I think you are describing the bi-elliptic transfer, where you raise the apoapsis above the target and then do a retrograde burn there to lower the periapsis. Technically this can save a bit of $\Delta v$ compared to a Hohmann transfer because the $\Delta v$ is spent more efficiently (the Oberth effect), but it also takes longer. If dv really isn't a problem you will always make the best time with a brachistochrone trajectory: accelerate continuously towards the target until you reach the halfway point, then flip around and deccelerate till you reach the destination at a standstill. This is really a generalised brachistochrone since you need to take into account the $1/r^2$ variation of gravity in the solar system. You can also take advantage of aerobraking at the target (if there is an acceptable atmosphere) to push the turnaround point even closer to the target and cut some time off the trip. Tweak as necessary - the idea still works. Porkchop plots are used to minimize the energy/$\Delta v$ requirement since in the real world lifting un-needed rocket fuel out of the Earth's gravity well is an expensive waste.
EDIT: I read an article from NASA describing work using genetic algorithms to optimize these sorts of continuously accelerating transfers. Sadly I don't have the reference any more, but its probably possible to find online.
EDIT 2: You can find a lot of information relating to real and fictional technologies from a hard sci-fi writing perspective at Atomic Rockets. If you are prepared to throw realistic $\Delta v$ requirements out the window then the only limitation on mission time is the acceleration your squishy human cargo is able to endure. For instance, if you pack your people in acceleration couches for the duration you can accelerate at, lets say, 3g long term (I'm not volunteering for this). An accelerate halfway/flip over/deccelerate mission takes (in the first approximation neglecting gravity) takes a time
$$ t = 2\sqrt{\frac{d}{a}} $$
to go a distance $d$ at an acceleration of $a$. To get a rough worst case figure take Earth and Mars at their furthest opposition $ d \approx 2.7 \mathrm{AU} $ so
$$ t_{max} \approx 65 \mathrm{hr}\ !!!$$
or if you only subject your passengers to 1g
$$ t_{max} \approx 113 \mathrm{hr}\ !!!$$
Clearly this is nothing like a real world mission profile. The reason is that the $\Delta v$ requirement for the 1g mission is
$$ \Delta v = a t = 4\times10^6 \frac{\mathrm{m}}{\mathrm{s}} $$
For the 3g mission it goes up to $6\times10^6 \frac{\mathrm{m}}{\mathrm{s}}$. You need a rocket engine with an exhaust velocity of the same order of magnitude if you want to achieve anything like a reasonable cargo/passenger capacity. This is, ahem, not currently feasible. But if you're willing to posit it then you hardly need to worry about complications like the sun's gravity - you can basically go in a straight line! (If you want to check my numbers, please do. I'm not entirely sure I believe them!)
The Kepler orbit of the Earth around the Sun is determined by two constants: the
specific orbital energy $E$ and the specific relative angular momentum $h$:
$$
\begin{align}
E &= \frac{1}{2}v_{r,\oplus}^2 + \frac{1}{2}v_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a},\\
h^2 &= r^2\,v^2_{T,\oplus} = \mu a(1-e^2),
\end{align}
$$
where $\mu = G(M_\odot + M_\oplus)$, $r$ is the distance Earth-Sun (at the moment of impact), $a$ is the semi-major axis, $e$ is the orbital eccentricity, $v_{r,\oplus}$ is the radial orbital velocity of the Earth, and $v_{T,\oplus}$ the tangential velocity. Now, suppose that a large asteroid collides with the Earth, with orbital velocity $(v_{T,A},v_{r,A})$ and mass $M_A$. Its relative velocity is then
$$
\begin{align}
v_{T,A}' &= v_{T,A} - v_{T,\oplus},\\
v_{r,A}' &= v_{r,A} - v_{r,\oplus}.
\end{align}
$$
We can express these relative velocities in terms of the total impact velocity $v_\text{i}$ and the impact angle $\theta$:
$$
\begin{align}
v_{T,A}' &= v_\text{i}\cos\theta,\\
v_{r,A}' &= -v_\text{i}\sin\theta,
\end{align}
$$
where I defined $\theta$ as in Fig. 1 of this article. So we obtain
$$
\begin{align}
v_{T,A} &= v_{T,\oplus} + v_\text{i}\cos\theta,\\
v_{r,A} &= v_{r,\oplus} - v_\text{i}\sin\theta.
\end{align}
$$
If we assume that the collision is central, that heat loss is negligible and that the debris remains gravitationally bound to the Earth, then conservation of momentum implies
$$
\begin{align}
M_\oplus\,v_{T,\oplus} + M_A\,v_{T,A} &= (M_\oplus+M_A)u_{T,\oplus}\\
M_\oplus\,v_{r,\oplus} + M_A\,v_{r,A} &= (M_\oplus+M_A)u_{r,\oplus},
\end{align}
$$
with $(u_{T,\oplus},u_{r,\oplus})$ the new orbital velocity of the Earth (and the gravitationally bound debris) after the impact. We get
$$
\begin{align}
u_{T,\oplus} &= v_{T,\oplus} + \frac{M_A}{M_\oplus+M_A}v_\text{i}\cos\theta,\\
u_{r,\oplus} &= v_{r,\oplus} - \frac{M_A}{M_\oplus+M_A}v_\text{i}\sin\theta.
