Math

The Maths behind ∆v and other such mysteries


{LINK REMOVED}http://www.mediafire.com/view/23ttj3bj2yq1ak9/KSP_Math_-_Orbital_Mechanics_-_draft_1.docx
{LINK REMOVED}http://www.mediafire.com/view/7547060fdhvrr5s/KSP_Math_-_rocketry.docx

forward

This is a recreation of my original WORD docs available from mediafire.
I do hope its understandable as the formatting options on steam are not really ment for maths which uses a lot of greek letters and subscripts and superscripts.

If you download the WORD documents view them in web format as some of the pics just won't fit on an A4 page

{LINK REMOVED}http://www.mediafire.com/view/23ttj3bj2yq1ak9/KSP_Math_-_Orbital_Mechanics_-_draft_1.docx
{LINK REMOVED}http://www.mediafire.com/view/7547060fdhvrr5s/KSP_Math_-_rocketry.docx




cheers
MAD

Orbital Mechanics, SOI & Conics

Orbital Mechanics
Calculating how an object moves in space under the effect of multiple gravitational bodies is very difficult, some say impossible and that’s not even factoring in General Relativistic effects.
So we simplify our multi-body problem into multiple two body problems - specifically you, your space craft orbiting a far more massive body such as a moon, planet or star.

SOI
Each massive body gravitationally dominates the local region of space around it, a region commonly called the Spheres of Influence (SOI).
The boundary of the SOI is where the gravitational forces between one body, say a planet and another even larger body, a star, balance. Inside the sphere the planets gravity dominates, outside of that sphere the suns gravity dominates and so we have moon spheres inside planetary spheres which are in turn inside a stars spheres of influence.
This is the important bit, while inside one objects SOI we ignore the effects of all other stellar bodies in the universe other than the primary body and your space ship.
If you want to read more on SOI try https://en.wikipedia.org/wiki/Sphere_of_influence_%28astrodynamics%29

In KSP the SOI for every star, planet and moon is listed on the KSP wiki http://wiki.kerbalspaceprogram.com/wiki/Category:Celestials

Conics
As stated above we simplify our multi-body problem into multiple two body problems; It’s what KSP does.
Inside each SOI we will orbit the primary body, usually in an ellipse; but depending on your velocity it could be one of the other conic trajectories, circle, ellipse, parabola and hyperbola.
The reason there called conics is because they represent different geometric slices through a cone.



Now to dispel one of Hollywood’s greatest myths, an object in space does not fall back to earth, it orbits the earth and will continue to orbit the earth ; ignoring solar wind, atmospheric drag etc without power, forever. The Moon has been orbiting the Earth for a few billion years so far. So if the Starship Enterprise in orbit around the earth suddenly lost power it wouldn’t just start to plummet into the ground.




So where in an elliptical orbit around Kerbin where Kerbin, represented as a blue circle acts as the one of the foci of the ellipses. As we move further away from Kerbin our speed decreases, as we convert kinetic energy into potential energy.
* The Apoapsis represents the point in our orbit furthest from Kerbin, it is where we have the lowest velocity.
* The Periapsis represents the point of the orbit closest to Kerbin, it is where we have the highest velocity.
The Apoapsis and Periapsis are directly opposite one another and the distance between them is the major axis, half of this is the semi-major axis ‘a’ remember this as you are going to be using it everywhere.

(There is also the semi-minor axis ‘b’ where for an ellipse


The orbital period T, the time it takes for one complete revolution is given by Keplers Third Law of planetary motion.


Our Speed anywhere on our ellipse may be calculated from the Vis-viva equation


With these two equations you can plot a coarse anywhere in the solar/kerbin system but more on that later.

For most orbital manoeuvres all we care about is finding the speed at either the Apoapsis or Periapsis and how much we need to modify it to get the trajectory we want. These modifications in speed is the Δν people talk about when they talk about space travel.
Δν is just mathematical jargon for change in velocity. The Delta symbol ‘Δ’ is used in mathematical notation the represent a change in something and in this case its velocity ‘ν’ because were lazy and don’t want to write Δvelocity all the time.
Now up unto this point I have been talking about speed and not velocity; velocity is just a speed in a certain direction, a vector.

Example 1 - How much Δν is required to move a satellite from low Kerbin orbit of 80km altitude to Geostationary Orbit

How much Δν is required to move a satellite from low Kerbin orbit of 80km altitude (green) to Geostationary Orbit (red).
figure 1


Low Kerbin Orbit (LKO) altitude of 80km equates to an orbital radius of 680km (680,000m) since the equatorial radius of Kerbin is only 600,000m.
(I pulled this info from the KSP wiki ) http://wiki.kerbalspaceprogram.com/wiki/Kerbin
(for comparison the Earth’s equatorial radius is 6,371km)

I should probably mention that Korbin’s atmosphere stops at 70km altitude (also from the KSP wiki) hence 80km for Low Kerbin Orbit

Geosynchronous orbit is a circular orbit whose period is exactly 1 day, 1 sidereal day which is the time taken for the planet to rotate 360 degrees in relation to the stars.
Not to be confused with a solar day which is the time it takes for the Sun to appear in the same position of the sky. They differ because the planet is also in orbit around the star.
When most people talk about days there talking about Solar Days, ie 365.25 Solar days per year

