Ouch ... where's my copy of BMW when I need it. Ok where to start.
How about basic orbital mechanics?
All orbits are conic sections ... ie. they look like a slice cut from a cone ... any one of a circle (degenerate elipse), ellipse, parabola or hyperbola. The orbits of the planets around the sun are roughly circular ... close enough for this discussion.
For a circular orbit, the orbital velocity is V = sqrt(mu/r)
For a heliocentric orbit, mu=1.32715E11 km^3/sec^2
Average Orbital Radius:
Earth: 1.496E8 Km
Mars: 2.274E8 Km
Orbital Velocities:
Earth: 29.78 Km/s (Ve)
Mars: 24.16 Km/s (Vm)
So these are the bounding cases for the orbit transfer ... you start from tangent to earth's orbit at earth's speed, and you need to end up tangent to Mar's orbit at mars speed. And the path between them has to be a conic section. And of course, timing is everything, but this is simply controlled by your launch time.
The most efficient path is called a Hohman transfer. It is simply an elliptical orbit tangent to earth & mars orbits ... in this case, a perigee radius equal to the radius of the earths orbit, and an apogee radius equal to the radius of mars orbit. Skipping a bit of math we can find that the velocity of this orbit at perigee (earth orbit) is 32.71 Km/s (V1), and the velocity at apogee (mars orbit) is 21.52 Km/s (V2).
This means that the minimum delta velocity (dV) required for the Hohman manuevers to rendevous with mars is (V1 - Ve) + (Vm - V2) = (32.71 - 29.78) + (24.16 - 21.52) = 5.57 Km/s
Add to this the dV required to escape earth's sphere of influece ... the hyperbolic excess velocity ... approximately 11 Km/s You'll have some hyperbolic excess velocity on the mars end also, but the current missions are scrubbing this off through aerobraking.
We can also find the time required to make this trip ... half the orbital period of the rtransfer orbit. P = 2 PI sqrt(a^3/mu)
a is the semi-major axis ... orbital radius for circular orbits. For elliptical orbits, its half the length of the major axis or (Re + Rm)/2
P = 516.6 days
P/2 = 258.3 days approximately 8.6 months
So I'm not sure where the 18 month figure came from ... maybe referring to the roundtrip time?
This is the lower bound on dV required ... and therefore fuel. There is an infinate number of other possible transfer orbits. Some will get you there considerably faster, but at the cost of more gas. And carrying more gas isn't cheap. Imagine planning to drive from NY to LA without stopping for gas ... so you tow a trailer with a couple drums of gas ... say you get 40MPG, and its a 3000 mile trip. So you have to carry 75 gallons of fuel ... about 480 lbs IIRC ... but towing 480lbs of fuels, plus the weight of the trailer your car will only get 35MPG ... no problem, you now carry 85 gallons, about 550 lbs ... but your trailer was only rated for 500 lbs, so you get a heavier trailer plus the fuel, now you only get 32 MPG ... etc., etc. That's the kind of thing we're getting into here.
That's all for tonight ... past my bedtime. Tomorrow ... the Rocket Equation, specific impulse, and how much more gas it would take to get there faster
One other bit ... exhaust velocity is not a constraint on final speed. As Hanny suggested, you could stand out on the hull of a ship moving at 0.1c chucking rocks off the back by hand and still be accelerating your spacecraft.