Student'ish. Engineering help request catenary under load
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Student'ish. Engineering help request catenary under load
Hello,
"Friend" of mine wants to install rope or cable based system to get groceries and a propane tank to his house from where he parks.
Goal: I want to recommend a rope or cable that can carry the load and (if possible) within his budget.
I Believe I need to calculate the load forces using a centenary formula to Pass Fail ropes and cables.
Firstly, is there any "catenary for dummies" I should read first? Ive spent some time reading the formula and I just cant see how it spits out "1400lbs is your maxim" from the inputs. That could be general ignorance.
Details: Load weight is About 100 LBS, strait line distance is about 1000ft, angle is about 22° climb and I'm guessing the rise for that would be 4500 ft.
I'm working with wags for now, but I intend to make the calculator in excel when Ive gotten the formula figured out.
As a wag, Ive been looking at a 1/2 rope with a maxim load of 300lbs, but i don't know the weight per foot.
I feel that I have insufficient information to make the calculator, maybe someone can help me ask the right questions.
"Friend" of mine wants to install rope or cable based system to get groceries and a propane tank to his house from where he parks.
Goal: I want to recommend a rope or cable that can carry the load and (if possible) within his budget.
I Believe I need to calculate the load forces using a centenary formula to Pass Fail ropes and cables.
Firstly, is there any "catenary for dummies" I should read first? Ive spent some time reading the formula and I just cant see how it spits out "1400lbs is your maxim" from the inputs. That could be general ignorance.
Details: Load weight is About 100 LBS, strait line distance is about 1000ft, angle is about 22° climb and I'm guessing the rise for that would be 4500 ft.
I'm working with wags for now, but I intend to make the calculator in excel when Ive gotten the formula figured out.
As a wag, Ive been looking at a 1/2 rope with a maxim load of 300lbs, but i don't know the weight per foot.
I feel that I have insufficient information to make the calculator, maybe someone can help me ask the right questions.
- Frederick_Law
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Re: Student'ish. Engineering help request catenary under load
"Details: Load weight is About 100 LBS, strait line distance is about 1000ft, angle is about 22° climb and I'm guessing the rise for that would be 4500 ft."
Your "friend" live on top of a 4500 ft cliff? And a 1000 ft rope is long enough?
Most building has 9ft ceiling. Assuming 10ft each floor. Your "Friend" is living on 450 floor.
Your "friend" live on top of a 4500 ft cliff? And a 1000 ft rope is long enough?
Most building has 9ft ceiling. Assuming 10ft each floor. Your "Friend" is living on 450 floor.
- Glenn Schroeder
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Re: Student'ish. Engineering help request catenary under load
I'm not an engineer, but I believe the first question should be to clarify that 4500 feet of elevation.AngryPictureDrawer2 wrote: ↑Mon Sep 12, 2022 1:03 pm I feel that I have insufficient information to make the calculator, maybe someone can help me ask the right questions.
"On the days when I keep my gratitude higher than my expectations, well, I have really good days."
Ray Wylie Hubbard in his song "Mother Blues"
Ray Wylie Hubbard in his song "Mother Blues"
Re: Student'ish. Engineering help request catenary under load
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I may not have gone where I intended to go, but I think I have ended up where I needed to be. -Douglas Adams
I may not have gone where I intended to go, but I think I have ended up where I needed to be. -Douglas Adams
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Re: Student'ish. Engineering help request catenary under load
well oops, I got my units wrong... derp says HIGHT is closer to 375ftGlenn Schroeder wrote: ↑Mon Sep 12, 2022 1:59 pm I'm not an engineer, but I believe the first question should be to clarify that 4500 feet of elevation.
thats a Wild @$$ Guess, Standing at point A looking at point B is a distance of 1000ft (Known), I guessed the angle to be about 22°
- Frederick_Law
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Re: Student'ish. Engineering help request catenary under load
1000 ft, half way across Niagara falls.
Feel like FBI will be here soon.
Feel like FBI will be here soon.
