In order to calculate the velocity, centripetal acceleration, and g’s felt by the rider for the first feature in
the coaster, a drop and an irregular loop, the following calculations were needed:
The first formula was used to calculate the velocity at the bottom of the first hill, before the cart entered
the first loop. The original formula provided to find the velocity at the bottom of the first hill was:
𝑚𝑣𝑓𝑖𝑛𝑎𝑙 2
𝑚𝑣𝑖𝑛𝑖𝑡𝑖𝑎𝑙 2
+ 𝑚𝑔ℎ𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =
+ 𝑚𝑔ℎ𝑓𝑖𝑛𝑎𝑙
2
2
However, since this doesn’t solve directly for 𝑣𝑓𝑖𝑛𝑎𝑙 , we can transform the formula to
𝑚𝑣𝑖𝑛𝑖𝑡𝑖𝑎𝑙 2 + 2(𝑚𝑔ℎ𝑖𝑛𝑖𝑡𝑖𝑎𝑙 ) − 2(𝑚𝑔ℎ𝑓𝑖𝑛𝑎𝑙 )
𝑣𝑓𝑖𝑛𝑎𝑙 = √
𝑚
When the values determined from the preliminary sketch were substituted into the formula, the
velocity at the bottom of the first hill was determined:
The values substituted are as follows:
g, of course, is 9.8, as it is a standardized value.
m is 0.0425 kg, which is the average of the two testing balls used in this roller coaster.
hinitial is 1.1 m, as this is the starting height of the launch hill.
hfinal is 0.5 m, as this is the bottom of the first hill before the cart enters the first loop.
vinitial is 0, because at the start of the launch hill, the ball is not moving. This enables one to cancel out
that term in the formula.
(0.0425)02 + 2(0.0425 ∗ 9.8 ∗ 1.1) − 2(0.0425 ∗ 9.8 ∗ 0.5)
𝑣𝑓𝑖𝑛𝑎𝑙 = √
0.0425
2(0.0425 ∗ 9.8 ∗ 1.1) − 2(0.0425 ∗ 9.8 ∗ 0.5)
𝑣𝑓𝑖𝑛𝑎𝑙 = √
0.0425
Once this is calculated, we get the final velocity of
𝑣𝑓𝑖𝑛𝑎𝑙 ≈ 3.43 𝑚⁄𝑠
Now, in order to calculate the g’s placed on the rider, one must first calculate centripetal acceleration.
The formula for this is 𝑎𝑐 =
𝑣2
,
𝑟
where v2 is the velocity at the bottom of hill one squared, and r is the
radius of the imaginary circle of derived from the imaginary circle that completes the rounded portion of
the launch hill.
𝑎𝑐 =
11.76
0.325
𝑎𝑐 ≈ 36.18 𝑚⁄ 2
𝑠
Now, to calculate g forces, simply divide the centripetal acceleration by 9.8 (the force of gravity) and add
or subtract 1 based on if it is positive or negative g’s. In this case, we will have to add 1 because it is a
situation with positive g’s.
𝑔′ 𝑠 =
𝑔′ 𝑠 =
𝑎𝑐
+1
𝑔
36.18
+1
9.8
𝑔′ 𝑠 = 4.69
The force placed on the rider at this point is 4.69 times the force of gravity. Generally, roller coasters
don’t exceed 5 g’s, so this is a good amount of g’s to place on the rider. The rider will survive this
section, and therefore it is safe to ride.
The second feature in this roller coaster is an irregular loop. The design strayed from a conventional loop
due to the amount of g’s it placed on the rider. An irregular loop will reduce the speed even more while
still keeping the sensation the same for the rider. To calculate if the irregular loop is safe, we need to
know the centripetal acceleration and the g’s. The velocity (3.43 m/s) coming into the loop will be the
same coming back out, as the two points are at the same height.
𝑎𝑐 =
𝑎𝑐 =
𝑣2
𝑟
11.76
0.325
𝑎𝑐 ≈ 36.18 𝑚⁄ 2
𝑠
Now, to calculate the g’s, follow the same procedure as used before, except subtract 1 this time because
the forces at the top of the loop are pushing against the force of gravity, not with it.
𝑔′ 𝑠 =
𝑔′ 𝑠 =
𝑎𝑐
−1
𝑔
36.18
−1
9.8
𝑔′ 𝑠 = 4.69
We see here that the centripetal acceleration in the first segment of the klothoid loop is the exact same
as at the bottom of the launch hill. This is because the same radius was used in the two centripetal
acceleration calculations, and since the bottom of the launch hill is just the beginning of the loop
feature.
However, a second centripetal acceleration calculation must be used to calculate the acceleration and
the g’s placed on the rider at the top of the loop. The same formulas are used, and the velocity remains
constant. The only difference is the radius used in the centripetal acceleration calculation.
𝑎𝑐 =
𝑎𝑐 =
𝑣2
𝑟
11.76
0.15
𝑎𝑐 ≈ 78.5 𝑚⁄ 2
𝑠
Now, we must calculate the g’s. In this case, we must subtract one, as the forces are negative g’s on the
rider.
𝑔′ 𝑠 =
𝑎𝑐
−1
𝑔
𝑔′ 𝑠 = 7
This exceeds the general maximum of most roller coasters, but it will not kill the rider. It will, however,
be met with a section immediately afterwards with minimal g’s, so the rider will survive this feature.
The next portion of the roller coaster is a corkscrew feature, which is really made up of five banked
turns. It is immediately after the irregular loop feature and will lead into a final hill.
First, one must determine a radius he or she deems suitable for the track. In this case, the radius used
was 0.25 m. Once one has the radius, then he or she must find the angle θ (the bank of the track) using
the equation tan(𝜃) =
𝑣2
𝑟𝑔
tan(𝜃) =
𝑣2
𝑟𝑔
tan(𝜃) =
11.76
0.25(9.8)
tan(𝜃) =
11.76
0.25(9.8)
tan(𝜃) = 4.8
Now we have found that the tangent of theta is 4.8. However, we need to solve for theta, as that is the
angle that the track must be banked at. One must take the inverse of the tangent function, which is
arctan (and also happens to be the cotangent).
tan(𝜃) = 4.8
cot(4.8) = 𝜃
𝜃 = 78.23°
This means that the banked curve will have a very steep angle, and that the cart will accelerate a great
amount.
Now, the centripetal acceleration must be calculated, using the previous velocity after the cart exits the
loop.
𝑣2
𝑎𝑐 =
𝑟
𝑎𝑐 =
11.76
0.25
𝑎𝑐 ≈ 47.08 𝑚⁄ 2
𝑠
However, now one doesn’t need to calculate g’s, as this is a linear acceleration, and won’t affect the
vertical force of gravity on a person.
However, every time the height of the corkscrew changes, the velocity changes also. The velocity and
angle calculations must be repeated for every new height.
0.4 m
𝑣𝑓𝑖𝑛𝑎𝑙 = 3.7 𝑚⁄𝑠
𝑎𝑐 = 54.76 𝑚⁄ 2
𝑠
tan(𝜃) = 145.04
cot(145.04) = 𝜃
𝜃 = 89.6°
0.3 m
𝑣𝑓𝑖𝑛𝑎𝑙 = 3.9 𝑚⁄𝑠
𝑎𝑐 = 62.6 𝑚⁄ 2
𝑠
tan(𝜃) = 614
cot(614) = 𝜃
𝜃 = 89.8°