The more friction between the wheel and cable the more torque winds up being applied to the wheel, making it accelerate faster. As long as the wheel turns freely with as little friction as possible in it’s own workings, the more friction between the cable and wheel (e.g. the heavier the load) the faster the wheel will turn due to more of the energy being transferred to torque. See Newton’s second law.
Nope, not at all. At that point because of the weight on the cable, he was going up hill. Friction had nothing (or very, very little) to do with it.
See, the cable is not perfectly taught, it sags and your weight makes the lowest point. The end of the run is up hill.
If you were hitting 50 MPH, then yeah, air resistance is going to be a big factor. All other things being equal, a 250-pound rider is going to reach a higher terminal velocity than a 150-pound rider, because of the relationship between a person’s mass/volume, linear size, and drag. This is true even before accounting for the rope geometry issues I described upthread: take a 150-pound rider and a 250-pound rider, put each on a skateboard at the top of a hill, and the 250-pound rider will hit a higher top speed.
What friction are you talking about? the rider is commonly suspended from the span rope via a carriage containing pulleys mounted on bearings - something like this. Friction losses in this device are minimal.
You won’t find me feeling sorry for him
I’ve seen the other side to being thin
Roll us both down a mountain and I’m
Sure the fat man would win!
This a great one! Let me see if I have it right.
Hypothetical #1) A catenary curve that is fixed, not of rope or chain. Point B lower than A, shape has a rapid fall, then to near level and then back up a little. No friction, no air resistance. Two bodies, different weights/shapes result in same acceleration, velocity, etc.
Hypothetical #2) Add in air resistance. Drag. Likely little change unless it is a very long distance or big big difference in drag. Cd of a 200 pounder vs a 100 pounder not much change.
Hypothetical #3) Friction on the line) Well yes, the force moving forward is some function of mass time gravity time sine of the angle overcoming the friction against the line along with drag. But again with bearings likely little difference in friction between those two weights and if anything the heavier will have more of each. Minimally.
Hypothetical #4) Keep out drag and friction, but add in that it is not a fixed line but a flexible one of some elasticity; the tension on the line that results from the weight changes the shape of the line. As explained, the curve actually travelled is very different for two objects of different weights. The heavier object falls at a steeper angle, at the bottom is exerting tension against the line which then has some elastic recoil that contributes to the force that keeps the object moving from the low point, below point B, to that final destination.
It would seem that the issues in Hypothetical #4 are the largest contributors to the difference between two differentially weighted objects.
Right?
I will admit that this is more than just a nitpick, it is an outright error. Ignorance fought.
Correct. If both bodies move along the exact same path under the influence of gravity, and there is no aero drag, they will experience the exact same acceleration/velocity profile.
Careful with your terms. Coefficient of drag (C[sub]d[/sub]) is a very particular engineering term, defined as the ratio of the actual aero drag force on the object, divided by the aero drag force on a flat plate of the same width. So yes, the coefficient of drag is approximately the same for both riders, since both riders are approximately the same shape (model them as a sphere, or possibly a cylinder, depending on orientation). But the heavy guy has greater drag (he’s wider) and greater weight. But weight increases faster than drag: for a given height, if he’s twice as wide, he’s got twice the drag force (because twice the frontal area) and four times the weight (because he’s twice as wide and also almost certainly twice as thick from front to back). Following the same fixed catenary curve you describe in H1, fat guy and thin guy will experience the same initial acceleration - but aero drag will reduce skinny guy’s acceleration before fat guy’s, and fat guy will hit a higher top speed.
I wasn’t even considering the linear elasticity of the line. If you’re using a steel wire rope with a lot of (initial) slack in it, the actual change in rope length due to different rider weights probably is not a huge factor. The catenary curve issues arise not from the stretch of the rope, but from the from the facts that:
-the rope can bend to assume different curves, and
-the rope has non-zero weight per unit length.
A similar effect happens when you have a heavy kid sitting on a skateboard and a skinny kid sitting on an identical skateboard and both start down a hill from rest. The heavy kid accelerates faster, presumably because the wind resistance acting on his greater mass has a lesser effect than a similar wind resistance acting on a smaller mass. There may also be an effect on the wheels from the greater mass that causes them to torque more.
It seems a trifle misleading to say that the wind resistance acts on a mass. It would be accurate to rephrase this as:
“The heavy kid accelerates faster because wind resistance acting on a body of greater mass has a lesser effect than similar wind resistance acting on a body of smaller mass.”
Probably more. That looks like steel cable with a diameter around 19mm/0.75". This (pdf) document says such cable would weigh around 1 lb per foot.
For most zipline setups, I think it’s safe to assume that:
- the pulley bearing is of good quality
- the mass of rotating parts is a very small fraction of the total moving mass
- there is negligible slip between the pulley and the cable
If these are valid, the energy that goes into torque is not significant.
