Air resistance on a sphere

Well I would think it would be proportional to the spheres cross-section (a circle) and its velocity...Im not 100% certain on that tho, but the question is kind of ambiguous...are we talking about the air resistance at a certain time or point in its fall (a force) or the total amount of force that acted on the sphere during its fall (work)....upon furthur review it looks ike the equation for force would be [.5(pi)(r^2)][C][D][v^2] where C is the drag coefficient of your sphere (I think its based on the material) and D is the density of air.

The resistence will be proportional to the square of the radius of the sphere and, at any given moment to the square of the velocity of the accelerating sphere. Beyond that it is impossible to answer your question with the information given. If the sphere is falling in a laminar, attached flow you will get one value. If there is a boundary layer separation, you will get another. The difference depends on the Reynolds number, which is (as I recall) the poroduct of density and diameter, divided by the kinematic viscosity of the fluid.

I bet there would be a lot of resistance. I would start out with a high number... like 20.

Scoates to the rescue!

Georgeob, how would density affect air resistance?

The density of the fluid through which the sphere fell would affect the resistence in two ways.

1. The resistence force on the sphere would be proportional to the density times the square of velocity times the square of the radius of the sphere.

2. The constant of proportionality would depend on whether the fluid flows smoothly around the sphere, remaining laminar and attached to the surface from front to back (unusual)

or the flow separated from the surface of the sphere, leaving a complex series of vortices in its wake.

Which outcome occurs, and the width of the turbulent wake in a separated flow, depends on a dimensionless constant, known as the Reynolds Number, which can be considered to be the ratio of the inertial forces to the viscous forces in the fluid near the sphere. Numerically the Reynolds Number equals the fluid density times the diameter of the sphere, divided by the kinematic viscosity of the fluid. At high values of the Reynolds Number the flow separates, leaving a high energy wake behind the sphere and significantly increasing the resistence force.

I just did some tests outside, and it looks like 20 may be a little low. It was kind of windy though, do you think that would make a difference?

I would suggest wind resistance is about 23-28 on a small sphere, and probably a lot more if the sphere is bigger, or if you're dropping it into particularly dense air, as Georgeob1 has already stated.

From http://muller.lbl.gov/teaching/Physics10/chapters2003/Appendix-Sept.11.htm


Unrequired calculation. For those who are interested, I'll calculate the terminal velocity for a sphere that has a radius of 10 cm. I'll assume the sphere has a density of 1 gm per cubic centimeter, i.e. it is similar to that of water. The physics equation for the force on an object depends on the shape of the object, and so we pick the sphere because the shape has a simple equation. For the sphere, the force is given by

F = (1/2)A r v2

In this equation, F is the force, A is the area of a circle (that's what the sphere looks like to the wind: A =p r2), r is the density of air = 0.001 gm per cubic cm, and v is the velocity.

The force of gravity is given by F = mg, where m is the mass and g is the gravitational constant. For typical physics units, we will use A in square centimeters, p in grams per cubic centimeter, v in centimeters per second. For these units, g = 980 = 1000 (approximately). The object falls faster and faster until the force of gravity equals that of the air. So we take the equation above and set it equal to F = mg. This gives

(1/2)A r v2 = mg

Now we substitute m = (volume)*(density), take the density of food = 1 gram per cubic centimeter, and use volume of sphere = (4/3)pr3. This gives the following equation:

(1/2)(p r2)r v2 = (4/3)pr3g

Plugging in the numbers, and solving for the velocity v, gives v = 5000 cm/sec. Using the fact that there are 3600 seconds in an hour, 100000 cm in a kilometer, and 1.6 km in a mile, we convert this velocity, and get that it is approximately equal to 100 miles per hour.

That's fast, but not too fast. If there air weren't there, an object falling from a height of 30,000 ft would reach a velocity close to 1000 miles per hour, i.e. ten times faster. (And at 10x the speed, it would carry 100x the energy. And if the ground is hard, most of that energy will go into crushing the food.)

Oops! Sorry Georgeob. I misread, and thought density was referring to the density of the sphere, instead of the fluid.

g...day's analysis is correct, except for his representation of the air resistence force on the sphere. The functional relationship is correct but there is a missing factor representing the wake turbulence behind the sphere as I explained above. In a real flow in air the missing quantity is the largest and dominant factor.

Here's one for you to calculate.
Sphere 36 ft in dia, assume strakes or fences at boundary layer for seperation
wind velocity at 75 mph
assume constant direction of wind fluid flow
temp 77 degree
wind pressure assume at 3,000 ft above sea level
What is the total force on the sphere (Ft lbs) and what is the X axis load (moment)?

The sphere diameter & flow velocity imply a low Reynolds number (in air) so the boundary layer flow is laminar. Still the result depends on where the strakes are located & their angle relative to the remote flow velocity vector - factors you have not specified.

The units of force are lbs. Ft. lbs are work or energy.

Stoke's Law--resistance of sphere's falling through an incompressible fluid at low Reynold's numbers. A special case solution to the navier-stokes equation.

http://en.wikipedia.org/wiki/Stokes'_law

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