Tension on cable & other engineering Q’s
I was at a local farm over the weekend with my family and some friends of ours. This is one of those places that, in the fall, will have hay rides, a maize maze, livestock demonstrations, a pumpkin cannon (used compressed air and launched the pumpkins about 8″ in diameter 2-300 yards) etc.
They had a zip line with a little seat where you could push a toddler about 75 feet. It used something like a beam trolley device, except the casters rode on the line vs. the flanges of a beam. A ~3/8″ steel cable was strung between two 6×6 posts with guy wires.
We (myself and my friend Seth) thought it was pretty cool and of course thought “I could build one of those”. Our discussion evolved from there until we came upon two problems that neither of us have a definitive answer for. Opinions, yes, but no real knowledge.
Let’s say you have a steel cable that is 100′ long and 8′ off the ground. If you place a weight (a 40 pound child) in the exact middle, we both think that the weight exerted on both ends of the cable is equal. Now, let’s move the child to the one quarter point, so you’re 25′ away from one end and 75′ from the other. The first question is- Is the force exerted on both 6×6’s equal, or is the post that the child is closest to under a larger load than the far post?
A somewhat related issue is the question of cable size with respect to the length of the zip line. If one were to determine the size of cable necessary for a 100′ line, would the cable size need to be increased if the line were to increase to 150′?
I know the dead load would increase, but I don’t think that to be significant. I think the longer cable would need to be stronger only because you would want to increase tension on the line to counteract the amount of deflection (probably not the right term for a flexible cable, but you know what I mean) beacause of the longer span. If deflection were not an issue, my belief is that the cable size could be the same size even though you have increased the span by 50%.
So, what do you think?
Replies
To get into the problem of the tensile load on the cable, think first of a cable between two supports almost zero apart. With the cable drooping down to just about 90 from horizontal, a load of W results in a tensile load of W/2 in each line ascending up to the mounting from the load, and each mounting is resisting that loading of W/2 which is "pulling" straight down.
Now think about tightening up the cable, but not all the way, and let's not consider a load on it yet.
So here is a concept for you. Consider a cable weighing nothing. Zero. Even though it weighs nothing, it can handle considerable axial load. Fix it to the supports on each tree, but allow it to droop a little. Just a little tiny bit.
Do the trees resist any pull, with a zero-weight cable that is not taut?
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"A stripe is just as real as a dadgummed flower."
Gene Davis 1920-1985
"Do the trees resist any pull, with a zero-weight cable that is not taut?"Yes. You have some tension, so the trees would have to resist that tension or the cable would come loose.
Jon Blakemore RappahannockINC.com Fredericksburg, VA
Think again. The cable is like a spider web fiber, lighter in fact, because its weight is zero. It is not taut, and has a miniscule amount of droop.
It is not like a chain or cable across your driveway. Those "fibers" have some weight to them, and it is the weight of the cable, fiber, or chain, and its "drape," or tautness, that is causing the tensile load at the connections. There is no one there doing any pulling except for Mr. Gravity.
Where'd we learn this stuff? Was it physics, in high school? Elementary mechanics, the statics part? Resolving forces into component vectors? Were we awake in class? Were we throwing spitballs, instead of paying attention? Does any of this matter?
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"A stripe is just as real as a dadgummed flower."
Gene Davis 1920-1985
"It is not like a chain or cable across your driveway. Those "fibers" have some weight to them, and it is the weight of the cable, fiber, or chain, and its "drape," or tautness, that is causing the tensile load at the connections. There is no one there doing any pulling except for Mr. Gravity."
Okay, I see your point. I can understand that the only load from a partially tensioned cable is the weight of the cable itself, which in your theoretical example, is nothing.
"Where'd we learn this stuff? Was it physics, in high school? Elementary mechanics, the statics part? Resolving forces into component vectors? Were we awake in class? Were we throwing spitballs, instead of paying attention? Does any of this matter?"
I have no idea what you're trying to say.
Jon Blakemore RappahannockINC.com Fredericksburg, VA
i am not getting it either, i must have been absent that day, or i smoked a joint that morning before class.
i am with florida on installing it though, i would tighten it up and test it with some dead weight to make sure it holds before you let the kids on it.
Why would a weightless cable droop?
BruceT
Why would a weightless cable droop?
Van der Walls force, static charge, magnetic field during sunspots, solar wind, aerodynamic drag, etc.
... sure glad some of youse guys dont design airplanes <G>
This sounded like something that would interest you.
Not a simple set of equations.