In situations where the changing shape of the rope isn’t a factor, the heavier object still ends up going quicker. I’ve tested this experimentally on a toboggan slope: By racing the four heaviest guys in my scout troop versus the single lightest, we managed to get somewhere in the vicinity of a 10:1 ratio of masses. The lighter guy started off about half a length ahead (presumably due to random variation in the operators starting us at slightly different times), but by the time the two toboggans cleared the starting position, the heavier one was already taking the lead. We also managed to set records for both the shortest and longest run of the day: The light guy had so little speed that he stopped just barely past the downhill part, while the heavy guys, having had negligible air resistance, nearly went out onto the road.
That said, the fact that the rope does change shape will also be relevant, and will also result in the heavy guy going faster. It’s an interesting question which of these two effects is more significant, and I’m not sure of the answer off the top of my head. I’ll maybe crunch some numbers on it a little later.
Oh, and on the catenary vs. parabola question: If the weight is distributed uniformly along the cable, as in a free-hanging cable, you’ll get a catenary (hyperbolic cosine). If the weight is distributed uniformly horizontally, as in a suspension bridge, then you get a parabola. And just for the sake of completeness, if the weight is proportional to the difference between the curve and the extreme point, as in an arched stone bridge, then you want a cycloid.
And just to bring this full-circle, a cycloid is also the solution to the brachistochrone problem: That is to say, if you have a curved path from point A down to point B, and you want an object traveling under gravity down this path to reach point B as quickly as possible, you want the path to be shaped like a cycloid. It’s also the solution to the tautochrone problem, meaning that masses released anywhere along the cycloid will all reach the bottom at the same time.
That’s exactly what I meant. Sorry for unclear phrasing.
I don’t get it. Do they really make ziplines so that the trolley has negligible friction? I would think this would get pretty dangerous, pretty fast.
Imagine a very tight zipline, 100 m long, with only a 10m difference in height between the endpoints. Let go of the empty trolley (at the high end, obviously).
If there is negligible resistance to the trolley’s movement along the cable, and the cable is tight enough not to form a significant, uh, catenary under the weight of the trolley, then we should expect that the trolley would cross the entire length of the cable accelerating downwards at 10 m/s². Which means it’ll do the whole traverse in about a second and a half, and will hit the catching station traveling at about 15 m/s downwards, which translates to about 150 m/s (= 540 km/h) along the axis of the cable.
My impression is that the mechanism of the trolley does have some resistance (friction may not be the correct word) which increases with speed, and eventually balances the pull of gravity. The pull (force) of gravity increases with mass, so this equilibrium will be reached later for a heavier passenger.
Does this make sense?
First thank you for explaining Cd to me.
Second, however, I think that the elasticity may play a role in the last phase of the run, depending on the sort of wire and the mass involved. Or at least I need to have my misunderstanding spelled out for me. (And help me out if I misuse terminology please.) What is the shape of the zipline curve at the instant the object is at its lowest point? My understanding is that at that instant it is two straight segments creating an angle AOB in which the object is the vertex of the angle. At that moment the length of the line is stretched slightly, is under some tension, and has some degree of potential energy in the form of elastic recoil. It is in fact under enough tension to offset the downward force vector of that mass at that velocity; the object continues to descend until they equal. The heavier the object and the greater its downward velocity at that point, the greater that downward force vector is at that point, the greater the elastic recoil. (Even if they are actually two separate catenary curves, not straight line segments, point O, the nadir, is determined by the upward force vector created by the tension of the segments equalling the downward force vector.)
In terms of the skateboard analogy, I do not think that the wind resistance is material. Friction is still the bigger issue preventing forward movement. But force forward is not only a function of g (acceleration)
Acceleration(in the horizontal plane) = (sineø * gravity) - (air resistance+friction)/mass
Right?
Problems:
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you cannot assume the cable is tight enough not to deviate from a straight line. The tension will exceed the strengh of the cable. This is why, for example, electrical/telephone cables on elevated poles all have some deliberate sag between support points.
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Acceleration will not be 10 m/s[sup]2[/sup]. if we do assume that the cable is infinitely tensioned so as to provide straight-line movement of the rider (which we can’t - see above), the acceleration of the rider will be 9.81 m/s[sup]2[/sup] times the sine of the angle of the cable with respect to the horizon. In your stated case of a 10m drop over a 100m-long cable, the sine is .1, so the acceleration is a steady 0.981 m/s[sup]2[/sup]; it will take 14.3 seconds for the rider to travel the length of the cable in this scenario, achieving a final speed of 14 m/s.
No - for reasons well stated by Machine Elf.
It’s important to note that speed is much lower than in true freefall due not to friction, but rather to the shallow angle of descent.
Close. It’s acceleration along the path on which the object is moving (which must start out below horizontal, else it would require some force other than gravity).
This is the same formula as in post #5.
The acceleration will be different, but conservation of energy dictates that for a given vertical drop, the final speed will be the same, since you have converted the
same amount of gravitational potential into kinetic energy. for a 10-meter drop, 14 m/s is the final speed (in the absence of aero drag), regardless of whether you’ve rolled down a 10m-tall hill, fallen 10m through the air, or zoomed along an infinitely-tensioned, 100m-long cable that dropped 10 meters from start to finish.
Got it. Close to the same formula. Sine is included. And the point is that for the zipline case the changing angle as a function of mass vs tension is probably the bigger issue and total drag (air resistance and friction combined) being of less effect because of mass less so.