New Mars Forums

I worked out the eqution of the ellips that has one focus of the sun and the other focus given by the coordinates (f1,f2)

It is:
a1 x^2+a2 x +a3 x y + a4 y +a5 y^2 =a6
a1=(4 d^2 - 4 (f1 -s1))
a2=s k (f1 -s1) - 8 d^2 s1
a3=-8 (f1-s1) (f2-s2)
a4= 2 k (f2 - s2) - 8 d^2 s2
a5 4 d^2 -4(f2-s2)^2
a6=(||f||^2-d^2)^2+(||s||^2-d^2)2-d^4

where
d is the sum of the distance from each foci to a point on the ellips
(s1,s2) are the coordinates of the sun
(f1,f2) are the coordinates of the second foci
k=||f||^2-||s||^2-d^2
||f|| denotes the maginitude of the postion vector giving the location of the second foci

I hope I understand this right.

Thus to get to earth L1 from LEO only requires a delta v of 6 m/s. However because systems exhibiting chaos are very sensitive to initial conditions some fuel is required to make corrections.

Thus more fuell is needed to correct for errors when going from LEO to earth L1 then is needed to go from LEO to L1.

Thus 450 m/s is more then sufficient fuel to go from LEO to earth L1.

The Cosmos in chaos

Region, opium, television, dreams and stories
Productivity, hard work and innovation.
A collision of contradiction,
An explosion
Joy, depression, sadness rejoice
The human equation
Dynamics
Unknown states
Chaotic results
And an end but
a beginning
Changing manifolds in fractal geometry
A beauty that is seen in a
Chaos of sound bites
The infinity of the continuum
Of ideas
Dwarfing the boxed
Ideas of our time
Woven by billions
The tapestry of time
The disjoint threads unite
Into a story
Stringing an order
A cosmos

I liked drinking carona  on the lieto deck by the pool, well sun tanning under the UV lights.

In such a ship you could:
-explore the entire solar system in luxory
-transport people between the L points of planets with miniamal energy
-learn how to live in space
-have a huge labratory for research into medical an material science.
-have a safe haven to escape to when exploring any planet

and the space station either rotates or we have another way to counter act zero g.

abstract from:
[http://www.gg.caltech.edu/~mwl/publicat … stract.pdf]http://www.gg.caltech.edu/~mwl....act.pdf
Paper at: [http://www.gg.caltech.edu/~mwl/publicat … ateway.pdf]http://www.gg.caltech.edu/~mwl/publicat … ateway.pdf

More on A simple example (1) (Finding the second Focus)

In the previous example the transfer orbit started where earths orbit crossed  the point (1,0) and arrived where the orbit of mars crossed the point (-1.5,0). The second focus was found to be at the point (-0.5,0). Because that the minimum d for which an elliptical solution will exist will be the [http://en.wikipedia.org/wiki/Hohmann_transfer_orbit]Hohmann transfer orbits (I think), when the two transfer orbit intersects the original orbit and the destination orbit are on opposite sides of the sun.

The transfer time is given by the equation,

t=pi*sqrt((ro+r2)^3/(8*Gm))
where:

r0 is the radius of the intial orbit
r2 is the radius of the final orbit
Gm is the gravitational parameter (3.986x10^5 km^3/s^2)

The [http://mathworld.wolfram.com/Apoapsis.html]apoapsis is the point of the transfer orbit furtherst away from the sun. Note this is also where the transfer orbit intersects the orbit of mars. The [http://mathworld.wolfram.com/Periapsis.html]periapsis is the point where the transfer orbit is closest to the sun. Note this is also where the transfer orbit intersects the orbit of earth.

