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We are building an outdoor batting facility and our construction company wants to know the tension on the three cables that will be used to support the net. It's a sttel building and I can't see this being an issue (most likely CYA), but does anybody know what tension the lines have to be to support the netting? Every coach I talked to just pulled the lines snug.
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To quantify "wicked, wicked tight":

The tension required to hold a single cable and a load which is uniformly distributed along its length is:

T=W*L/(8*h) where

T = tension
W = total weight of cable and netting for one cable
L = span of the cable
h = sag at the center of the cable

So for a single cable, assuming the cable and netting weight 115 lbs, the span is 70 feet, and the cable is allowed to sag 4" (0.3333 feet), the tension is 3019 lbs. If you have three parallel cables, the load on each wall holding the cable anchors would be 9000 lbs. Obviously the cable needs to be more than strong enough to stand the tension.

Too much tension? Allow more sag.

Once you and the construction companay have decided on an acceptable sag, resist the temptation to tighten the cables in order to reduce the sag.
quote:
Originally posted by 3FingeredGlove:
To quantify "wicked, wicked tight":

The tension required to hold a single cable and a load which is uniformly distributed along its length is:

T=W*L/(8*h) where

T = tension
W = total weight of cable and netting for one cable
L = span of the cable
h = sag at the center of the cable

So for a single cable, assuming the cable and netting weight 115 lbs, the span is 70 feet, and the cable is allowed to sag 4" (0.3333 feet), the tension is 3019 lbs. If you have three parallel cables, the load on each wall holding the cable anchors would be 9000 lbs. Obviously the cable needs to be more than strong enough to stand the tension.

Too much tension? Allow more sag.

Once you and the construction companay have decided on an acceptable sag, resist the temptation to tighten the cables in order to reduce the sag.

Excellent formula! Since we are getting into the engineering of it, the cable should be sized greater than the calculated value of T. Perhaps 2T.
quote:
T=W*L/(8*h) where

T = tension
W = total weight of cable and netting for one cable
L = span of the cable
h = sag at the center of the cable

So for a single cable, assuming the cable and netting weight 115 lbs, the span is 70 feet, and the cable is allowed to sag 4" (0.3333 feet), the tension is 3019 lbs. If you have three parallel cables, the load on each wall holding the cable anchors would be 9000 lbs. Obviously the cable needs to be more than strong enough to stand the tension.



quote:
the cable should be sized greater than the calculated value of T. Perhaps 2T.




I promise to bring the snacks if you guys promise to set up the cage & do the math part,...ok? Big Grin crazy
Last edited by shortstopmom
quote:
Originally posted by 3FingeredGlove:
To quantify "wicked, wicked tight":

The tension required to hold a single cable and a load which is uniformly distributed along its length is:

T=W*L/(8*h) where

T = tension
W = total weight of cable and netting for one cable
L = span of the cable
h = sag at the center of the cable

So for a single cable, assuming the cable and netting weight 115 lbs, the span is 70 feet, and the cable is allowed to sag 4" (0.3333 feet), the tension is 3019 lbs. If you have three parallel cables, the load on each wall holding the cable anchors would be 9000 lbs. Obviously the cable needs to be more than strong enough to stand the tension.

Too much tension? Allow more sag.

Once you and the construction companay have decided on an acceptable sag, resist the temptation to tighten the cables in order to reduce the sag.


like i said, wicked wicked tight. Wink
quote:
Originally posted by 20dad:
quote:
Originally posted by 3FingeredGlove:
To quantify "wicked, wicked tight":

The tension required to hold a single cable and a load which is uniformly distributed along its length is:

T=W*L/(8*h) where

T = tension
W = total weight of cable and netting for one cable
L = span of the cable
h = sag at the center of the cable

So for a single cable, assuming the cable and netting weight 115 lbs, the span is 70 feet, and the cable is allowed to sag 4" (0.3333 feet), the tension is 3019 lbs. If you have three parallel cables, the load on each wall holding the cable anchors would be 9000 lbs. Obviously the cable needs to be more than strong enough to stand the tension.

Too much tension? Allow more sag.

Once you and the construction companay have decided on an acceptable sag, resist the temptation to tighten the cables in order to reduce the sag.


like i said, wicked wicked tight. Wink


And if the fancy formulas don't work make sure you have plenty of duct tape handy
Last edited by jerseydad

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