Easiest way nowadays to calc all the stress is a high-falutin' FEA program (Finite Element Analysis)
Roebling's original formula was something like:
Page 1419 of 1934 Mark's handbook:
t1=1/2*t2 + sum(l* w* sin a +1/2 n W1 sin a) -f2 sum(L8w*cosa+0.5*n*W1 cos a)
t2 and a have their own long expressions.
No wonder the ig eng. firms in the 1890s thru the 30's had rooms full of young (mostly female) mathematecian just cranking out calcualtions all day long.
When I built my kids a long pulley slide, just pulled some 3/8" wire rope between a couple of 44" dia trees and figured the loads like Frenchy <G>
I can only speak from vast experience. I've installed lots of zip lines, did one last weekend in Colorado actually. I used scrap yard 7/16" stainless cable wrapped around 2 pines trees about 250 apart. I used a come-a-long to take some of the slack out and 3 cable clamps on each end, a small snatch block for pulley, 3 feet of chain and a round disk of plywood for a seat. It will pin your ears back! The one before I had a spool of power pole guy wire I used. No idea what the breaking strain on any cable I've ever used but I've never had one break either and I'm going back 40 years.
Florida,
I set up one zip line for my kids that uses 300 feet of cable and a cool little aluminum trolley made by petzl (I think a rock climbing gear mfg). They hang from a t handle and I also have a cross bar hanging from a rope under the tbar handle for them to sit on if necessary.It starts out with the cable about 8 feet off the ground with no load on it and ends about six feet off the ground but under load the kids feet drag for the last fifty feet if they don't hold them up.The ground has a gentle slope between the two trees so they get moving along about as fast as I can run without getting painfully out of breath.I stretched the 5/16 stainless cable as tight as I could with a come along and it has virtually no sag until loaded.My problem is that the kids don't always drag there feet and slam into the cable clamps at the end of the run with more force than I am comfortable with.Have you figured out some rule of thumb for building a bit of sag into the cable so it runs uphill at the end and stops on its own?Does the zip line work better if the beginning is relatively high and starts off with a fast drop? I tried to avoid this by putting a lot of tension on the cable with the goal being to make the acceleration a bit less dramatic for the younger kids. Any insight of tips would be helpful as it works ok right now but stopping usually involves an adult at the end acting as a bumper for the kid on the zipline.Thanks,
Karl
Sounds great! I've had to fuss with every zip line I've put up to get the stopping point just right. The one I did last weekend I had to move the cable 3 times to get it right. This was the first one I did that wasn't on relatively flat ground so it was a bit fussier than most. I started on the uphill end with the cable about 12 feet up the tree. On the low(downhill) end I put the cable about 20 feet up so it wouldn't be too fast but ended up having to go to 25 feet with more sag so the rider would stop where we wanted.
I think removing the stop block was a good idea. My DIL made an early ride of the current line. I was at the low end and when I saw her coming realized she was about to break the sound barrier. I made a grab at her but it was pointless. To say that she wasn't prepared for the stop block would be a gross understatement. When the pulley hit the stop she flipped straight up, flew off the seat and fell 8 feet between a tree and a small stump. She got up speaking very harshly at me which was my impetus for slowing it down.
The previous one I made I put as much tension in the cable as I could without pulling the tree over since the rider dropped off in the water so stopping wasn't an issue. Lots of tension makes them fast!
Funny you should mention the Petzal pulley as we picked one up from a mountaineering shop and installed it right before I left. It worked great and isn't nearly as heavy as a snatch block. While we were there we got a kids climbing harness as a safety backup for small kids. We clipped them to the chain and didn't have to worry about them falling off the seat.
I also had a real brainstorm on this one. I've always used a ladder to get up to the jumping off point. Since this one was in a real upscale neighborhood I was concerned that the neighbors might now see a ladder as an asset. Went to a sporting goods store and picked up a 2 man ladder tree stand. It gave us the ladder for access and a great place to safely stand to mount the seat as well as a place for an adult to help the smaller kids. You can hardly see it against the tree.
Florida,
Thanks for the thorough reply.Your story about your DIL slamming into the cable end and doing a flip is disturbingly familiar. Last time we had a group of kids over, one daredevil 7 year old girl decide to take one last ride unattended by an adult. She had the other kids give her a good shove to get her speed up early in the run and it was about twenty feet before the end of the run that I looked over and saw her on it. The trolley slammed into the clamps and she did a complete rotation of a full flip with her body completely outstretched and landed flat on her back. Fortunately my original attempt at a stop bumper was a small mountain of wood chips dumped in the path of the zip line. It had been knocked and trampled down to the point that it no longer worked as a bumper but it did provide a nice soft cushion for her to land on.She hopped up and said "WOW, that was fun". After checking her out to make sure she was indeed healthy and well and seriously admonishing her and the other kids I ran into the house to change my shorts.I really want to tune the thing so it is safe and easy to load (not outrageously high), fast enought to be fun yet still safe and self stopping without killing the fun of the last half of the ride.Even though this property has 6 acres and a hundred or more good trees it was surprisingly difficult to find an unobstructed path between two substantial trees with a good slope between them.Karl
try as you might you will never find someything kids like to do that is completely safe (well video games I supposes are reasonably safe)
however you might try to outthink them..
