# Example Problem

## Calculations: Example

Let's see how we can predict the maximum speed of a rider using the information from Figure 5. For this example, we will assume a rider weight (including trolley and harness) of 578 N (or equivalently, 130 pounds).

### Step 1: Determine the Distance Traveled

Please find the horizontal distance to the lowest point and vertical drop values from Figure 5 and insert them into the boxes below. Click on the "Calculate" button to see the result.

 Vertical Drop Horizontal Distance 2 2 That's correct! $\mathrm{Distance}\left(d\right)=\sqrt{{214}^{2}+{16}^{2}}\phantom{\rule{0ex}{0ex}}d=\sqrt{45796+256}\phantom{\rule{0ex}{0ex}}d=214.597m\phantom{\rule{0ex}{0ex}}d\approx 215m$

### Step 2: Approximate Acceleration

First let’s find sin(θ). It can be calculated by dividing the opposite leg (vertical drop) by the hypotenuse (distance (d)). Enter the two values into the boxes below to calculate sin(θ). For simplicity, enter whole numbers only.

 Vertical Drop That's correct! $\mathrm{sin}\left(\theta \right)=\frac{16m}{215m}=0.0746$

Next, we’ll use the chart below to determine loss. The loss values have been derived through experimentation on the WVU Canopy Tour. In this example the rider is 130 pounds. Find the approximate value of loss in the chart and use it in the box below.

So, for a rider of 130 pounds, acceleration can then be calculated as: (enter the values of sinθ and loss below)

$\mathrm{Acceleration}\left(a\right)=g×\mathrm{sin}\left(\theta \right)-\mathrm{loss}$

 sin (θ) That's correct! $a=9.81×0.0746-0.45\phantom{\rule{0ex}{0ex}}a=0.732-0.45\phantom{\rule{0ex}{0ex}}a=0.282\frac{m}{{s}^{2}}$

### Step 3: Calculate Maximum Velocity

Since the values of all variables are now known, the maximum velocity (speed) of a 130 pound rider can be determined. Enter the values into the boxes below to find out.

 Acceleration (a) That's correct! ${V}_{\mathrm{max}}=\sqrt{\left(2×\left(0.282\right)×\left(215\right)\right)}=11.0\frac{m}{s}$ Thus, knowing the distance of the zip line to the lowest point in the catenary curve, the vertical drop, and the weight of the rider, you can predict the maximum speed on the zip line. In this example, a 130 pound rider will have a maximum acceleration of 11.0 m/s or 24.7 mph (since the unit conversion from m/s to mph is: 1m/s = 2.24mph).