Gravitational slingshot maximum I have recently read an article about gravitation slingshot assist used by Voyagers 1-2, and was thinking on why this hasn't been used for travel between solar and other systems.
I mean sligshot can be done as many times as it is necessary to get speed of lets say half the speed of light that would allow to travel to Alpha Centauri in ~10-20 years can it not? There must be a flaw in my thinking that 3 or 4 planets can be re-used to get to necessary speed otherwise it would already have been done (drawing below). Even if planets would align differently I should always be able 'find' the planet that would allow me to jump to one that is closer to the sun, and repeat the acceleration again and again.

What maximum (theoretical) speed could be achieved using planets of solar system as sligshot and how much would this speed wary from planetary alignment and what realistic speed could be achieved?
UPDATE: To be more specific on the second part of the question Lets say craft weight's 500kg at starting speed of 30,000 km/h initially it slings around Mercury (radius 2440km), Venus (radius 6052 - 300 (atmosphere) = 5750 km), and Earth (radius 6378 - 300(atmosphere) = 6050km) until diameter of planets is to wide to not to crash craft on surface. Then it flies to the moons of Saturn - Titan (radius 5150km), Rhea (1527km), Lapetus (1470km), Dione (1123km), Tethys (1062km), Enceladus(504km), Mimas (396km) and starts slinging there until diameter is to wide too. What approximate maximum speed could it get to leave the solar system?
 A: One can get an order of magnitude estimate of the maximum speed attainable by gravitational slingshots without doing any real calculation. 
The 'rough physics' reasoning goes as follows:
The gravitational field of the planets used for slingshots needs to be strong enough to "grab" the speeding spaceship. As a planet cannot "grab" a spaceships moving faster than the planet's escape velocity, it is impossible to slingshot a spaceship to speeds beyond the planetary escape velocities. 
So no matter how often our solar's system planets line up and no matter how often you manage to pull off a perfect gravitational slingshot, you are practically limited to speeds not exceeding roughly the maximum escape velocity in the solar system (i.e. 80 km/s or 0.027 % of the speed of light, the escape velocity of Jupiter).
(Note: by working with well-defined trajectories one can refine the above argument and get all the numerical factors correct.)
A: The faster you go, the less velocity you theoretically can gain from a gravity assist.
The reason for this is that the faster you go the harder it is to bend the orbit. To proof this we have to use the patched conics approximation, which means that while within a sphere Kepler orbits can be used. The sphere can be simplified to be infinitely big, since the bending of the actual patched conic will hardly be affected by this. While the eccentricity is low (equal or greater than one, since it will have to be an escape trajectory) the trajectory will be able to be bend 360° effectively reversing the relative velocity of the space craft with the celestial body, so the change in velocity would be twice that relative velocity, which is also the theoretical maximum gain. When the eccentricity increases this angle decreases. This angle can be derived from the following equation:
$$
r = \frac{a(1-e)^2}{1+e\cos(\theta)}
$$
where $r$ is the distance from the spacecraft to the center of mass of the celestial body, $a$ is the semi-major axis, $e$ is the eccentricity and $\theta$ is the true anomaly. The semi-major axis and eccentricity should remain constant during the trajectory, so the radius would only be a function of the true anomaly which is by definition equal to zero at periapsis and therefore the maximum amount of bending will be roughly twice the true anomaly at $r=\infty$, which means
$$
\theta_{\infty} = \lim_{r \to \infty} \cos^{-1}\left(\frac{a(1-e)^2-r}{er}\right) = \cos^{-1}(-e^{-1})
$$
When the eccentricity gets really high this angle will become 180°, which means that the trajectory is basically a straight line.
There multiple ways to alter the eccentricity. In this case the relevant variables would be:


*

*The hyperbolic excess velocity, $v_\infty$, which will be equal to the relative velocity at which the spacecraft "encounters" the celestial body, with this I mean that the sphere of the celestial bodies is very small compared to the scale of the orbits of the celestial bodies around the sun, thus the relative velocity can be approximated with the difference of orbital velocity relative to the sun, approximated with a Kepler orbit at a encounter between the two when using an trajectory ignoring the interaction between them.

*The height of the periapsis, $r_p$, which is basically limited by the radius of the celestial body (surface or outer atmosphere).

*The gravitational parameter of the celestial body, $\mu$.


$$
e = \frac{r_p v_\infty^2}{\mu} + 1
$$
The gravitational parameter is just a given for a specific celestial body, since a lower eccentricity is desirable, therefore the periapsis should be set to its lower bound, the radius of the celestial body. This way the eccentricity is only a function of hyperbolic excess velocity and thus the relative velocity of the spacecraft with the celestial body.
Using a little bit more maths it can be shown what the change in velocity would be after such a close gravity assist. For this I use a coordinate system with a unit vector parallel to the direction of the relative encounter velocity, $\vec{e}_{\parallel}$, and a perpendicular unit vector, $\vec{e}_{\perp}$:
$$
\Delta \vec{v} = -v_\infty \left(\left(\cos{\left(2\theta_\infty \right)}+1\right)\vec{e}_{\parallel} + \sin{\left(2\theta_\infty\right)}\vec{e}_{\perp}\right) = \frac{2{\|\vec{v}_\infty\|}}{\left(\frac{r_p v_\infty^2}{\mu} + 1\right)^2} \left(\sqrt{\frac{r_p v_\infty^2}{\mu}\left(\frac{r_p v_\infty^2}{\mu}+2\right)}\vec{e}_{\perp} - \vec{e}_{\parallel}\right)
$$
$$
{\|\Delta \vec{v}\|} = \frac{2\mu v_\infty}{r_p v_\infty^2 + \mu}
$$
When plotting these values for Earth, so $\mu = 3.986004\times 10^{14}\frac{m^3}{s^2}$ and $r_p = 6.381\times 10^{6}m$ (I used the equatorial radius plus the altitude at which atmospheric effect can be neglected, 300 km), you would get the following results:

If you want an as high as possible velocity, then you want that this change in velocity would be in the direction of your velocity around the sun. If you have enough time and the orbit is eccentric enough that it crosses multiple orbits of celestial bodies then there are a lot of possibilities, but as soon as you have an escape trajectory from the sun you basically pass by each celestial body at most one more time.
If you just want to get an high as possible velocity you might want to get closer to the sun in an highly eccentric orbit, since its "surface" escape velocity is $617.7 \frac{km}{s}$.
A: You are all thinking too hard about this.  The slingshot effect is all about frame of reference.  Relative to the body you are approaching, the entrance velocity increase must equal exit velocity decrease or you violate simple laws of physics (i.e. gravitation). From the solar system perspective you will have a net gain in velocity if you approach a planet from the right direction, else you will have a net speed decrease after exiting.  The theoretical maximum velocity increase on exit is therefore a function of the speed of the host (slingshot) body in the frame of reference and the vector of approach.
