When designing our cannon, we had to think about several things. One of our main concerns centered on how we would achieve our desired angle, but other problems such as stability, the ability to stand up on its own, and overall strength of design were important. After looking at several different designs, we decided that we would use the two tennis-can tubes to do a cross with one tube being the base, and the other being the barrel that would fire the ball.
We chose this design because this would have a stable base, the firing tube would be long enough to have good accuracy, and the angle would be easy to adjust. The gas law that our project involves is Gay-Lussac's Law. This supports our project because Gay-Lussac's law states that pressure and temperature are directly proportional if the volume remains constant. When we ignite the ethanol, this will cause a combustion reaction, which will result in an extreme release of heat. This, in turn, increases the pressure, and the pressure will reach a point where it exceeds how much is needed to blow the nerf ball out of the tube. Because the ethanol has a very low vapor pressure, it will most likely saturate the air in the chamber, allowing for rapid ignition, and the nerf ball will launch almost instantly after the ignition. We are a using 99% mixture of ethanol, and the molecular formula for ethanol is C2H5OH. When combusted with oxygen you get C2H5OH + O2 which goes to C2H5OH + O2,à CO2+H2O, then you balance the equation you get C2H5OH + 3O2,à 2CO2+3H2O.
We chose this design because this would have a stable base, the firing tube would be long enough to have good accuracy, and the angle would be easy to adjust. The gas law that our project involves is Gay-Lussac's Law. This supports our project because Gay-Lussac's law states that pressure and temperature are directly proportional if the volume remains constant. When we ignite the ethanol, this will cause a combustion reaction, which will result in an extreme release of heat. This, in turn, increases the pressure, and the pressure will reach a point where it exceeds how much is needed to blow the nerf ball out of the tube. Because the ethanol has a very low vapor pressure, it will most likely saturate the air in the chamber, allowing for rapid ignition, and the nerf ball will launch almost instantly after the ignition. We are a using 99% mixture of ethanol, and the molecular formula for ethanol is C2H5OH. When combusted with oxygen you get C2H5OH + O2 which goes to C2H5OH + O2,à CO2+H2O, then you balance the equation you get C2H5OH + 3O2,à 2CO2+3H2O.
A preliminary idea we had before we decided on the one below (if unable to see images click them)
This was the design we ended up choosing for our final cannon (if unable to see images, click them)
We wanted to find the angle that would launch the ball the farthest, so we used the formula h = -16t2+v0t+h0. To find v0, you need to use the formula (speed (ft/sec))cos(launch angle). We had to guess what the speed in feet per second would be, but we should still find the best angle because if the speed differs it will affect the distance, but the angle should still be the best. This formula finds how high the ball will be fired, and the height directly affects the distance. When we put this formula in a calculator, we found out that a 43 degree angle would give us the highest launch. We decided to double check this so we found a website that would do the math for us. When we plugged in the formula we got a 45 degree angle would give us the best launch. We decided to go with the 43 degree angle because the extra forward force would help compensate for wind.
Here is a math problem that is similar to the problem we had to solve:
A cannonball is shot upward from the upper deck of a fort with an initial velocity of 192 feet per second. The deck is 32 feet above the ground.]
1. How high does the cannonball go? __________ (Remember you are looking for specific part of the vertex.) The cannonball goes 608 feet high.
2. How long is the cannonball in the air? ___________ (Remember you can use the quadratic formula.) The cannon ball is in the air for 12.16 seconds
1. How high does the cannonball go? __________ (Remember you are looking for specific part of the vertex.) The cannonball goes 608 feet high.
2. How long is the cannonball in the air? ___________ (Remember you can use the quadratic formula.) The cannon ball is in the air for 12.16 seconds
To solve this problem you have to use the formula h = -16t2+v0t+h0. v0 is the initial velocity, so that is 192feet/second. h0 is the initial height. which is 32 feet, and t is time. When you plug in the formula you have height=-16t^2+192t+32. You can now put this into a calculator if you substitute t for x. When you put the problem into a calculator you will get a parabola. After getting the parabola in a calculator you solve for the vertex, and the x-intercepts. The vertex is 608 feet high at about 6 seconds, and the realistic x intercept is about 12 seconds. This problem is very similar to the problem that we had to solve for ,except we involved launch angle in ours.
Your procedure is very precise and I'm glad you included the formula that was used because it helps the reader comprehend on what you are plugging the variables into.
ReplyDeleteyou did a great job explaining, balancing the chemical equation, and solving the math problem
ReplyDeleteThe only think that I would improve is the inclusion of a possible graph for the math problem. When I try to click the image, I still can't view them. Otherwise, very good explanation.
ReplyDeleteI agree with Isai, you did an excellent job explaining it. Although the text was dark and a little harder to read, keep up the excellent work!
ReplyDelete