Working out Delta V for Kerbin Orbit - Problem with my maths?

[yurivonkerbin]

Hi all,

I'm working out some of the basic calculations for a ground-to-orbit insertion for Kerbin as one of my first challenges in KSP and I appear to be getting something wrong. Currently I am using the following calculation from the normally very useful 'Atomic Rocket' page on designing realistic space vehicles in science fiction:

ÃŽâ€_vo = sqrt[ (G * P_m) / P_r ]

where:

• ÃŽâ€_vo = deltaV to lift off into orbit or land on a planet from orbit (m/s)
  
• G = 0.00000000006673 or 6.673e-11 (gravitational constant)
  
• P_m = planet's mass (kg)
  
• P_r = planet's radius (m)
  
• sqrt[x] = square root of x
  

Now, the KSP wiki says you need a Delta V of 4500 m/s for a stable orbit (mentioned on the Kerbin entry in paragraph 3) so I wanted to see if I could work it out myself and thus be satisfied my maths was correct. I worked it out by hand initially and was a little puzzled that I seemed to be coming out at a figure only around half of the 4.5km/s mentioned above. So I put together an Excel spreadsheet. As you can see below, the planetary mass and planetary radius are taken from the Kerbin entry on the wiki, and the gravitational constant is what I understand to be the standard one used in RL and in KSP;

EDIT: Please note that 'Gm/Pr' should read 'GPM/Pr' to confirm where that calculation comes from.

As you can see, my final calculation is still coming out only around half of the 4,500 m/s Delta V mentioned on the KSP wiki. Can anyone suggest what I'm doing wrong here?

Many thanks for the assistance, I'd really like to be able to work this out.

[Red Iron Crown]

I think you've only calculated what your orbital speed would be at an orbital radius of 600km, which is not possible (you'd hit things).

[yurivonkerbin]

Ah, so the Planetary Radius should be the Orbital Radius I am aiming for?

[Red Iron Crown]

Not quite. The orbital radius should be PlanetaryRadius + OrbitalAltitude. So for the typical 70km orbit, the orbital radius is 670km.

Even then, that equation just gives you the horizontal speed required in orbit at a specified altitude, it does not consider the gravity losses and aerodynamic losses to get there. Computing those is non-trivial, I'm not sure how it's done. I think the 4,500m/s figure is derived empirically.

Edit: I misunderstood your question. Yes, where you are putting planetary radius you should be putting desired orbital radius.

Edited June 23, 2014 by Red Iron Crown

[Yasmy]

1) What you have calculated is the orbital velocity at surface level, if Kerbin did not have an atmosphere.

For orbital velocity at some radius above the surface of the planet, add the orbital altitude to the planetary radius: v = sqrt(G Pm / (Pr + altitude))

2) A rotating planet has a non-zero surface velocity. Depending on which direction you head into orbit relative to the surface velocity, the cost to orbit is increased or decreased by something less than or equal to the surface velocity. This is why people pitch over east to orbit Kerbin. It reduces the cost of orbital insertion.

If you did this calculation (orbital velocity +/- component of surface velocity along orbital inclination) for an airless planet or moon, you would roughly get the correct delta-v to orbit. To achieve an orbit around a body with an atmosphere there are two other costs:

3) Atmospheric losses: Atmospheric drag is proportional to the atmospheric density and the square of your velocity. You have to expend a lot of delta-v to push though the atmosphere.

4) Gravity losses: Any fuel spent burning radially away from a planet is lost delta-v. Until you pitch over, almost all of your delta-v is lost to gravity (and atmospheric drag on a body with an atmosphere). To orbit, you have to expend delta-v perpendicular to the body's radius.

Gravity losses also apply around airless bodies. Any delta-v spend going up to avoid mountains, or to give your engines time to get to orbital velocity is delta-v lost to so-called gravity drag.

It depends on your flight profile, but roughly speaking, gravity drag and atmospheric drag for Kerbin orbit are both around 1 km/s.

