Physics of Beer Pong
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Don’t assume that this game is not a blood thirsty sport either. Alcohol is related to 100,000 deaths annually in America, usually occurring after a night’s game of beer pong (Sage). I will rely heavily upon the laws of physics to explain in great detail the appropriate method to overcome your opponent's game, forcing them to streak across the neighborhood. The rules of beer pong vary between locations and can cause great strife amongst rivaling teams. The number one rule is that the house that is hosting the tournament controls the rules. The house rules must also be announced at the beginning of the tournament. The official rules located on Wikipedia may give you a good foundation for the general rules. Alterations in the game usually form from city traditions. In Valdez, Alaska, a place well known for its intense game of beer pong, there is a common tradition of using quotes from Family Guy and BASEketball to distract the other player. Usually after a few beers these jokes miraculously become funny. Another beer pong tradition is to force the losing team to streak publicly. The website PongRules allows people to post their own rules around the world (PongRules). Remember, beer pong is governed by the drinkers for the drinkers. There are two types of table measurements players may use. The official rules state that the table must be eight feet long, at least two feet wide, and stand four feet tall. However, it is more commonly played on a ping pong table where the height reaches 30 inches (Organization). For the purposes of this project, I will be deriving my calculations using the dimensions of a ping pong table. The regulation ping pong ball consists of being 3star, 40mm in diameter, and weighing 2.7 grams. Now, at each end of the table ten cups must be sorted in a triangular position similar to bowling pens but with the rims touching like so: Usually three 12ounce beers of your choice are divided equally among the 10 cups. How to Cite this Page
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"Physics of Beer Pong." 123HelpMe.com. 21 Feb 2017 <http://www.123HelpMe.com/view.asp?id=153474>.
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Two teams of two will be separated on both sides of the table face to face. Two regular size ping pong balls are used along with a cup of water to wash away any filth that may collect on the wet balls.
To start the game, one player on each team will toss the ball simultaneously into the opponent’s cup while staring into their enemy’s eyes. The first person to make it into their opponent’s cup starts off the game with both balls. The basic objective is to make your ping pong ball into one of your opponent’s cups. If this is achieved your opponent is not allowed to proceed with his shot until he finishes that cup of beer and sets it aside. If both players make a shot, then the opponents must drink up and pass their balls back. Also, if both players make it into the same cup, then the game is over and the rest goes towards the loosing side. Another advantage is the bounce shot. If you are able to make it into your opponent's cup after a bounce on the table then they must chug two cups. But be aware, once a bounce shot is engaged, the opponents are welcome to swat it out of the way. If your opponent is down to 6, 4, 3, or 2 cups, then a reformation may be called and must be aligned appropriately in these formations: (http://en.wikipedia.org/wiki/Beer_pong) Also keep in mind that verbal psycheouts can be used to break your opponent's concentration. Keep in mind, variations of these rules are common. Now you know the rules and are ready to begin your first step towards sobriety or alcoholism. 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 soontobe 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 xx0 = (vo cos θ0)t and the vertical motion equation is vy2 = (vo sine θ0)2 – 2g(yyo). By combing these two equations we are able to produce y = (tan θ0)x – gx2/(2(vo 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, vo, 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 ti89. But first we will simplify it the best we can: y = (tan θ0)x – gx2/(2(vo cos θ0)2) y2vo2 cos2 θ = 2vo2 * tan θ *cos2 θ * x – gx2 2y(vocos θ2 = 2vo2 * sin θ*cos θ * x – gx2 2y(vocos θ)2 = x*vo2 * 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 ti89. 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. Now I am going to show you how to humiliate your opponent. The opponents first cup has already been drank and there are nine left. If you skimmed over the rules section you may have missed the bounce shot. If the bounce shot is accomplished, the opponent must drink two cups, but he may also attempt to swat the ball out of the way. First, shift yourself perpendicular to the table and line your shot parallel to your opponent's cup. The distance between you and the middle cup should be 2.3 meters. You must still throw at a speed of 6 m/s, this won't be hard unless you have already had to drink a few cups. Now we must find the angle at which we need to toss. This simple answer lies within some simple equations. We will use The Range equation to calculate the first distance (see the graph above) is d1 = ((vo2 * sin(2θ))/g). This equation is derived by eliminating t and combining vectors. Now d2 is derived from the horizontal motion equation: (vo*cosθ)t However, we also need the time, which can be solved for using the vertical motion equation, h=v*t+1/2 * at2 (Halliday & Resnick). Since theta is the angle desired we must change the velocity to equal vo * (sinθ), and by setting this equation equal to zero and acceleration