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| In [[celestial mechanics]] '''Lambert's problem''' is the [[boundary value problem]] for the [[differential equation]]
| | How to find apartments:<br>Do you plan to visit Paris for a week or so? Have you discussed where you will stay once you arrive?<br>In most areas choosing an apartment is hard, yet in Paris, you have access.<br>First when choosing an apartment you need to look at the neighborhood you plan to be in along with your budget. Are you going to stay one night or several months?<br><br>There are many apartments available in various areas around Paris. The rates depend [http://tinyurl.com/nqbkz6z uggs on sale] the number of people that is going to stay there and how long you will be staying. If you are traveling alone, why not choose a Bed & Breakfast? A comfy apartment or bed and breakfast would be nice to have giving you your freedom, romance, and your own private view.<br><br>The [http://tinyurl.com/nqbkz6z ugg boots sale] sites and tours are always going on for you to fill in those empty days and nights that you have free. Enjoy any one of them whenever you'd like.<br>The most popular apartments are near [http://tinyurl.com/nqbkz6z ugg boots sale] the Seine River. You have all the privacy of your own home and the view is right outdoors. The island apartments on the river have a small population, relaxing settings making a person not wanting to leave. Of course, if you want to leave the island the boats are always making their rounds and you can hop on one to be off for a good time in the city.<br><br>On the Ile Saint Louis Island, there are only 6,000 people and the most romantic spot in Paris. Enjoy the daily life of the old capital with its 17th century buildings and fantastic views all around you. If you decide not to leave the island, the groceries stores and shops are open all the time.<br><br>The Luxe Apartments is something you may want to consider as well. These apartments are all non-French people wanting to live the Parisian ways. The sizes and cost varies considering how many people are staying there.<br><br>Living in luxury, enjoy the view, and relax is the way to live anywhere.<br>The France Lodge suggest when you are looking for an apartment it be furnished and equipped. When traveling to another country it is hard to take all that you will need especially if you are staying for a long period. The bed and breakfast is nice for one or two people, it is efficient and relaxing, again the views are fabulous.<br><br>[http://tinyurl.com/nqbkz6z http://tinyurl.com/nqbkz6z] The sizes of the apartments in Paris range from Studios to 5 rooms and [http://tinyurl.com/nqbkz6z ugg boots sale] there is a solution for all budgets. If you need assistance with finding your knew temporary home away from home check in with an information office and they will be able to assist you.<br>Keep in mind when you are apartment hunting that you probably want to be near where you are going to be most of the time. You might want to consider the bus stop too; sometimes walking very far isn't too pleasant. Check in with your [http://www.dailymail.co.uk/home/search.html?sel=site&searchPhrase=local+travel local travel] agent and see if they can help you make you choose the right apartment and location to fit your budget.<br><br>Sometimes these places have a lot more information than one thinks.<br>Hunting for an apartment in Paris is not going to be easy. There are so many things to think about. The area I think would be the hardest to choose because of the beauty of the views and you really want to see everything. Just remember [http://tinyurl.com/nqbkz6z ugg outlet] your going to be there awhile so you'll have time to do that site seeing you want to do. |
| | |
| :<math> \ddot {\bar r } = -\mu \cdot \frac {\hat r } {r^2}\ \ </math> | |
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| of the [[two-body problem]] for which the [[Kepler orbit]] is the general solution.
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| The precise formulation of Lambert's problem is as follows:
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| Two different times <math>\ t_1 \ ,\ t_2\ </math> and two position vectors <math> \bar r_1 = r_1 {\hat r}_1 ,\ \bar r_2 = r_2 {\hat r}_2\ </math> are given.