Edit: triple ninjaed!

Edited June 23, 2014 by Yasmy

[yurivonkerbin]

1) What you have calculated is the orbital velocity at surface level, if Kerbin did not have an atmosphere.

For orbital velocity at some radius above the surface of the planet, add the orbital altitude to the planetary radius: v = sqrt(G Pm / (Pr + altitude))

Gotcha, I'll edit my calculator.

2) A rotating planet has a non-zero surface velocity. Depending on which direction you head into orbit relative to the surface velocity, the cost to orbit is increased or decreased by something less than or equal to the surface velocity. This is why people pitch over east to orbit Kerbin. It reduces the cost of orbital insertion.

Okay, so I need to pitch east on an orbit over Kerbin and subtract the surface velocity from the Delta V I need. The surface velocity; would that be the Sidereal Rotational Velocity listed on the Kerbin wiki page?

If you did this calculation (orbital velocity +/- component of surface velocity along orbital inclination) for an airless planet or moon, you would roughly get the correct delta-v to orbit. To achieve an orbit around a body with an atmosphere there are two other costs:

3) Atmospheric losses: Atmospheric drag is proportional to the atmospheric density and the square of your velocity. You have to expend a lot of delta-v to push though the atmosphere.

4) Gravity losses: Any fuel spent burning radially away from a planet is lost delta-v. Until you pitch over, almost all of your delta-v is lost to gravity (and atmospheric drag on a body with an atmosphere). To orbit, you have to expend delta-v perpendicular to the body's radius.

Gravity losses also apply around airless bodies. Any delta-v spend going up to avoid mountains, or to give your engines time to get to orbital velocity is delta-v lost to so-called gravity drag.

It depends on your flight profile, but roughly speaking, gravity drag and atmospheric drag for Kerbin orbit are both around 1 km/s.

Okay, so I have to add 2000 m/s to my final Delta V to take account of Gravity and Atmospheric losses. How is that actually worked out? I.e., how would I work that out if I was taking off from another planet with an atmosphere but different qualities of that atmosphere to Kerbin?

Many thanks; I appreciate all of the assistance so far.

[Yasmy]

Okay, so I need to pitch east on an orbit over Kerbin and subtract the surface velocity from the Delta V I need. The surface velocity; would that be the Sidereal Rotational Velocity listed on the Kerbin wiki page?

Yes. So for Kerbin, it should cost about 2*174 m/s extra to insert into a retrograde orbit, or just 174 m/s extra to insert into a polar orbit. Note that the approximately 4500 m/s to Kerbin orbit that many people quote already assumes launching east.

Okay, so I have to add 2000 m/s to my final Delta V to take account of Gravity and Atmospheric losses. How is that actually worked out? I.e., how would I work that out if I was taking off from another planet with an atmosphere but different qualities of that atmosphere to Kerbin?

I'm not aware of any simple approximate formulae, though a simple ballpark approximation should be possible. But you can't get better than approximate for a problem this complex. Otherwise there are two routes to go:

1) Full simulation

2) Trial and error

A few people on the forums have written their own KSP physics simulators to calculate the costs to orbit through an atmosphere. The answer depends on the details of your rocket and trajectory. They calculate delta-v by integrating the equations of motion as the rocket ascends.

On the other hand, many people have lifted off from every landable body in KSP many times, and through trial and error have created delta-v maps as guidelines.

See, for example, the delta-v map on wiki cheat sheet page.

Edited June 24, 2014 by Yasmy

[GoSlash27]

*edit* looking at that equation, it wouldn't be able to approximate a DV requirement to land or launch, even from an airless body. It looks like that just gives an orbital velocity for a known altitude from a planetary center.You'd need to know the rotational period of the body, it's radius, it's mass, and your orbital altitude (assuming a prograde orbit) to be able to calculate a DV. It would be your orbital velocity minus the planet's rotational velocity.