If one was to look up at an object in geosynchronous orbit it would appear to be in the same spot of sky all the time. We put communication satellites into geosynchronous orbit because they appear to hold a fixed position, we can point our satellite antennas at them without having to use complicated tracking mechanisms.
From the KSP Wiki 1 Sidereal Day is 21,549.425183s
https://en.wikipedia.org/wiki/Sidereal_time

From Kepler’s 3rd law
For a circle r = a ie the radius of the circle is also its semi-major axis hence
We know the desired orbital period T, hence solve for the semi-major axis 'a'


have I lost you yet? Yout probably asking what is this 'u' thingy is?
'µ' is the 'Standard gravitational parameter' and its not a 'u', its a funny Greek letter 'mu' 'µ'
I swear that mathematicians and physicist just throw in all these Greek letters to confuse people.
Anyway the 'μ' of a celestial body is the product of the gravitational constant G and the mass M of the body and it is unique for every body in the solar / kerbin system.
https://en.wikipedia.org/wiki/Standard_gravitational_parameter
Don’t mix up the µ for Kerbin with the µ for Kerbol (the Star) and make sure your working in the correct units.
I will try to keep to the SI system and use meters and seconds.
Also when dealing with such huge numbers don’t get too fixated on decimal places too much.

refer to figure 1, given that we now know the radius for the starting and finishing orbits we can calculate the semi-major axis a of our ellipse. ae being shorthand for the semi-major axis of the ellipse


Now using the vis-visa eqn we can calculate the velocity at the Apoapsis and Periapsis for both the circular orbits and the elliptical orbits


Periapsis
r = 680,000m
a = 2,071,667m
ve = 2,946.6 m/s
vc = 2,278.9 m/s
Δν1 = ve – vc = 667.7m/s (ie the elliptical velocity - the circular velocity)
ie we have to accelerate to prograde by 667.7m/s to transition from our circular orbit to obtain an elliptical orbit which will take us to geostationary

Apoapsis
r = 3,463,334m
a = 2,071,667m
ve = 578.5 m/s
vc = 1009.8 m/s
Δν2 = vc - ve = 431.3m/s
Ie we have to accelerate to prograde by 431.3m/s to transition from our elliptical transfer orbit to obtain a circular orbit at geostationary

Δνtotal = Δν1 + Δν2 = 1,098.9 m/s

Just as a note, if you were returning from Geostationary orbit to 80-km LKO and weren’t performing aero-braking the Δν would be exactly the same except that you will be burning to retrograde.



Solar day = 21,600 s
Solar year = 9,203,544.6 s
Sidereal day = Solar day * Solar year / (Solar year + Solar day)

Side Note: Sidereal Day

A solar day it the time it takes for the Sun to appear in the same location in the sky.
A sidereal day is the time it takes for stars to appear in the same location in the sky.
They differ slightly because the Earth (planet) revolves around the Sun.


Usually we just measure that time period with an acurate clock but if you want you can calculate one given the other.

In KSP
1 solar day = 6 hrs (21,600 s)
1 solar year = 9,203,544.6 s (426.09 days)

In 1 solar day the planet will have moved: 216,000 / 9,203,544.6 * 360 = 0.84 degree around the star
hence 1 sidereal day is shorter by 0.84 degrees.
ie 1 sidereal day is (360 - 0.84) / 360 of a solar day.

Instead of thinking in degree you can think in time
0.84 / 360 * 216,000 s = 50.693512 s
Hence a sidereal day is shorter than a solar day by 50.693512 s

In KSP
1 sidereal day is 21,549.425183 s long


side side note
A solar day it the time it takes for the Sun to appear in the same location in the sky. (given that you yourself havent moved in longatude. If you have moved the time it takes for the sun to appear in the same position will vary depending on how much you have moved.
If you have an accurate clock you can measure how long it takes for the sun to appear in the same location in the sky and use that time to calculate your change in longatude from your original position.
18th century navigators used that method to calculate a ships position while at sea.
A sextant is a device for measuring angles very precisly, typically the angle of the sun (or stars) above the horizon.

1 knot is a unit of speed equal to one nautical mile per hour.
A vessel travelling at 1 knot along a meridian travels approximately one minute of arc (ie 1/60th of a degree) of geographic latatude in one hour. [60 nautical miles per degree of longatude]


https://en.wikipedia.org/wiki/Marine_chronometer
https://en.wikipedia.org/wiki/Sextant
https://en.wikipedia.org/wiki/Longitude

Side note - Placing multiple satellites into geostationary

If you want to place satellites into geostationary orbit you probably want to place several of them so you can communicate with locations anywhere on the globe.


Question: How do you ensure that your satellites are placed evenly in their orbit ?
ie for 3 satellites their seperated by exactly 120 degrees.
ie for 4 satellites their seperated by exactly 90 degrees.

Answer: Get the bus to drop them off. 8-)

Q: How do you fit an elephant in your fridge ?
A: Open the door and place the elephant in your fridge, close the door. easy 8-)

Q: How do you fit a girrafe in your fridge ?
If you were going to say ' Open the door and place the girrafe in the fridge ?' that would be silly.
A: Open the door, remove the elephant and then place the giraffe in your fridge and close the door. 9-).