Put one end high enough to get a good run and have the other end low enough to start dragging the kids before coming to the stop.
That way no matter how much they push each other or how fast they get going the ground slows them down before they slam into the end..
The force isn't equal, but the tension in the cable is the same on both sides of the load. However, as the load moves from one post to another, the tension changes. (I think)
Without being confusing (hopefully) the force is part magnitude (amount) and part direction. The direction and magnitude are changing constantly from post to post.
Stupid me would build what I believed to be plenty strong enough based on my best readings on the subject.. then I'd put some test weights on and check it out..
Once I achieved 200% of my weight I'd feel comfortable enough to put my own butt on the line.
I know the more brillant will sit down with pen and paper and do all those calculations but those of us with a deficiet in math skills need to find other ways. If I spend $50.00 more on bigger materials than is needed so what I've saved more than that without hiring someone to do the calculations for me..
Jon, I had a nice reply all typed out but lost it...suffice it to say, your last paragraph is correct, it's all about deflection, and the only difference the length makes is more dead load.
The tension on the cable is a affected by 2 things: The weight of the child and the angle that the cable makes as it is weighted down. With the kid in the center each cable makes the same angle as it sags. With the kid over to one end the angle changes. The steeper the angle (in this example that's the short side) the greater the tension in the cable on that side. The ratio of the tensions is equal to the ratio of the angle cosines.
So, does anyone other than Mike have an opinion about whether the tension on the cable changes with respect the position of the load? And what about the cable size needed if you increase than span by 1.5?
Jon Blakemore
RappahannockINC.com Fredericksburg, VA
"does anyone other than Mike have an opinion about whether the tension on the cable changes with respect the position of the load?"
I don't think it does. But I have no clue how to prove it.
Bumpersticker: If it weren’t for the kids, this would be a Mercedes!
If you and another guy were stretching a 100' rope with a 100lb weight on it, and the weight was an inch away from the other guy, who do you think would be carrying more weight then?
You're talking about weight at the anchor points. We were talking about the tension in the rope. Two entirely different things.
Error, mistake, slip, blunder, or lapse? I prefer to think of it as an unfortunate juxtaposition of circumstances.
This was his original question:
The first question is- Is the force exerted on both 6x6's equal, or is the post that the child is closest to under a larger load than the far post?
I agree that the tension IN the cable doesn't change, but the way it's resolved at each end has to do with the geometry of how much the cable sags, as a previous poster noted.
Okay, let's say you and I have a 50' rope with a 10 lb. weight and we're pulling it extremely taught.When the weight is in the center, I think all would agree that the load is the same.But what about moving the weight 5' towards me. Would you have any less weight on your end?
Jon Blakemore RappahannockINC.com Fredericksburg, VA
Yes. It would be imperceptible, but as it moved closer to the end it would become more pronounced.
Draw yourself a vector load diagram and figure it out.
It's easiest to see when you treat the cable as if it were a spider web fiber . . . ultra strong with no weight of its own.
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"A stripe is just as real as a dadgummed flower."
Gene Davis 1920-1985
The less sag in the cable the more load on the cable for a given weight.
with a near flat cable a 80 pound kid can put a lot of load on the cable.
Dont try to get the cable flat/
Edited 11/6/2008 9:38 pm ET by edwardh1
To calculate the load set up a triangle. The sag of the cable at the center (say 2 inches) is one leg corresponding to 1/2 the load weight , the horizontal distance to the center point where the load hangs is the second side and the actual cable length from one side to the load is the hypotenuse. Calculate the angles then use the 1/2 the kids load weight and calculate for the load on the cable. it will be surprisingly high for a slight sag in the cable. more sag is better
Edited 11/6/2008 9:45 pm ET by edwardh1
edwardh1,
Your calculations work only if the pulley is tied off to each end of the cable. Since the cable passes thru the pulley and the pulley is free to roll, the tension is the same on both sides of the load.
tension is always the same in the cable as the cable connects to itself
My home made zip line kid carrier
- use pulleys made for wire rope, not fabric rope
- pulleys should be a little loose in the carrier
- two pulleys align the pulleys, stop wobble and are WAY faster than one pulley
- run cable near ground on one end to let kids drag legs to stop
Yes, changing the position of the load will change the tension in the cable.
