Convinced that Nitro hits max speed at bottom of second drop

[Blackhand1001]

I've always thought that Nitro actually hits its top speed of 80mph at the bottom of its second drop. The bottom of the first drop is elevated about 15 feet in the air. I checked the elevation under the bottom of the first drop and the elevation of after the second drop and theres about a 20ft difference in elevation. This drop literally skims the ground. That means the bottom of this drop is about 35ft lower than the first. This would also explain why the picture was originally located after the second drop.

Anybody else have thoughts on this.

[AV Matt]

Is the height of the ground the same under both drops?

[AV Matt]

T's definitely possible to have a higher speed on the second drop if the combination of the initial speed at the top of the second hill plus the potentialspeed from the height of the second drop alone is greater than the two values for the first drop. If we know the heights of the two hills and are fairly certain about length of drop for both hills, we can calculate a theoretical max speed at the bottom of both hills. 

[Bubba Z]

My head hurts AV. All I know is to squeeze  your buttcheecks in the helix.

[B-mac]

This certainly is more possible than the speed of 97 mph Kings Dominion advertised for Intimidator 305 before the trims were added to the first drop. If I remember right it's only possible for it to hit 93 mph max :B

[Blackhand1001]

Is the height of the ground the same under both drops?

The elevation under the second drop is about 20 ft lower.

[AV Matt]

Just to clarify...

Not the vertical distance between the track and the ground, but the actual height above sea level of the ground under both drops is what I'm asking about.

Let's say both drops went all the way to the ground. But if the bottom of the second drop was 20 feet higher above sea level than the bottom of the first drop, that's the same as the track being 20 feet off the ground if the land was completely flat under both drops. If the terrain varies at all, you have to account for that.

The relative distance from the highest point of the ride  to the lowest point is what matters, not how close the track is to the ground below it. For instance, the midcourse brake on Voyage is only 5-10 feet above the ground, but it's actually 100 feet higher than the end of the ride; that's why it goes so damn fast all the way back to the station.

[Bubba Z]

I thought the voyage had L.I.M's.

[Millennium Force 44]

LOL nice Bubba

[Blackhand1001]

Just to clarify...

Not the vertical distance between the track and the ground, but the actual height above sea level of the ground under both drops is what I'm asking about.

Let's say both drops went all the way to the ground. But if the bottom of the second drop was 20 feet higher above sea level than the bottom of the first drop, that's the same as the track being 20 feet off the ground if the land was completely flat under both drops. If the terrain varies at all, you have to account for that.

The relative distance from the highest point of the ride  to the lowest point is what matters, not how close the track is to the ground below it. For instance, the midcourse brake on Voyage is only 5-10 feet above the ground, but it's actually 100 feet higher than the end of the ride; that's why it goes so damn fast all the way back to the station.

Both the height above the ground and the elevation above sea level are lower below the second drop. The track is about 15-20 feet closer to the ground and the elevation is about 20ft lower as well. Together that means about a 35-40ft difference in track height.

[AV Matt]

Ok. That's what I was getting at.  Now I'm sure one of you coaster stats guys knows the height of the first two hills...

[Blackhand1001]

Ok. That's what I was getting at.  Now I'm sure one of you coaster stats guys knows the height of the first two hills...

Yup. It makes sense especially since thats where the onride phone was for the first several years.

[AV Matt]

Why would that be any indication? The on-ride photo for Son of Beast was taken at the base of the lift hill.

[Bubba Z]

Coaster stat guy !!!! the train drops 215 feet (66 m) at a 68 degree angle, reaching its top speed of 80 miles per hour (130 km/h), then up a second 189-foot (58 m)-tall (58 m) hill and then diving down to the left through a 161-foot (49 m) airtime hill. After the moment of weightlessness, Nitro enters its unique element, the Hammerhead turn! 

[AV Matt]

Based on that info, we're not sure what the overall drop of the second hill is. Based on the first drop info, here's a little calculation. We're going to neglect frictional losses and calculate the theoretical maximum speed the train could reach in a 215 ft free-fall.

U = P + K  (Total Energy = Potential Energy + Kinetic Energy)

Since energy is conserved (constant), the total energy at the top of the hill = the total energy at the bottom of the hill. Thus:
P1 + K1 = P2 + K2 (Potential at the top of the hill + Kinetic at the top of the hill) = (Potential at the bottom of the hill + Kinetic at the bottom of the hill).

Potential Energy = mgh (mass of train x gravitational acceleration constant x height above the ground)

Kinetic Energy = 1/2mv^2 (One half x mass x velocity_squared)

g=9.8m/s^2 (9.8 meters per second_squared) <-- this is a constant value of gravity pulling down on objects that are near the surface of the Earth. Note, the hill would have to be more like a mile high for the value of g to be affected very much.

Reminder: "_1" is the value at the top of the hill, while "_2" is the value at the bottom of the hill/drop.

(mgh)_1 + (1/2mV^2)_1 = (mgh_)2 + (1/2mV^2)_2

m = mass of the train. Since m appears in all four energy quantities, they all cancel out, so it doesn't matter what m equals.

Factor out the m:

(gh)_1 + (1/2V^2)_1 = (gh)_2 + (1/2v^2)_2

(gh)_1 - (gh)_2 + (1/2v^2)_1 = (1/2v^2)_2 = (speed)_2

Stats:
distance of drop = 66 meters
speed at bottom of drop = 130 kilometers per hour

Let's assume the height of point 1 is 66 meters and the height of point 2 is at zero. Let's also assume the train doesn't have any initial speed at the top of the hill since we're not sure what it is. In actuality the train is probably moving somewhere between 5-10 mph at the top of the chain lift. But that is a small factor in the final result. So...

(9.8m/s^2 x 66m) - (9.8m/s^2 x 0) + (1/2 x 0^2) = (1/2 x v^2)
Simplifies to: (9.8m/s^2 x 66m) - 0 + 0 = (0.5 x v^2)

646.8 (m/s)^2  = 0.5v^2

1,293.6 (m/s)^2 = v^2
36 m/s = v
36 m/s x (1km/1,000m) x (3,600s/h) = 129.6 km/h

So ignoring friction and lift hill speed we come up with a top speed within 0.4 km/h of the advertised speed.

[Edwardo]

...at the bottom of the first hill. ;-)

I gots a headache now.

[Ogolo]

Now lets move on to some quantum physics.

[AV Matt]

Now I have a headache.   

[Steve]

e=mc2

[B-mac]

I understood everything that was in Matt's post, glad to see high school wasn't a total waste

[Edwardo]

I understood it I just hate math :-).