Answer: Get the bus to drop them off.
for a 3 sat network the bus has a period of 2/3 of geostationary period.
for a 4 sat network the bus has a period of 3/4 of geostationary period.
(note: there are other solutions that work but I am trying to keep things simple)
That way each orbit of the bus equates to 120 degrees / 90 degrees at geostationary.
Geostationary Period is equal to one Sidereal Day ie 21,549.425 s

Scott Manley has a nice video setting up 3 geostationary satellites.
https://youtu.be/Z3FHOQFVV0k

Since Scott video covers a 3 sat network I will do 4

We need to find the bus's ellipses semi-major axis 'a'

T = 3/4 * Siderial day = 3/4 * 21,549.425183 s = 16,162.069s
Standard Gravitational Parameter, ukerbin = 3.5316000×10^12 m3/s2
Solving for the Semi-major axis a = 2,858,919.28m



Bus orbital characteristics for a 3/4 Sidereal day
Apoapsis = 2,863.33 km altitude ie Geostationary
Periapsis = 1,654.50 km altitude


* Get to orbit
* circularize at approximatly 1,654.5 km
* Using the Mun as reference set your inclination to 0.0 degrees, ie over the equator
(It's more efficient to change inclination when travelling at lower orbital velocities but if you want to change your inclination at low kerbin orbit then do so)
* circularize at 1,654.5 km (V = 1,251.6m/s)
* plot a transfer orbit where Apoapsis = 2,863.33 km ie geostationary orbit
If you like place the periapsis directly over the KSP the 2nd and 3rd satellite will be +/- 45 degees longatude thus giving the KSP two satelites to choose from (just in case one breaks down)
* execute the transfer
* every time you reach Apoapsis (geostationary) release one of your sattelites and circularize it.
Δv to circularize it will be ~ 113.1 m/s
Orbital velocity at geosync is 1,009.8 m/s
Try to get your orbital period to be 16,161.319s (5 Hours, 59 minutes and 8.425s)


'Remote Tech' is a mod dealing with realistic satallite communication.
https://youtu.be/L42s75UmYec
http://ryohpops.github.io/kspRemoteTechPlanner/
http://forum.kerbalspaceprogram.com/index.php?/topic/75245-105-remotetech-v169-2015-11-10/

for 3 sats
Bus orbital characteristics for a 2/3 Sidereal day
Apoapsis = 2,863.33 km altitude ie Geostationary
Periapsis = 1,222.70 km altitude

Example 2 - How long will the transfer take?

We have already calculated 'a' (the semi-major axis) for our ellipse, just plug it into Keplers third law to get our orbital period; our transfer time is half a period.



Orbital Period T = 9,970s
Transfer time is half that ie 9,970/2 = 4,985s --> 1hr, 23m, 5s

This is the time for an object to esentially drift along an elliptical path using only two short burns to perform the maneuver.



Earth to Mars

Hohmann transfers and double impulse maneuvers

This type of manoeuvre is called a Hohmann transfer https://en.wikipedia.org/wiki/Hohmann_transfer_orbit
It is one of the most energy efficient ways of moving around in space, it is definitely the easiest to calculate and execute since all of your manoeuvring occurs at the Apoapsis or Periapsis. It is what most KSP player do.
In the game you don’t have to calculate anything, just orient your craft to prograde or retrograde and burn until your ellipse grows to where you want to go, then follow the ellipse to the next burn.

Changing your orbit is most efficient when done at Apoapsis or Periapsis but you don’t have to do it then……
https://en.wikipedia.org/wiki/Orbital_mechanics

A Hohmann transfer from a low circular orbit to a higher circular orbit


If you’re prepared to spend extra fuel you can reduce your travel time
Generic two impulse transfer


Note: How all our transfers are part of an ellipse
If you really wanted to calculate it you could use the vis-viva eqn to calculate your speed at any portion of the ellipse, then use the geometric identity x2/a 2 + y2/b2 = 1 to work out the component vectors at that point and do the same for you transfer ellipse yada yada
And calculate your Δν’s that way, but I am for sure not doing it now.

If you want you can follow these links which talk of such things
https://books.google.com.au/books?id=2U9Z8k0TlTYC&pg=PA349&lpg=PA349&dq=orbital+mechanics+ion+engines&source=bl&ots=CEUWJKnECp&sig=wLe7n6v-3mqUpzVahae-yP6KuQ4&hl=en&sa=X&ved=0ahUKEwjZwbG70dTJAhVDk5QKHaXbDuMQ6AEITDAI#v=onepage&q=orbital%20mechanics%20ion%20engines&f=false
Chapter 6, section 6.6

Bi-elliptical transfers

https://en.wikipedia.org/wiki/Bi-elliptic_transfer
There is one other maneuver you should be aware of, the Bi-elliptical Transfer
In certain situations, require less delta-v than a Hohmann transfer maneuver.
It is easy enough to work out as its just two hohmann tranfers one after another but boy does it take a long time to perform.

Patched Conics

Inside each SOI we will orbit the primary body, usually in an ellipse; but depending on your velocity it could be one of the other conic trajectories, circle, ellipse, parabola and hyperbola.
What happens when we want to go to stuff outside of Kerbins SOI, you know like other planets and such?
Say we want to go to Duna.
1) First you have to leave Kerbins SOI on a hyperbolic trajectory
2) This puts us into the Star Kerbols SOI where if we have done our sums right we will travel along an elliptical orbit which will intercept Dunas SOI
3) Once within Dunas SOI we will swing past Duna on a hyperbolic trajectory. If were planning on staying there we will need to slow down somewhat in order to be captured and remain in Dunas SOI.
So we have transitioned into three different SOIs and in each SOI essentially doing a different conic trajectory, our entire trip is essentially a patchwork of different conic sections ie Patched conics
It’s an approximation. It’s a very good approximation of a very difficult mult-body problem.
Our task is to ensure that as we transition from one SOI to another we are doing the right speed and direction such that we get to where we want to go.
https://en.wikipedia.org/wiki/Patched_conic_approximation

