In general, an elimination that gives the less-substituted product in excess is said to follow the "Hofmann rule" (as opposed to the Zaitsev rule). What's important is why this happens. There are several reasons that will drive a system to give the Hofmann product: Steric bulk of the base (tert-butoxide gives more than ethoxide). However, remember that smaller bases are also better nucleophiles and will do more substitution. Branching in the substrate. This may create steric hindrance for the base approaching the hydrogen that leads to a Zaitsev product. Slow down that reaction, and the production of the Hofmann product becomes a larger part of the outcome. Poor leaving groups. This pushes us to a transition state with less C-X cleavage, less double bond character, and less of the thermodynamic influence from substitution of the incipient alkene. Of course, the eponymous "Hofmann Elimination" is the best example, and can be used to create non-thermodynamically preferred alkenes. This actually involves a very different mechanism. An excess of iodomethane does an SN2 reaction repeatedly on the nitrogen of an amine, with base neutralizing the HI byproduct. Then silver oxide precipitates AgI leaving hydroxide as the counterion. This basic anion associates with the positively charged nitrogen, leading to syn elimination (predominantly).

We can see the propensity for syn elimination and the trans product by looking at the structure of the trimethylammonium cation: Look at the H-C-C-N Dihedral Examine the ESP map Translucent ESP map	And the structure of the trans-cyclooctene product: Black background White background Spacefilling model Wireframe Ball & Stick
Note that NO anti-periplanar hydrogens are oriented correctly with respect to the trimethylammonium leaving group, and that there is one syn hydrogen oriented correctly--this will lead to creation of the E double bond. As a further note, E-cyclohexene is the smallest cycloalkene that can be isolated with a trans double bond.