 2016 by dafidge9898

[Pecan]

3 hours ago, Sans Solo said:

...The Tsiolkovksy Rocket Equation...

... assumes instantaneous velocity change, = infinite TWR.  You don't have it, so won't get a 'perfect' burn.  It's effectively impossible to calculate accurate burns through an atmosphere and in every other case it's best to add 10% (or so) anyway, just as a margin for error.

[FullMetalMachinist]

19 minutes ago, Pecan said:

4 hours ago, Sans Solo said:

...The Tsiolkovksy Rocket Equation...

... assumes instantaneous velocity change, = infinite TWR.

No, no, that's not it. The maneuver node code in KSP assumes instantaneous velocity change. The Tsiolkovsky equation doesn't care at all about TWR. 

22 minutes ago, Pecan said:

You don't have it [infinite TWR], so won't get a 'perfect' burn.

That is the issue with the node predictions where the lower your TWR, the further from the "ideal" result of the node you will be. But it has nothing to do with how much dV your ship has. 

@Sans Solo, for your problem, it sounds like you're doing the math right, just that the atmosphere is tripping you up, as others have said. If you don't want to rely on a mod to do the math for you, then I would say just ignore the atmosphere part of the journey when you do your calculations. 

Pretend that you always start in low orbit, and do all the math from there, using the vacuum numbers. Then take the resulting craft, and stick it on top of a lifter that is capable of getting the whole thing into low orbit. 

[wumpus]

16 hours ago, Pecan said:

... assumes instantaneous velocity change, = infinite TWR. 

It certainly accounts for fuel useage over time, so it isn't really an instantaneous issue.  It does ignore gravity losses (such as going to orbit from Kerbin), so the bits when TWR<<2 is going to "cost" more than twice the delta-v the rocket equation gives you.  Once you are going effectively sideways (and your TWR is more a matter of getting into orbit before falling back down) even this doesn't matter.  There is also the aero drag on the vessel, which is much harder to calculate (note how long it took Squad to even compute an effective instantaneous drag needed for a more accurate KSP).

Edited July 26, 2016 by wumpus
really stupid writing fixed

[OhioBob]

As others have said, ascending through the atmosphere during launch will result in gravity and drag losses.  There is no easy way to compute what these losses will be but, for a launch from the surface of Kerbin, the losses are usually about 1000-1200 m/s.  That is, to attain a final orbital velocity of 2300 m/s will require a launch vehicle Δv of about 3400 m/s (computed using vacuum Isp).

Even after attaining orbit, you will still find that the Δv you add during a maneuver is not the same thing as the final speed minus the initial speed.  This is because during a maneuver your spacecraft is either increasing or decreasing in altitude, and this change in altitude results in a change in velocity.  For example, say you are performing a propulsive maneuver to raise the apoapsis of your orbit to intercept Mun.  Typically this is about a 860 m/s Δv maneuver.  As you burn the engine the spacecraft speeds up and its altitude begins to increase.  Because it is gaining altitude, some of the spacecraft's kinetic energy is being converted to potential energy, resulting in a loss in velocity.  So while you are adding kinectic energy during the engine burn, some of that is being lost as a result of the increasing altitude.  You will find that the difference between your final and initial speeds is less than the amount of Δv added.  For instance, the initial velocity might be 2280 m/s, you add 860 m/s Δv, and your velocity at engine shutdown is 3100 m/s.  We see that 3100 - 2280 = 820 m/s.  Don't worry about this, it is normal.  In this example we "added" 860 m/s Δv even though it might not appear like it when we observe the before and after speeds.

[Zhetaan]

@Sans Solo:

Most likely, you are seeing a difference between your results and predictions because of a combination of cosine losses (very minor, unless you accidentally started by thrusting in the wrong direction) and gravity losses, that simply indicate in this case that you've changed orbits with a non-instantaneous burn ... which you already knew and others here have said.

There are three major sources of loss that I know of:  drag, gravity, and cosine.

Drag is atmospheric loss--a problem on launch but not on the Mun.  There's no easy way to figure it because it depends very much on both the design of the rocket and on the skill of the pilot.