Kerbin to Duna


Computationally start with our heliocentric (that’s Greek for Sun Centered) transfer around the Star Kerbol. We know that we are going to be travelling in an elliptical orbit because our orbital velocity is well below Kerbols escape velocity
Just like a Hohmann transfer we want one end of our ellipse, the periapsis in this case to rest on Kerbins orbit and the Apopsis of the ellipse to rest on Dunas orbit.
Well that for an energy efficient elliptical transfer, once again there is nothing stopping you going for a quicker, less fuel efficient double impulse transfer but for computational ease were going to stick with a Hohmann style transfer.
Technically this is a heliocentric transfer because there are masses at one or more of it ends but hopefully you will understand me more when I say Hohmann like transfer.
Just like a Hohmann transfer we can calculate the elliptical velocities at Periapsis and Apoapsis. These will be the velocities we will require when we depart Kerbins SOI or enter Dunas SOI. As for direction we will be departing along Kerbins prograde ie the direction Kerbin is orbiting around the Star.

If we were going to an inner planet we would depart along Kerbins retrograde because we are trying to reduce our orbital velocity around the star Kerbol

Knowing the velocity required as we leave Kerbins SOI allows us to calculate the velocity required to depart Kerbin. This velocity is commonly known as the “Ejection Velocity”
It is the velocity at Periapsis of a hyperbolic which leaves Kerbin.
Our first Δν impulse will be the difference between out ejection velocity and our orbital velocity, everything after this point will be unpowered trajectories until we need to break at the other end around Duna, and that will be our second Δν impulse.

Now this leads to a couple of choices, do we travel along Kerbins orbital plane and do a minor correction to transition to Dunas orbital plane or do we shoot off at some slightly odd angle slightly offset from our orbital plane in order to hit Duna ballistic.



The first requires a slight trajectory correction part way through our manoeuvre; the second requires a slight vector change at departure which can incur an additional small Δν expenditure at either end.


Once we get to the other end there are quite a few options open to us
1. Decelerate a lot and go into circular capture around Duna at some radius.
2. Decelerate a bit less and go into an elliptical capture around Duna. The largest possible ellipse has an Apoapsis which just stops shy of Dunas SOI, this is called a “Marginal Capture”
3. Decelerate go into an elliptical orbit which we circularize at some radius.
4. Since Duna has an atmosphere you can aero brake and decelerate in Dunas atmosphere and land
5. Or aero brake somewhat and go into orbit around Duna
Aero braking can save you a lot of that fuel you were going to use decelerating and hence reduce or eliminate your Δν requirement at your destination. Just don’t burn up by accident.
As a general rule it is usually more efficient to decelerate as close to the planet as possible.

going into a circular capture around Duna at 5,000km

Going to the Mun

Going to the Mun doesn’t require us to leave Kerbins SOI but it does require us to enter the Muns SOI and remember once inside some ones SOI you treat everything as a two body problem in this case just you and the Mun.
1) Getting to the Mun is easy, just plot an elliptical trajectory which takes us to Mun orbit. Just arrainge things so that the Mun is actually there when you get to Apoapsis.
2) Once you enter the Mun’s SOI you will accelerate towards the Mun on a hyperbolic trajectory. I don’t know if I would call it good or bad luck if you were too accurate and that trajectory passed thru the Mun itself.

Once again computationally we calculate the velocity of our kyperbolic periapsis given that we know our velocity as we entered the Muns SOI.
Our first Δν impulse is that required to change our LKO into an elliptical one that reaches Mun orbit and the second Δν impulse is orbital deceleration so that we are captured by the Mun. Aerobraking is out of the question since the Mun has no atmosphere. Actually I think the Mun has lots of atmosphere, craters, arches, the occasional monolith or crashed flying saucer.

a trip to the Mun

Hyperbolics

A circle, your rocket orbits at a constant speed. Vc
An ellipse, your rocket ship loses speed the further it gets away from the planet and eventually comes back to perform an orbit.
A parabola, your rocket ship loses speed the further it gets away from the planet but it never returns. Its velocity at infinity is zero. The velocity at Periapsis is known as your ‘Escape velocity’.
A hyperbola, your rocket ship loses speed the further it gets away from the planet but it never returns. Its velocity at infinity is some speed greater than zero. This speed is known as the ‘Hyperbolic excess velocity’ ν∞


If your entering or exiting a SOI you probably going to be travelling on a hyperbolic trajectory.



Your velocity at Periapsis (the closest point of your orbit to the planet) is lowest for a circle and as it increases your orbit becomes more elliptical, then parabolic and finally hyperbolic.
Your velocity for any given radius ‘r’ is given by the Vis-Viva equation





For r = SOI we can use the Vis-viva equation to solve fo a directly and once you know that you can calculate the velocity at periapsis ie our Ejection Velocity.


Once you know your ejection velocity it’s a simple matter of determining the Δν impulse required to send you to out of Kerbins SOI and all the way to Dunas SOI

Example 3 - Calculate the Δν to get from Kerbin to Duna

Ok let put it all togeter and calculate the Δν to get from Kerbin to Duna
From the KSP wiki
aKerbin = 13,599,840,256 m (circular)
aDuna = 20,726,155,264 m (slightly elliptical) for greater accuracy you will need to calculate Dunas exact radius for the transfer window you plan to use.