equal to g we can finish this by writing 0 = vo*(sinθ)*t+1/2*gt2 –h By using the quadratic formula the only possible equation without a negative time is Time = 1/g (vo*sinθ + (vo2*sin2θ + 2gh)1/2) So after substitution into the d2 equation the distance between d1 & d3 becomes d2 = (vo*cosθ)* 1/g (vo*sinθ + (vo2*sin2θ + 2gh)1/2). Now the last distance equation seems difficult but is just simple equations we have used previously. Since theta is the desired variable, we must solve for theta again. By using tangent we are able to find the angle at which the ball is coming in from, but we need to break it down into X & Y components. So vx = vocosθ, the common vector equation. Since gravity acts upon the velocity in the y direction we can form the equation vy = vosinθ + g*t, and we already have the equation for t, so just substitute it in. This allows us to somewhat find θ, now θ = tan1(vosinθ + ((vo*sinθ + (vo2*sin2θ + 2gh)1/2)))/ vocosθ, after the time substitution is included then gravity is eliminated. To calculate the range we just use the d1 equation, ((vo2 * sin(2θ))/g). But now we must take into effect the energy lost in the bounce. If we use the gravitational potential energy equation (mass * gravity * change in height), we can find the potential energy lost. But better yet, let’s just use the ratio between the two! After numerous tests on a ping pong table I used the final height divided by the initial height to prove that a loss of 22% occurs after the first bounce. So we use the range equation and multiply µ into it, which is equivalent to 78%. So our equation becomes ((µ*vo2 * sin(2θ))/g). Now we also have an equation for theta that we solved for previously, so let’s shove that in there also. Now the last distance d3= ((µ*vo2 * sin(2* tan1(vosinθ + ((vo*sinθ + (vo2*sin2θ + 2gh)1/2)))/ vocosθ))/g). Awesome, we have all three equations now. We now need to add them all up and solve for the total distance, D = d1 + d2 + d3: D = ((vo2 * sin(2θ))/g) + (vo*cosθ)* 1/g (vo*sinθ + (vo2*sin2θ + 2gh)1/2) + ((µ*vo2 * sin(2* tan1(vosinθ + ((vo*sinθ + (vo2*sin2θ + 2gh)1/2)))/ vocosθ))/g) So depending on your height, table dimensions, speed, gravity, and theta, it can all be calculated with this equation. This equation is very similar to the golden ratio in that it has an endless amount of possibilities + beer, so I like to refer to it as the Platinum Equation. We will now substitute all of our values into this Platinum Equation to find out which values of theta will work best: 2.3 = ((62 *sin(2θ))/9.8) + (6*cosθ)*1/9.8 (6*sinθ + (62*sin2θ + 2*9.8*.88)1/2) + ((0.78*62 *sin(2*tan1(6*sinθ + ((6*sinθ + (62*sin2θ + 2*9.8*.88)1/2)))/ 6*cosθ))/9.8) θ = 79.66° & 64° I am not going to bust out my mad trig skills, but instead I will resort to using the solve function on my ti89 to solve for theta. After waiting about 8 minutes for the calculator to solve for the results we finally end up with θ = 79.66° & 64°. This shows that if you toss the ball at an angle of 79.6° to the horizon it will bounce and land straight into the opponents’ cup. Also, if you throw it downwards at an angle of 64 to the horizon, it will land into the cup. I set the height equal to 0.88 meters because this is the height after the height of the cup is subtracted. Sir Isaac Newton will now demonstrate the results of our calculations. An alternative to this problem allows us to find a target to aim for. If we add the equations d1 and d2 we can find the distance in which the ball needs to travel in order to bounce successfully. If we substitute 79.5° into theta we are able to produce a distance of 1.46 meters. I will not go into great detail outlining the rest of the shots. If a cup is 4” to the left, then just move 4” to the left. If it is a greater distance, then sub your own values into the Platinum Equation and you will have the correct angle. The rest of the game is common sense, but depending on how much you have consumed, your accuracy will decline similar to ex does on a coordinate plane. After reading this article you may think you have what it takes to become a beer pong champion. But similar to other sports practice is essential. The fact that you must throw at a constant velocity and a perfect angle is almost impracticable. There are many other uncertainties that must be taken into consideration. I did not include air resistance because this variable changes depending on air density and temperature, and the calculations stump my calculator. But your own uncertainties have the potential to cover this up. You might toss at 6.2 m/s causing you to overcome air resistance. After a few games, measuring the perfect distances will also be hard to achieve. With physics these problems and uncertainty values can be calculated but the work to do this is too labor intensive for myself. My point of this experiment was to show the reader what your mind takes into account while playing a sport. Our mental ability is amazing in that it can analyze the problem mentally and output physically. And with practice, our minds are able to analyze the problem and produce results with more accuracy. The problem presented demonstrates how the cognitive mind interprets wonderful calculations and physically produces outstanding results. Works Cited Halliday & Resnick. "Fundamentals of Physics." Resnick, Halliday &. Fundamentals of Physics extended 8th edition. Danvers, MA: John Wiley & Sons, Inc., 2008. 5875. Halls, Steven B. Men height. 10 November 2003. 16 November 2007 Organization, Beer Pong. Beer Pong Constitution. 2005. 18 November 2007 PongRules. Pong Rules. 2007. 20 November 2007 Sage, Church of the Divine. Annual Causes of Death in America. 21 November 2007 
