# Thread: Rearrange Tower Guy Tentioning Equation

1. ## Rearrange Tower Guy Tentioning Equation

Hello. My name is Jim.
I am a new member to the Math Help Forum. I am a retired old (octogenarian) electrical engineer specializing in communications, specifically microwave propagation, who has forgotton too much of what I used to consider fun. I would appreciate your help in solving for the following:

WL
T = ---- * SQRT(L^2 I^2 H^2 - 2IH) (The ^ indicates superscript as in squared.)
2I
This is a catenary equation used to calculate the tension of a tower guying cable from the visual sag in the cable. It is based on the length of the base distance from anchor to tower (L), the height of guy cable attachment point on the tower (H), the weight and length of the guy cable, contained in (W), and the tension (T) in the cable for the amount of sag as defined by the intercept distance (I). The intercept distance (I) is that distance to a point on the tower below the guy attachment height, as visually observed on the tower using a surveyor's transit telescope attached to the guy cable at the anchor, to establish the guy starting angle and amount of sag resulting after tensioning.

I'm trying to solve for the individual values of I, H, and L by rearranging the tensioning equation.

Your help in restoring this ol' fogies appreciation for mathamatics would be sorely welcome.

2. ## Re: Rearrange Tower Guy Tentioning Equation

Help...
An error "Message from webpage" states "Your submission could not be processed because the token has expired. Please reload the window"

I don't what this means nor what I need to do to proceed. Please help reprocess my posting.

Thanks
jhhayes

3. ## Re: Rearrange Tower Guy Tentioning Equation

Ohhhh. I see an error on my part (don't get old...):
The second term of the guy tensioning equation should read SQRT(L2 + I2 + H2 - 2IH): L, I & H summed not multiplied)
What do I need to do to proceed?
jhhayes

4. ## Re: Rearrange Tower Guy Tentioning Equation

Originally Posted by jhhayes
Hello. My name is Jim.
I am a new member to the Math Help Forum. I am a retired old (octogenarian) electrical engineer specializing in communications, specifically microwave propagation, who has forgotton too much of what I used to consider fun. I would appreciate your help in solving for the following:

T = (WL/2I)* SQRT(L^2+ I^2+ H^2 - 2IH) (The ^ indicates superscript as in squared.)

I've added the "+"s here. Is that what you meant? Also I have written WL/2T. Internet Explorer does not respect begininning spaces.

This is a catenary equation used to calculate the tension of a tower guying cable from the visual sag in the cable. It is based on the length of the base distance from anchor to tower (L), the height of guy cable attachment point on the tower (H), the weight and length of the guy cable, contained in (W), and the tension (T) in the cable for the amount of sag as defined by the intercept distance (I). The intercept distance (I) is that distance to a point on the tower below the guy attachment height, as visually observed on the tower using a surveyor's transit telescope attached to the guy cable at the anchor, to establish the guy starting angle and amount of sag resulting after tensioning.

I'm trying to solve for the individual values of I, H, and L by rearranging the tensioning equation.

Your help in restoring this ol' fogies appreciation for mathamatics would be sorely welcome.
I assume you mean arrange for each of I, H, and L in terms of the others. You cannot solve a single equation for three different variables.
T = (WL/2I)* SQRT(L^2+ I^2+ H^2 - 2IH)
To solve for I, start by multiplying both sides by 2I/WL:
2IT/WL= sqrt(L^2+ I^2+ H^2- 2IH. Square both sides: 4I^2T^2/W^2L^2= L^2+ I^2+ H^2- 2IH.
Subtract 4I^2T^2/W^2L^2 from both sides: 0= I^2- 4I^1T^2/W^2L^2- 2IH+ L^2+ H^2= (1- 4T^2/W^2L^2)I^2- (2H)I+ (L^2+H^2)= 0

That is a quadratic equation of the form aI^2+ bI+ c= 0 with a= (1- 4T^2/W^2L^2), b= (-2H), and c= (L^1+ H^2). Put those into the quadratic formula: $I= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$

To solve for H do much the same thing.
Starting from 2TI/WL= sqrt(L^2+I^2+ H^2- 2IH), above, square both sides:
4T^2I^2/W^2L^2= L^2+ I^2+ H^2- 2IH
subtract both 4T^2I^2/W^2L^2 from both sides to get H^2- 2IH+ (L^2+I^2- 4T^2I^2/W^2L^2)= 0
Again that is a quadratic equation, ax^2+ bx+ c= 0 with a= 1, b= -2I, and c= L^2+ I^2- 4T^2I^2/W^2L^2.

T = (WL/2I)* SQRT(L^2+ I^2+ H^2 - 2IH)
To solve for L, start, again, at
2IT/WL= sqrt(L^2+ I^2+ H^2- 2IH) and square both sides:
4I^2T^2/W^2L^2= L^2+ I^2- 2IH. Now mutiply both sides by L^2:
4I^2T^2/W^2= L^4+ (I^2- 2I)L^2 and then subtract 4I^2T^2/W^2 to get
L^4+ (I^2- 2I)L^2- 4I^2T^2/W^2= 0
Notice that we have only even powers of L.
If you let X= L^2, you have X^2+ (I^2- 2I)X- 4I^2T^2/W^2= 0 which is, again, a quadratic equation with a= 1, b= (I^2- 2I), and c= (-4I^2T^2/W^2).

Once you have used the quadratic formula to solve for X, take the square root to solve for L.

5. ## Re: Rearrange Tower Guy Tentioning Equation

Good show, HallofIvy.

I was able to complete the solution for values I and H. But I'm having difficulty with that for L.
I don't understand your explanation of "If you let X= L^2....Once you have used the quadratic formula to solve for X, take the square root to solve for L."

I have attached a PDF of my work in MathCAD so as you can see if my approach is correct?

I really do appreciate your efforts to bail me out on this.
Thanks so much.

jhhayes

6. ## Re: Rearrange Tower Guy Tentioning Equation

My appologies again, HallofIvy.

I see an error in my last equation of the MathCAD paper.
Attached is a revised PDF showing my correction to the formula.
But the resulting number is still in difference to the previous calculations, i.e.,
the result shoule be 112 ft. but the calculation produces 159 ft.

Where have I errored?

Thanks
jhhayes