The semi-major axis of our transfer ellipse = (a Kerbin + aDuna) /2 = 17,162,997,760 m

Using the vis-viva equation we can calculate the orbital velocity for both Kerbin and Duna as well at the transfer ellipses velocity at Apoapsis and Periapsis
Since we are inside Kerbols SOI the Standard gravitational parameter µKerbol = 1.1723328×1018 m3/s2



VKerbin = sqrt(1.1723328×1018 / 13,599,840,256) = 9,284.50 m/s
VDuna = sqrt(1.1723328×1018 / 20,726,155,264) = 7,520.84 m/s

VPeriapsis = sqrt(1.1723328×1018 * (2/13,599,840,256 - 1/17,162,997,760)) = 10,202.85 m/s
VApoapsis = sqrt(1.1723328×1018 * (2/20,726,155,264 - 1/17,162,997,760)) = 6,694.78 m/s

Kerbin SOI: Since were already orbiting Kerbol (the star) at 9,284.50 m/s and we need to be doing 10,202.85 m/s to transfer to Duna then the velocity we need to exit Kerbins SOI is (10,202.85 - 9,284.50) = 918.35 m/s
Duna SOI: Likewise the velocity we enter Dunas SOI is (7,520.84 - 6,694.78) = 826.05 m/s

Leaving Kerbin

Find a for r = SOI ie 84,159,286m
µKerbin = 3.5316000×1012 m3/s2
Vr = 918.35 m/s


a = -4,650,267.67 m

Yes its a negative number, the semi-major for all hyperbolics are negative. Don't think about it too much or you will give your self a head ache. It's a math thing.

We will assume we are departing Kerbin from low Kerbin orbit (LKO), say 80km altitude
Kerbin’s equatorial radius is 600km, hence or departure orbit has a radius of 680km (680,000 m)
µKerbin = 3.5316000×1012 m3/s2
Our circular orbital velocity around Kerbin: vc = sqrt(µ/r) = sqrt(3.5316000×1012 / 680,000) = 2,278.9 m/s ( I can confirm this easily enough in KSP)

Now that we know a using the vis-viva eqn calculate the Ejection Velocity at 80km altitude
Vr = sqrt (3.5316000×1012 (2/680,000 + 1/4,650,267.67 )) = 3,338.64 m/s
Since were already orbiting kerbin at 2,278.9 m/s the Δν for our first impulse is

Δν1 = vr – vc = 3338.64 - 2,278.9 = 1,059.71 m/s

A quick check here from OLEX’s website http://ksp.olex.biz/ , he calculates 1055m/s.
There is no real way of ascertaining who is more accurate or what assumptions Olex has made (that is unless he reads this and want to let me know) or how many decimal places he is working with but the values are so close that it makes little difference especially when dealing with such huge numbers as µ

Arrival at Duna
According to the KSP wiki Dunas atmosphere is only 50km high hence if we go for a 60km circular capture around Duna
Dunas equatorial radius is 320km hence r = 380km (380,000 m)
µDuna = 3.0136321×1011 m3/s2
Hence vc = sqrt(3.0136321×1011 / 380,000) = 890.5 m/s

Now using the vis-viva eqn we can calculate our hyperbolic parameter a given that our velocity is 826.05 m/s as we enter Dunas SOI 47,921,949 m
µDuna = 3.0136321×1011 m3/s2


a = -449,942.67

Our velocity at Periapsis r = 380,000 m
Vr = sqrt (3.0136321×1011 (2/380,000 + 1/449,942.67)) = 1,501.97m/s

Since circular velocity is 890.5 m/s so we only have to change our velocity by 1,501.97m/s - 890.5 m/s = 611.4 m/s

ie Δν2 = 611.4 m/s
olex calculates 612.61m/s; close enough for government work

The total Δν required to go from Kerbin to Duna would be 1,059.7 m/s + 611.4 m/s = 1,671.1 m/s
Plus whatever Δν requirement we would need for orbital inclination changes



Marginal Capture at Duna
Rather than going into a circular capture at Duna how much Δν do we need for a marginal elliptical capture
This is an ellipse where Periapsis is just above Dunas atmosphere and its Apoapsis just below Dunas SOI; and bigger then it would never come back to Duna
From the KSP wiki Dunas SOI is 47,921,949 m, we shall set our Apoapsis to 10km short of this ie 47,911,949 m
Our marginal capture ellipse
Periapsis = 380,000 m
Apoapsis = 47,911,949 m
aellipse = (47,911,949 + 380,000)/2 = 24,150,974.5 m
µDuna = 3.0136321×1011 m3/s2
from the Vis-viva eqn the orbital speed at Periapsis = sqrt(3.0136321×1011 *(2/380,000 – 1/24,150,974.5)) = 1,254.4 m/s

In order to obtain a marginal capture we need to reduce our velocity by 1,501.97 – 1,254.4 = 247.57 m/s
ie Δν2 = 247.6 m/s

since all other elliptical captures with a Periapsis at 60km altitude lie somewhere between these two extremes one can think our Δν2 lies somewhere between these to values
Δν2 = 247.6 m/s to 615.6 m/s
This assumes we are not performing aero-breaking



Some in game screen captures of a trip from Kerbin to Duna

Departure from Kerbin

Elliptical transfer to Duna

We are intercepting Duna early, oops I am early by 28 days, the optimal time to depart should have been day 230

Just outside Dunas SOI

Capture at Duna


In game pic of aerobraking without a heatshield, coming back hyperbolic.
Who said comming back steep was bad?. Periapsis set to 2km

Orbital Inclination

https://en.wikipedia.org/wiki/Orbital_inclination
Most objects, planets, moons, in a solar system tend to orbit on the same plane about the Sun. This is why when you look up into the sky it appears that all the planet follow the same path through the sky, the ecliptic.
There are minor variations in orbital planes, for example Mars has an inclination 1.85 degrees. In KSP Duna is inclined 0.6 degrees.
Calculating the Δν required to change from one plane to another



For small inclination changes Δν = 2V sin(ɵ/2) where V is the elliptical velocity where the orbital planes meet.

