Legacy:Travel Weariness

From Ring of Brodgar
Revision as of 02:01, 28 October 2010 by 130.89.162.247 (talk) (→‎Quoting Loftar)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigation Jump to search

Traveling via teleporting or crossroads gives travel weariness, which can be removed by smoking tobacco, drinking alcoholic beverages or sleeping in a bed. Sleeping logs you out, and removes some weariness over time, while you're logged out. The amount of weariness lost depends on bed type, bed quality and time logged out.


How Travel Weariness Works

When your Travel Weariness "buff" hits 100%, you won't be able to quick travel, quite simply. The one exception is that you can always quick travel to your Hearthfire, although it will still cost travel weariness. The amount of weariness you can take is based on agility. Higher agility means you'll be able to travel farther. The amount of weariness gained when traveling is determined, non-linearly, by distance. Traveling the same distance with many small hops is preferable to one or a few large hops.

You can still choose to log in at your Hearthfire, but it will cost Travel Weariness, calculated from your last position in game.

Quoting Loftar

"With the original n00bish 10 Agility, you'll be able to travel about half a supergrid in one hop. The cost incurred by one hop scales as O(n^{3/2}), and the amount of weariness a character can handle scales linearly with agility."

"A character has α · 500 "travel points" , where α is his Agility; and a travel costs δ^(3/2) / 1000 such points, where δ is the distance travelled, measured in units 11x11 of which make up a tile.

Indirectly, that means that the cost, as a fraction of one's total travel weariness, is δ^(3/2) / 500000α, or approximately σ^(3/2) / 13705α if σ is the distance measured in tiles. It also means that, using your whole TW in one hop, you would be able to travel (13705α)^(2/3) tiles; but, obviously, further than that if you travel in several hops, such as 3 · (13705α / 3)^(2/3) ≈ (23738α)^(2/3) tiles in three hops. In general, it means that, by using n hops instead of 1, you get n / (n^(2/3)) = n^(1/3) times further."

This means the minimum agility required for the jump, when measured in tiles, is given by:

α= δ^(3/2) / 13705

The minimum agility is also visible on http://havenmap.student.utwente.nl/Preview/ by clicking New Ruler, and putting the markers on the appropriate locations.

Solutions

Some helpful things to reduce your travel weariness are: