If it wasn't for your design, I wouldn't have thought of an alternative for that part of the track. I wanted to make sure the crater was sized appropriately. So I went back to Google Earth, and used the measurement tool to find out. It's roughly 33 ft. across, which is more or less 10 meters. Which got me thinking... what are the dimensions of the tiles anyway? I've heard anything from 8 to 10 feet. As soon as I know the dimensions of the tiles, I'll be able to create a more accurate crater.
Time for some serious number-crunching.
Here's how I worked it out:I built a coaster lift hill to see how many squares of supports (up the side) a given height would generate, and started with 100 feet. Conveniently, each square up the side is equal to each square of the rct land tiles – the total area is the same. You can see this by comparing the horizontal squares and the vertical squares: they have an identical criss-cross pattern, and therefore an identical size. The problem with rct lift hills is that the entrance and exit ramps take up a different amount of space. To compensate for this , I brought down the lift hill until it looked like it took up a whole-number multiple of squares (remember the bottom one is only half the height of the subsequent ones) and then built a flat section to test my idea.
At 90 feet, the structure has 8 full-size squares and 2 half-size squares, equal to 9 full-size squares. Now the calculation is a simple 90/9 = 10, so the side of each rct tile is 10 feet. So, how big is the crater in rct2 terms? Well, 33/10 = 3.3 squares. That’s not a very useful number because the crater is going to be more or less round (using curved edges).
To calculate the circumference of the crater, we use pi * diameter. For the real crater, this is 3.142 * 33 = 103.7 feet (approx). For our rct2 crater, it is 3.142*3.3 = 10.37 squares. Because we need 4 rct curves to produce a round crater, we need each curve to take up 10.37/4 = 2.6 squares.
So there you have it. Your rct tile is 10 feet square, the crater is 3.3 squares across, and the total circumference is 10.37 squares, with each curved piece needing to take up 2.6 squares. The good news is that in practise you don’t need to worry about the 0.3 of a square. Simply build a 3-square path, and another 3-square path intersecting that to form a cross. Build a straight wall at each end and a curved wall next to each of them. Delete the paths, and hey presto, there’s your 30-foot crater.
Of course, you then have to find out how high the real crater is and budget your land accordingly. As an experiment, I built the crater at 25 feet high. To create a smooth slope down to 0 feet, it requires 5 squares long and 3 squares wide from each end of the cross. For the intervening slopes, you need 10 whole squares plus 5 triangles = 12.5 squares. The total area you will have to budget for a 25 ft cone is therefore (12.5*4) + (15*4) = 50+60 = 110 squares, and the cone itself has a diameter of 13 squares at its widest point. However, the rct land tool does not create a smooth curve at the base, but instead a large hexagon. The result is a huge circumference of (5*4)+ (3*4) = 20+12 = 32 squares, so be careful. Your volcano may be a lot bigger than you think.

The modest 25-foot cone. There's not much space inside, but it has a large footprint!

EDIT: I just had a thought. You said that the maximum land height is 120 feet, so let's pretend you build a volcano that high. How big will the circumference of the hexagon be? We know how big a 25 ft volcano is, so 120 / 25 = 4.8 which is how many of my volcanoes will fit into yours. Therefore, the total circumference will be 32 squares times 4.8 volcanoes = 32*48 = 1,536 squares. That's an impractical number to visualise, so how wide is the 120 ft volcano across? Answer: 13*4.8 = 62.4 squares. We'll round that up to 63 squares for rct practicality - it's better to budget 1 square over than 1 square under. If you went for a 100 ft cone, to comfortably contain your 90 foot drop, this works out to 90/25 = 3.6 therefore 13*3.6 = 46.8 or 47 squares. That produces a hexagonal circumference of 32*3.6 = 115.2 or 116 squares to budget. Wow. That's a lot of land.
So now you can work out exactly how big the footprint of your volcano will be, depending on the height, by comparing it to mine. The scary thing is that this is the calculation for the actual cone, without taking the caldera into account. My guess is that you will be able to fit a pretty big rollercoaster into that volcano, whatever size you pick.
Edited by Woodpecker, 12 May 2009 - 02:48 PM.