Drag coefficient of 2012 Game Piece

[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.

[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

[Ian Curtis]

Originally Posted by 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.

I knew someone would bring up the drag crisis, that's what I get for keeping it simple I suppose...

To remain at the low spot requires pretty spectacular attention to surface roughness. We tunnel tested several models as part of one of my aero labs, and indistinguishable (by fingers) differences in surface roughness will knock you out of the crisis. Seeing as these are low quality foam balls produced in the thousands by robots, I don't think there is much to worry about.

(Where did that chart come from? I've never seen Cd v. Re with how the drag crisis moves as a function of roughness before, that's a nice chart!)

[Michael Hill]

Originally Posted by Ian Curtis

I knew someone would bring up the drag crisis, that's what I get for keeping it simple I suppose...

To remain at the low spot requires pretty spectacular attention to surface roughness. We tunnel tested several models as part of one of my aero labs, and indistinguishable (by fingers) differences in surface roughness will knock you out of the crisis. Seeing as these are low quality foam balls produced in the thousands by robots, I don't think there is much to worry about.

(Where did that chart come from? I've never seen Cd v. Re with how the drag crisis moves as a function of roughness before, that's a nice chart!)

The chart is in A Brief Introduction to Fluid Mechanics. I'm really not sure if drag is entirely predictable just because robots may rough them up a bit. Unfortunately, drag might be a real issue too. At 5 m/s, the drag force is right at around 1 N (and weight is 3.1 N). That's pretty non-negligible.

[Michael Hill]

The best solution may just be to have a look-up table to be honest. I've already derived equations of motion, and you end up with a system of second-order nonlinear equations which would have to be solved numerically. It's something probably a little too advanced to teach high-schoolers how to do, and it may also be a little too much for the CPU to handle (with any reasonable accuracy).

For the curious:
m * p_dd = -K * sqrt(p_d^2 + h_d^2) * p_d
-m * h_dd = m*g + K * sqrt(p_d^2 + h_d^2) * h_d

where:
m = mass of ball
p = x-distance
h = negative y-distance (coordinate frame unit vector pointed down. positive h is downward)
K = rho * S * C_D/2
rho = density of air
S = Cross-sectional area of ball
C_D = Drag Coefficient
?_d = ? dot (as in first time-derivative)
?_dd = ? double-dot (as in second time-derivative)

It can also be expressed in terms of Speed, V, and flight path angle, gamma as a single order system of nonlinear eqations.

m * V_d = -m * g * sin(gamma) - K * V^2
-m * V * gamma_d = m * g * cos(gamma)

Expressing that way is "prettier," but is less useful.