Ballistic Trajectory

Ballistic Trajectory

Of course you don’t have to don’t have to perform a mid-course correction if you travel to your destination via a ballistic trajectory.
This required two inclination changes, one at your origin and one at your destination. You can use the above formula to calculate the Δν for each the inclination change but the actual trajectory is so dependent on the exact planetary positions there is no general solution so I will skip this bit all together.

Phase Angles and Launch Windows

For a minimum energy transfer one must leave Kerbin on a specific date, otherwise when you get to Duna orbit, Duna won’t be there.
Phase angle is the angle between source and destination planet relative to the star at the time of launch.
For a Hohman style transfer we are travelling literally from one side of the solar system to the other.
Our destination is 180 degrees from our launch position.



We can use Kepler’s 3rd law of planetary motion to calculate the orbital period of the transfer orbit as well as well as our planets.
Calculating the phase angle is fairly straightforward.
If your transfer takes X days then we just roll back the destination positions of all the planets X days to find their starting position.

Example 4 – find the Phase angle for a trip from Kerbin to Duna

µKerbol = 1.1723328×1018 m3/s2
aKerbin = 13 599 840 256 m (circular)
aDuna = 20 726 155 264 m (slightly elliptical)
aTransfer = 17,162,997,760 m



TKerbin = 9,203,545 s (426.1 days)
TDuna = 17,315,400 s (801.6 days)
TEllipse = 13,048,005.44 s  TTransfer = TEllipse/2 = 6,524,002.72 s (302.0 days)

So what fraction of a Duna orbit is 302 days?
302 days / 801.6 days * 360 degrees = 135.6 degrees
Phase angle is 180 – 135.6 degrees = 44.36 degrees

OLEX http://ksp.olex.biz/ says 44.36 degrees 8-)

Now there is a degree of error in this calculation since it is based on the regular motion around a circular orbit and Duna’s orbit isn’t perfectly circular.



The eccentricity for a perfect circle is 0
The eccentricity for Duna is 0.05 (data from KSP Wiki) so the error is pretty trivial,
In practice you typically have about ± 5 day window of launch opportunity.

Example 5 - What’s the lead angle to get to the Mun

µKerbin = 3.5316000×1012 m3/s2
a mun = 12,000,000 m (circular)
a transfer ellipse = (aperiapsis + aapoapsis)/2 = (680,000 + 12,000,000)/2 = 6,340,000 m



T mun = 138,984 s (38.6 hours)
T transfer ellipse = 53,374 s  TTransfer = TEllipse/2 = 26,686.89 s (7.4 hrs)
In the 7.4 hrs it takes to get to mun orbit, how far will the mun have moved in its orbit given that the Muns orbital period is 38.6hrs? You can uses keplers equn to calculate the Muns orbital period also.

7.4 hrs / 38.6 hrs * 360 gegrees = 69.1 degrees


Just for reference if your going to Minmus your lead angle is 65.1 degrees and it takes 9 kerbin days to get there.

Ejection angle

KSP is really nice as it shows you departure trajectories but without such niceties how does one calculate where in our orbit to perform our first burn “Ejection Burn” for interplanetary travel?

In essence how does one calculate ones ejection angle η?



From the above picture η = 180 - ɵ
Don't ask me what the funny 'n-like' symblol is, I will have to look it up myself.
In our case it is the Ejection Angle.
There is probably some wierd sub-atomic partical which uses the same symbol.
Damn, now I am going to have to look it up...... its called 'eta' and there are eta (η) and eta prime meson (η′)
I wonder if you can use them as a cataylist for fusion?

For a hyperbolic a < 0 but for the following trigometric identities we are just using its magnitude ie |a|
Tan(ɵ) = b/a
ɵ = ATan(b/a)
or
Cos(ɵ) = a/f
ɵ = ACos(a/f)
= ACos(a/(a + rc)) rc is the radius of our initial circular orbit

Example - Ejection Angle

for our tip from Kerbin to Duna ie Example 3, what is our ejection angle ?

In example 3 we have already calculated the semi-major axis of our departure hyperbola from Kerbin as
a = -4,650,267.67 m
rc = 80km altitude ie 680,000 m
ɵ = ACos(a/f) = ACos(4,650,267.67 /( 4,650,267.67 + 680000))
= ACos(0.8724)
= 29.26 degrees
η = 180 - ɵ = 150.74 degrees

Olex http://ksp.olex.biz/ says 151.28 Degrees, eh close enough for government work.

Summary

Rocketry

∆v

Delta-V (∆v)
There are two common usages of the term DELTA-V

ΔV - that which a space craft is capable of performing.
ΔV - that required to perform a manoeuvre or chain of manoeuvres.

Calculating the ΔV of a space craft.

Every action has an equal and opposite reaction.
A rocket burns fuel producing hot gasses which shoot out one way and in doing so, push the rocket the other way.