But once you factor in atmospheric drag (something that cannot be done with calculus), you end up with a completely different number.

Anywho... the process would be to calculate orbital velocity at your chosen altitude. Rp+A. Your equation looks fine to do that part. Next, you would work out surface rotation velocity by radius and period Vs= 2pi r(m)/p(sec). Then Vo-Vs would give you your DV. If you were to correct for inclination, you would multiply Vs by the cosine of your ascention, assuming that 0* ascention is aligned with the equator prograde.

Gravity losses are generally minor. That'd be dependent on the sine of your pitch vector which applies 1 local G of acceleration downward, regardless of what your engines are generating. Since the bulk of an approach or departure are tangent to the surface, gravity losses don't account for much. Not something that you can compute, since your approach profile determines it.

The big killer is atmospheric drag, which cannot be mathematically solved. It must be either simulated or gathered from empirical results.

Best,

-Slashy

Edited June 24, 2014 by GoSlash27

[Specialist290]

In addition to what's already been said, you may find this page to be most enlightening on the subject.

[yurivonkerbin]

Okay, progress so far:

Using good ol' fashioned pencil, paper and scientific calculator, I have so far determined that I need a Delta V of approximately 4136 m/s minimum to reach Kerbin Orbit. This is based on the total of the following (with numbers rounded up to the nearest m/s);

- 2,279 m/s gross Delta V for orbit operations (before corrections of atmo, grav, safety)

- 422 m/s for gravitational drag

- 610 m/s for atmospheric drag (standard addition for Earth's atmo drag, according to sources)

- 1,000 m/s safety margin

- Minus 175 m/s sidereal rotation of Kerbin for a due east launch direction

This roughly corresponds with the 4,500 m/s approx figure on the wiki, so I seem to have worked that out okay.

My next problem is getting the thing into orbit in terms of fuel;

- In my calculations, the RT-10 that I was planning to use as my launch stage (I am using the career mode and want to use only the parts I have available as I progress with research) has a thrust of 250,000 newtons and my current craft weight (RT-10 included) is coming out at 4,648kg. This results in an acceleration of 53.79 m/s (roughly 5.5 gees) based on the formula

Acceleration = Thrust / Craft Mass

.

- This gives me a take-off duration of approximately 43 seconds based on the formula

Take off Duration = Delta V for Orbit (m/s) / Craft Acceleration (m/s)

based on the power of the RT-10 and the assumption that I use the Gross Delta V (2,279) in this calculation, not the total Delta V with all the grav, atmo and safety corrections.

- However, the RT-10 empties the tank in approximately 28 seconds, leaving me 15 seconds short of the required burn time that (if I have worked this out correctly) I require at that engine thrust to get into orbit.

So, do I therefore need a second stage? This is obviously going to increase the weight of the craft so I will have to do some re-calculation, but would I be looking at needing an upper stage based on a Liquid Fuel Engine (I believe the LV-T30 is available as a default part at the start of the career mode)?

I really appreciate all the help so far on this; I was never really any good at maths when I was at school - despised it, in fact - but KSP has actually got me interested in having a go at learning and working this out for myself, so it's rather important to me to be able to calculate the majority of this to get a real sense of achievement for my first orbital accomplishment.

Many thanks, fellow Kerbanauts!

[rkman]

- 422 m/s for gravitational drag

- 610 m/s for atmospheric drag (standard addition for Earth's atmo drag, according to sources)

- 1,000 m/s safety margin

Just for reference, realistic gameplay numbers are

~1400m/s gravity loss

~800m/s drag loss (Kerbin's atmosphere is nothing like Earth's)

i give myself ~150m/s safety margin.

Not to discourage, but you're not the first to try this, and the maths required might be a bit more involved than you have anticipated: http://forum.kerbalspaceprogram.com/threads/46194-I-need-someone-help-me-do-some-math-for-launch-optimization

[yurivonkerbin]

Just for reference, realistic gameplay numbers are

~1400m/s gravity loss

~800m/s drag loss (Kerbin's atmosphere is nothing like Earth's)

i give myself ~150m/s safety margin.

