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galactus · Sep 17, 2010, 06:57 AM

Many of us have seen the modeling of a passengers height on a Ferris wheel with respect to time, but how about a double Ferris wheel?

I got to thinking about this, and here is an example where I made up the dimensions. If it is not realistic, my apologies. Just used to show the method.

Suppose a double Ferris wheel has a bar length of 60 feet, and the height from the ground to the center of the bar is 50 feet, The wheels have a diameter of 20 feet and revolve once every 10 seconds. The bar revolves once every 20 seconds.

Like a single wheel, but we need two expressions. One for the wheels and one for the bar.

The radius of the bar is 30 feet and it revolves once every 20 seconds, so we have

$-30cos(\frac{2{\pi}t}{20})$

The wheels revolve once every 10 seconds and have a radius of 10 feet.

So, we have $-10cos(\frac{2{\pi}t}{10})$

We also have to add on the height of the bar from the ground.

So, in all we have $h(t)=-30cos(\frac{{\pi}t}{10})-10cos(\frac{{\pi}t}{5})+50$

This gives the height of the passenger with respect to time for the passenger on the bottom wheel after they get on the ride at t=0.

I thought this was interesting so I posted it.

Unknown008 · Sep 17, 2010, 08:03 AM

I wasn't quite understanding the 'double' Ferris wheel at first, but then the equations helped me understand what you were talking about. It's interesting :)

ebaines · Sep 17, 2010, 10:13 AM

You'd probably want to pick a different value for the ratio of the angular velocity of the two wheels. By having the two angular velocities in the ratio 2:1 the riders never get to reach the maximum possible height - the highest they ever get is 50+30-10 = 70m. Make the second wheel turn at ω=10π/3 and the passengers will get to reach a max altitude of 50+30+10 = 90m, for a more impressive view!