As the rocket burns fuel it accelerates. As the rocket burns fuel it also gets lighter.
A lighter rocket will accelerate more for the same thrust.

So under constant thrust
the 1st second of operation you may change our velocity by Δv1 (m/s)
the 2nd second we change our velocity by Δv2 but since our rocket is slightly lighter you accelerate a little bit more ie Δv2 > Δv1
the 3rd second we change our velocity by ΔV3 where Δv3 > Δv2

The total change in velocity Δv = Δv1 + Δv2 + Δv3 + …..

By the process of differential calculus we can derive a general formula for how much our velocity has changes for the entire burn ie
http://www.braeunig.us/space/propuls.htm
https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

Isp

Isp or Specific Impules

Most high school science students are familiar with ‘F = ma’ but what if a force is applied over a period of time. Well that’s called ‘Impulse’
An Impulse may be short as in a golf club striking a golf ball or may be long as in a rocket engine.

Thrust profiles for various solid rocket cores


Rockets outside of KSP rarely produce constant thrust so the ‘Total Impulse’ is the integral of the thrust (force) over time.

Total Impulse = ∫ F dt (N.s)

Hence Impulse is the ‘average’ force (Thrust) for a given time and it's equal to the change of momentum.

(Isp) Specific Impulse (s) = Total Impulse (N.s) / Propellant mass (kg)

Specific Impulse, measured in seconds is useful for measuring an engines efficiency.
Think of it as the number of seconds a rocket can generate a given thrust per kg of fuel consumed.
The higher the specific impulse, the lower the propellant flow rate required for a given thrust, and in the case of a rocket, the less propellant needed for a given delta-v

http://www.asa-houston.org/technical/Rocket%20Motor%20Specific%20Impulse.htm

The strongest parameters influencing ISP are the amount of chemical energy in the fuel, the efficiency of combustion, and the shape of the exhaust nozzle / spike.
So two engines burning the same fuel won’t necessarily have the same efficiency.
The Exhaust nozzle is usually designed for maximum efficiency with a particular pressure (altitude) in mind depending on their role and stagging.
You can even have different models of engine which are identical except for the length of their nozzle
That’s why most engines list an Isp for sea level and for vacuum as they may differ considerably. In KSP while in the assembly building right click on the engine

effect of pressure on exhaust


As for fuel Hydrogen / Florine is the most energetic but since no one wants to work with Florine we will settle for second best Hydrogen / Oxygen.
Liquid Hydrogen/Oxygen engines require expensive cryo pumps and special handling so a lot of rockets use chemicals which are liquid at room temperature.
They may be toxic or corrosive and most certainly explosive but that’s rocket science for you.
https://en.wikipedia.org/wiki/Rocket_propellant

Typical Exhaust Velocity
Liquid Oxygen + liquid hydrogen 4462 m/s
Liquid Oxygen + Kerosene (RP-1) 3510 m/s (please note that the kerosene you buy from most places is mostly water)
Nitrogen tetroxide N2O4 + Hydrazine N2H2 3369 m/s

Since we’re talking about Exhaust velocities and Specific impulse


Hence our rocket equation can be written as


This is far more useful for KSP since the stats for a rockets Specific Impulse is listed for every rocket (right click on the part while in the assembly building)
And also available from the KSP wiki http://wiki.kerbalspaceprogram.com/wiki/Parts


The “Nerv” has an Isp of 800! OMG I’ll use it for everything…..
It also only uses ‘Liquid Fuel’ so I can ditch all that heavy ‘Liquid Oxidiser’, it’s so awesome!
NO
It also weighs 3 tons! so if you were paying attention when I said m0/m1 as in


you will understand that it is only awesome when pushing heavy loads around. For lighter stuff use smaller engines like the ‘terrier’ for example.
Isp is only a measure of engine efficiency but Δv lets you know how far you can go.

Then there is the Ion Engine with an Ispv of 4,200s !!! (Now say it with a disappointed tone in your voice) “With only 2kN thrust”.
Cheer up, just stack 20 of them in a row and you can start to go places, places far far away that no other rocket can get to, ever !



Here is a link to a Scientific American artical on engines
check out VASIMR (Variable Specific Impulse Magnetoplasma Rocket)
http://en.wikipedia.org/wiki/Variable_Specific_Impulse_Magnetoplasma_Rocket

And some much older articles
Scientific American, Electric propulsion in space, 1961
{LINK REMOVED}http://www.mediafire.com/view/ab8xb78q2p6s51w/Electric_Propulsion_in_Space.pdf

TWR

Thrust to Weight Ratio is what gets you into space.
You need a TWR > 1 to lift off Kerbin.
I generally design by rocket with a TWR of 1-6 to 1.7; Too high and your rocket tends to burn up on launch, too low and you waste lots of fuel during Launch.

TWR = Trust/Weight (pretty obvious one would think?)

