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Tower math ?


By Old F, Section Homebrewed Electricity
Posted on Wed Dec 31st, 2003 at 05:02:32 PM MST
counting my toes and pulling my hair   :  )

Tower math ?

I am crunching some numbers for tower loading and I have run in to a snag.
I have work out the thrust load of the wind generator at the top of the tower
And the tower drag load.

Now here is the snag the example I am working  from says to add the thrust load
and tower drag load together and divide by 10 this come from the base being 10 foot
square to come up with the lifting force on the two legs opposite the pivot point.

This example is for a four sided tower and I have a three .  My base triangle is six foot
on a side do I divide by six to come up with the lift force or is there more to it than that.  I am thinking there is.      

Old F
Tower math ? | 5 comments (5 topical, 0 editorial)

Re: Tower math ? (none / 0) (#1)
by kell on Wed Dec 31st, 2003 at 06:15:12 PM MST
(User Info)

The tower in the example has attachment point(s) 10 feet from the axis of pivoting. Yours has an attachment point 5.196 feet from the axis of pivoting, if I did the geometry correctly. Now if in the example dividing by 10 gives you the total lifting force at the attachment point(s), then for your own tower, dividing by 5.196 will give the total lifting force at the attachment point. But one point (the triangle vertex) will take all the force. The example tower has two feet attached to the ground, or a ten foot connecting rod attached to the ground or however it's built. And I hope your load figures are in foot-pounds, so that when you divide by feet, the final result comes in pounds, which are a unit of force. Only way it would make sense. Foot pounds are a measure of torque. Like if you apply 50 lbs force to a lever at a point 30 feet from the fulcrum you get 1500 ft lbs of torque. If the lever sticks out 10 feet on the far side of the fulcrum you get 1500 ft lbs divided by 10 feet = 150 pounds force at the far end of the lever. Why can't I get carriage returns on this posting?


Re: Tower math ? (none / 0) (#2)
by TomW on Wed Dec 31st, 2003 at 07:44:46 PM MST
(User Info)

Kell;
You asked:
Why can't I get carriage returns on this posting?

The answer is because you need to set your comment posting format preferences to "auto format" and you likely have it set to "html formatted".

In case your not sure where that is it is between the Preview and Post buttons just below the text entry window. You can also set it in your user preferences.
Cheers.

TomW

"Education consists mainly of what we have unlearned."--Mark Twain
[ Parent ]



Re: Tower math ? (none / 0) (#3)
by Old F on Wed Dec 31st, 2003 at 09:12:03 PM MST
(User Info) http://www.oldf.homestead.com

Thanks Kell

The force is measured in foot pounds. Should have said that in the first post.
With that snag out of the way things are starting to make sense.
And the numbers are falling in to place better than I expected.
Tho I will add an extra 15 or 20 % as a safety factor.
 Looks like I am well on my way of meeting  my goal

A 30 foot tower and  from the of  looks things now  a free stander

Rated for an 8 foot diameter blade set in a 60 mph wind an supporting up to 200 pound
of machinery.
Now off to gather some  more data on strength of  materials  

Kell many thanks till your better paid

Old F

Riding the learning curve and loving it : )

[ Parent ]



Re: Tower math ? (none / 0) (#4)
by monte350c on Wed Dec 31st, 2003 at 11:42:25 PM MST
(User Info)

Hi OldF,

Here's a page that helped me a bit, it's pretty basic but explains the concepts!

http://www.asme.org/education/precollege/esp/act5sr1.htm

Hope this helps,

Ted.



Re: Tower math ? (none / 0) (#5)
by Old F on Thu Jan 1st, 2004 at 06:55:36 AM MST
(User Info) http://www.oldf.homestead.com

Thanks Ted

Every bit helps

Old F

[ Parent ]



Tower math ? | 5 comments (5 topical, 0 editorial)
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