Not to discourage, but you're not the first to try this, and the maths required might be a bit more involved than you have anticipated: http://forum.kerbalspaceprogram.com/threads/46194-I-need-someone-help-me-do-some-math-for-launch-optimization

Oh wow, I was really off, wasn't I?

Thanks for the link; I will take a browse through it and see what I can discern.

[Pecan]

...- This gives me a take-off duration of approximately 43 seconds based on the formula Take off Duration = Delta V for Orbit (m/s) / Craft Acceleration (m/s) based on the power of the RT-10 and the assumption that I use the Gross Delta V (2,279) in this calculation, not the total Delta V with all the grav, atmo and safety corrections...

I think I've got a bit lost somewhere. Fine, you want to work out all the maths yourself and have determined that, indeed, you need about 4,500m/s to launch.

So why were you planning on using an RT-10, knowing it only gives 2,279m/s? Clearly you need another stage but the one thing not immediately available in career mode is a decoupler :-( If you want to SSTO from the very start use a T-30 - which is one of the game's all-around best engines - and a stack of fuel tanks.

Alternatively, if you really want to use the RT-10, you'll need the decoupler and so second-science-stage Basic Rocketry, which is simple enough. Since this still needs all the mass of the T-30 it is heavier and more complex than just a single-stage though.

[yurivonkerbin]

I think I've got a bit lost somewhere. Fine, you want to work out all the maths yourself and have determined that, indeed, you need about 4,500m/s to launch.

So why were you planning on using an RT-10, knowing it only gives 2,279m/s?

Sorry, just to clarify; the 2,279 m/s is the minimum Orbital Delta V I would need to get into orbit (minus corrections for grav, atmo, sidereal), regardless of engine type. The RT-10 is just the first solid booster available so I naturally thought of using it. I suppose if the RT-10 only burns for 28 of the 43 seconds I need than it's giving me (4137 m/s / 43 * 28 =) 2,413.25 m/s, which is 58% of total m/s I require.

However, if it makes more sense to use the T-30 with a stack of tanks, I will rework my calculations on design to use that. I suppose the T-30 is also better given the ability to throttle and cut-off a liquid engine as well. Thanks for the advice.

[Mr Shifty]

So, do I therefore need a second stage? This is obviously going to increase the weight of the craft so I will have to do some re-calculation, but would I be looking at needing an upper stage based on a Liquid Fuel Engine (I believe the LV-T30 is available as a default part at the start of the career mode)?

It's fantastic that you're working this out yourself. One of the great things about KSP is getting to apply all those math skills you never thought you'd use. It sounds like you're starting to notice the tyranny of the rocket equation, which causes your total fuel required to increase exponentially with payload weight. (As your payload weight increases, you need more fuel to lift it into orbit, but that fuel adds weight, which means you need even more fuel, etc etc.) There are some rough rules of thumb that have mostly been determined empirically:

- With decent staging and rocket design, you can count on a mass fraction (payload weight/fuel weight) of about 10% to lift stuff to orbit. So if your payload weighs 10 tons, you'll need at least 100 tons of fuel. (Very good rocket designers can add a few percent to this number, but it's extremely difficult to get mass fractions above about 15%.)

- There's also an upper limit to the total amount of dV you can stuff onto a rocket with the vanilla engines currently in KSP, and it's somewhere less than 20 km/s. Getting even 10 km/s onto a KSP rocket can be a challenge.