| |
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| Find the solution <math> \bar r(t)</math> satisfying the differential equation above for which
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| :<math> \bar r(t_1)=\bar r_1</math>
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| :<math> \bar r(t_2)=\bar r_2.</math>
| |
| | |
| ==Initial geometrical analysis==
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| [[Image:Lambert Fig1.png|thumb|right|300px| | |
| Figure 1:
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| <math> F_1 \ </math> : The centre of attraction
| |
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| <math> P_1 \ </math> : The point corresponding to vector <math> \bar r_1\ </math>
| |
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| <math> P_2 \ </math> : The point corresponding to vector <math> \bar r_2\ </math>
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| ]] | |
| [[Image:Lambert Fig2.png|thumb|right|300px|
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| Figure 2:
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| Hyperbola with the points <math> P_1 \ </math> and <math> P_2 \ </math> as foci passing through <math> F_1 \ </math>
| |
| ]]
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| [[Image:Lambert Fig3.png|thumb|right|300px|
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| Figure 3:
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| Ellipse with the points <math> F_1 \ </math> and <math> F_2 \ </math> as foci passing through <math> P_1 \ </math> and <math> P_2 \ </math>
| |
| ]]
| |
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| The three points | |
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| ; <math> F_1 \ </math> : The centre of attraction
| |
| ; <math> P_1 \ </math> : The point corresponding to vector <math> \bar r_1\ </math>
| |
| ; <math> P_2 \ </math> : The point corresponding to vector <math> \bar r_2\ </math>
| |
| | |
| form a triangle in the plane defined by the vectors <math> \bar r_1\ </math> and <math> \bar r_2\ </math> as illustrated in figure 1. The distance between the points <math> P_1 \ </math> and <math> P_2 \ </math> is <math> 2d \ </math>, the distance between the points <math> P_1 \ </math> and <math> F_1 \ </math> is <math> r_1 = r_m-A \ </math> and the distance between the points <math> P_2 \ </math> and <math> F_1 \ </math> is <math> r_2 = r_m+A \ </math>. The value <math> A \ </math> is positive or negative depending on which of the points <math> P_1 \ </math> and <math> P_2 \ </math> that is furthest away from the point <math> F_1 \ </math>. The geometrical problem to solve is to find all [[ellipse]]s that go through the points <math> P_1 \ </math> and <math> P_2 \ </math> and have a focus at the point <math> F_1 \ </math>
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| The points <math> F_1 \ </math>, <math> P_1 \ </math> and <math> P_2 \ </math> define a [[hyperbola]] going through the point <math> F_1 \ </math> with foci at the points <math> P_1 \ </math> and <math> P_2 \ </math>. The point <math> F_1 \ </math> is either on the left or on the right branch of the hyperbola depending on the sign of <math> A \ </math>. The semi-major axis of this hyperbola is <math> |A| \ </math> and the eccentricity <math> E\ </math> is <math> \frac{d}{|A|}\ </math>. This hyperbola is illustrated in figure 2.
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| Relative the usual canonical coordinate system defined by the major and minor axis of the hyperbola its equation is
| |
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| :<math>\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 \quad (1)</math>
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| with
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| :<math>B = |A| \sqrt{E^2-1} = \sqrt{d^2-A^2} \quad (2)</math>
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| For any point on the same branch of the hyperbola as <math> F_1 \ </math> the difference between the distances <math> r_2 \ </math> to point <math> P_2 \ </math> and <math> r_1 \ </math> to point <math> P_1 \ </math> is
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| <math> r_2 - r_1 = 2A \quad (3)</math>
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| For any point <math> F_2 \ </math> on the other branch of the hyperbola corresponding relation is
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| <math> s_1 - s_2 = 2A \quad (4)</math>
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| i.e.
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| :<math>r_1 + s_1 = r_2 + s_2 \quad (5)</math>
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| But this means that the points <math> P_1 \ </math> and <math> P_2 \ </math> both are on the ellipse having the focal points <math> F_1 \ </math> and <math> F_2 \ </math> and the semi-major axis
| |
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| :<math>a = \frac{r_1 + s_1}{2} = \frac{r_2 + s_2}{2} \quad (6)</math>
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| The ellipse corresponding to an arbitrary selected point <math> F_2 \ </math> is displayed in figure 3. | |
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| ==Solution of Lambert's problem assuming an elliptic transfer orbit==
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| First one separates the cases of having the [[orbital pole]] in the direction <math> \bar r_1 \times \bar r_2\ </math> or in the direction <math> -\bar r_1 \times \bar r_2\ </math>. In the first case the transfer angle <math> \alpha </math> for the first passage through <math> \bar r_2</math> will be in the interval <math>\ 0 < \alpha < 180^\circ </math> and in the second case it will be in the interval <math> 180^\circ < \alpha < 360^\circ </math>. Then <math> \bar r(t) </math> will continue to pass through <math> \bar r_2</math> every orbital revolution.
