I am wanting to use an aluminum pole (as a Jin Pole) to lift the mast on my sailboat into place. My mast is about 30 ft. long and weighs between 80 – 100 lbs. The aluminum pole would be stood upright and attached to fore and lateral stays to hold it in place vertically. It would be fitted with a pulley near the top end with a line running down and tied around the mast (near its middle) for lifting. The other end of the line (free end) would be run downward from the pulley thus lifting upward on the mast when pulled down. The mast would be lifted up until high enough to set it in the mast holder at its base (about 18 ft up where the lifting line is attached to the mast). My question is how large (diameter) aluminum pipe/tube would I need to lift the 100 lb. mast vertically if I used 3/16 in. wall thickness for the aluminum pole? (Assume it will be 6061 T6 strength aluminum). The pole will need to be an overall length of about 19 ft. if this matters. Distance between the lifting pulley and where the line is attached to the mast to start the lift will be about 18 ft.

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    $\begingroup$ This question would be vastly improved by a diagram $\endgroup$ – Jonathan R Swift Oct 12 '18 at 4:22
  • $\begingroup$ Hi Jonathan, Thanks much for taking the time to look at this. I cannot provide a diagram (at least easily) but to simplify, just think of this as a vertical aluminum pole, held in place vertically by three guy wires. At the base of the pole is a 100 lb. weight that will be lifted straight up by running a line from the weight up to the top of the pole, through a pully, and back down to the ground where a person will pull down on the line to lift the weight. How large a diameter pipe will be needed if it has 3/16 in walls (6061 T6 aluminum)? $\endgroup$ – Jim T Oct 12 '18 at 16:49

Assuming the aluminum pole is attached to the mast at 90 degree angle with ropes attached to the end of the mast and front stay, we have a triangle of approximately 60 ft base and 18 ft height. The vertical load on the pole initially when it is upright is

$$P = 2\times(100/2) (30/18) = 166.66\text{ lbs}$$ Now say as a first guess we assume a 2.5 in diameter aluminum pole and check for Euler's critical column load:

$$p_{critical} = \frac {\pi^2EI}{L^2}$$

where $L =18\text{ ft}$, $E = 10000\text{ ksi}$, and

$$I = \frac {\pi(D^4- d^4)}{64} = 0.5136\text{ in}^4$$


$$Pc = \frac{\pi^2 (10000000\times 0.5136)}{(12\times 16)^2} = 1068\text{ lbs}$$

This is an acceptable answer even though it is much more than 166.7 lbs we rough calculated, considering a factor of safety of 3 and the fact that in a windy day with a rocking boat loads can increase by several folds.

  • $\begingroup$ Thanks Kamran. Your help is greatly appreciated. $\endgroup$ – Jim T Oct 13 '18 at 16:52
  • $\begingroup$ For the record, how did you get to P=166 lbs? The given equation is ambiguous. (2100/2)/(30/168)=(2100*168)/(2*30)=5800. (2100/2)*(30/168)=187.5. 2100/(2*30*168)=0.208... And where did that 2100 come from? $\endgroup$ – Wasabi Oct 15 '18 at 16:34
  • $\begingroup$ @Wasabi , i apologise for the ambiguity. I use my cellphone to answer the questions without the benefit of Latext or calculator. In a nutshell if we assume an isosceles triangle with base at 60ft and height of 18ft we get the force on the pole as 30/18=1.66 times the force of 50 lbs on each end of the base of triangle. 1.67times 50 times 2 is 167 lbs force on the pole. Do me a favor and correct my equation to show the clear arithmetic. Thanks. $\endgroup$ – kamran Oct 15 '18 at 19:37

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