- Thrust is an important part of the calculation. As others have pointed out, you lose quite a bit of dV to gravity as you ascend. With more thrust, you can dedicate more of your acceleration to increasing your speed, rather than resisting gravity. This is simple trigonometry: increasing orbital speed requires thrust that is parallel to the planet's surface. Resisting gravity requires thrust that is perpendicular to the planet's surface. Since gravity is roughly constant during ascent (it decreases only slightly), the perpendicular component of thrust that you need is fixed. (It is equal to g.) BUT, if you accelerate too fast, you'll start to lose too much dV to air resistance. dV lost to air resistance does not depend on your ascent angle, but does depend heavily on your altitude and speed. The math is too much for me (with a graduate degree in engineering), but it's been shown that a Trust-to-Weight Ratio (TWR) of 2.0 is optimal for atmospheric ascents. TWR=2 gives you the maximum thrust without losing excess energy to air resistance.

- Likewise, it has been shown, using a bunch of maths (and also empirically) that the optimal speed during ascent is terminal velocity. Terminal velocity changes with altitude. There's a table on the wiki that shows TV values for various heights. I personally use an add-on called Kerbal Engineer Redux, which shows current terminal velocity, as well as a percent atmospheric efficiency (i.e. your speed/current TV.)

- Note that accelerating straight up never helps you with orbital speed. The only reason you accelerate upwards with launching is to punch through the thick lower layers of atmosphere as quickly as possible, so that you can increase your speed without violating the terminal velocity rule. This is why there's a strange 'gravity turn' rule in KSP, where you burn straight up for 8-10km, then start to turn your vessel (usually east to take advantage of Kerbin's spin) to burn more parallel for orbital speed. On an airless world, you'd gain enough altitude to not hit mountains, then burn sideways as much as possible, keeping your altitude constant, until you reach orbital velocity. On atmospheric worlds, you have to keep enough vertical thrust to raise your trajectory above the atmosphere.

[Pecan]

...However, if it makes more sense to use the T-30 with a stack of tanks, I will rework my calculations on design to use that. I suppose the T-30 is also better given the ability to throttle and cut-off a liquid engine as well. Thanks for the advice.

Exactly so. You're welcome, and well done for doing all the working-out the hard way :-)

[GoSlash27]

The equation you really need to plan your launch vehicle is this one:

ÃŽâ€V= 9.81Isp*ln(Mw/Md)

where Isp is the specific impulse of your engine in seconds, ln denotes a natural logarithm, Mw is the mass of your rocket with fuel and Md is the mass of your rocket without fuel.

As you have already surmised, what you can do is determined by how much ÃŽâ€V you can generate.

This equation is in the form of "this much rocket will make this much ÃŽâ€V". A much more useful version would say "in order to make this much ÃŽâ€V, you need this much rocket". I will show the derivation of such an equation below. If you want to derive it yourself, don't highlight the following text...

Mw/Md is what the rocket surgeons refer to as the "mass ratio", which I shall refer to as Rm.

Since the mass of fuel and tanks you need are wholly dependent on the mass ratio, first order of business is to rewrite the main equation to solve for mass ratio.

ÃŽâ€V=9.81*Isp*ln(Rm)

ÃŽâ€V/(9.81*Isp)=ln(Rm)

e^(ÃŽâ€V/9.81*Isp)=Rm

Next is to define which parts of our mass define the numerator and denominator of our mass ratio.

Mt= Mass of our tanks fully loaded

Me=Mass of our engines

Mp= Mass of our payload

and Rfe= the ratio of of the mass of our full tanks to empty. You would simply take the loaded mass of your tank and divide it by the empty mass. Thankfully, all small and large diameter liquid tanks work out to "9".

Expressed this way, Rm=(Mp+Me+Mt)/(Mp+Me+Mt/Rfe)

We need to rewrite this to solve for Mt.

Rm=(Mp+Me+Mt)/(Mp+Me+Mt/Rfe)

Rm(Mp+Me+Mt/Rfe)=Mp+Me+Mt

Rm(Mp+Me+Mt/Rfe)-Mp-Me=Mt

but we have an "Mt" on the wrong side of the equation, so...

