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I need help.
Don’t want to bore people fartless, so I’ll not go into detail here, but I need someone who’s a bit decent with maths/trig to do what I’m told is simple. (Might be, but not to me.)
I know there are a lot of learned people here, so: I have to put a steel wire rig across a road to take a banner. The banner will not be central on the wire rig but needs to be pretty horizontal when in place.
If the banner was central even I could do it. TBH, I haven’t taken spot levels yet, but I can probably get the end fixings level but that would put the banner out of level because it not central on the catenary curve.
It’s the off centre bit that flummoxes me.
To get the banner off centre but reasonably level means that one end has to be fixed higher than t’other, so what I need to know at the end is how much higher.
I meant ‘are the ends of the catenary wire at the same level’ ? If yes then its fairly simple - decide where along the catenary you want to put the banner, then adjust the hangers to get it level . That spreadsheet one of our pleasant forummers kindly pointed out should sort it should it not ?
Diagrams I can’t do, the scanner has packed up. Doesn’t everything when you need it most?
Right stats:
27.5 total span between fixings.
Natural catenary droop for span (without over tensioning/stretching the cable) is, to all practical purposes, 500 mm.
If end fixing were level, that would obviously put the “perigee” mid span.
I need it 8 metres from one end.
So how much higher must one end be than t’other to give me that?
Looks like simple proportions thing to me but as you raise one end, the span gets longer. Not sure if that is critical.
NumNum You are right, up a bit/down a bit would be one way with a lightly weighted free running pulley on the line, but ther’s a fair bit of weight and therefore tension using 5mm stainless steel wire rope and it’s across a TfL buses route. So not a lot of cock up margin.
To create an aysmmetric catenary with its lowest point 8 metres into a 27.5 metre span: The nearest anchor will be 0.32m above the lowest point of the cable, the furthest anchor will be 0.52m above the lowest point. This assumes a vertical sag of 0.04 metres for every metre of span, consistent with the symmetric model.
Because the sag is negligible, this suggests a lightweight cable, so I can’t see the extra metre or so having much of an impact. That reduces it to a simple ratio problem.
Disclaimer: I take no responsibility for the safety of passengers in open-top buses, who may or may not be garroted as a direct result of my calculation.
Scorch You’re a gent and that’s a cup of tea and bun owed.
You’re right, it is a calculus problem, and that’s exactly why it was beyond me. (Never could get to grips with that, which is why I took the spherical trig. option at GCE.)
It’s not I don’t trust your workings, but the next step in to run a check using a scale model. Grand term for some bits of balsa glued together and a length of cotton.
I’d hate to be the first South East London mass murderer. Imagine the headline “Londonbiker’s grisly mayhem with a village festival banner”
If nothing else, it would be seriously embarrassing.
Funny thing is… After I’ve done the maths, Oldguy has done the model, the method statements have been signed, safety permits copied in triplicate… it will come down to a guy saying “Up a bit your end…”
Jetstream My normal spirit level is about half an inch in a large highball tumbler and I fail to see how that aids calculating a catenary curve.
Scorch Oldguy hasn’t done the model yet. I wouldn’t trust that anyway, at 1:100 scale, to be that accurate either. (You have to scale the weights and charateristics of the materials to do that.) Simply checks the practical might accord with the theory.
Next problem is tensioning the cable without winch anchor points available.
