The formula is t= (S^2*w)/8d where t= tension, s= span, w= weight per ft and d= sag.
There is more to it than that of course. Call your local Okonite cable rep and get a copy of EHB-90 "Engineering Data for Copper and Aluminum Conductor Electrical Cables". There is a very good writup on pages 45-49. If you can't get the book go to the Okonite Website and ask them to send you a copy of the book or those pages.
has a "Power Cable Manual" that they will send you that has a write up on sag calculations, They also have a form you can enter you parameter on and they will calculate sag it on their computer and fax ( maby Email by now )you back a solution .
Like BJC says, there is a lot to this, but it can be done. I normally use AlcoA's Sag10 program (it does, however, cost money). From time to time, though I have had to put this stuff into spreadsheets, so it is workable...just not as easy as the slick software package.
Wind pressure in lbs/ft^2 is calculated using
Pw = 0.00256*(Vw)^2
Vw = Wind speed in miles per hour
Wind load per unit length is equal to the wind pressure multiplied by the conductor diameter.
Using the same units, Fw comes out in lbs/ft
Fw = Pw * (Dc + 2t)/12
Dc = conductor diameter (inches)
t = ice thickness, if you get ice in your area (inches)
I have to give BJC and kraigb stars for the excellent posts. The S.C. Okonite plants are in my area and I have done work for them. The cable plant is quite impressive in terms of cleanliness and the processes used to produce their product.
Suggestion: Reference:
1. Wadhwa C. L., "Electrical Power Systems," Second Edition, John Wiley & Sons, 1991
Reference 1 Chapter 7 Mechanical Design of Transmission Lines
includes a good treatment of sags, including wind, ice, and you could possibly consider conductor bundles. However, to leave it up to a specific software is often required.
mtroche, please try this. You will be glad to find the sag.
Keep a stop watch on your left hand and a dry buzz stick on the right. Hit the conductor by the stick and start counting time. Let the stick rest on the conductor. When you receive the third wave return, stop the watch. The time noticed by the watch be squared and multiply it by 0.0372.
The answer is the sag in meters.
By using the sag tension formula, I know nobody who measures it. Sorry for this word.
Another way to measure is to put a mark on the tower at location approximately the sag value from the insulator position. Get the position on the other tower which is in line with the mark on the tower and the tangent to the sagging conductor. Get average of the two values. It is approximately the sag. This can be exactly calculated by considering the sagging conductor as a parabola.
Sorry for correction
The above sag is in fact in feet. It needs correction.
The correct formula in meter for:
First return wave sag = 1.006*T^2
Second return wave sag = 0.2515*T^2
Third return wave sag = 0.1118*T^2
T is time in seconds
SooryaShrestha is right that not too many line crews actually measure sag by finding the midpoint and measuring down from attachment point height using targets on the poles (this method is one that we teach our line personnel, but we don't really use it on distribution). More often than not, we use the return wave technique described in SooryaShrestha's post.
There are times when transmission lines are installed to very exacting initial tensions calculated specifically for a particular installation. In those cases, dynamometers are used to set the initial tension, and targets (or return wave) may be used to verify the correct sag.
I have seen Southwire's reference books, and I wholeheartedly agree with BJC that they are worth having.
Suggestion:
Reference 1 indicates that:
1. Ice loading produces:
wi=pi x rho x (2 x r x t + t**2), in kg/m
where
wi is weight of ice per unit length of conductor
r is radius of the conductor in meters
t is thickness of the ice in meters
rho is the density of ice, e.g. 910 kg/m**3
Please, notice that this does not account for icicles that can potentially become very heavy. Then safety factor has to account for them, e.g. 5 or more since the normal safety factor may be 2 only.
2. Wind loading produces:
wa=2 x (r + t) x p, in kg/m
where
wa is wind loading per unit length
p is the wind pressure in kg/m**2
t is ice coating thickness in meters
r is radius of conductor
Total loading W=sqrt[(w + wi)**2 + wa**2], in kg/m
where w is weight of conductor downward
Loading factor q=W/w
