## Trig/Algebra Question - Sailboat

So not being a math whiz I have a problem that I was hoping you all could help me with. I'll start with referring you to the following website which shows the parts of the sailboat that I'm trying to figure out.

http://www.marinespars.com/image002.jpg

My rig is a little less complex than this. I lost my mast (the big pole in the centre) last year and have found a replacement mast. The problem is I don't have the spreaders or rigging (specifically the upper/outer lower shrouds). My problem is how long should my spreaders be?

Constraints:
- The spreaders have to be 5 degrees from the horizontal (so they tilt slighly upwards and are not perpendicular to the mast as the picture shows)
- The angle from where the upper shroud and lower outershroud connect at the spreader must bisect (the angle below and above where the upper and outer lower shroud connect to the spreader must be the same)

Measurements
- Height of the mast (30 feet)
- Height from the bottom of the mast to where the spreaders connect (15 feet)
- Horizontal distance from the bottom of the mast to where the outer lower shroud connects (4 feet)
- Vertical distance from the bottom of the mast to where the outer lower shroud connects (1 foot)

The reason why the distance from the bottom of the mast to where the lower outer shroud connects is become my mast sits on top of a cabin which is higher than the deck of the boat (by 1 foot).

Solve: Solve for the length of the spreader
Length of the upper shroud and outer lower shroud

Now these are basic measurements so if you could give me a formula/process on how to solve if these dimensions were to change that would be awesome.
 PhysOrg.com mathematics news on PhysOrg.com >> Pendulum swings back on 350-year-old mathematical mystery>> Bayesian statistics theorem holds its own - but use with caution>> Math technique de-clutters cancer-cell data, revealing tumor evolution, treatment leads
 Recognitions: Gold Member Science Advisor Staff Emeritus If I am understanding correctly, dropping a perpendicular line from the point where the shrouds meet the spreader to the deck forms a right triangle with the outer shrould being the hypotenuse. If we call the angle at the spreader "$\alpha$" and the distance from the foot of the perpendicular "x", we have $tan(\alpha)= x/15$ or $x= 15 tan(\alpha)$. Further, drawing a perpendicular from the point where the inner shroud is attached to the deck to the first perpendicular gives a second right triangle where the angle at the spreader is also $\alpha$, the height is 14, and the base is 4- x. That gives $tan(\alpha)= (4-x)/14$ or $4- x= 14 tan(\alpha)$. Dividing one equation by the other eliminates "$tan(\alpha)$ leaving $$\frac{4- x}{x}= \frac{14}{15}$$. We can solve that to get $x= 60/29$, about 2.07 feet. Now, the right triangle for which the outer shroud is the hypotenuse has one leg of 15 feet and the other of 2.07 feet. By the Pythagorean theorem, the length of the outer shroud is $$\sqrt{15^2+ 2.07^2}= \sqrt{225+ 4.3}= \sqrt{229.3}= 15.14$$ or about 15 and 1 and 1/2 inches. We also have that 4- x= 4- 2.07= 1.93 feet. That gives a gives a right triangle with height 14 feet and base 1.93 feet. Bh the Pythagorean theorem again, the length of the inner shroud is $$\sqrt{14^2+ 1.93^2}= \sqrt{196+ 3.72}= \sqrt{199.72}= 14.13$$ or about 14 feet and 1 and 1/2 inches. The length of the spreader will be that 1.93 feet of 1 foot and 11 inches. Now, the "15 feet" and "14 feet" heights you give do NOT take into account that 5 degrees from the horizontal of the spreader- but doing so only changes the lengths by a fraction of an inch so is not really important.

 Similar discussions for: Trig/Algebra Question - Sailboat Thread Forum Replies Classical Physics 3 General Physics 1 General Physics 16 Introductory Physics Homework 1 Calculus & Beyond Homework 8