**PART I---basic kinematics **

For those of you who don’t want to slog through the mathematics necessary to do this calculation:

THE BOTTOM LINE IS THAT ICE, DEBRIS OR ANYTHING BREAKING OFF THE WIND TURBINE BLADES (including the blades themselves) CAN IMPACT A POINT ALMOST 1700 FEET AWAY FROM THE BASE OF THE TURBINE.

**WHAT WE KNOW:**

RADIUS OF BLADE: OVER 100 FEET

ROTATIONAL SPEED: UP TO 1 REVOLUTION EVERY 3 SECONDS (OR ABOUT 20 REV/MIN)

**PRELIMINARY RESULTS:**

ROTOR TIP SPEED:

IN ONE REVOLUTION, THE BLADE TIPS SWEEP OUT A CIRCLE WHOSE RADIUS IS OVER 100 FEET. THIS DISTANCE IS 2*PI*R OR ABOUT 628 FEET. IF IT TAKES 3 SECONDS TO MOVE THIS DISTANCE, OUR SPEED IS 628/3 FEET PER SECOND. THIS IS ABOUT 210 FEET/SECOND OR 150 MPH.

When you do the mathematics in detail, you find that launching the fragment horizontally is NOT the worst case scenario for maximum horizontal range. (LAUNCHING FROM THE TOP OF THE TURBINE (horizontally) YIELDS A RANGE OF SLIGHTLY MORE THAN 1000 FEET.)Instead, this maximum distance occurs when debris is released with the blade at a 45 degree angle from the vertical.

Imagine the blade at 45 degrees from its vertical position. At this point, the

projectile will be launched about 70 ft. from the horizontal position of the hub.

(This is 100 times the cosine of 45 degrees). Also, it will be about 70 feet higher (vertically) than the hub. (Again, we assume that the blades are 100 ft. in length). Thus, the vertical distance it has to fall is 300 feet (hub height) plus 70 feet (vertical distance that the piece of ice, or whatever, is from the hub).

Now, the range for this projectile is:

R= v**2/g (that's "v squared divided by "g", the gravitational acceleration).

This is the range to come back down to the ORIGINAL vertical height. So after this distance, it is BACK at 370 feet off the ground.

R=( 210 ft/sec x 210 ft/sec)/(32ft/sec/sec). or about 1400 ft.

Now, at this position, (neglecting air resistance), its vertical velocity is the

same as when it was launched (except that it's now going DOWN instead of up).

So, the vertical velocity is about 140 ft/sec. (210 x .7 or v cos 45)

The extra time it takes to fall to the ground from this height is:

s= v times t + 1/2 g times t squared.

SO, 370=140 t + 16 t**2

Solving for t, we get about 2.5 seconds. In 2.5 seconds the increase in the range is:

v(horizontal) times t or 140 x 2.5 or about 350 feet.

Thus, the TOTAL range of a projectile is: 1400 + 350 = 1750 feet. From this we subtract the 70 feet that the projectile was behind the hub when it was launched, and you end up with 1680 feet for the horizontal range from the base of the hub.

**PART II---comments on inclusions of drag coefficients and risk assessment**

1) Friction is NOT a fundamental force. What this means in practice is that any attempt to take into account air resistance in a description of ice throw can be fraught with model dependent errors. The drag coefficient usually quoted in wind developers' "papers" of 1.0 is totally inappropriate for the study. Variability in the Reynolds number is completely ignored. They assume a perfect ice cube of size = 4 inches. Then, they assume it always tumbles. But these are chunks of ice that are forming on propeller blades! Ice that forms on propeller blades tends to be shaped like propeller blades. And they can be QUITE aerodynamic (as are the blades). Any models employing ice cubes are at best, useless, and at worst, deceptive.

Interestingly, when the chunks become larger, the Reynolds number increases, and viscosity becomes less important. What this means is that the larger, more dangerous chunks will tend to travel further than small ones.

Moreover, the study of "harvested" ice that is subjected to wind tunnel testing is likewise demonstrably without merit. The procedure is to break off chucks of ice, make molds, and then subject them to wind tunnel tests. But real ice melts. It changes shape. It becomes smoother. The "studies" ignore this, instead adopting a drag coefficient = 1.0 This is close to the drag that a half a tennis ball (say) would present if it were thrown into the wind with the open "cup" catching the wind at all times! A rather silly assumption, and one that is totally inappropriate.

