Coaster Physics Questions, Please Help!

I'm trying to figure out "what" the radius is.

When dealing with something like a coaster, THIS doesn't offer much help.

This is the situation...

Vertical Drop is 64.5 meters (211.6141 feet). Creating a MGPE of 35.5676 m/s (79.5625 mph).

The G's at the bottom of the drop will be less than or equal to (at max) 3.5 g's.

Assume that this is a single-car Dive Coaster, to eliminate the length issues in the equation.

After doing the math I got a r(radius) of 51.599 meters (167.2910 feet).

So my question is this. What exactly is the radius? Is it the length of the turn in this picture (below), or what?

i1099.photobucket.com/albums/g399/ShippudenMan/RAD_zpsffad8c3c.jpg

[not drawn to scale]

I know that drop curves (radi) aren't perfect quarter circles, so I'm just going to assume to use a 3:5 ratio for the curve.

(3 is vertical distance and 5 is the horizontal distance)

Although with a single-car dive coaster, I'think a quarter circle is reasonable.

I will most likely be posting more questions like this on this thread. I'll appreciate the help.

I cannot see your picture, but your calculations are somewhat correct. At the lowest point in the valley, you want a vertical force of 3.5 G.

The formula for that is G*g = v^2/r . Since gravity is in the same direction as the vertical force at that point, the track can only account for 2.5 Gs.

So that only gives you the radius at the bottom of the valley. If you need to know what the rest of the radii should be, you'll need some more advanced

math.

Since the G-force is related to radius, speed and track angle, I guess you'll need to solve or iterate some kind of differential equation.

I cannot see your picture, but your calculations are somewhat correct. At the lowest point in the valley, you want a vertical force of 3.5 G.

 

The formula for that is G*g = v^2/r . Since gravity is in the same direction as the vertical force at that point, the track can only account for 2.5 Gs.

So that only gives you the radius at the bottom of the valley. If you need to know what the rest of the radii should be, you'll need some more advanced

math.

 

Since the G-force is related to radius, speed and track angle, I guess you'll need to solve or iterate some kind of differential equation.

The radii at the bottom of the valley or the "pull out" is all I really needed to know.

BTW, the formula I used is;

r=V²/Acentripetal

The correct formula for the radius, assuming zero air resistance and friction, would be as follows.

Symbol list:

a = acceleration (units = m/s^2)

g = acceleration due to earth's gravity (9.81 m/s^2)

v = velocity

r = radius

When we are talking about the bottom of a vertical drop, a = g + [(v^2)/r], but the units of acceleration here are in meters per second. When we algebraically rearrange the equation to solve for r, we get r = (v^2)/(a-g). But if you notice that if you want to calculate the radius of the track throughout the whole bottom of the drop, that is extremely difficult to calculate: Assume that you instantly change from 0 to 3.5 G's (which will cause an extremely nasty pump, by the way) in order to keep at a constant 3.5 G's, rearranging our initial equation slightly will show that a-g = (v^2)/r. This means that if you want the centripedal acceleration to be a constant 2.5 G's, r will have to continually increase so that v^2 / r = 2.5*(9.81 m/s^2) at all times, so r will have to increase in a quadratic manner as time increases.

But then if you want to get rid of that nasty pump, then you will have to vary how the acceleration changes from 0 - 3.5 G's in either a set amount of time or a set distance (depending on whichever makes the equations easier.) Though I know the basic mathematical concepts behind calculating the radius of the track as it varies in time (you will need to learn Differential Equations, which I learned last year as a sophomore in Mechanical Engineering) but the actual numbers and calculations required are absolutely ridiculous (dare I say impossible) without a computer, which is why coaster designs until the late 1980's were always more or less very rough and in the shape of regular geometric designs, because it was simply beyond the scope of human capability.

And this is why Newton is so powerful, because all you have to do is enter in how you want the G's and roll angle to vary in a certain time frame, and then it will compute those values for you.

Why do you wish to learn this in the first place? Are you trying to solve a (sounds like high school) physics problem, are you trying to build something in No Limits, or are you just curious about the bare-bones basics of the mathematics behind calculating the shaping of a roller coaster's track layout?