The First Cup

This is where you get your game on. Make sure you have a good stance to rely on, take some marshal arts classes if you need help on your posture. Now stand directly in front of the center of the table with your back in an upright position, but stay relaxed. Align your dominate hand parallel to your opponents cup, and pull your hand downards in a tossing motion while keeping your body in a comfortable position.

We will begin by tossing the ball into the front cup. The distance should be exactly 2.1234 meters away. This measurement is calculated by using an 8 foot ping pong table while taking into consideration the distance from the end of the table to the center of the first cup.

The height of the person is an extremely important factor that needs to be taken into consideration. I will be using the height of an average American male (5’9”) which is equivalent to 1.7526 meters (Halls). If we subtract this from the initial height of a ping pong table, we are left with 0.99 meters or about 1 meter. So now we are standing a meter above the ping pong table and can now place our position into a coordinate plane.

To make this soon-to-be equation even more complicated, we must also take into consideration the height of the cup, which is 0.12 meters. It is important to take this into consideration because otherwise the ball will be hitting the side of the cup instead of directly in it. To simply this calculation, we will subtract the height of the cup from the final height of the American, leaving us with 0.88 meters.

Now we are able to find the correct angle needed to make this cup shot. If we use theta as our changing variable then our velocity must be constant. This is where practice comes in handy. Through my data calculations, I figured that I threw about 6 meters per second. This seems to be a common speed for beer pong. If your measurement is different, then just use my simple equations to figure out your own ideal angle.

Now we are about to figure out my winning angle. Since we are using two different coordinate directions we need to combine the horizontal motion and vertical motion of our object. The horizontal motion equation for a projectile is x-x₀ = (v_o cos θ₀)t and the vertical motion equation is v_y² = (v_o sine θ₀)² – 2g(y-y_o). By combing these two equations we are able to produce y = (tan θ₀)x – gx²/(2(v_o cos θ₀)²) . Where y is the height in the vertical direction, θ₀ is the initial angle which we will be solving for. In the horizontal direction, x is the distance between you and the cup, and g becomes the acceleration of Earth which is a constant 9.8 m/s². The initial velocity, v_o, will be approximately 6 m/s depending on your throw. After a few drunken weekends you should be able to toss at a constant velocity. Unless you are a math genius, this trajectory equation I am about to solve will need to be done using a scientific solving calculator like a ti-89. But first we will simplify it the best we can:

y = (tan θ₀)x – gx²/(2(v_o cos θ₀)²)

y2v_o² cos² θ = 2v_o² * tan θ *cos2 θ * x – gx2

2y(v_ocos θ² = 2v_o² * sin θ*cos θ * x – gx²

2y(v_ocos θ)² = x*v_o² * sin(2θ)-gx²

Even though this equation is not much different than the one we started out with, it will make it a little easier to solve for θ using your ti-89. We now have all the variables so let’s start plugging in some numbers:

2*0.88(6*cos θ)² = 2.1234*6² * sin(2θ)-9.8*2.1234²

63.36*cos²θ = 76.44*sin(2θ)-44.2

cos²θ = 1.21*sin(2θ)-0.7

cos²θ - 1.21*sin(2θ) = -0.7

Solve for the equation above:

Solve (cos²θ - 1.21*sin(2θ) = -0.7,x)

And you get:

θ = 68.01° & 44.44°

These are the angles at which you can throw to make your ball into the first cup. So pull out your protractor and measure the angle before you give it a toss.