\end{align}
$$
So the orbital energy and angular momentum will have changed into
$$
\begin{align}
E' &= \frac{1}{2}u_{r,\oplus}^2 + \frac{1}{2}u_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a'},\\
h'^2 &= r^2\,u^2_{T,\oplus} = \mu a'(1-e'^2).
\end{align}
$$
(the change in $\mu$ is negligible). Right, let's plug in some numbers. Suppose we start with a circular orbit, with a radius equal to the present-day semi-major axis:
$$
\begin{align}
\mu &= 1.32712838\times 10^{11}\;\text{km}^3\,\text{s}^{-2},\\
r &= a = 1.49598261\times 10^{8}\;\text{km},\\
e &= 0.
\end{align}
$$
For a circular orbit, it follows that
$$
\begin{align}
v_{T,\oplus} &= \sqrt{\frac{\mu}{r}}= 29.785\;\text{km}\,\text{s}^{-1},\\
v_{r,\oplus} &=0\;\text{km}\,\text{s}^{-1}.
\end{align}
$$
The impact velocity of an asteroid will always be at least equal to the Earth's escape velocity $11.2\,\text{km/s}$, which is the speed it takes up as it falls into the Earth's gravitational potential well. The article that I already linked to states that typical asteroid impact velocities are in the range of $12-20\,\text{km/s}$. In theory, the impact velocity can be as large as $72\,\text{km/s}$ in the case of a head-on collision, when the Earth and the asteroid have opposite orbital velocities, thus a relative velocity of ~$60\,\text{km/s}$, augmented with the escape velocity as the asteroid falls into our gravitational potential well. This is very unlikely for asteroids, but it is possible for comets.
So, let us assume a typical impact velocity $v_\text{i}=16\,\text{km/s}$,
a mass $M_A = 0.1M_\oplus$ and an impact angle $\theta=45^\circ$. We find
$$
\begin{align}
u_{T,\oplus} &= 30.813\;\text{km}\,\text{s}^{-1},\\
u_{r,\oplus} &= -1.0285\;\text{km}\,\text{s}^{-1},\\
E' &= -411.87\;\text{km}^2\,\text{s}^{-2},\\
h'^2 &= 2.1248\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\
a' = -\frac{\mu}{2E'} &= 1.61109\times 10^8\;\text{km},\\
e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0788,\\
r_\text{p} = a'(1-e') &= 1.48411\times 10^8\;\text{km},\\
r_\text{a} = a'(1+e') &= 1.73807\times 10^8\;\text{km},
\end{align}
$$
with $r_\text{p}$ and $r_\text{a}$ perihelion and aphelion. Evidently, the influence on the Earth's orbit is substantial.
In the case of a direct-from-behind collision, we get $\theta=0^\circ$, $v_\text{i}=11.2\,\text{km/s}$, so that
$$
\begin{align}
u_{T,\oplus} &= 30.803\;\text{km}\,\text{s}^{-1},\\
u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\
E' &= -412.72\;\text{km}^2\,\text{s}^{-2},\\
h'^2 &= 2.1234\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\
a' = -\frac{\mu}{2E'} &= 1.60778\times 10^8\;\text{km},\\
e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0695,\\
r_\text{p} = a'(1-e') &= 1.49598\times 10^8\;\text{km},\\
r_\text{a} = a'(1+e') &= 1.71958\times 10^8\;\text{km}.
\end{align}
$$
As expected, the radius at impact has become the perihelion, and the change in eccentricity is lowest.
And just for fun, let's try the worst-case scenario: $\theta=180^\circ$, $v_\text{i}=72\,\text{km/s}$:
$$
\begin{align}
u_{T,\oplus} &= 23.239\;\text{km}\,\text{s}^{-1},\\
u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\
E' &= -617.10\;\text{km}^2\,\text{s}^{-2},\\
h'^2 &= 1.2086\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\
a' = -\frac{\mu}{2E'} &= 1.07530\times 10^8\;\text{km},\\
e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.391,\\
r_\text{p} = a'(1-e') &= 0.65462\times 10^8\;\text{km},\\
r_\text{a} = a'(1+e') &= 1.49598\times 10^8\;\text{km},
\end{align}
$$
so that the radius at impact has become the aphelion, and the change in eccentricity is highest. Although I wonder how much would be left of the Earth after such an apocalyptic event...
If the collision isn't central, then part of the energy will be transferred to the axial rotation of the Earth, which should reduce the effect on the orbit. But that will be more difficult to quantify.
