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).

Measurements for the last zip line on the WVU Canopy Tour. — **Figure 6:** This figure shows a 255 meter zip line that is hung between two trees. The launch platform is 16 meters high and the landing platform is 11 meters high. The vertical drop from the launch point to the lowest point on the line is 16 meters and the horizontal distance is 214 meters. The drawing is not shown to scale.

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.

$Distance (d) = \sqrt{horizontal {distance}^{2} + vertical {drop}^{2}}$

**Calculate the Distance Traveled**
Horizontal Distance	²
Vertical Drop	²
Distance
	That's correct! $Distance (d) = \sqrt{214^{2} + 16^{2}} d = \sqrt{45796 + 256} d = 214.597 m d \approx 215 m$

Step 2: Approximate Acceleration

$Acceleration (a) = g \times \sin (θ) - loss remember : g = 9.81 \frac{m}{s^{2}}$

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.

$\sin (θ) = \frac{vertical drop}{distance (d)}$

**Calculate the sin(θ)**
Vertical Drop
Distance
sin (θ) =
	That's correct! $\sin (θ) = \frac{16 m}{215 m} = 0.0746$

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.

Graph showing loss according to rider weights in pounds. An outline is included to indicated the loss value of 0.45 loss per meter for a 130 pound rider. — **Figure 7:** Loss per Meter According to Rider Weight in Pounds

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

$Acceleration (a) = g \times \sin (θ) - loss$

**Calculate the Acceleration**
sin (θ)
Loss
Acceleration =
	That's correct! $a = 9.81 \times 0.0746 - 0.45 a = 0.732 - 0.45 a = 0.282 \frac{m}{s^{2}}$

Step 3: Calculate Maximum Velocity

$Maximum Velocity (V_{\max}) = \sqrt{2 \times a \times d}$

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.

**Calculate the Maximum Velocity**
Acceleration (a)
Distance (d)
Max Velocity =
	That's correct! $V_{\max} = \sqrt{(2 \times (0.282) \times (215))} = 11.0 \frac{m}{s}$ 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).