Gravity loss is the effort you need to put into preventing falling down to the surface--inasmuch as orbit consists of going sideways relative to the direction of gravity such that you miss the ground, you encounter gravity losses during the parts of the burn when your rocket is not burning at a ninety degree pitch.  That's easier to figure out, but not on the fly.  It would be possible, for example, to take off from the Mun's surface, pitch over to horizontal immediately, and attempt to make orbit that way.  Assuming no terrain obstacles (we can call hitting a mountain a special case of drag losses), only the most insanely powerful of rockets would be able to add horizontal speed quickly enough to make orbit before the vertical component (gravity) pulled it back to the ground.  Therefore, the rocket needs to devote some thrust to countering that vertical component.  The key to it is that the most efficient ascent is the one that only devotes exactly the required thrust to keeping your vessel--well, aloft seems entirely the wrong word for airless bodies--but without unnecessary thrust additions (wasting fuel) or subtractions (causing you to lose altitude that you have to make up again ... wasting fuel).  For an easier example, imagine a hypothetical craft that always has a TWR of exactly 1.  It can float, but it cannot make orbit, because all of its thrust goes to keeping it off the ground and none goes to adding orbital energy.  Contrariwise, such a craft already in orbit is much more capable, because it comes loaded with the orbital energy; there's no risk of falling into the ground, so you can afford to take your time with less powerful-but-more-efficient rockets.  This one is mostly dependent on pilot skill; to minimise the loss, look up constant-altitude approaches and landings.

Cosine loss is the part of the thrust vector not directed through the rocket's centre of mass.  This component may be utterly necessary to your survival (using gimballed engines to avoid a special case of drag losses, for example), or it may be a convenience.  A common source of cosine loss is in unbalanced or asymmetric rockets.  The stock shuttles (and the real Space Shuttle) have high cosine losses because the engines have to point in an odd direction to compensate for the big, engineless fuel tank's effect on the centre of mass.  You can encounter it also in situations where you have an oddly-oriented part as a control point (such as a surface-mounted docking port) such that when you point it prograde, your engines point elsewhere.  You can throw away an entire tank of fuel and go nowhere; for an exaggerated example, imagine a booster attached sideways to a pod on the pad.  You can point prograde, thrust all day, and never go up; that's cosine loss.  This one is dependent mostly on rocket design, though it's completely unavoidable unless you design your rocket with no gimbals, asymmetric thrusters, unbalanced payloads, control surfaces, or RCS jets.  Reaction wheels are okay.

[Spricigo]

Well,  there is 3 points to address there.  

 

First : Tsiolkovksy Rocket Equation only account for the variation of velocity caused by propulsion.  Was well explained above: there's drag and gravity to take in account.

Secound: we are dealing with vectors. We need to take in account direction and orientation. If your exhaust  vector  it's not perfectly lined with your velocity vector part of the effect will be change in direction not intensity. (particularly common during less than perfect gravity turn and engine ginbaling to counter ship wobbling) 

Third: maneuvering takes time,  and conditions may change during that time. 

[Red Iron Crown]

Steering losses need to be considered as well. Any propellant expended when not pointing exactly prograde or retrograde will have a lesser effect on vessel speed. For an extreme example, put SAS in Normal+ mode and burn, your craft will consume delta-V but speed won't change at all.

[Pecan]

16 hours ago, FullMetalMachinist said:

No, no, that's not it. The maneuver node code in KSP assumes instantaneous velocity change. The Tsiolkovsky equation doesn't care at all about TWR. 

It takes something like 850m/s dV from LKO to Ap at Mun altitude.  Please try this with a constant burn at a TWR of 0.1.  Moving kind of blurs the results, doesn't it?  The equation doesn't care about thrust, per se, but it answers the question "what does it take to get from here to there" and, unfortunately for accuracy, we are never at "here" for most of the burn.

[Sans Solo]

16 hours ago, FullMetalMachinist said:

No, no, that's not it. The maneuver node code in KSP assumes instantaneous velocity change. The Tsiolkovsky equation doesn't care at all about TWR. 

That is the issue with the node predictions where the lower your TWR, the further from the "ideal" result of the node you will be. But it has nothing to do with how much dV your ship has. 