Thrust in kN is listed for every engine in KSP. 1 Newton = 1 kg . m/s2 (It’s a measure of Force, the physics kind, not the Skywalker kind). 1kN = 1000 N
Mass in kg (Kilograms) listed for every part in KSP. 1 ton = 1000kg’s
Weight in N (Newtons) Say what? Yes Newtons.
Weight = Mass * local gravitational constant g
You may say you weigh 80kg but from a true physics perspective they actually weigh 80 *9.80665 = 784.5 Newtons, [Don’t try to think about it]
On Mars where the local gravity is only 0.376 of an earth ‘g’ you weight 0.376 * 784.5 = 285 Newtons. You may weigh 285 N but your Mass is still 80kg]

When in space TWR is mostly irrelevant since rocket efficiency and Δv is what you should be primarily concerned about. TWR or should I say Thrust to Mass since in space you have no weight, will determine how long a particular manoeuvre will take to complete.
Low TWR means that it will take longer to perform a manoeuvre, so long that some time critical manoeuvres may become impossible.
High TWR means that your burn time is much shorter, it may mean that due to human limitations in pressing buttons you’re not able to control your burn duration with sufficient accuracy to do fine manoeuvres.

If I need a TWR > 1 to lift of Kerbin, how much do I need to lift off the Mun?
Ideally you need a TWR > 1 because your weight is a local thing but if you using KER (Kerbal Engineer Redux) it calculates everything relative to Kerbin.
According to the KSP wiki http://wiki.kerbalspaceprogram.com/wiki/Mun, the Mun has a local gravity of 0.166g.
Hence to lift off the Mun you will need a TWR > 0.166.
That’s just to lift off, if you want to do it in a timely fashioned then you may want a bit more, say : TWR = 1.65 * 0.166 = 0.274
Since the Mun has not atmosphere you don’t have to worry about burning up on take-off so there is no real upper limit other than control issues.

Burn Time

How long will it take me to perform a manoeuvre? The KSP Navball shows you it, but how do you calculate it?
It’s not difficult, it’s just messy

Burn time (s) = Fuel required (kg) / Propellant flow rate (kg/s)

Fuel required (kg)
That is the fuel required for a manoeuvre with a particular Δv
m1 ( the final mass) = m0 ( the starting mass) – the mass of the required fuel (mf)
Substatute into the rocket equation and we get after a bit of manipulation



With a bit of substitution and manipulation



Propelant flow rate (kg/s)
One usually doesn’t calculate flow rate, one usually just measures it in the lab, but in KSP you can’t do that.
What you can do is determine it given a rockets Isp

Propellant flow rate (kg/s) = Thrust (N) / Exhaust Velocity (m/s)

------------------------------
As of KSP v1.1 they actually give you the consumption rate of each engine as well as the capacity of each tank.
To calculate Maximum Burn Time just divide one by the other.

Maximums Burn Time = Sum of all your tanks / Engine flow rate

[FL-T400 when they say a tank has 220 units of fuel and 180 units of oxidiser their not talking about litres, actually I have no idea what unit their talking in as it seems to be just a derivative of the weigh i.e. 900kg of liquid oxidizer /5  220 units of liquid oxidizer. Just for reference liquid oxygen is 1.14kg/litre]

Example - Confirming Kerbal Engineer Redux calculations

[/previewicon]


Personal note to the person / persons who wrote Kebal Engineering Redux, thanks for the great add-on.

Example - Can I get home?

Can I make it back home safely? or How much ΔV have I got left?




and that pretty much agrees with KER



In KSP propellant weighs 5kg/unit
We can read the remaining units of propellant straight of the resources tab.

Rocket Design

Staging and such

1 - late staging .......... ∆v = 5,598m/s, TWR 1.59
2 - early staging ........ ∆v = 5,307m/s, TWR 2.38
3 - Cross Fed ............ ∆v = 5,593m/s, TWR 2.37 ............ Best of both
4 - Fuel pods ............. ∆v = 6,406m/s, TWR1.75


Asparagus 4 for when you want to go places....

Even more for asparagus 6

Lighter than a 2-man pod is two 1-man pods.
Pilot and scientist team making multple biome hops is a great way to generate science.
Have you scientist collect all the science and reset the equipment between each survay.
Dock in space to refuel, return to Kerbin and re-enter with just the two pods.




Talking about Torque - getting the most out of your thrusters

Be warned thrusters are heavy, I generally only put them on craft which require lateral thrust for docking purposes.
For 50kg a small inline reaction wheel can generate 5 KN.m
For 100kg a Advanced Inline Stabilizer can generate 15 KN.m
And for 200kg the Advanced Reaction Wheel can generate 30 kN.m


Rocket stability



It looked flat on approach......

Links

Some really good online calculators
http://ksp.olex.biz/
https://alexmoon.github.io/ksp/


Orbital Mechanis
http://www.braeunig.us/space/orbmech.htm

Rocketry
http://www.braeunig.us/space/propuls.htm



My stuff
{LINK REMOVED}http://www.mediafire.com/view/23ttj3bj2yq1ak9/KSP_Math_-_Orbital_Mechanics_-_draft_1.docx
{LINK REMOVED}http://www.mediafire.com/view/7547060fdhvrr5s/KSP_Math_-_rocketry.docx
{LINK REMOVED}http://www.mediafire.com/view/1f88dgh0ke3744d/KSP_maths.png

[MangledMango]

Kerbal Space ProKilograms Of Meth In My System

[pinky binky]

kerpal space gram

[chb62455]

why did you inflict such horrific pain upon yourself?

[ant]

sir, this is a wendys

[VGER]

Dang, you are dedicated!!!!!

[Fritz]

what

[pp_extended]

if i knew half of this when i was in ap physics i wouldve gotten a much better grade lol

[Kevdog457]

No lie, I'm gonna use this to pass my next few classes in aerospace engineering. This is better than a textbook, who would've thought?? On STEAM? Thx to whoever did this!

[WolfyST]

ABSOLUTE CHAD

[Nervous]

simple math