Simpler way to calculate Force to raise a Tilt Tower

Discussion in 'Antennas, Feedlines, Towers & Rotors' started by WA5DX, Aug 11, 2015.

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  1. WA5DX

    WA5DX Ham Member QRZ Page

    Does anyone have a simpler way to calculate the Force (pounds) required to raise a tower that pivots at the base on the ground?

    In other words, if you have a tower lying on the ground and it weights 86 pounds, and is 39'-11"tall, and it pivots at the base and when you "walk it up" the "lifting point" will always be 6-foot which is your lifting height.

    The formula below is what I am using now and the 173.99 pounds would be at a 45 degree angle. At 75 degrees the (F) force would be 97.62 pounds.

    Thanks in advance for your help, WA5DX

    Force Formulas to calculate the required force to raise a Tilt-Up Tower:

    This Formula is based on a fixed pivot point at or near the base of the tower.

    Formula is based on the total weight is evenly distributed over the entire height of the tower.

    Formula is: U x M = C ( / L ) = f2 x S = F

    Example: 22.17 x 67.49 = 1496.25 / 6.08 = 246.10 * .707 = 173.99 Pounds

    H = distance from vertical to top of tower in Feet = 28.25
    L = Lower Section, lifting point in Feet from Tilt base = 6.08

    S = COSIN of angle = 0.707
    T = mass total weight of tower in pounds = 86.00
    U = Upper Section length in Feet = 22.17

    lp = percent of total height of tower lower section = 0.22
    up = percent of total height of tower upper section = 0.78
    a = weight of lower section in pounds = 18.51
    M = weight of upper section in pounds = 67.49
    C = U x M in pounds = 1496.27
    f2 = down force applied at L = 246.10
    F = force required to lift tower at L =
  2. K8ERV

    K8ERV QRZ Member QRZ Page

    Just have a drone pick it up for you.

    TOM K8ERV Montrose Colo
    AK5B and K0UO like this.
  3. N7EKU

    N7EKU Ham Member QRZ Page


    It doesn't look so bad. Have you tried just putting it in a spreadsheet? I could probably give it a try latter today if you want.


  4. WA5DX

    WA5DX Ham Member QRZ Page

    Hi Mark, actually I wrote this in my Quattro Pro spread sheet. I only have to enter the values for "H", "L", "S" and "T" and everything else is calculated to give me "F". I just have not been able to find a simpler Formula. I no longer hand calculate the COSIN and just use my Casio scientific calculator.

  5. N1VAU

    N1VAU XML Subscriber QRZ Page

    Hey, thanks for that formula, I need it right about now!
  6. N7EKU

    N7EKU Ham Member QRZ Page

    Too fast for me WA5DX!

    Will Quattro Pro do algebraic functions? OpenOffice and Excel will both do it.


  7. WA5DX

    WA5DX Ham Member QRZ Page

    Hi Mark,

    Yes it does.
  8. WA5DX

    WA5DX Ham Member QRZ Page

    Clayton here is a drawing showing each "Value" and where it is referenced, hope this will help. Note: this shows the antenna/tower at a 45 degree angle and "F" will have less weight as the angle increases. Formula.gif
  9. DK7OB

    DK7OB Ham Member QRZ Page

    I'm just playing with masts here too, and so I'm interested in this problem.

    I dug out the remnants of my classes in technical mechanics from > 30 years ago and tried to find a formula myself.

    We can calculate with the weight of the mast concentrated in its center of gravity, which is at half the mast length for a mast with evenly distributed weight.

    I came to this formula:

    F = G / 2 * l * s / (s*s + h*h)

    l = mast length
    G = mast weight
    h = height of point of attack of lifting force from ground
    s = ground distance of point of attack of lifting force from pivot.
    F = force needed to hold the mast in this position

    See attached hand sketch for details.

    Units are arbitrary (m, ft, N, lbs), but don't mix.

    The formula passes simple plausibility checks (like lifting the mast from ground with h=0).

    Also attached is a Libre Office spreadsheet if you want to play around (mast.txt, rename to mast.ods after download)


    Attached Files:

    WA5DX likes this.
  10. WA5DX

    WA5DX Ham Member QRZ Page

    Good morning Wolf,

    Thank you for this formula and it is much simpler than mine. Running your formula compared to mine using a length of 40-feet and a weight of 86 pounds here is what I found:

    at 15 degrees Yours = 70.69 pounds V mine = 23.69 pounds (note 1 )
    at 30 degrees Yours = 123.01 pounds V mine = 116.74
    at 45 degrees Yours = 143.33 pounds V mine = 173.99 ( note 2 )
    at 60 degrees Yours = 126.14 pounds V mine = 161.92
    at 75 degrees Yours = 77.06 pounds V mine = 97.62

    Note 1, your number looks a bit high at this angle (mine looks a bit light) because most of the weight should still be on the ground at this lifting point of 22-feet and 9-inches from the pivot point. Note 2 at 30 degrees your number is very close to mine. At 45 degrees we are close percentage wise, at 60 degrees mine shows a lot more weight but we get real close again at 75 degrees. I guess some day when I have the time and materials that I will build a scale model to test both formulas, which would be a fun project.

    I really appreciate your help and hopefully this will be of some help to other Hams.

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