The tension in cable that sags dues to its weight only (dead load) will vary over its length. It will be highest at the anchor points and lowest at the middle (where it is horizontal).
Changing the span and/or the cable length will change the shape of the cable and have a definite effect on the tension. Whether or not you need to increase the cable size if you increase the span by 1.5 would have to be calculated based on the weight of the cable, the length of the cable (or the sag) and the span.
1) Assuming the cable sags, the tension in the cable does change in relation to the position of the load.
2) The span has no real effect of the required cable cross sectional area (size) other than the additional load of the weight of the greater (longer) cable mass.
The tension in the cable changes as a function of position only if the "pulley" is fixed to both ends of the cable. Since it is a pulley it isn't fixed, but rather free to roll then the tension is equal on both sides of the pulley and the pulley will roll along the cable until it reaches equilibrium. (though it might oscillate for a while depending on how the momentum in the system)http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0012/pulleyFBD.gif
Here's some text I copied to support the argument above. Note that the tricky language is the "non stretching rope". That makes it easy to calculate but doesn't really exist in any form. (though steel cable probably comes as close as anything)Note that the tension force, T, on each of the blocks is of the same magnitude. In any nonstretching rope (the only kind of rope you’ll encounter on SAT II Physics), the tension, as well as the velocity and acceleration, is the same at every point. Now, after preparing ourselves to understand the problem, we can begin answering some questions.
Tim,Would it be correct to say that, as the amount of sag in the cable increases, the difference in the load between the two points also increases?In other words, the theoretical infinitely taught cable has no sag and the load is the same no matter what.If the cable is still very taught, with 6" of maximum sag, the load would be slightly more as you travel towards one pole because the angle of the load is changing.If the cable had 6' (12x increase) of sag, the load would increase even more because the angle increases even more?
Jon Blakemore RappahannockINC.com Fredericksburg, VA
No, I don't beleive that is true except possibly at the ends.
Understand the forces that remain constant, those that change with position and how they change.
Constants: the load (L) and the equal (L/2) reactions (the proper term for the loads on the supports).
At any given point, the tension in the cable, if we are assuming a single frictionless pulley supporting the load on a continuous cable, is constant on both sides if the system were in equilibrium.
What is not possible, without external force being applied to the load, is for the system to be in equilibirium with the load at any place other than the center of the cable (another important assumption is the the supporting ends are at the same elevation). If the load is held in another position than the middle BY an externally appplied force, then the force required hold the load off center is reflected in the changing tension in the cable. Greater force is required to hold the load farther from the center. Consequently, the tension in cable increases as well.
If you had a theoretically, infinitely taut horizontal cable, that had no sag, it would be able to support no verticle load.
What decreases with incresed sag id the horizontal reactions. If you tied one end of a rope to a tree, ran it through a tire and pulled on the other end, as you pulled the rope tighter (and therfore had less sag) it would take more force. While the weight of the tire dies not change, the angle through which you exert the forcee does and less of the force has a vertical component.
A very saggy cable will exert the same downward force (with the same load) as a taut cable, but will exert much less force horizontally on the supports, and vice versa.
For your 2nd point, increasing the span increases the length of cable, hence the dead weight. This increase in dead weight increases the tension in the cable. At some point, the tension will become greater then the cable tensile strength and it will fail under it's own weight. So you increase the diameter of the cable. That, of course, increases the dead weight and you must again check to make sure the maximum tension is still below the tensile strength of the cable.
If the cable is pulled taut enough that absolutely NO sag occurs, the tension in the cable is infinite - really!
Forrest
Would it be correct to say that the load on a theoretical cable that is tensioned infinitely would be equal on both ends regardless of where the load was applied?
Jon Blakemore RappahannockINC.com Fredericksburg, VA
yes...the tension on both ends would be the same...that is to say, infinite. In other words, unaffected by the load hanging from it :-)
Yes. The reality is that the tension in a cable for a zip line is MUCH greater than the vertical "load" of the person on it. Having tested with mine, I have to seriously crank on the come-along to get the cable up. With a 200 lb adult on the center of a pretty darn taut 200' length, my 3000# come-along won't pull it one more click tighter, even with both hands on the end of the lever.
Forrest
Just to bring zip lines up again. The boy came up with this deal where the zip line rider's seat is attached to a rope that's attached to a wagon that's ridden in by a trusting sister holding a flag.
Helmets seems a good idea.
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Forrest
Edited 1/23/2009 8:47 pm ET by McDesign
Here it is from a physics textbook.
Thank you for that. The greater the theta, the greater the load, as seen in the diagram.
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"A stripe is just as real as a dadgummed flower."
Gene Davis 1920-1985