@Sans Solo, for your problem, it sounds like you're doing the math right, just that the atmosphere is tripping you up, as others have said. If you don't want to rely on a mod to do the math for you, then I would say just ignore the atmosphere part of the journey when you do your calculations. 

Pretend that you always start in low orbit, and do all the math from there, using the vacuum numbers. Then take the resulting craft, and stick it on top of a lifter that is capable of getting the whole thing into low orbit. 

@FullMetalMachinist Thanks for the answer, yeah I really want to know the whole mathematics of it before I get to use the aid mods, because I'm planning to use this to pursue rocketry as a hobby and probably as a profession too.

So I understand that the calculations would work better for orbit and factoring aerodynamics, gravity loss and vectors would be tedious. But do you have any formula I could use so that I may be able to launch my vehicle to orbit in the most efficient way? Any tips?

[OhioBob]

Quote

So I understand that the calculations would work better for orbit and factoring aerodynamics, gravity loss and vectors would be tedious. But do you have any formula I could use so that I may be able to launch my vehicle to orbit in the most efficient way? Any tips?

@Sans Solo, there is no single formula that can do this.  There is a reason the term "rocket science" is used to mean a very complex and difficult task.  I think that most people figure it out by extensive experimentation.  I studied the problem by writing a computer simulation and performing repeated simulated launches until I found what I thought was an ideal design.  The simulation involved many formulas; way more than I care to explain.

There are also different ways to define "most efficient".  If by most efficient you mean getting to orbit for the lowest Δv, then that's very different from getting to orbit for the least cost per unit mass of payload.  Generally speaking, trying to minimize Δv results in a bad design; you end up with a big overpowered rocket that's a lousy payload delivery system.  I generally lean toward maximizing payload unit cost, which results in a very design in terms of TWR.  I ended up boiling my rocket design method down to just a few simple rules that I think work pretty well.  I describe it here:

And below is a thread in which I compared different optimization methods, i.e. least Δv versus payload efficiency.

 

 

[Spricigo]

2 hours ago, Pecan said:

It takes something like 850m/s dV from LKO to Ap at Mun altitude.  Please try this with a constant burn at a TWR of 0.1.  Moving kind of blurs the results, doesn't it?  The equation doesn't care about thrust, per se, but it answers the question "what does it take to get from here to there" and, unfortunately for accuracy, we are never at "here" for most of the burn.

Actually the equation don't even consider position also.  It only says how much [ship velocity] (a vector )  change based in [propellant exhaust velocity]  (also a vector) . 

 

The problem with long burn it's  that either

-exhaust velocity it's not lined up with ship velocity,  resulting in part of it affecting ship velocity direction. 

-exhaust velocity is lined up with ship velocity at all times but ship velocity is changing direction,   so exhaust velocity it's not constant either. (part of the early burn will cancel out part of the later burn) 

 

Now you should notice that either way there will be some 'wasted delta-V'.  And maybe you are asking why.  The reason it's that gravity is still there and you still need to figth it's effects. As stated above,  TRE don't account for gravity effects. 

[Sans Solo]

Quote

there is no single formula that can do this.  There is a reason the term "rocket science" is used to mean a very complex and difficult task.  I think that most people figure it out by extensive experimentation.  I studied the problem by writing a computer simulation and performing repeated simulated launches until I found what I thought was an ideal design.  The simulation involved many formulas; way more than I care to explain.

@OhioBob Thanks for the data and the nugget of criticism. I may have constructed my question poorly. Though I'm no stranger to rocket science. I was asking in terms of dV to fuel consumption and how to expend the dV efficiently in terms of flying the spacecraft. Though I appreciate the data.

[OhioBob]

Below is a nice summary and explanation of the losses from the book Orbital Mechanics, Theory and Applications by Tom Logsdon.

Spoiler

 

Quote

I was asking in terms of dV to fuel consumption and how to expend the dV efficiently in terms of flying the spacecraft.

I would say that for most orbital maneuvers you can use your Δv most efficiently by minimizing steering losses and by taking advantage of the Oberth effect.