Ice is NOT like this. While the developers tout their results as being representative (decidedly untrue), they ignore MacQueen's 1983 study that concluded that a maximum range of 800m (about 2500 feet) was quite possible. Indeed, even a range of 2 km. (over one mile) was conceivable. They discount this because he "assumed that the ice 'fragments' were actually large flat slabs weighing perhaps 80 kg." Actually, he was modeling BLADE throw, another issue that seems to be ignored despite the fact that within the past year there has been at least one documented instance of this; an entire rotor blade broke off from the hub. (Wethersfield, N. Y.) Incidentally, as near as I can see, the MacQueen study is the ONLY peer reviewed analysis for throw possibilities. The rest are calculations done by wind company employees and/or consultants.

2) I never claim my calculation to be anything other than a maximum calculation of distance, beyond which you don't have to worry. I am not usually accused of being conservative in many ways, but when it comes to human life, I suppose I am. Why worry when you can just adopt my calculation and not be concerned at all with tragedy in the future? Moreover, any risk assessment data is useless, since the calculations are not assuming an appropriate model to begin with! I remember when de-icing airplane wings was said to be unnecessary, posing no risk to public safety, and only after tragedy struck is it now "de rigeur" to do it (and do it carefully

and thoroughly). All other mumbo jumbo is exactly that unless they get the basic physics right.

3) If you are going to invoke air, and air resistance, it is the height of deception not to include the effects of lift. Frisbees fly far. Why? Because of air. If you throw a frisbee facing the direction of travel, it will travel only a few feet. The drag coefficient is probably of order unity in this situation. If you sling it in the direction of travel, it goes very far. And rime ice, of course, will likely be shed from the rotors in a similar aerodynamic fashion. The wind companies assume that the fragment will tumble; thin blades of ice may not (as a frisbee does not, when properly oriented).

4) Throughout these discussion, the wind developers have been groping for setback distances that they can live with. They started with 1.5 times the ground-tip distance.: about 150m. As near as I can tell, this was just pulled out of a hat. (In physics, we call this a "toy model".) Then, the distance was increased to 200m. in several " papers." Now, in a recent " paper", 400m is quoted. They are getting closer to my original number! What about data? If you refer to the Atlantic Wind Test Site memo of March 27, 2002, they state:

*"Summary---Following the moderate wind icing event at AWTS on March 27, 2002, fragments of ice, large enough to cause injury, have been observed being thrown from the turbine blades. Concerns over dangers of flying ice are legitimate. In 15 m/s winds, ice was observed to travel approximately 200m."*

Instead of recognizing this, companies present a figure from an utterly useless and anecdotal "questionnaire" that purports to show that ice throw is "unlikely" more than 100 meters from the site. This figure is a completely misleading representation of "data" that has been bandied about for years by developers.

5) In the beginning, the claim was that the rotor sensors would stop the blades because of ice buildup. Now, even the papers put forward by the companies admit their error here. They state: "...rime ice formation appears to occur with remarkable symmetry on all turbine blades with the result that no imbalance occurs and the turbine continues to operate." Another failure of their initial assumptions and models.

In conclusion, there are some problems with wind turbines that have unavoidable consequences. Birds will die, bats will die. In these scenarios you NEED to adopt a risk analysis study. But here, YOU CAN ELIMINATE THE ENTIRE PROBLEM, if you just adopt a conservative value for your setbacks.

REFERENCE: J. F. MacQueen, et. al, IEE Proceedings, Vol. 130, Pt. A, No. 9, pp. 574-586 (1983).

If you have any questions, just e-mail or call. It would be a lot easier to explain this if I could write it on a piece of paper, but I hope you can picture

this adequately.

Professor Terry Matilsky matilsky@physics.rutgers.edu

Department of Physics and Astronomy Tel: 1-732-445-5500 ext. 3876

Rutgers University Fax: 1-732-445-4343

Piscataway, N. J. 08854

*Source:*
http://xray.rutgers.edu/~ma...