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Re: Student'ish. Engineering help request catenary under load
buhahha that gave a much more reasonable tension in my goal seek.
weight per foot = 0.063
height of lowest point = 15ft
known vertical = 375
known horizontal = 927.024
tension = 78.71 (units, hopefully lbs/foot)
weight per foot = 0.063
height of lowest point = 15ft
known vertical = 375
known horizontal = 927.024
tension = 78.71 (units, hopefully lbs/foot)
- Frederick_Law
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Re: Student'ish. Engineering help request catenary under load
thanks for that, that is way higher than what Ive been getting with catenaryFrederick_Law wrote: ↑Mon Sep 12, 2022 2:36 pm https://www.balancecommunity.com/pages/ ... calculator
- bentlybobcat
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Re: Student'ish. Engineering help request catenary under load
1/2" rope SWL 300 pounds?
What is it, wet paper?
A descent 1/2" rope should have a SWL in the 1000-1500 pound range for cheap poly up to 3000-4000 for something high performance.
For design start here:https://www.omnicalculator.com/math/catenary-curve
1000 foot between the parking lot and car? Wow.
What is it, wet paper?
A descent 1/2" rope should have a SWL in the 1000-1500 pound range for cheap poly up to 3000-4000 for something high performance.
For design start here:https://www.omnicalculator.com/math/catenary-curve
1000 foot between the parking lot and car? Wow.
Bent
- Frederick_Law
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Re: Student'ish. Engineering help request catenary under load
yep, homie lives in a yurt in the middle of the woods. summer time he can drive all the way up but in winter it becomes too muddy and he has to park much lower down and walk up. if you think the numbers are bad now its much worse walking the path to his place.
Ive been following this sites instruction: https://engineerexcel.com/catenary-cabl ... -in-excel/
So here is where I'm at: not a lot of confidence.
using a catenary formula I have the below inputs
1/2" rope weight = 85LBS with 9500LBS tensile strength
load = 100 lbs (not sure how to apply load so I added it to the per foot rope weight)
distance = 1000ft
vertical =261
horizontal = 965
sag = 47ft
after goal seek iterations I came up with 408.95 Lbs of tension
does this sound even remotely close to reality?
Re: Student'ish. Engineering help request catenary under load
If it were me, I would take my best swag, build in about 250% overkill, because it is just a guess, and then iterate during installation.
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I may not have gone where I intended to go, but I think I have ended up where I needed to be. -Douglas Adams
I may not have gone where I intended to go, but I think I have ended up where I needed to be. -Douglas Adams
- Frederick_Law
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Re: Student'ish. Engineering help request catenary under load
Get homie to buy a 4X4 LOL
You don't want a sagging line. The load will be bouncing up and down.
Your calculation is just hang the line free and find tension on both end.
Got nothing about pulling it tight.
You'll need 2 strong posts to hold the line.
Which when it snap, you got 1000ft line cutting through everything in its path.
Try hang the tank on a shorter line and see how it goes.
You got 2 months LOL
You don't want a sagging line. The load will be bouncing up and down.
Your calculation is just hang the line free and find tension on both end.
Got nothing about pulling it tight.
You'll need 2 strong posts to hold the line.
Which when it snap, you got 1000ft line cutting through everything in its path.
Try hang the tank on a shorter line and see how it goes.
You got 2 months LOL
Re: Student'ish. Engineering help request catenary under load
You can calc this by placing the point load in the worst case situation, which will be approximately between the 1/3 and 1/2 mark on the low side.
Also, the rope stretch is going to be an issue. I would recommend looking into Galvanized Aircraft Cable, the stretch is very low, and the strength to weight ratio is high.
Realistically, the challenge is the lateral load at the mounting points for the ends of the cable, this is exacerbated by the long cable length requiring a fairly high tension to keep it from sagging too much and allowing your load to drag. As the cable tension increases, and your cable approaches a straight line, the forces at the ends approach infinity.
It might be worth splitting the run in half to downsize the components, sag, and resultant forces, assuming you've got some big trees to anchor on.
Also, the rope stretch is going to be an issue. I would recommend looking into Galvanized Aircraft Cable, the stretch is very low, and the strength to weight ratio is high.
Realistically, the challenge is the lateral load at the mounting points for the ends of the cable, this is exacerbated by the long cable length requiring a fairly high tension to keep it from sagging too much and allowing your load to drag. As the cable tension increases, and your cable approaches a straight line, the forces at the ends approach infinity.
It might be worth splitting the run in half to downsize the components, sag, and resultant forces, assuming you've got some big trees to anchor on.