Friction on the ball

Has anybody measured the coefficient of friction for the balls this year?

CoF describes the “grippiness” between two surfaces. What material would you like measured interacting with the ball?

  1. Carpet, such as used for the FIELD.

  2. Whatever is the grippiest stuff you can find that we can put on moving parts of our robot.

Oh alright. That makes sense. What I was really wondering about is the CoF between the ball and the air and whether air resistance would be worth accounting for with such a large, light, and rough ball.

Air is a fluid, so there isn’t “friction” per se as there aren’t two distinct surfaces. Instead, the question is whether the resistance to motion from viscosity or drag is significant.

To determine that you first need a value called the Reynolds number:

Re = (v * L)/(k.v.)

Where v is velocity of the object, L is the characteristic dimension of the object, and k.v. is the kinematic viscosity of the fluid (which is dynamic viscosity normalized by density). For air at 70 F, the kinematic viscosity is 1.64 x 10E4 ft^2/s. For a sphere, the characteristic dimension is the diameter.

If the Re is much less than 1, then the viscous forces dominate. If it’s much larger than 1, then the turbulent and drag forces dominate. While it depends on velocity, we can solidly say that the flow is always turbulent for the ball in this year’s game, as the velocity where Re = 1 corresponds to something in the 10E-4 ft/s range. If the ball were travelling that slowly, then it’s either at the top of an arc or on the ground. If it’s at the top of an arc it would only be in that speed range for a very brief time and we can reasonably ignore the change in the dominant force. If it’s on the ground, then carpet friction is the dominant force.

For a turbulent flow, the drag is modeled as:

Drag = (1/2) * § * (v^2) * (Cd) * (A)

Where p is the density, v is the velocity, Cd is the drag coefficient, and A is the cross sectional area. Density of air at sea level at 70F is about 0.15 lbm/cu ft. The Cd is about 0.5 for a sphere with Reynolds number of 10E5 (you need to look a a VERY non-linear chart to determine this). The cross sectional area is the area of the circle that is the maximum cross section of the sphere, or roughly 3.14 sq ft. Pick a velocity and plug in the numbers and you’ll get a result with units of (lbm * ft)/(s^2). Divide by 32 (that’s g in ft/s^2) to get pounds of force.

Using these nominal numbers:

Drag (in lbf) = 0.0036797 * v^2 (v in ft/s)

**Taking a typical example using 20ft/s as the speed, the drag force on the ball is about 1.47 pounds in a direction opposite the motion of the ball. **Whether that is significant is a judgement for you to make. Also note that this drag is non-linearly dependent on velocity. So, that 1.47 pound number at 20ft/s is roughly 0.36 pounds at 10ft/s. If you want to model the path of the ball in the air, you need to simulate it or use some seriously sophisticated math.

If you think in terms of launching the ball toward a goal, the drag is largest (and possibly significant) right after launch when speed is high, and then drops quickly as the speed drops.

Keep in mind that this is idealized, as it doesn’t account for surface texture or spin. Spin typically increases drag, and surface texture can either increase or decrease it. But, as an estimate, the drag from the air at a 20 ft/s launch is about a pound and a half.

Hope that helps.

Ok, what about a CoF between the ball and say a kit wheel?

Anyone have a good way of measuring that?

-Mr. Van
Coach, Robodox

One of the most basic ways is to drag a wheel across the ball surface with a scale attached to measure the force.

You would have to be creative for fixturing. I would use 3 wheels (or whatever object you want to measure) to get 3 even points of contact. Fixture them to hold them steady, and add weight to get an even number to calculate so the total (including the wheels) is 10 or 15 pounds.

Take the fixture of wheels, remove the fabric from the ball and place it on a table. Make sure the fabric is held secure. Connect the scale to the fixture of wheels and drag. The number read, before the fixture of wheels skids, is the static force. The number read while the wheels skid is the kinetic force. Take the readings and divide them by the 10 or 15 pounds (whatever the fixture weighs with the wheels), and that is your coefficient of friction. The higher the coefficient of friction, the more grip there is.

I am sure if I missed something, it will get corrected for you. :slight_smile:

Hope that helps!

In a calculation of the force required to shoot the ball from the line between the alliance zone and the white zone into the high goal, could I, in an effort to avoid using differential equations, assume that the drag is a constant vector pointing from the point at which the ball passes the plane of the goal to the point of release?

I am trying to see if drag is a significant factor, and I would like to calculate the force required to shoot the ball with drag taken into account.

I ignored air resistance, I would be more concerned with frictional losses of poorly loaded mechanisms than air resistance. Plus, I didn’t want to dust the cobwebs off the fluids book either.

did you see this thread?

Also, you should probably take a look at Impulse and its relation a change in momentum. Force is not the only thing to think about.Here is one link

I concur. This is either an energy or momentum problem.

If you do it as a momentum problem, you need to determine the impulse necessary (m * v = F * (delta t)) to get the ball going fast enough to be on a trajectory to score or get over the truss. So, large force for short time or small force for long time.

From a spatial perspective, energy provides a more convenient model ((1/2) * m * v^2 = F * d). So, apply a large force over a short distance or a smaller force over a long distance.

Either way, you need to account for the aforementioned friction and drag losses. If you pick a spring that stores just enough energy to give the ball the velocity for your desired trajectory, you will be short. You’ll be losing energy to friction, drag, and other un-modeled inefficiencies. So, you can get a good guess with some high school physics, but to refine that requires simulation or testing or both.

For students interested in the physics, math, and computer simulation aspects:

*- Free-body force diagram

  • Derivation of differential equations of motion

  • C pseudo-code for the difference equations for 2nd order numerical integration

  • Includes drag and spin

  • Example graphs

I created an interactive plotter which takes these equations of motions into account, including the drag forces. Might be interesting for some to get a feel for the motion. You can change the initial parameters such as speed and drag coefficient to see what affect that has.