EL DRAGO GRAND OPENING
Another One by 6ix Flags
Welcome to 6ix Flags; the thrilling and adrenaline rushing theme park which is based in the heart of Toronto. We now present to you: El Drago, new for the summer of 2016!!
Starting Height: 10 feet
Duration: 100 seconds
Height Requierments: 50"
Max Speed: 195km/h
Manufacture: 6ix Flags
Our Plan & Difficulties
For our rollercoaster, we knew that we needed to have the range reach up to a maximum of 300 feet, and start at a minimum of 10 feet. Therefore, for our first equation, we decided to use a linear equation, which met these requirements. We based our first equation on how a real life rollercoaster works. If you look at a real rollercoaster, the highest point occurs at the first drop, and it takes a gradual time on a straight path to get there. So we used a linear equation that reached the max of 300 feet to model this. After the linear equation, we needed to include the other six equations, and to do this we modeled our rollercoaster based on something that people would actually want to ride on.
Some difficulties we faced were the restrictions on the domain. At first, we used guess and check to find our equations. We guessed the “a” value, and we guessed the vertical and horizontal shifts, so to get each equation to intersect, we had to restrict the domain to 10 decimal places. We realized that this method did not actually involve any thinking, so we re-did our graph. This time, we solved for the “a” (or “c” and “d”) values by using a point from the previous equation and plugging in the values.
General description of functions
To get this equation we had to solve for the “a” value. We got the “b” value because we knew that the rollercoaster needed to start at a minimum of 10 feet, since the “b” value is the vertical translation, we made it 10. To find the “a” value or the slope, we plugged in the point we wanted the equation to pass through. We needed it to go to 300 feet (y value), and we wanted it to happen in 20 seconds (x value), so we subbed in the point (20,300) into the equation y=ax+b, and the “a” value was 14.5.
We needed this equation to intersect the previous equation, and this is why we vertically shifted it 300 units upward, and horizontally translated it 20 units to the right. Next, we chose 2 as our stretch value as it looked the most appropriate in the scenario of a rollercoaster model. Also, the “a” value had to be negative, as we needed a reflection over the x-axis.
For this equation we started by picking our “a” value. We made this value -12 because we wanted it to be reflected over the x-axis, and to have a slight curve. Next, we shifted this equation 99 units up as it had to intersect the previous equation, which ended at 100 feet. Finally, we found the appropriate “d” value for this function by plugging in a point from the previous graph (20,300) that this function needs to go through.
For this exponential function we decided to use base 2, as we did not want the curve to be too steep (realistic rollercoaster curve). Also, we shifted this graph up by 89 units in order for it to connect to the end of the previous logarithmic curve. Finally, we again solved for the “d” value by plugging in a point from the previous equation.
For this sinusoidal function we translated our equation 115 units up in order to reach our previous equation. Then, we figured out our “a” value by finding a max and min that will give us 115 when added together and divided by 2. These two numbers were 141 and 89. Then to find the amplitude (a) we subtracted the two numbers and divided by two (26). Next, we included a horizontal stretch by a factor of 2, as we wanted our curve to be more stretched out. Additionally, we solved for “d” value by subbing in (40,115.11) because this function needs to intersect the one before it. Finally, we let domain of this function be only 10 units long, as we only wanted half a cycle of this sinusoidal function.
For this equation we first shifted our equation 53 units to the right (so it connects with the previous function), and had a negative “a” value, as we needed this curve to open downwards. We then solved for the “c” value by plugging in a point from the previous graph. Also the time restriction is very important to this equation, as we needed the curve to be very small, which is why it has been restricted to just 3 units.
For this function, we needed something with a slight curve, so we decided to use a log function. We knew the “a” value vertically stretched the function, so we made it 3, because leaving it at 1 did not provide enough of a curve. We also made the “a” value negative since we needed the function to go from y=∞ to y=-∞. Since the last function ended at y=84 after the time restriction was placed, we shifted this 83 units up. Then we used the point (55, 84.15), plugged it into this equation and solved for the “d” value to ensure it intersected with the previous function.
In this part of the rollercoaster, we needed the path to travel upwards, but not too fast. So we decided to use a degree 6 polynomial, since we knew that the base around the vertex would be a little wider than a regular quadratic, or degree 4. To make the base even wider, we made the “a” value as small as possible, 0.01. Because the time restriction stopped the previous function at x=60, we made the “d” value -60, since we knew this would put the vertex 60 units to the right. To find the correct “c” value for the function, we subbed in the point (60, 80.8), which was a point that the previous graph “ended” at.
For this rational function, we first shifted it 71 units to the right so that it meets the previous function. We then added a horizontal compression by a factor of ½, as we did not want the curve to be too steep. Finally, we solved for the “c” value by getting a point from the previous function, which this equation must also pass.
This is almost the end of our rollercoaster, so we needed to start making the functions descend to reach the minimum of 10 feet again. To do this, we made the “a” value negative so the parabola would open downward, and we also made the “a” value tiny which compressed the function vertically, so the curve wasn’t as steep. The last function ended at x=78 seconds, so the “d” value was -78 which made the vertex shift 78 units to the right. To find the “c” value, we subbed in a point which the last function passed through, (78, 80.44), and solved to get 80.4383198. This moved the parabola 80.4383198 units up, and made the vertex at (78, 80.4383198).
This is our final function, and we decided to use a linear because most rollercoasters end by going down a little. Most rollercoasters end off at the same height that they start at, so we knew ours had to finish at a height of 10 feet. This is why we shifted the function 10 units up (C value). We also knew that the final time had to be 100 seconds, so we shifted the function 100 units to the right (D Value). To find the correct “a” value which would intersect this function with the previous one we took a point that the last function crossed, (95, 31.95), and subbed it into this equation.