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| In case <math> \bar r_1 \times \bar r_2\ </math> is zero, i.e. <math> \bar r_1 </math> and <math>\bar r_2\ </math> have opposite directions, all orbital planes containing corresponding line are equally adequate and the transfer angle <math> \alpha </math> for the first passage through <math> \bar r_2</math> will be <math> 180^\circ </math>.
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| For any <math> \alpha </math> with <math>\ 0 < \alpha < \infin </math> the triangle formed by <math> P_1 \ </math> , <math> P_2 \ </math> and <math> F_1 \ </math> are as in figure 1 with
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| :<math> d = \frac{\sqrt{ {r_1}^2 + {r_2}^2 - 2 r_1 r_2 \cos \alpha}}{2} \quad (7)</math>
| |
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| and the semi-major axis (with sign!) of the hyperbola discussed above is | |
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| :<math> A = \frac{r_2 - r_1 }{2} \quad (8)</math>
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| The eccentricity (with sign!) for the hyperbola is
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| :<math> E = \frac{d}{A} \quad (9)</math>
| |
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| and the semi-minor axis is | |
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| :<math>B = |A| \sqrt{E^2-1} = \sqrt{d^2-A^2} \quad (10)</math>
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| The coordinates of the point <math> F_1 \ </math> relative the canonical coordinate system for the hyperbola are (note that <math> E </math> has the sign of <math> r_2 - r_1 </math>)
| |
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| :<math>x_0 = -\frac{r_m}{E} \quad (11)</math>
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| :<math>y_0 = B \sqrt{{ \left(\frac{x_0}{A}\right) } ^2 - 1} \quad (12)</math> | |
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| where
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| :<math> r_m = \frac{r_2 + r_1 }{2} \quad (13)</math>
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| Using the y-coordinate of the point <math> F_2 \ </math> on the other branch of the hyperbola as free parameter the x-coordinate of <math> F_2 \ </math> is (note that <math> A </math> has the sign of <math> r_2 - r_1 </math>)
| |
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| :<math>x = A \sqrt{1+ {\left(\frac{y}{B}\right)}^2} \quad (14)</math> | |
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| The semi-major axis of the ellipse passing through the points <math> P_1 \ </math> and <math> P_2 \ </math> having the foci <math> F_1 \ </math> and <math> F_2 \ </math> is
| |
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| :<math>a = \frac{r_1 + s_1}{2} = \frac{r_2 + s_2}{2} \ = \frac{r_m + E x}{2} \quad (15)</math>
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| The distance between the foci is | |
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| :<math> \sqrt{ {(x_0 - x)}^2 + {(y_0 - y)}^2} \quad (16)</math>
| |
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| and the eccentricity is consequently
| |
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| :<math>e = \frac { \sqrt{{(x_0 - x)}^2 + {(y_0 - y)}^2}} {2 a} \quad (17)</math>
| |
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| The true anomaly <math>\theta_1 </math> at point <math> P_1 \ </math> depends on the direction of motion, i.e. if <math> \sin \alpha</math> is positive or negative. In both cases one has that
| |
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| :<math>\cos \theta_1 = -\frac{(x_0+d) f_x + y_0 f_y}{r_1} \quad (18)</math>
| |
| | |
| where
| |
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| :<math>f_x = \frac{x_0 - x }{\sqrt{{(x_0 - x)}^2 + {(y_0 - y)}^2} } \quad (19)</math>
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| :<math>f_y = \frac{y_0 - y }{\sqrt{{(x_0 - x)}^2 + {(y_0 - y)}^2} } \quad (20)</math>
| |
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| is the unit vector in the direction from <math> P_2 </math> to <math> P_1 </math> expressed in the canonical coordinates.
