Need some formulas from the nerds here

[Science Abuse]

I need to test some materials, but before I do that, I need to know how much stress I need to test them to. It's a mast, so i need to know (@sea level):

x wind speed over y sail area = ? Lbs of force

 

and

 

X material (calculate from weight per foot) that is y feet long produces an equivilent ? lbs force at the tip when held perpendicular to gravity.

 

Basicaly I want to see at what windspeed my home made masts will deflect perminently. There's lots of variables, I know, but takeing them out means I get to over engineer my stuff, no harm in that.

 

http://img112.imageshack.us/img112/7127/formulas2yf.png

[Berto]

delete?

[Berto]

For your bottom situation, you can use my book if you want or look up online leaf spring or cantilever beam

 

stress=(6P/bt^2)*x

P=transverse end load (force at length Y, the tip of the beam)

b=beam width

t=beam thickness

x=distance from the free end of the beam.

 

if trying to find max stress, x=L=Y (in this case)

 

 

the wind force stuff i'll have to try to remember it should be really easy but i forgot how to correlate wind speed to force.

 

here is a start

 

http://en.wikipedia.org/wiki/Wind_power

 

the chart on that link gives you Watts/m^2 if you know the area of the sail calculating the power and force is easy. and if gives you a range of wind speeds

[Science Abuse]

I haven't taken a math course in which ^ was used....what's it mean?

 

As for wind stress, I just had to go down the hall here. Turns out is something our engineering dept uses quite often for our Windload/Hurricane doors. Should'a known.

[Gergwheel1647545492]

^ is a carot or carrot, so m^2 is the same as m squared or m to the second power

[Science Abuse]

Ok now the hard part. I've got PSF @ sea level when perpendicular to the wind. Now I'm looking for something to give me the thrust conversion for varying angles relative to the wind. Boom @ 20deg to wind = minus X lb side force and + X lb forward force. I might be able to find this ins aeronautical mathmatics, since it'd also be used for pitch angles of propellars.

[Berto]

look up the use of sine and cosine and it should give you how to break it down into components

 

just remember at 45 degrees cosine and sine are equal so you have the force separated equally into x and y

[Science Abuse]

look up the use of sine and cosine and it should give you how to break it down into components

 

just remember at 45 degrees cosine and sine are equal so you have the force separated equally into x and y

Oh snap, I should've known that! But I'm looking for forward force. That would give me lateral force, and the force passed out the rear. The force transferred to the sail/vehicle wont be equal to 1/2 PSF, would it?

 

edit: Maybe it would. The fastest Sailboats do 2x windspeed, and the fastest land sailers do 4x wind speed, with very narrow sails.

http://www.windjet.co.uk/images/stories/land/iron_duck_side.jpg

 

Astonishingly enough, the fastest ever recorded speed of a wind powere craft was set back in 1938 by a canvas sail and a stern steer. As far ahead as we've come, we've yet to surpass that. (he had a 70mph wind, though)

http://www.windjet.co.uk/images/stories/ice/deuce.jpg

[Guest HyperBurn]

um.. The angle of the dangle is inversely poportional to the heat of the beat.