Hersheypark 04/21/2013 - Working Up Courage and Questionable Legality

There is a 20 foot tall grassy hill that you want to gleefully ride you wagon down. There are two dirt paths to the top of the hill. It has a steep and not-so-steep dirt incline to get to the top. The not-so-steep incline is a longer walk, but its easier to pull you wagon up. The steeper incline is harder to pull your wagon up, but it's not as long of a walk to the top because it is more direct.

Either incline you wish to frolic up, you are storing potential energy. Either way you go, it takes the same amount of energy to pull your wagon to the top of the hill. One is more direct and steeper, and one is less direct and not as steep. Either way you go, you will have the same amount of stored energy to use, which is riding you wagon gleefully down the hill.

That's all I know about it. Hope my stupid analogy helped a little. :)

Yes it did. I kinda was thinking that, but I didn't want to expand any further on my lame physics knowledge. ;)

So to nerd out for a second:

Work = Force x distance (actually, integral but they're the same in this case)

The work required to lift the train to a certain height by any path is the same. Since the distance to go straight up is shorter, the force must be larger.

Another approach is to calculate the component of the normal force in the y-direction, which is a fairly simple trig. setup.

Fchain = cos(angle) x Fg

If the lift is vertical, it requires approximately 1.414 times the force to lift the train as a 45 degree lift.

My brain hurts now. Heh...

Put less simply, (1^2 + 1^2)^1/2 = 2 ^1/2....

I will take your word for it, Gator. :D

ApolloAndy said:
So to nerd out for a second:

Work = Force x distance (actually, integral but they're the same in this case)

The work required to lift the train to a certain height by any path is the same. Since the distance to go straight up is shorter, the force must be larger.

Another approach is to calculate the component of the normal force in the y-direction, which is a fairly simple trig. setup.

Fchain = cos(angle) x Fg

If the lift is vertical, it requires approximately 1.414 times the force to lift the train as a 45 degree lift.

I love mathematics (In fact I made it to the state finals in a highly respected math competition, am on an engineering team at our school that got 4th in the nation.) Your math is right, but forget the rounding. Use the exact answer which is the square root of two.

So basically, if I'm understanding this correctly, if the force needs to be greater to lift the train up a vertical lift, that means the engine or whatever running the chain lift would have to be more powerful than a less vertical lift, but the energy required would be the same amount total if the lift hill was at a lesser angle? Am I learning new things? Or is this still beyond me?

Fahrenheit still feels very heavy to me on that, but I guess that's just gravity for ya. Lol

I don't claim to be any good at math or know anything about it, but yes, I think you got it right.

Yes. That's right. Which is one of the reasons I assume the train is so small on any coaster with a vertical lift. I'm not sure what type of motor would be required to lift a 36 passenger train up a a vertical lift, but I imagine it wouldn't be cheap.

In fact, I'm still curious why more companies aren't going with double station/short train (Cheetah Hunt style) to save on support structure, lift motor, etc.

Bad capacity probably. Fahrenheit's line moves like molasses.

Yes, but that's because it has a short train and a single station. Cheetah Hunt and TTD both have decent capacity in spite of having short trains because they have double stations. See also: Storm Runner, Kingda Ka, Sheikra, Maverick (I assume, I've never been).

Fahrenheit needed *a* station solution...like StormRunner, Maverick, Millennium Force....something that would move trains in and out faster. I've been in that line, and it is truly painful (and I used to be a fan of Gwazi, so I know something about lines that don't move).

Why anyone would be wasting time with Fahrenheit when Storm Runner is right across the way is beyond me.

As "similar" as those rides are, they landed a world apart on my favorites list. Fahrenheit is a marginally better than "meh" looper, whereas Storm Runner is a must ride multiple times every visit.

Fahrenheit feels a little...lightweight to me in terms of thrills (but the trains feel heavy up the lift...hehe). Storm Runner will always hold such a great place in my heart, but I also really love it over Fahrenheit as well. It still is interesting to me that Intamin can have such a difference from coaster to coaster in terms of re-rideability, comfort, thrill, and smoothness, even in very similar coasters (not comparing Fahrenheit and Storm Runner in this case). Even looking at two kind of "basic" patterns like TTD and Kingda Ka, TTD is by far the better ride, when they should be pretty much exactly the same. I never know what to expect from them.

The power of the motor has nothing to do with it. With the right gearing, pullies or whatever, you could lift the train with a hand crank. Granted, it would be super slow, but you could do it.