Drag coefficient of 2012 Game Piece

[yarudl]

This year shooting is going to play a huge role in the game and I was wondering if anyone has calculated the drag coefficient for the game pieces yet? This would be a great value to be able to incorporate to increase accuracy. Any help is appreciated.

[sanddrag]

Anyone live near one of those indoor skydiving places?

[davidthefat]

Originally Posted by sanddrag

Anyone live near one of those indoor skydiving places?

There's one in Hollywood.

[yarudl]

I think over the next week I'm going to work with my physics teacher and the bleachers at our school to do an experiment.

[team698]

The indoor skydiving is a dead end because the force of drag is directly proportional to the speed of the game piece. (I can't imagine that the speeds will approach that of skydiver at terminal velocity) The coeffient of drag can be determined mathmatically, look up the equation, or experimentally if you can find a room whose air density is similar to the playing field and if you know the velocity that your game piece will be launched at.

[davidthefat]

Just do experimentation. Just forget about the drag and actually launch it and compare it to the prediction and the real result to find the margin of error and you are good to go.

[shuhao]

Wait.. the drag coefficient of the game piece?

I thought it can be assumed 0 as it's the "drag coefficient" (not sure if right there, but c1 * v * r + c2 * v^2 * r^2.. with the first term ~= 0) of air that matters (c1 = 3.1e-4, c2=0.85 at STP)

[Ian Curtis]

If you decide to include drag in your calculations, you turn the algebraic equations of projectile motion into differential equations and differential equations are complicated. This is because while the acceleration of the ball in simple projectile motion depends only on the time since launch, if you include drag it depends upon the velocity, and velocity also depends on acceleration -- see how it might get complicated?

It is pretty easy to calculate the magnitude of the force though, and check what the magnitude is.

Drag Force=.5*density*velocity^2*Area*Cd

For a sphere, a Cd of .5 is a pretty good. Entire books have been written on drag, but FIRST robots tend to operate pretty well on the back of a napkin.

The projected area of the ball should be about .34 square feet. That is, the area of a circle with a diameter of 8".

The density of air at sea level in the standard atmosphere is .00237 slugs/ft^3.

Velocity will obviously depend, but I'd say 30 ft/s is probably a pretty decent upper bound.

If you do the math, you'll find that the force of drag on the ball at that speed would be about .36 lbs, and it'll go down fairly quickly, if the ball is traveling at 15 ft/s, it goes down to .09 lbs.

[sanddrag]

Originally Posted by team698

The indoor skydiving is a dead end because the force of drag is directly proportional to the speed of the game piece. (I can't imagine that the speeds will approach that of skydiver at terminal velocity) The coeffient of drag can be determined mathmatically, look up the equation, or experimentally if you can find a room whose air density is similar to the playing field and if you know the velocity that your game piece will be launched at.

Sorry, during my fluid mechanics lectures on drag I must have been designing a gearbox in my mind. FRC season will do that.

[shuhao]

Originally Posted by Ian Curtis

If you decide to include drag in your calculations, you turn the algebraic equations of projectile motion into differential equations and differential equations are complicated. This is because while the acceleration of the ball in simple projectile motion depends only on the time since launch, if you include drag it depends upon the velocity, and velocity also depends on acceleration -- see how it might get complicated?

It is pretty easy to calculate the magnitude of the force though, and check what the magnitude is.

Drag Force=.5*density*velocity^2*Area*Cd

For a sphere, a Cd of .5 is a pretty good. Entire books have been written on drag, but FIRST robots tend to operate pretty well on the back of a napkin.

The projected area of the ball should be about .34 square feet. That is, the area of a circle with a diameter of 8".

The density of air at sea level in the standard atmosphere is .00237 slugs/ft^3.

Velocity will obviously depend, but I'd say 30 ft/s is probably a pretty decent upper bound.

If you do the math, you'll find that the force of drag on the ball at that speed would be about .36 lbs, and it'll go down fairly quickly, if the ball is traveling at 15 ft/s, it goes down to .09 lbs.

Is the viscosity calculation equally correct here?

[Aren Siekmeier]

In air, anything at FIRST relevant velocities is at such a low Reynolds number that viscosity is not a concern (inertial drag hardly is).

[shuhao]

yeah but im using that to calculate drag ... feel like it's easier... not sure if it is completely correct, though.

[Chris Hibner]

Originally Posted by team698

The indoor skydiving is a dead end because the force of drag is directly proportional to the speed of the game piece. (I can't imagine that the speeds will approach that of skydiver at terminal velocity) The coeffient of drag can be determined mathmatically, look up the equation, or experimentally if you can find a room whose air density is similar to the playing field and if you know the velocity that your game piece will be launched at.

It is actually not a dead end. The question is what is the coefficient of drag (not drag force).

Fdrag = 1/2 * rho * A * v^2 * Cd

You can easily find out what rho is at your indoor skydiving place by using the temperature, ambient pressure, humidity, and a standard air chart.

A is easy

Fdrag is the weight of the ball as long as you can get it to hover.

v can be read form the indoor skydiver operator controls (and you can measure it inside the tunnel).

Then all you have to do is solve the equation for Cd.


While it is theoretically possible using the skydiving place, the are a lot of books that have characterize the Cd of spheres with many surface textures. I'm sure the wind tunnel would be unnecessary.

[Michael Hill]

I just did some back of the envelope aerodynamic calculations for the game piece this year and realized we're RIGHT on a near asymptote when trying to predict drag on a ball. (Hey, I'm bored at work with nothing else to do.)

I estimated the Reynolds number to be around 120000 (assuming around 10 m/s) and the relative roughness factor to be around 4.9 *10^-3 (I assume 1 mm dimples with an 8 inch ball, best I can estimate without having the game piece in my posession).

Then I found this little jewel (see attachment):

As you can see, the roughness (epsilon/D) line for 5*10^-3 has a nice little vertical right around Re=120000.

This doesn't take into account any spin, mind you (and I really don't feel like getting into Magnus Effect forces right now).

Enjoy!

P.S. In my opinion, the dodgeball couldn't be any more imperfectly designed if consistancy is desired.

[Michael Hill]

I should also note that 120000 is a rather high Reynolds number. 10 m/s is extremely fast for this kind of game. When I was doing the calculations, I was still stuck in Aim High mode, but looking back, Re should probably be about half that, and a C_D of 0.5 should be a decent estimation.

If you want an even better estimation, you can use C_D = 0.09015*log(Re) + 0.06924

Where:
C_D = Drag Coefficient
Re = Reynolds Number = rho*V*D/mu
rho = Density of Air
V = Velocity
D = Diameter of Ball
mu = Dynamic Viscosity

This is fine as long as the Reynolds number stays below about 8*10^4