Don't involve math, it doesn't really add anything here. Also, that cart is going to weigh more than 50 lb unless it's barely accommodating one person, but again it shouldn't matter.
I would probably let someone move it at their speed roughly. Even though you don't need to move to pull the rope, it still gives a rough approximation. So yeah, I'd probably so no faster than 30' per round. That means a full 2 minutes to cross 600 ft. Seems reasonable enough to me.
| 2 people marked this as a favorite. |
Exactly as fast as it needs to be to make for a good story.
Do you want to have an encounter while the gondola is being hauled up the zip line? Will the civilians need multiple trips to get everybody to safety? Will it be fun to have an encounter before all of the civilians are safe?
Be the GM. Make a decision. There is no rule in the book for pulling a 50# gondola up a zip line. Also, there are lots of variables. How well oiled is the windlass? How big is the take-up spool? How tight are the trolley bearings? How strong is the guy operating the windlass? How much head or tail wind?
Forget all this junk.
Here's how I would do it:
The hobgoblins are 25 minutes away. You have 60 citizens and the cart can take 6 of them each trip. The gondola trolley seems old and poorly maintained, and so is the windlass - old and rusty. In its current state, it travels down at about 30 feet per second (roughly 10 MPH) and returns at 5 feet per second. It takes about 30 seconds to get 6 people on it and 30 seconds to get them off.
So, 30 seconds to load, 20 seconds to zip down, 30 seconds to unload, 120 seconds to pull it back. 230 seconds per round trip, about 3 and a half minutes. That means you can do 6 round trips (plus one more down trip) before the hobgoblins get here. That's only 42 of your 60 civilians. You're going to have to fight!
Note: Observant players might notice that the machinery is not it peak operating conditions. This is when I'd allow relevant skill checks. Oil the trolley and windlass, file off some of the rust from the moving parts, do whatever good engineers do. If they succeed, I would double the travel time:
So, now it's 30 seconds to load, 10 seconds to zip down, 30 seconds to unload, 60 seconds to pull it back. 130 seconds per round trip, just a bit over 2 minutes. That means you can do all 10 trips before the hobgoblins get here, plus an 11th trip down if the PCs want to ride it last. You don't have to fight!
That way, clever or interested PCs who solve a problem can get everybody to safety without the encounter. But if they don't try to solve it, or fail, they have to fight the hobgoblins before they save everybody.
And yes, I made up all those numbers, except for a quick google search to find out how fast zip lines typically go. This one might be a little bit too fast for a 6-man gondola, but hey, it's a game, let's have a wild ride!
Exactly as fast as it needs to be to make for a good story.
This. Convenient, eh? :)
Without knowing the mechanics involved it is impossible to say. DM Blake's speed of plot is a good one.
I disagree slightly with some of the other posts suggesting walking speed, having seen how slowly a hand (or electric) winch winds in a trailer boat or off road vehicle. An easy answer would be 5' per round. Which is still faster than the given real world examples but is easy to manage in gameplay.
If there was a counterweight system it could be significantly faster and the windlass system is replaced with a braking system to control speed. In this scenario the counterweight is slightly heavier than the cart but less than the cart plus a notional minimum load.
Maths answer:
Some assumptions
For ease of maths, 1kg=10 newtons=2 lbs and 1.5 metres = 5 feet
The cart's mass is 100lbs per occupant capacity (from Internet)
An average labourer can output an average of 50-150 watts per day (Internet again). Lets assume the power output of a character is 10 watts per point of strength.
The coefficient of friction between the cart's pulley and the zip line is 0.1 and the zip line is perfectly horizontal.
The mechanical efficiency of the windlass is 10% taking into account the condition of the windlass and more significantly the amount of total human power that can be directed into winding.
Mass to wind=Mass of cart×friction as the cart's weight is supported entirely by the zip line = 100 lbs × 0.1 = 10 lbs = 5Kg = 50N per person capacity
Winding power = STR x Power per STR x Mechanical efficiency. Assuming 10 STR = 10×10×.1=10 watts
Power=Newtons × metres / seconds
Speed = power/newtons = 10/50= 0.2 m/s = 6m per rd = 20 feet per round.
So an average strength character can move a one person cart at 20 feet per round along a perfectly horizontal zip line.
The formula can be generalised as
Speed per round = 20×(Total strength applied/10)/carrying capacity. If the cart has to wound up an incline then the vertical lift per round has to be factored in and will significantly slow the winding speed.