# Calculate the lifting weight

coltonk
TL;DR Summary
How do I calculate how much force is needed to lift a horizontal metal tube to a vertical position.
how do I calculate how much force is needed to lift a horizontal metal tube to a vertical position. The metal tube is 25 feet long, weighs 300lb and and pivots 5 foot from the bottom. I will be using a hand winch and the cable attaches at the bottom of the 5 foot end of the tube.

Welcome to PF.
So you want to raise a tilting mast for a flag or an antenna. This is not a new problem. The answer will depend on what assumptions we make.
Does the fulcrum rest at zero height or five-plus feet above the ground?
Does the mast start horizontal or tilted through the fulcrum with head of mast on the ground ?
How much mass is at the head of the mast?
How much counter-ballance weight is there at the bottom?
Maybe you could attach a side elevation diagram of your setup.

Lnewqban
coltonk
This is for an antenna. The fulcrum is 5 foot from the bottom of the tube and the fulcrum is 5-1/2 foot about the ground. It has no counter-balance weight except for the 5 foot after the fulcrum. I am not concerned with the added weight of the antenna right now because I will over-size the hand winch. In the future, I will have a 25 foot long 3" square tube inside the 25 foot long 4" square tube. The weight of the 4" square tube is 9.44lb/ft and the 3" square tube is 6.88lb/ft.

I am trying to make sure my current hand winch is capable.

Does the winch wire pass through a pulley at the base of the stand, 5½ feet below the fulcrum? That way the winch can be fixed to the stand at a convenient operating height.

coltonk
yes it does

Here is a quick estimate based on several angular approximations.

The bottom 5' is balanced by the next 5' on the other side of the fulcrum.
The unbalanced mass is only the top 15', which has a centre of mass, CoM.
That CoM is 5' + ( 15' / 2 ) = 12.5' from the fulcrum.
The CoM weighs 300 * ( 15' / 25' ) = 180 Lbs.
Downwards lever force needed on lower end = 180 * ( 12.5' / 5' ) = 450 Lbs.
But wire is at a diagonal angle of about 45° so we must multiply by √2.
Tension in wire will be about 450 * 1.4142 = 636.4 pounds tension.

An accurate value would need to take into account all angles during the raising of the mast.

I will have a 25 foot long 3" square tube inside the 25 foot long 4" square tube. The weight of the 4" square tube is 9.44lb/ft and the 3" square tube is 6.88lb/ft .
25 * ( 9.44 + 6.88 ) = 408. lbs NOT 300. lbs as originally specified.
Tension in wire will be closer to 636.4 * ( 408 / 300 ) = 865.5 lbs.

Mentor
This is for an antenna. The fulcrum is 5 foot from the bottom of the tube and the fulcrum is 5-1/2 foot about the ground. It has no counter-balance weight except for the 5 foot after the fulcrum.
If the fulcrum and ground supports can tolerate the extra weight of a counterweight at the bottom of the antenna mast, that would seem to be a big improvement in the forces required to raise (and lower) the mast...

If the fulcrum and ground supports can tolerate the extra weight of a counterweight at the bottom of the antenna mast, that would seem to be a big improvement in the forces required to raise (and lower) the mast...
The extra weight of counterweights would be more than compensated by a lower tension requirement in the winch wire, so the fulcrum and support stand must be built very strong in the first place, especially if it has no counterweights.

I'll describe a picture I have in my mind.

One pintle bar could be attached across the mast tube at the fulcrum, or there could be two pintles, welded externally either side of the outer tube.

A strong-back tension bar or rod, offset about 6” at the fulcrum, ends attached at the bottom of the mast and 5 feet above the fulcrum, would prevent the outer mast tube bending at the critical pintle weld while tilting.

Near the top of the stand would be two U-shaped journals to support the mast pintles.
The winch would first be used to raise the mast fulcrum, to place the pintles in the journals. That would need a temporary short jury pole at the top of the stand above the journals.
A bolt would pass through the top of each 'U' journal, above the pintle, to prevent a pintle lifting out of it's journal in a cross-wind, or during some unfortunate tilting upset.

Near the bottom of the mast, a horizontal round cross-bar could carry flat weights on each side to counter-balance the mast.

How will an inner tube be raised if it is inside the outer tube ?