The [http://mathworld.wolfram.com/SemimajorAxis.html]semi major axissemi major axis of the transfer orbit is half the distance between the apoapsis and periapsis.
a=||(1,0) – (-1.5,0)||/2=2.5/2

The [http://mathworld.wolfram.com/SemiminorAxis.html]semi minor b axis can be found by using simple triganomatry. At the semi minor axis each foci is d/2 away and the distance between the foci is ||(0,0)-(-0.5,0)||=0.5. The resulting triangle formed is an [http://mathworld.wolfram.com/IsoscelesTriangle.html]isocialise triangle. Half of an isocialise triangle is a [http://mathworld.wolfram.com/RightTriangle.html]right angle triangle. In this right angle triangle the hypotensis is d/2, the distance from one foci to the cener of the ellipse is 0.25. Thus the third side can be found using the [http://mathworld.wolfram.com/PythagoreanTheorem.html]Pythagorean Theorem.
b=sqrt((d/2)^2-(0.25/2)^2)=(1/2)*sqrt(6.5) ~ 1.27

A simple example (1) (Finding the second Focus)

I’d like to work out a simple example first. For now assume mars and earth are in the same plane and have circular orbits. The distance earth is from the sun is 1 AU (astronomical unit) and mars is approximately 1.5 AU. For simplicity let the initial point of the earth be at (1,0) in rectangular coordinates and the let the transfer orbit take intersect the orbit of mars at the point (-1.5,0).

Recall the second focus can be found with the intersection of the two circles:
||f-e||=d-||s-e||
||f-m||=d-||s-m||
e is the position vector of the earth (1,0)
m is the position vector of where the transfer orbit will intersect the orbit of mars (-1.5,0)
f is the second focus the transfer ellipse.
The distance between where the transfer orbit intersects the orbit of mars and earth is 1-(-1.5)=2.5.
The two circles will just touch each other when:
d-||s-e||+d-||s-m||=2.5
which implies
d=(2.5+||s-e||+||s-m||)/2=(2.5+1 +1.5)/2=2.5

When two circles just touch each other the intersection will be on a line between there two centers. More specifically it will be at d-||s-e||=1.5 away from the earth and d-||s-m||=1 away from where the transfer orbit intersects the orbit of mars. The coordinates of the intersection are therefore (1,0)+(-1.5,0)=(-1.5,0)+(1,0)=(-0.5,0)

Hence the location of the second focus is (-0.5,0)

silly me. There is always an elliptical trajectory. There just isn't one for every value of d (the some of the distance from each foci).

Here are the coordinates of  mars from the [http://nssdc.gsfc.nasa.gov/space/helios … 0054659957]website:

You submitted the following name/value pairs:

planet = 02
resolution = 60
start_year = 2004
start_day = 1
stop_year = 2006
stop_day = 1
MARS coordinates:

YYYY DDD   AU     ELAT   ELON     HLAT   HLON   HILON
2004   1   1.475    0.04   51.08      3.07  116.12  335.44
2004  61  1.552    1.01   82.82      0.12   16.68    7.07
2004 121  1.618   1.63  111.70     -2.61  274.37   35.82
2004 181  1.658   1.85  138.72     -4.59  170.30   62.81
2004 241  1.664   1.67  164.98     -5.56   65.57   89.14
2004 301  1.635   1.14  191.61     -5.37  321.25  115.89
2004 361  1.576   0.32  219.75     -3.94  218.39  144.09
2005  55   1.500  -0.65  250.44    -1.33  117.96  174.72
2005 115  1.428  -1.51  284.44      1.96   20.74  208.57
2005 175  1.385  -1.85  321.39      4.74  286.58  245.47
2005 235  1.391  -1.43  359.36      5.61  193.61  283.57
2005 295  1.442  -0.45   35.75       4.21   99.06  320.08
2005 355  1.517   0.61    68.99      1.47     1.18  353.27

I thought I’d post some relevant links:

[http://nssdc.gsfc.nasa.gov/space/helios … 0054659957](1) At this link you can give the time and date and it will calculate the position of any planet.

[http://www.marsacademy.com/traj.htm](2) The mars academy gives some equations to help calculate delta v’s and travel times.
Unfortunately they do not derive anything.

[http://en.wikipedia.org/wiki/Interplanetary_travel](3) I thought this site was interesting. It is a free online encyclopedia.
It talks a little bit about with new computational power how some of the old
orbital transfer solutions have been reevaluated