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Marworth

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  1. So I was writing my original post in a rush and didn't think to reference the rules I was working backwards from. The following is from UV 55-56: However, a rowed Vehicle's movement depends not just on its inches of Swimming, but on the strength of the rower(s). A rowed Vehicle's maximum movement equals the maximum of it's inches of Swimming or the rower's inches of Leaping (see below), whichever is less. To determine how fast a rower can row a watercraft, consult the Strength Table (Hero System 5th Edition, Revised, page 34) and find out how far the rower can make a running broad jump forward Leap with his STR. Subtract 1" for each point of Size the water Vehicle has. If the final result is 0" or less, the rower isn't strong enough to move the vehicle. If the result is 1" or more, that's how fast the rower can move the vessel, to a maximum of the inches of Swimming it's purchased. If the rowing inches exceed the Swimming inches, that usually indicates too many rowers or a poor vessel design. Sometimes, an operator has more than one rower -- a slave galley in a fantasy setting might have a crew of hard-muscled slaves chained to the oars, for example. In this case, add the weight all the rowers can lift together (don't add their STRs, just the weight the can lift). Then compare this total to the Strength Table to determine the "group STR" of the rowers. Use that STR as described above. (Alternately, if all the rowers are the same STR, for every x2 rowers, add 5 STR.) If you read the above passage, it clearly outlines the math to determine how fast a group of rowers can row a ship, or with a little basic algebra, how strong a group of rowers must be to row a ship of a given Size at a given number of Swimming inches. Referencing the 5E core, we know that the Leaping inches referred to above are equal to the STR in question divided by 5. So from the above passage the equation for how fast a group of rowers can row a ship is as follows: STR / 5 - Size = Inches of Swimming If we rearrange this we get: (Inches of Swimming + Size) * 5 = STR required to row the ship at that speed In my above question I referenced the Trireme as built on UV 61 which has Size: 12, Swimming (rowed): 4", and 170 oars (meaning there are up to 170 rowers) So again, using our rearranged equation and plugging in the numbers we get: (4+12) * 5 = 80 STR The lifting STR of 80 STR is 1,600,000kg If we assume all the rowers have the same STR, then they all contributed the same amount to the group lifting capacity used to determine the group STR. It therefore follows that if you divide the lifting capacity of the group, by the number of members of the group, you would get their individual lifting capacity. 1,600,000kg / 170 rowers = 9412kg If we reference the Strength Table again we find that this lifting capacity falls between the values for a STR of 40 and 45, which is completely insane. For further edification, applying the same math to a few other rowed vehicles in UV you get the following results: Viking Longship (UV 62) [size: 13, Swimming: 4", Oars: 60] (13+4)*5=85 Lifting Cap for STR 85: 3,200,000kg 3,200,000kg/60 oars= individual STR 55-60 Canoe (UV 61) [size: 0, Swimming: 6", Oars: 2] (0+6)*5=30 Lifting Cap for STR 30: 1,600kg 1,600kg/2 oars= individual STR 25 So by this, the average human, who is at most STR 10, cannot even come close to rowing the smallest and most basic of boats. It in fact would take TWO someones 8 times as strong as an average human to row a simple rowboat, something that I am fully capable of, on my own, as a human who is most definitely not STR 10. Now I'll ask again... Am I missing something, or after 4 years of working with the HERO System have I finally found a rule that doesn't make sense?
  2. So there isn't a 6E vehicle book, so I'm trying to use the 5E one. Here's my problem, unless I'm missing something huge, based on the rules given for rowing speeds in the book, to row the example trireme at it's given rowing speed of 4", you would need a rowing crew with an average STR of 43. Rowing Movement: 4" Size: 12 Oars: 170 4+12 = 16" STR needed to make a running broad jump forward of 16": 80 Lift Capacity of STR 80: 1.6ktons = 1,600,000kg 1,600,000kg / 170 oars = 9412kg lift capacity per oarsman STR needed to lift 9412kg: 43 What did I miss?
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