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Name: Bradon
Status: Student
Grade: 6-8
Location: MA
Country: United States
Date: January 2007


Question:
I am doing a science project on which shape is the most aerodynamic. I plan to use a wind tunnel and test a square, oval, circle, and triangle and mark which object moves the furthest. How do I account for the difference in the weight of the objects?



Replies:
Hi Bradon,

I am a little puzzled by your question. To begin with, objects tested in wind tunnels, are held stationary and the air is allowed to flow over them. Since they are held fast, and cannot move, I am puzzled by your question about testing them to see "how far they move". They should not move at all in a normal wind tunnel test.

Without knowing how you plan to orient the objects with respect to the air flow, it is hard to comment on any results. If you plan to hold the objects perpendicular to the air flow, then you should find the force exerted on each shape by the wind, to be simply in proportion to their total area. For example, if you hold a circular disk, and a square plate perpendicular to the wind, and the circular disk has a diameter equal to the length of one side of the square plate, then because the square plate has more area, you should expect that there will be more force exerted on it.

Regards,

Bob Wilson


Brandon-

I wish I knew more about your "moves the furthest". There are usually a bunch of ways to do an experiment, and what looks intuitive before your eyes, sometimes is not familiar from a distance. I suspect sliding will not work very well, if that is the move you mean. I am hoping you mean "tilts the furthest" on a pendulum or string. That way, given the measured pendulum tilt in degrees and weight of the object, you can figure out the force generated by the air-drag.

If hanging straight down (no wind) is zero degrees, then drag force F = weight x sine(angle). If you do not know sines and cosines yet, it will be close enough to use sine(angle) = displacement / pendulum_length as long as the angle is less than 30 degrees (displacement < length/2) If the displacement gets larger than that, you'll need to reduce it somehow, by making the object heavier, by making it smaller but the same weight, or by reducing the wind speed for all the objects.

Then figure out the frontal-projection area of each object, (the area of its silhouette, as you look at it from the front.) and divide its drag-force by that. (Force / area = pressure.) That is the frontal drag-pressure of that object.

The objects can be of different masses and slightly different sizes, and the frontal drag-pressures will still be comparable. Then one shape can be shown to have less drag than another.

I hope the wind speed is the same for all the objects. If your object blocks more than 1/4 of the wind-tunnel's cross-section area, it will force the air to speed up to squeak by, or reduce the amount of air going through. Then different shapes might experience different wind-speeds, which would not be good for a comparative experiment.

Suppose you are not using pendulum tilt. If your wind is emerging from a diverging cone or widening rectangular tunnel, and you place your object on thin wire-legs on the floor in the widening zone, then yes, the object that slides furthest has the highest ratio of drag_force/weight. In this case I think you have little choice but to make the mass of each object proportional to its frontal area, to make them comparable. Such use of sliding friction can be rather imprecise, though. Might not work too well.

Jim Swenson


Bradon,

Since you plan on measuring the distance moved, all objects should be of equal weight. Ideally, the objects should be suspended in such a manner as to make their weight irrelevant, such as by a string from which they can swing freely. In the case of the string, instead of the distance travelled, you'd measure how far back each object is pushed.

Ryan Belscamper



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