Telescoping tubes always seem to have a problem with bending and corrosion. Is it possible to place the secondary extension tube external to the main mast ?
That might also make pintles, ballast weights and any external guys easier to attach.

berkeman and Lnewqban
coltonk
25 * ( 9.44 + 6.88 ) = 408. lbs NOT 300. lbs as originally specified.
Tension in wire will be closer to 636.4 * ( 408 / 300 ) = 865.5 lbs.
I know the weight specified was based on just the 4" tube but ultimately it will also have the 3" tube. I appreciate the help! It helps quite a bit. I believe my hand winch is rated at 2500lb, so it should be adequate to raise and lower both tubes.

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Lnewqban
Homework Helper
Gold Member
I know the weight specified was based on just the 4" tube but ultimately it will also have the 3" tube. I appreciate the help! It helps quite a bit. I believe my hand winch is rated at 2500lb, so it should be adequate to raise and lower both tubes.
Consider the forces on the anchors and structure supporting the hand winch.

coltonk
Let me see if I understand this correctly.
The short side of the fulcrum is 1/5 the total length of the tubes and the long side is 4/5 the total length. If the center of mass weight of the long side of the fulcrum is 865.5lb then it would take a hand winch with the capacity to pull 5 times the center of mass weight of the long side of the fulcrum (5 x 865.5lb)

Mentor
If the center of mass weight of the long side of the fulcrum is 865.5lb
Wait, you have a 25' antenna mast that weighs 865 pounds? I must be misunderstanding what you are saying... Can you upload a sketch of your design as you envision it currently?

Lnewqban
Let me see if I understand this correctly.
NO.
The 865.5 lbs would be the tension in the winch wire supporting the double RHS tube.

The 25 foot long double tube weighs only 408 lbs.
Divide the 25' tube into three regions.
The 5' on one side of the fulcrum, balances the near 5' on the other side.
That leaves only 15' of unbalanced mast, which weighs 408 lbs * ( 15' / 25' ) = 244.8 lbs.
That unbalanced mass is centred 12.5' from the fulcrum.
Down force on foot of tower needed will be 244.8 lbs * ( 12.5' / 5' ) = 612 lbs.
The wire is at about 45°, so tension in wire will be about 612 * √2 = 865.49 lbs tension.

berkeman
coltonk
Thanks Baluncore,

Ive never tried to calculate anything like this before and I appreciate all your help.

Mentor
The 25 foot long double tube weighs only 408 lbs.
Wow, that's a heavy mast for only 25' long. I've helped put up 50'+ masts at Field Day that were only a couple hundred pounds (we used guy-lines affixed near the top to pull them up and then to brace them to the ground).

@coltonk -- What kind of antenna are you putting on top? Will it also have a rotor motor associated with it? How are you going to affix the bottom of the mast in place once the mast is vertical (pin? other?)?

Similar antennas from Field Day 2016 courtesy of SBARA.org:

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coltonk
berkeman,
The mast is support by two 4" x 1/4" thick C-channels embedded in 6000lb of concrete. The C-channels are on each side of the 4" tube and extend up about 6 feet with a bolt at about 5.5' above the concrete running thru the C-channels and tube to act as a fulcrum to raise and lower the tubes. There is a 3" tube inside of the 4" tube and will be raised up about 20 feet above the 4" tube by a second hand winch. I will have a hexbeam on the top of the 3" tube but at this time will not have a rotor but I will have a leg coming off the antenna support pipe with a cord to the ground so I can manually move the hexbeam if needed. I live in Oregon and am told that the heaxbeam facing east/south-east should cover move of the USA. Will most like only turn the hexbeam when I want to talk to places west of me like japan. The attached drawing is an engineered drawing and my design is patterned after it.

#### Attachments

• Tilt Tower.pdf
427.8 KB · Views: 74
berkeman
Here is the winch wire tension in lbs for different angles of elevation.
Pully on ground is 6" beyond foot of mast, so wire is finally at 45°.
Top of mast (truk) starts at ground with -16° slope. It rises through 0° to 90°.
Wire tension:
mast truk  wire
angle hgt tension
deg   ft   lbs
-16  0.0 1000.2
-15  0.3  984.8
-10  2.0  946.0
-5   3.8  906.2
0   5.5  865.5
5   7.2  823.8
10   9.0  781.1
15  10.7  737.5
20  12.3  692.9
25  14.0  647.5
30  15.5  601.1
35  17.0  554.0
40  18.4  506.2
45  19.6  457.8
50  20.8  408.8
55  21.9  359.4
60  22.8  309.6
65  23.6  259.5
70  24.3  209.3
75  24.8  159.1
80  25.2  109.0
85  25.4   58.4
90  25.5    0.0

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