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| If <math> \sin \alpha</math> is positive then
| |
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| :<math>\sin \theta_1 = \frac{(x_0+d) f_y - y_0 f_x}{r_1} \quad (21)</math> | |
| | |
| If <math> \sin \alpha</math> is negative then
| |
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| :<math>\sin \theta_1 = -\frac{(x_0+d) f_y - y_0 f_x}{r_1} \quad (22)</math>
| |
| | |
| With
| |
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| *semi-major axis
| |
| *eccentricity
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| *initial true anomaly
| |
| | |
| being known functions of the parameter y the time for the true anomaly to increase with the amount <math> \alpha</math> is also a known function of y. If <math>t_2 -t_1 </math> is in the range that can be obtained with an elliptic Kepler orbit corresponding y value can then be found using an iterative algorithm.
| |
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| In the special case that <math> r_1 = r_2 </math> (or very close) <math> A = 0 </math> and the hyperbola with two branches deteriorates into one single line orthogonal to the line between <math>P_1</math> and <math>P_2</math> with the equation
| |
| | |
| :<math>x = 0 \quad (1')</math>
| |
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| Equations (11) and (12) are then replaced with
| |
| | |
| :<math>x_0 = 0 \quad (11')</math>
| |
| :<math>y_0 = \sqrt {{r_m}^2 - d^2} \quad (12')</math>
| |
| | |
| (14) is replaced by
| |
| | |
| :<math>x = 0 \quad (14')</math>
| |
| | |
| and (15) is replaced by
| |
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| :<math>a = \frac{r_m + \sqrt {d^2 + y^2}}{2} \quad (15')</math>
| |
| | |
| ==Numerical example==
| |
| [[Image:Lambert Fig4.png|thumb|right|300px|
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| Figure 4: The transfer time with
| |
| | |
| : ''r''<sub>1</sub> = 10000 km
| |
| : ''r''<sub>2</sub> = 16000 km
| |
| : ''α'' = 120°
| |
| | |
| as a function of ''y'' when ''y'' varies from −20000 km to 50000 km. The transfer time decreases from 20741 seconds with ''y'' = −20000 km to 2856 seconds with ''y'' = 50000 km. For any value between 2856 seconds and 20741 seconds the Lambert's problem can be solved using an ''y''-value between −20000 km and 50000 km
| |
| ]]
| |
| | |
| Assume the following values for an Earth centred Kepler orbit
| |
| | |
| *''r''<sub>1</sub> = 10000 km
| |
| *''r''<sub>2</sub> = 16000 km
| |
| *''α'' = 100°
| |
| | |
| These are the numerical values that correspond to figures 1, 2, and 3.
| |
| | |
| Selecting the parameter ''y'' as 30000 km one gets a transfer time of 3072 seconds assuming the gravitational constant to be <math> \mu</math> = 398603 km<sup>3</sup>/s<sup>2</sup>. Corresponding orbital elements are
| |
| *semi-major axis = 23001 km
| |
| *eccentricity = 0.566613
| |
| *true anomaly at time ''t''<sub>1</sub> = −7.577°
| |
| *true anomaly at time ''t''<sub>2</sub> = 92.423°
| |
| | |
| This ''y''-value corresponds to Figure 3.
| |
| | |
| With
| |
| | |
| *''r''<sub>1</sub> = 10000 km
| |
| *''r''<sub>2</sub> = 16000 km
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| *''α'' = 260°
| |
| | |
| one gets the same ellipse with the opposite direction of motion, i.e.
| |
| | |
| *true anomaly at time ''t''<sub>1</sub> = 7.577°
| |
| *true anomaly at time ''t''<sub>2</sub> = 267.577° = 360° − 92.423°
| |
| | |
| and a transfer time of 31645 seconds. | |
| | |
| The radial and tangential velocity components can then be computed with the formulas (see the [[Kepler orbit]] article)
| |
| | |
| :<math> V_r = \sqrt{\frac {\mu}{p}} \cdot e \cdot \sin \theta \ </math>
| |
| | |
| :<math> V_t = \sqrt{\frac {\mu}{p}} \cdot (1 + e \cdot \cos \theta).</math>
| |
| | |
| The transfer times from ''P''<sub>1</sub> to ''P''<sub>2</sub> for other values of ''y'' are displayed in Figure 4.