Best Answer
From a previous revision to this answer:
I started on that long drive six months ago but then I never completed the answer. This question recently briefly reappeared on the front page. It's well past time to make my answer complete.
What are porkchop plots?
When the goal is to send a spacecraft from the Earth to another planet, it's not enough to reach the target planet's orbit. The vehicle has to meet the target planet itself. The amount of energy needed to accomplish this varies widely depending on departure and arrival dates. A porkchop plot is a graphical interplanetary mission planning tool that depicts as a contour plot the required energy as a function of departure date and arrival date. The energy needed by the launch vehicle is key in determining the feasibility of such a mission. A mission plan that requires more energy than a launch vehicle could possibly provide is not feasible. In addition to feasibility, a porkchop plot aids in planning key mission operations and in planning the optimal trajectory between the two planets. The plot shown in the question and replicated below show launch energy. It does not show the change in energy needed at arrival. Other porkchop plots do show this as a second set of contour lines.
The original post asks three key questions:
Before answering the above, it will help to explain how a porkchop plot is constructed.
How are porkchop plots constructed?
As a mission planning tool, a porkchop plot makes certain simplifying assumptions with regard to reaching the target planet. Later on, more detailed analyses address those simplifying assumptions. The key simplifying assumptions used in making a porkchop plot are the patched conic approximation and impulsive maneuvers.
These assumptions reduce the problem to one of finding Keplerian orbits about the Sun that take the spacecraft from the vicinity of the Earth to the vicinity of the target planet in the requisite amount of time. Finding such transfer orbits is the subject of Lambert's problem. A number of such transfer orbits might exist. I’ll denote the angle subtended by the departure point, the Sun, and arrival point as $\theta$, with $\cos \theta = \frac{\vec r_1 \cdot \vec r_2} {r_1 r_2}$. The principal value of this angle will be between 0° and 180°, inclusive. For now I’ll ignore cases where $\theta$ is 0° or 180°. This means that the transfer plane is well-defined and that the number of solutions is finite.
Lambert's problem does not have closed form solutions; a number of iterative techniques have been developed to find solutions. One solution, “the short way”, or “Type 1” transfers, has the change in true anomaly equal to $\theta$ as described above. Another solution, the “long way”, or “Type 2” transfers, has the change in true anomaly equal to 360°-$\theta$. Other solutions may exist as well. For example, one way to transfer from Earth to Mars in 2.5 years is to make more than a complete orbit during the transfer. Porkchop plots typically only show the Type 1 and Type 2 solutions, and typically only show at most one of these two solutions for a given pair of departure and arrival dates. If one of the two solutions is close to optimal, the other solution will inevitably follow a retrograde path and thus will involve huge expenditures of energy. There’s no reason to show these highly sub-optimal solutions.
Why is there a gap in porkchop plots?
The plot can be cleaned up further by removing cases where the better of the two solutions still involves huge energy expenditures. Huge energy expenditures are obviously going to result when the transfer time is very short or very long. A not so obvious place where this happens is when the angle subtended between the line from the Earth and Sun at departure and the target planet and Sun at arrival is nearly 180°. That the Earth and target planet have slightly different orbital planes means that the transfer plane will be nearly orthogonal to the planetary orbital planes when the transfer angle is close to but not equal to 180°. This makes the approach of having a maneuver at departure and a maneuver at arrival extremely expensive for those transfers that are close to 180°. Removing those very expensive transfers from view is what creates the gap in the porkchop plot.
This excessive cost for near 180° transfers is to some extent an artifact of the approach used to create a porkchop plot. Adding a third maneuver enables the use of much smaller in-plane maneuvers at the start and end, with a small plane change somewhere along route. There’s a problem, however. This mid-course plane change would necessarily mean thrust from the spacecraft itself. This is undesirable. Tspacecraft itself provides very little of the energy with the two burn approach. The energy for the Earth departure comes from the launch vehicle, and in the case of Mars, most of the energy for Mars arrival comes from aerobraking. It’s better to fold that plane change into the maneuvers at departure and arrival so as to keep the thrust needed by the spacecraft down to a minimum.
Why don’t they use an optimal Hohmann transfer?
An optimal Hohmann transfer doesn’t exist. Hohmann transfers are in-plane maneuvers that transfer from one circular orbit to another that share a common orbital plane. Planetary orbits are slightly inclined with respect to one another and are elliptical rather than circular. Generalizing the concept of a Hohmann transfer to that of a 180° transfer, in most cases that 180° transfer doesn’t exist. When it does, that the orbits are not coplanar and that the orbits are not circular means that this 180° transfer is no longer optimal.
Why are there two local energy minima?
There aren’t just two local energy minima. There are a countable infinite number of local minima. A porkchop plot only shows the first two. The other solutions take even longer than Type 1 and Type 2 transfers and are more sensitive to errors.