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Calculating wind turbine tower loads

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Posted by Curbie on February 23, 2009, 6:02 pm
For guyed towers there are two points of ground contact (support), 1)
the tower base and 2) the guy wire anchors.

Is the loading on the base only vertically downward from weight of the
mast and turbine, or does horizontal forces from wind velocity place
upward forces from the guy anchors that translate into increased
downward forces on the base?



Posted by vaughn on February 23, 2009, 6:29 pm

   The base force is almost entirely downward. and may greatly exceed the
weight of the mast and turbine because some wind forces get translated
downward, and because of the static strain of tight guy wires.  Many guyed
radio towers actually come to a point where their base contacts the
foundation, and are only secured there by a pin which merely serves to keep
the base from slipping off the foundation.

     Exactly correct.

 Vaughn   (I am not a structural engineer, and don't play one on the

Posted by Curbie on February 23, 2009, 7:14 pm
 If four guy anchor/wires are properly placed at 0, 90, 180, & 270
degrees apart and the wind coming in from 45 degrees is the horizontal
wind load evenly dispersed between anchors/guys at 0 and 90 degrees
and the base?

Or maybe more properly asked: given the same arrangement, if the wind
coming in from 0 degrees is the horizontal wind load evenly dispersed
between anchor/guy at 0 and the base?

On Mon, 23 Feb 2009 18:29:22 GMT, "vaughn"

Posted by daestrom on February 23, 2009, 11:37 pm

In the case where the wind is coming 'from' the same direction as a guy,
that guy will have to carry all the wind force and the other guys will have
no force at all (except for any 'pre-load' tension in them).

The tension in the guy will be *more* than the wind force alone.  If the guy
is at a 45 degree angle, the tension in the guy will be about 1.414 (

The force on the base will be equal to all the weight plus an amount equal
to the wind load (if the guy is at 45 degrees).

So a 100 lbf wind load on a tower/turbine that presents a 50 lbf windload
will have a guy wire tention of 141 lbf and the tower base will be
supporting 150 lbf.

hope this helps


Posted by Curbie on February 24, 2009, 2:43 am
 When calculating horizontal wind loads do you just treat the drag
coefficient of the "swept area" as a flat plate x Betz x safety

Calculating Drag        
Wind speed (MPH)    80.00               MPH
Rotor Radius (R)        10.00               Ft.
Wind Speed (V)        =MPH * ((5280 / 60) / 60)   Ft./Sec.
Cross Sectional Area (S)    =PI() * (R ^ 2)           Ft.^2
Drag Coefficient (Cd)    1.28               Flat plate
Air Density (rho)        0.002377           Lbs. Sec.^2
/ Ft.^3 at sea level
Drag (D)        =Cd * rho * v ^ 2 * S / 2       Lbs.
Betz            59.3               Percent
Force             D * Betz               Lbs.

Thanks to daestrom great explaination, just what I needed. (speadsheet
Thanks to vaughn for reply and help.

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