| |
| | |
| ==Practical applications==
| |
| | |
| The most typical use of this algorithm to solve Lambert's problem is certainly for the design of interplanetary missions. A spacecraft traveling from the Earth to for example Mars can in first approximation be considered to follow a heliocentric elliptic Kepler orbit from the position of the Earth at the time of launch to the position of Mars at the time of arrival. By comparing the initial and the final velocity vector of this heliocentric Kepler orbit with corresponding velocity vectors for the Earth and Mars a quite good estimate of the required launch energy and of the maneuvres needed for the capture at Mars can be obtained. This approach is often used in conjunction with the [[Patched Conic Approximation]]. This is also a method for [[Orbit determination]]. If two positions of a spacecraft at different times are known with good precision from for example a [[GPS]] fix the complete orbit can be derived with this algorithm, i.e an interpolation and an extrapolation of these two position fixes is obtained.
| |
| | |
| ==Open source code to solve Lambert's problem==
| |
| [http://www.mathworks.com/matlabcentral/fileexchange/26348-robust-solver-for-lamberts-orbital-boundary-value-problem From MATLAB central] | |
| | |
| [http://sourceforge.net/projects/keptoolbox/ PyKEP a Python library for space flight mechanics and astrodynamics (contains a Lambert's solver, implemented in C++ and exposed to python via boost python)]
| |
| | |
| [[Category:Orbits]]
| |
| [[Category:Conic sections]]
| |
How to find apartments:
Do you plan to visit Paris for a week or so? Have you discussed where you will stay once you arrive?
In most areas choosing an apartment is hard, yet in Paris, you have access.
First when choosing an apartment you need to look at the neighborhood you plan to be in along with your budget. Are you going to stay one night or several months?
There are many apartments available in various areas around Paris. The rates depend uggs on sale the number of people that is going to stay there and how long you will be staying. If you are traveling alone, why not choose a Bed & Breakfast? A comfy apartment or bed and breakfast would be nice to have giving you your freedom, romance, and your own private view.
The ugg boots sale sites and tours are always going on for you to fill in those empty days and nights that you have free. Enjoy any one of them whenever you'd like.
The most popular apartments are near ugg boots sale the Seine River. You have all the privacy of your own home and the view is right outdoors. The island apartments on the river have a small population, relaxing settings making a person not wanting to leave. Of course, if you want to leave the island the boats are always making their rounds and you can hop on one to be off for a good time in the city.
On the Ile Saint Louis Island, there are only 6,000 people and the most romantic spot in Paris. Enjoy the daily life of the old capital with its 17th century buildings and fantastic views all around you. If you decide not to leave the island, the groceries stores and shops are open all the time.
The Luxe Apartments is something you may want to consider as well. These apartments are all non-French people wanting to live the Parisian ways. The sizes and cost varies considering how many people are staying there.
Living in luxury, enjoy the view, and relax is the way to live anywhere.
The France Lodge suggest when you are looking for an apartment it be furnished and equipped. When traveling to another country it is hard to take all that you will need especially if you are staying for a long period. The bed and breakfast is nice for one or two people, it is efficient and relaxing, again the views are fabulous.
http://tinyurl.com/nqbkz6z The sizes of the apartments in Paris range from Studios to 5 rooms and ugg boots sale there is a solution for all budgets. If you need assistance with finding your knew temporary home away from home check in with an information office and they will be able to assist you.
Keep in mind when you are apartment hunting that you probably want to be near where you are going to be most of the time. You might want to consider the bus stop too; sometimes walking very far isn't too pleasant. Check in with your local travel agent and see if they can help you make you choose the right apartment and location to fit your budget.
Sometimes these places have a lot more information than one thinks.
Hunting for an apartment in Paris is not going to be easy. There are so many things to think about. The area I think would be the hardest to choose because of the beauty of the views and you really want to see everything. Just remember ugg outlet your going to be there awhile so you'll have time to do that site seeing you want to do.