Wire sag calculation Catenary?

[Machinist1]

Hi all!

I have an odd question that i thought somebody with a degree could help with. I am looking to find the formula to describe wire sag over a given length and given horizontal tension and given wire diameter. I believe the formula is a catenary but am unsure how to apply it to my situation.

I have a .024" (inch) wire diameter.
i have a horizontal tension of 30 lbs
i have a distance between level centres of 30ft

help would be greatly appreciated.

sorry if this is in the wrong category, its my first day :P

PS: i have a chart from a website i found but the description is suspect, here is the link.
http://www.millwrightmasters.com/School/tight_wire_sag_chart.htm

 

[Bill Simpson]

Will this do what you want?

http://www.spaceagecontrol.com/calccabl.htm

You need the weight of your cable per unit length and I don't know what your cable is made of.

 

[Machinist1]

Will this do what you want?

http://www.spaceagecontrol.com/calccabl.htm

You need the weight of your cable per unit length and I don't know what your cable is made of.

right sorry, forgot to mention the wire is steel with a weight of 491lbs/cuft. so .284lbs/cuin (.001542lbs per linear ft)

and according to that website, the lowest point on the curve is .005". and i see on the chart i provided that the sag is .031" (lets say the chart is proven but uses a .016" wire, the spaceagecontrol website does not come close to the chart on the millwright website).

 

[PhanthomJay]

Hey, that Millwright website sag calculator was based on info I had provided in response to a question posted some time ago on these forums! It was based on the parabolic approximation to the catenary curve, which is pretty darn accurate when sag to span ratios are small.

Anyway, the two charts are just about identical. For .016 inch diameter steel cable strung to 30 pounds in a 30 foot span, the sag is , per the Millwright site, .031 inches. And per Mr. Simpson's spaceage site, the sag is .002551 feet, which is also .031 inches to 3 sig figures. I think you forgot to convert feet to inches.

When sag-span ratios are less than about 10 percent, the parabolic approximation
$sag = wL^2/8T_H$ is very good to 3 significant figures, where L is the horizontal span in feet, w is the weight of the cable in pounds per foot, $T_H$ is the horizontal tension at the lowpoint of the cable, in pounds, and the sag is in feet. Otherwise, use the spaceage site.

 

[Machinist1]

thank you Jay. i think i might have been getting mixed up with the catenary equation because it asks for weight per unit length. so i assumed it meant the full weight of the wire over 30ft span. not just the length of 1 foot of wire.