Student'ish. Engineering help request catenary under load

AngryPictureDrawer2
Posts: 19
Joined: Wed May 04, 2022 1:43 pm
Answers: 0
x 12
x 14

Student'ish. Engineering help request catenary under load

Unread post by AngryPictureDrawer2 »

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. grumph
User avatar
Frederick_Law
Posts: 1944
Joined: Mon Mar 08, 2021 1:09 pm
Answers: 8
Location: Toronto
x 1634
x 1467

Re: Student'ish. Engineering help request catenary under load

Unread post by Frederick_Law »

"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. oa
Burj Khalifa-01.jpg
User avatar
Glenn Schroeder
Posts: 1518
Joined: Mon Mar 08, 2021 11:43 am
Answers: 23
Location: southeast Texas
x 1755
x 2126

Re: Student'ish. Engineering help request catenary under load

Unread post by Glenn Schroeder »

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. grumph
I'm not an engineer, but I believe the first question should be to clarify that 4500 feet of elevation.
"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"
User avatar
SPerman
Posts: 2042
Joined: Wed Mar 17, 2021 4:24 pm
Answers: 14
x 2215
x 1865
Contact:

Re: Student'ish. Engineering help request catenary under load

Unread post by SPerman »

image.png
-
I may not have gone where I intended to go, but I think I have ended up where I needed to be. -Douglas Adams
AngryPictureDrawer2
Posts: 19
Joined: Wed May 04, 2022 1:43 pm
Answers: 0
x 12
x 14

Re: Student'ish. Engineering help request catenary under load

Unread post by AngryPictureDrawer2 »

Glenn 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.
well oops, I got my units wrong... derp says HIGHT is closer to 375ft

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°
User avatar
Frederick_Law
Posts: 1944
Joined: Mon Mar 08, 2021 1:09 pm
Answers: 8
Location: Toronto
x 1634
x 1467

Re: Student'ish. Engineering help request catenary under load

Unread post by Frederick_Law »

1000 ft, half way across Niagara falls.
Feel like FBI will be here soon.
AngryPictureDrawer2
Posts: 19
Joined: Wed May 04, 2022 1:43 pm
Answers: 0
x 12
x 14

Re: Student'ish. Engineering help request catenary under load

Unread post by AngryPictureDrawer2 »

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)
User avatar
Frederick_Law
Posts: 1944
Joined: Mon Mar 08, 2021 1:09 pm
Answers: 8
Location: Toronto
x 1634
x 1467

Re: Student'ish. Engineering help request catenary under load

Unread post by Frederick_Law »

AngryPictureDrawer2
Posts: 19
Joined: Wed May 04, 2022 1:43 pm
Answers: 0
x 12
x 14

Re: Student'ish. Engineering help request catenary under load

Unread post by AngryPictureDrawer2 »

thanks for that, that is way higher than what Ive been getting with catenary
User avatar
bentlybobcat
Posts: 64
Joined: Tue Sep 21, 2021 8:43 am
Answers: 0
x 3
x 48

Re: Student'ish. Engineering help request catenary under load

Unread post by bentlybobcat »

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.
Bent
User avatar
Frederick_Law
Posts: 1944
Joined: Mon Mar 08, 2021 1:09 pm
Answers: 8
Location: Toronto
x 1634
x 1467

Re: Student'ish. Engineering help request catenary under load

Unread post by Frederick_Law »

bentlybobcat wrote: Tue Sep 13, 2022 8:54 am 1000 foot between the parking lot and car? Wow.
And 300 ft up.
AngryPictureDrawer2
Posts: 19
Joined: Wed May 04, 2022 1:43 pm
Answers: 0
x 12
x 14

Re: Student'ish. Engineering help request catenary under load

Unread post by AngryPictureDrawer2 »

Frederick_Law wrote: Tue Sep 13, 2022 9:40 am And 300 ft up.
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?
User avatar
SPerman
Posts: 2042
Joined: Wed Mar 17, 2021 4:24 pm
Answers: 14
x 2215
x 1865
Contact:

Re: Student'ish. Engineering help request catenary under load

Unread post by SPerman »

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.
-
I may not have gone where I intended to go, but I think I have ended up where I needed to be. -Douglas Adams
User avatar
Frederick_Law
Posts: 1944
Joined: Mon Mar 08, 2021 1:09 pm
Answers: 8
Location: Toronto
x 1634
x 1467

Re: Student'ish. Engineering help request catenary under load

Unread post by Frederick_Law »

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
User avatar
tsmith
Posts: 35
Joined: Wed Dec 08, 2021 11:36 am
Answers: 0
x 99
x 30

Re: Student'ish. Engineering help request catenary under load

Unread post by tsmith »

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.
Post Reply