Rm(Mp+Me+Mt/Rfe)-Mp-Me=Mt

RmMp+RmMe+RmMt/Rfe-Mp-Me=Mt

commutative

RmMp-Mp+RmMe-Me+RmMt/Rfe=Mt

Distributive corollary

Mp(Rm-1)+Me(Rm-1)+RmMt/Rfe=Mt

Distributive corollary

(Mp+Me)(Rm-1)+RmMt/Rfe=Mt

rewrite that a bit...

(Mp+Me)(Rm-1)+Mt(Rm/Rfe)=Mt

and move it back over

(Mp+Me)(Rm-1)=Mt-Mt(Rm/Rfe)

Reverse distributive

(Mp+Me)(Rm-1)=Mt(1-Rm/Rfe)

(Mp+Me)(Rm-1)/(1-Rm/Rfe)=Mt

We can tidy it up a bit by replacing "1" in the denominator with "Rfe/Rfe"

(Mp+Me)(Rm-1)/(Rfe/Rfe-Rm/Rfe)=Mt

(Mp+Me)(Rm-1)/((Rfe-Rm)/Rfe)=Mt

And since division by a fraction is the same thing as multiplying by it's inverse...

(Mp+Me)(Rm-1)*(Rfe/(Rfe-Rm))=Mt

so

Rfe(Mp+Me)(Rm-1)/(Rfe-Rm)=Mt

coupled with the reverse of the ÃŽâ€V equation above, you can substitute it in.

Rfe(Mp+Me)(e^(ÃŽâ€V/9.81*Isp)-1)/(Rfe-e^(ÃŽâ€V/9.81*Isp))=Mt

You now have the power to determine exactly how much fuel you need to make a rocket generate a needed ÃŽâ€V given the engine specs, tank specs, and payload you need to move.

This also applies to staging, where you treat subsequent stages as payload for earlier stages. The design process is to start at the end of your mission and work backwards.

I need to put a capsule in orbit that can support a Kerbal and return to Kerbin. That will weigh this much. It needs to have solar panels and the ability to dock, so it needs RCS junk, batteries, etc. And we'll deorbit the heavy stuff and let it burn up so our Kerbal has just the parachute and reentry shield, so add in a decoupler. The whole orbital vehicle will weigh this much.

That's payload. So how much to get this from exiting the atmosphere to orbit? That's this much ÃŽâ€V. This engine looks good for that, so plug it in. That creates a stage with such-and-such tanks and such-and-such engines, for a total mass of this much. How much ÃŽâ€V

to get it from the gravity turn to space? Should I break it down into multiple stages?

Rinse and repeat until you've designed all the way back down to the pad. Remember that saving a little weight at the end saves a whole lot of weight at the beginning, which is critical.

Good luck and remember that failed launches are really just discoveries of what doesn't work!

-Slashy

Edited June 28, 2014 by GoSlash27

[GoSlash27]

Oh, I almost forgot the other half of the problem!

Not only do you have to generate the ÃŽâ€V for each phase of the launch, but you also have to do it rapidly enough to not waste it fighting gravity and aerodynamic drag. You can generate 4,500 M/s ÃŽâ€V, but if you do it at less than 1 G, you'll never leave the pad. Likewise, if you generate it at the wrong acceleration for a phase, you'll waste a lot of it fighting gravity and drag instead of establishing your orbit.

So to that end, you want enough engine to generate the proper acceleration for each phase of the flight.

The important equation here is Ga=T/(Ca*Mt)

where Ga= Gs of acceleration, T is the thrust in kilonewtons, Ca is the local acceleration at 1 G, and Mt is the total mass of the vehicle you're moving. Naturally, you'll want to rewrite this as we've done above to show how much thrust you need to generate your necessary acceleration.

There's no consensus on this, but the conventional wisdom is you need 2 Gs for the boost phase (pad to gravity turn), 2 Gs tapering down to 1G for the transstage phase (gravity turn to apoapsis), and .5 G or even less from apoapsis to orbit.

Best,

-Slashy

Edited June 27, 2014 by GoSlash27